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Analytic Model For The Design And Selection Of Electronic Digital Computing Systems

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Analytic Model For The Design And Selection Of Electronic Digital Computing Systems

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Content This dissertation has been microfilmed exactly as received 6 6-5495 SCHNEIDEWIND, Norman Floyd, 1928- ANALYTIC MODEL FOR THE DESIGN AND SELECTION OF ELECTRONIC DIGITAL COMPUTING SYSTEMS. University of Southern California, D.B.A., 1965 Economics, general University Microfilms, Inc., Ann Arbor, Michigan Copyright "by Norman Floyd Sebneidewind 1966 ANALYTIC MODEL FOR THE DESIGN AND SELECTION OP ELECTRONIC DIGITAL COMPUTING SYSTEMS by Norman Ployd Schneidewind Dissertation presented to the PACULTY OP THE GRADUATE SCHOOL OP BUSINESS ADMINISTRATION UNIVERSITY OP SOUTHERN CALIFORNIA in partial fulfillment of the requirements for the degree DOCTOR OP BUSINESS ADMINISTRATION January 1966 This dissertation, written by NORMAN FLOYD SCHNEIBBWIND under the guidance of the Faculty Committee, and approved by all its members, has been presented to and accepted by the Faculty of the Graduate School of Business Administration in partial fulfillment of the requirements for the degree of DOCTOR OF BUSINESS ADMINISTRATION Date £ £ & £ Approvedt y d , ACKNOWLEDGEMENTS I wiah to express my appreciation to the following individuals and organizations for the assistance and information which was provided to me during the preparation of this dissertation: The Dissertation Committee comprised of Dr. Richard Beckwith, Chairman; Dr. Gerald Fleischer and Dr. C. Michael White for suggestions which resulted in strengthening the dissertation. Mr. George Wood, Consultant,Representative, International Business Machines Corporation, Los Angeles, California, for providing technical and financial data on the IBM System 360. Computer Sciences Laboratory, University of Southern California, for providing computer time on the Honeywell 800. Honeywell Electronic Data Processing for the use of its Advanced Linear Programming System. Mty wife, Verene Schneidewind, for typing the pre liminary draft and for encouraging my efforts throughout the entire Doctoral program. Mrs. Marian Kost, for editing and typing the final draft. Finally, I wish to thank Dr. Richard Williamson, Associate Dean, Graduate School of Business Administration, for his assistance in obtaining a fellowship which per mitted me to pursue a portion of the Doctoral program on a full-time basis. — N.F.S. iv TABLE OF CONTENTS Page LIST OF TABLES........................................ vi LIST OF ILLUSTRATIONS ................................ xi INTRODUCTION.......................................... 1 PART I. HARDWARE MODEL Chapter I. DEVELOPMENT OF THE MODEL ............. 6 II. APPLICATION OF THE HARDWARE MODEL............ ^3 INPUT DATA FOR HARDWARE MODEL.......................... 75 PART II. HARDWARE-SOFTWARE MODEL III. DEVELOPMENT OF THE HARDWARE-SOFTWARE MODEL . . 95 IV. APPLICATION OF THE HARDWARE-SOFTWARE MODEL . . 115 INPUT DATA FOR HARDWARE-SOFTWARE MODEL................. 177 PART III. REPLACEMENT ANALYSIS V. DEVELOPMENT OF REPLACEMENT ANALYSIS..........200 VI. APPLICATION OF REPLACEMENT ANALYSIS..........23*f SUMMARY AND CONCLUSIONS............ 296 BIBLIOGRAPHY .......................................... 303 v LIST OP TABLES Table Page 1. Technological Coefficient Matrix.................^7 2. Processing Time Matrix (65,000 Characters of Memory, 2 Input-Output Channels, and 1 Tape Controller) • • • . • • • • • . ................. *f8 3* Integer Programming Method..................... 53 *f. Parametric Programming Method............ . . • 57 5. Discrete Programming Method. 59 6. Computer Unit Utilization (Minutes)...............66 7. Daily Computer Schedule............................69 8. Specifications of Magnetic Tape Units.............87 9* Reciprocal of Effective Transfer Rate of Magnetic Tape Units.......... 88 10. Specifications of Printers .............. 90 11. Reciprocal of Effective Transfer Rate of Printers...................................... 91 12. Reciprocal of Effective Transfer Rate of Card Readers ......................92 13. Processor Memory Cycle........................... 92 1*+. Rental Rates . 93 15. Ratio of Compiler to Assembler Characteristics • 118 16. Annual Disbursements, Rental System..............121 17. After-Tax Equivalent Uniform Annual Costs, Assembly Language Rental System (Interest Rate = 5 / 0 ......................... 123 vi Table 18. 19. 20. 21. 22. 23. 2 k. 25. 26. 27. 28. 29. 30. 31. Page After-Tax Equivalent Uniform. Annual Costs, Compiler Language Rental System (Interest Rate = 5fo).......................................12k Purchase-Rental Schedule .............. 126 Computer Maintenance Cost Schedule ....... 128 Annual Disbursements, Purchase-Rental System . . 129 Straight Line Reduction in Salvage Value .... 130 After-Tax Study, Depreciation and Investment- Tax Credit, Purchase-Rental Systems (Income Tax Rate = 50 Per Cent)............ 133 After-Tax Equivalent Uniform Annual Costs, Purchase-Rental System (Interest Rate = 5%).............. 138 After-Tax Study, Depreciation and Investment Tax Credit, Assembly Language-Rental vs. Compiler Language-Purchase-Rental............... 1^-5 After-Tax Equivalent Uniform Annual Costs, Assembly Language-Rental vs. Compiler Language-Purchase-Rental (Interest Rate = 5^). ......... .............Ik9 After-Tax Equivalent Uniform Annual Costs, Compiler Language-Purchase-Rental (3?m = 1.20, Interest Rate = 5%) • . . . . . . . . . . . . . 158 After-Tax Equivalent Uniform Annual Costs, Assembly Language-Rental vs. Compiler Language-Purchase-Rental (Interest Rate * 10fO . . . . ............................. 160 After-Tax Equivalent Uniform Annual Costs- Reduced Salvage Value, Assembly Language- Rental vs. Compiler Language-Purchase-Rental (Interest Rate = 5%)................ 168 Comparison of Methods of Depreciation (Pive Year Life, Interest Rate -5%) .............. 171 Estimated Programmer Production Rates. ..... 178 vii Table Page 32. Estimated Annual Programmer Costs............ 178 33. Annual Computer Time Required for Assembly , Language Programming, .,,,, ........ 180 3*f. Annual Cost of Assembly Language Programming. • 180 35. Annual Computer Time Required for Compiler Language Programming............ 183 36. Annual Cost of Compiler Language Programming. • 183 37. Annual Program Maintenance Cost, Assembly Language (G-m = .05)............ 185 38. Computer Time Required for Program Maintenance Assembly Language (K = .25) • • . 186 39. Annual Program Maintenance Cost, Compiler Language (Gm = .05, Fm = 1.10;.................188 **0, Computer Time Required for Program Maintenance Compiler Language (K » .25, Pm = 1.10). • . • 189 *+1. Object Program Execution Time, Assembly Language, Ca = .03 (JOB 3 Constraint = 7,200 Seconds)..................................190 *+2. Object Program Execution Time, Compiler Language, Ga = .03, F+ = 1.10 (JOB 3 Con straint = 7,200 Seconds). ................... 193 *+3. Technological Improvement of Several Computer Characteristics .............. 20*f Mf. Compound Annual Rates of Decrease in Cost/ Performance Ratios— Forecasted Compound Rates of Price and Rental Decrease.............222 ^5. Decline in Salvage Value of Several Computer Systems.......... 229 lf6. Equivalent Uniform Annual Costs, Constant Rate vs. Compound Rate of Salvage Value Decrease (Six Year Life)................ 5 • . 233 *+7. Schedule of Costs of Defender and Four Purchased Challengers ..... ............. 235 viii Table Page **8. Defender Salvage Value Schedule (First Cost = $668,000) 238 4-9. Challenger 1 Salvage Value Schedule (First Cost s $4-58,000).............................. 239 50. Challenger 2 Salvage Value Schedule (First Cost = $1+17,000).......... 24-0 51. Challenger 1 Capital Recovery Cost (First Cost = $4-58,000) (Interest Rate = 5 Per Cent) 24l 52. Defender Capital Recovery Costs (Interest Rate = 5 Per Cent) ............ 24-2 53* Defender-Challenger 2 Capital Recovery Cost (First Cost = $4-17,000; (Interest Rate = 5 Per Cent) .......... 243 54-. Depreciation and Investment Tax Credit, Challenger 1 vs. Defender (One More Year) and Challenger 2 (Income Tax Rate = 50 Per Cent)........... 24-5 55. After-Tax Equivalent Uniform Annual Costs, Challenger 1 vs. Defender (One More Year) and Challenger 2 (Interest Rate = 5 Per Gent) .............. 24-7 56. Challenger 3 Salvage Value Schedule (First Cost = $379,000).............................. 253 57. Defender-Challenger 3 Capital Recovery Cost (First Cost = $379,000; (Interest Rate = 5 Per Cent) 254 58. Depreciation and Investment Tax Credit, Challenger 1 vs. Defender (Two More Years) and Challenger 3 (Income Tax Rate = 50 Per Cent)................... 256 59. After-Tax Equivalent Uniform Annual Costs, Challenger 1 vs. Defender (Two More Years) and Challenger 3 (Interest Rate = 5 Per Cent) 258 60. Equivalent Uniform Annual Cost of Disburse ments of Defender (Interest Rate = 5 Per Cent) ............................267 ix Table Page 6l. Equivalent Uniform Annual Cost of Disburse ments’of Challenger 2 and Challenger 3 (Interest Rate * 5 Per Cent) ..... . . . 268 62# Comparison of Costs for Two Rates of Price Decrease Challenger 1 (7 Year Life) vs. Defender (One More Year) and Challenger 2 (6 Year Life) and Challenger 2A (6 Year Life). ....... ....................... 27k 63. Schedule of Costs of Defender and Pour Rented Challengers ................ ..... 276 6*f. Defender Capital Recovery Cost (Interest Rate * = 10 Per Cent)......................... 277 65. Differences in Income Tax Disbursements, Challenger 1R vs. Defender (One More Year) and Challenger 2R (Income Tax Rate = 50 Per Cent)................... 278 66. After-Tax Equivalent Uniform Annual Costs, Challenger 1R vs. Defender (One More Year) and Challenger 2R (Interest Rate « 10 Per Cent) ................... 279 67. Differences in Income Tax Disbursements, Challenger 1R vs. Defender (Two More Years) and Challenger 3R (Income Tax Rate = 50 Per Cent) 282 68. After-Tax Equivalent Uniform Annual Costs, Challenger 1R vs. Defender (Two More Years) and Challenger 3R (Interest Rate = 6 Per C e n t ) ................... 283 69. Equivalent Uniform Annual Cost of Disburse ments of Defender (Interest Rate = 10 Per Cent) .......... 287 70. Equivalent Uniform Annual Cost of Disburse ments of Challenger 2R and Challenger 3R (Interest Rate = 10 Per Cent)................ 289 71. Comparison of Costs for Two Rates of Rental Decrease Challenger 1R (7 Year Life) vs. Defender (One More Year) and Challenger 2 (6 Year Life)/Challenger 2RA (6 Year Life) • 293 x LIST OF ILLUSTRATIONS Figure 1, Application Flow Chart 2. After-Tax Equivalent Uniform Annual_Costs Rental Systems (Interest Rate = 3. After-Tax Equivalent Uniform Annual Costs Purchase-Rental Systems (Interest Rate 5%)........ ........................... 5. 6. 7. 8. 9. 10. 11. 12. After-Tax Equivalent Uniform Annual Costs Assembly Language-Rental vs. Compiler Language-Purchase-Rental (Interest Rate = 5$) . . . . . . . . . . . . . . After-Tax Equivalent Uniform Annual Costs Assembly Language-Rental vs. Compiler Language-Purchase-Rental (Interest Rate = 10$)............................ Comparison of Capital Recovery Costs and Total After-Tax Equivalent Uniform Annual Costs, Compiler Language Systems (5% Interest Rate vs. 10$ Interest Rate) . • Central Processor Technological Improvement, Cost/Performance Ratios (Data of Table ^3) • Magnetic Tape Unit Technological Improvement, Cost/Performance Ratios (Data of Table *f3) . Card Reader Technological Improvement, Cost/Performance Ratios (Data of Table *f3) • Printer Technological Improvement, Cost/Performance Ratios (Data of Table ^3) » Page *f5 12? 1^3 l5*f 165 Replacement Analysis Time Chart, Replacement Analysis Flow Chart, 166 206 207 208 209 215 216 xi Figure Page 13. lb. 15. 16. 17. 18. 19. Forecasted Purchase Price of Compiler Language Challengers Based on 50 Per Cent and 100 Per Cent of Historical Technologi cal Progress (Data of Table 19 and Table 221 * Forecasted First Year Rental of Assembly Language Challengers Based on 50 Per Cent and 100 Per Cent of Historical Technologi cal Progress (Data of Table 16 and Table ¥*■) 225 After-Tax Equivalent Uniform Annual Costs of Challenger 1 (Purchased Compiler Language System) and Defender (One More Year) Followed by Challenger 2 (Purchased Compiler Language System) for 5 Per Cent Interest Rate (Data of Table 55) .......... 252 After-Tax Equivalent Uniform Annual Costs of Challenger 1 (Purchased Compiler Language System) and Defender (Two More Years) Followed by Challenger 3 (Purchased Compiler Language System) for 5 Per Cent Interest Rate (Data of Table 59)••••••••••• 262 Defender Capital Recovery Costs for 5 Per Cent Interest Rate (Data of Table 52)............. 26*+ Capital Recovery Cost of Challenger 2 (Data of Table 53)? and Challenger 3 (Data of Table 57) (Purchased Compiler Language Systems) for 5 Per Cent Interest Rate............ 265 Ca •' - — - — Data of Table 51)? Defender (One More Year) Followed by Challenger 2 (Purchased Compiler Language System— Data of Table 53) and Defender (Two More Years) Followed by Challenger 3 (Purchased Compiler Language System— Data of Table 57) for 5 Per Cent Interest Rate . . . . . • • ................. 266 Defender Equivalent Uniform Annual Cost of Disbursements for 5 Per Cent Interest Rate (Data of Table 60) 269 xii Figure 21. Equivalent Uniform Annual Cost of Disburse ments of Challenger 2 and Challenger 3 (Purchased Compiler language Systems) for 5 Per Cent Interest Bate (Data of Table 6l) 22. Equivalent Uniform Annual Cost of Disburse ments of Challenger 1 (Purchased Compiler Language System— Data of Tables 55 and 59)» Defender (One More Tear) Followed by Challenger 2 (Purchased Compiler Language System— Data of Table 55) and Defender (Two More Tears) Followed by Challenger 3 (Pun- chased Compiler Language System— Data of Table 59) for 5 Per Cent Interest Rate. . * 23* After-Tax Equivalent Uniform Annual Costs of Challenger 1R (Rented Assembly Language System) and Defender (One More Tear) Followed by Challenger 2R (Rented Assembly Language System) for 10 Per Cent Interest Rate (Data of Table 6 6 ) ........... 2k, After-Tax Equivalent Uniform Annual Cost of Challenger 1R (Rented Assembly Language System) and Defender (Two More Tears) Followed by Challenger 3R (Rented Assembly Language System) for 10 Per Cent Interest Rate (Data of Table 6 8 ) . . . . . .......... 28^ 25. Defender Capital Recovery Costs for 5 Per Cent Interest Rate (Data of Table 52) and 10 Per Cent Interest Rate (Data of Table 69) . . . . 285 26. Defender Equivalent Uniform Annual Cost of Disbursements for 10 Per Cent Interest Rate (Data of Table 69).............................. 288 27. Equivalent Uniform Annual Cost of Disburse ments of Challenger 2R and Challenger 3R (Rented Assembly Language Systems— Data of Table 70) for 10 Per Cent Interest Rate . . . 290 28. Equivalent Uniform Annual Cost of Disburse ments of Challenger 1R (Rented Assembly Language System— Data of Table 66), Defender (One More Tear) Followed by Challenger 2R 270 271 xiii Figure Page (Rented Assembly Language System— Lata of Table 66) and Defender (Two More Years) Followed by Challenger 3R (Rented Assembly Language System— Lata of Table 68) for 10 Per Cent Interest Rate.....................291 xiv INTROnJCTIOET Historically, it has been possible to base computer selection on an analysis of predicted performance with respect to representative applications. This method of selection is feasible when the number of applications and the number of computer systems which must be evaluated is small. The application and development of computer systems has increased so rapidly that a selection based on an enum eration of possible solutions is no longer feasible. A mathematical model is developed for the selection of electronic digital computer systems. The model may be applied to the selection of entire systems or to the selec tion of individual components for an existing installation. Optimization is achieved by minimizing the total cost of installing and operating the proposed system over its assumed life, subject to satisfying all time constraints which are imposed on the proposed system. The cost3 to be minimized include any savings (negative costs) resulting from the installation of the proposed system. The model provides management with a recommendation based on those variables which can be quantified. Qualitative factors, such as the reliability and reputation of the manufacturer, 1 would also be considered by management before making a decision. Qualitative factors are important when the quan titative analysis does not produce a significant difference in alternatives. The discussion consists of three major parts: Part I - Hardware Model, Part II - Hardware-Software Model and Part III - Replacement Analysis, The Hardware Model is a deterministic model which uses the method of integer programming. The model can only be applied to the selection of systems for either single program or multi-program batch processing applications; it cannot be applied to applications in which inputs occur randomly, Gost per period is minimized subject to con straint equations which stipulate the times in which jobs must be performed. The constraint equations are a summa tion of the product of the reciprocal of effective transfer rates and technological coefficients. Effective transfer rates and memory requirements are based on the use of an assembly language. An exact solution to the integer pro gramming model is obtained by using the Gomory Method, and approximate solutions are also obtained by using parametric linear programming or discrete programming. The dual solu tion is used to indicate critical job(s), where the criti cal job is one which causes a change in the optimal solu tion if the time constraint or technological coefficients change. 3 Optimal solutions are obtained for each time period. Optimal solutions change from period to period due to an increase in the amount of data to be processed, an increase in the amount of memory required or a decrease in the appli cation time constraints. The Hardware-Software Model is used to obtain an optimal solution in each time period for the selection of hardware and software. Optimal period costs are converted to an equivalent uniform annual cost. The best available hardware-software system (present challenger) is determined on the basis of the lowest minimum equivalent uniform annual cost. The Replacement Analysis is used to detezmiine whether the existing system (defender) should be replaced by the present challenger. Replacement is indicated when the minimum equivalent uniform annual cost of the present challenger is less than the minimum equivalent uniform annual cost of every combination of additional defender service, followed by its successor, which can occur within the assumed life of the present challenger. A reduction in processing time may be a major reason for the original installation of a computer or for the replacement of a computer with a faster model. When this is the case, the proposed system (challenger) time constraints will be less than the existing system (defender) time constraints, nil order to compare defender and chal lenger on an equal service basis, estimated future defender costs should be adjusted to reflect the same set of time constraints as those chosen for the challenger. In the case of an existing computer system, it is necessary to determine additional future equipment costs which will be incurred in order to process within the time constraints established for the challenger. If the defender is a cler ical system, the additional future personnel costs which will be incurred in order to process within the challenger time constraints should be determined, A relationship is shown to exist between the tech nical and financial variables involved in the selection of computer systems. The use of a compiler language is favored when the computer system is purchased and the mini mum attractive rate of return is low. An assembly language is favored when the computer system is rented and the mini mum attractive rate of return is high. PART I HARDWARE MODEL CHAPTER I DEVELOPMENT OP THE HARDWARE MODEL Present Methods There has long existed a requirement for a metho dology designed to select the optimal computer system required to satisfy the time constraints of one or more applications. An optimal system is defined as one which will process one or more applications within specified time intervals and at minimum cost. The present methods of computer selection may be divided into three major classes as follows; 1. Analysis of Representative or Benchmark Problem1 This method employs the concept of using sample task(s) chosen to be representa tive of the totality of tasks to be performed. 3y trial and error calculations, a search for an optimal solution is performed. A sampling procedure must be resorted to in this method, since the number of enumerations grows rapidly 1 Myron B. Solo. "Selecting Electronic Data Pro cessing Equipment," Datamation. (1958), 6 with an increase in number of applications. 2 2. Computer Aided Analysis The task of finding a solution by enum eration can be made more manageable by employ ing computer programs which use a library of hardware and software data for many manufac turer’s equipment. Many solutions can be obtained by trying different combinations of components for various manufacturer’s equip ment. This method is essentially a mechaniza tion of the manual trial and error procedure and does not guarantee a solution which maxi mizes overall effectiveness. 3 M- 3. Numerical Scoring Methods * Scoring methods of many varieties have been used which have as their objective the quantification of various hardware and software ^Donald Herman, Fred C. Ihrer, "The Use of a Com puter to Evaluate Computers," American federation of Infor mation Processing Societies Proceedings. Vol. 25. 196M-. ^Solomon Rosenthal, "Analytical Technique for Automatic Data Processing Equipment Acquisition," American Federation of Information Processing Societies Conference Proceedings. Vol. 2$. 196M-. * 1 'E. 0. Joslin, "Cost Value Technique for Evaluation of Computer System Proposals," American Federation of Information Processing Societies Conference Proceedings. Vol. 25, 196M- attributes into a numerical score which pur ports to represent the overall effectiveness of the system. This method does not maximize overall effectiveness because the assignment of points to individual attributes does not ade quately measure the interdependence of attributes in satisfying the time constraints of many applications. When many applications, with diverse processing characteristics, must be performed by the same components a resource allocation procedure is called for. The appli cation of this procedure will result in the selection of components which will satisfy the constraints of all applications at minimum cost. Operating Environment The model to be described is applicable to batch processing systems, wherein input data is collected in advance of processing and the sequence and scheduling of computer runs is known in advance. The model is not appli cable to systems where random inputs occur and the sequence of computer processing is dependent upon the nature of the random inputs. The following definitions and characteris tics serve to define the operating environment! 1. A complete computer application is referred to as a job. Jobs are processed on a scheduled basis according to a known frequency. A self-contained phase of a job is referred to as a run. She runs comprising a given job are processed in sequence. There is only one sequence feasible for a given job. Input data for a given job is collected in advance of the processing of that job. Data which is generated after the beginning of com puter processing for a given job, is included in the input for the next processing cycle. The individual computer units operate as general purpose devices, A mode of operation is defined as a type of data processing opera tion. Examples of modes of operation are file processing, file maintenance, file searching, sorting, merging, collating, computing, print ing, card reading and card punching. One or more modes of operation are possible in each run. In a given run, concurrent input, computing and output operations may take place. It is also possible to have no concurrent operations. In this type of computer system all operations occur serially. 10 6. (Through multi-programming, two or more runs of different jobs may he processed concurrently. The above operating environment con stitutes the batch processing of data by the most advanced hardware-software systems presently available. Considerations in the Development of the Model The processing time of a given computer unit in a particular mode of operation is a function of both equip ment and application parameters. As an example, for a mag netic tape unit engaged in file maintenance of a multi-reel file, where tape rewinding is involved, the following equipment and application parameters determine file pro cessing time: Equipment Application Forward Speed File Size Ke-wind Speed Fraction of File Which is Alphabetic Recording Alphabetic Density Fraction of File Which is Decimal Recording Density Decimal Inter-Block Gap Block Size Tape Length Reel Change Time An effective transfer rate can be computed from the above parameters. For a magnetic tape unit, in a file maintenance mode of operation, the effective rate is the average rate of transferring information over the entire file. This throughput rate reflects the delays caused by 11 interblock gaps, rewinds, and reel changes. The effective rate is to he distinguished from the instantaneous rate, which is equal to speed multiplied by density. Effective rates for magnetic tape units can also be calculated for other modes of operation, such as sorting. Effective trans fer rates can be computed for all types of computer units and for all modes of operation. The reduction of many equipment and application parameters to a single effective transfer rate for each unit and mode of operation simplifies the task of con structing a model to represent the times of performing data processing jobs. If effective transfer rates are not used, the number of variables in the model is so large that a solution cannot be obtained. The time of performing a batched processed job can be expressed as a summation of the product of technological coefficients and reciprocal of effective transfer rates of the computer units. There will be one or more dominant units in each run, depending upon the number of input- output channels and degree of simultaneity. The dominant units are those which govern the time in which a given run will be completed. Other units, which are not dominant, must obey the constraints that their processing times are less than or equal to those which are dominant. In order to formulate a problem as a set of linear constraints, an 12 assumption must be made as to which component(s) is (are) dominant. However, in most cases, the relationship is obvious. In a system in which no concurrent operations can take place, all unit processing times are dominant and run time is the sum of all unit processing times. When concur rent operations are permissible, run time is the sum of the dominant unit times. The non-dominant units must perform their processing tasks in a time less than or equal to the dominant units (s) time. For some units, there is a value of effective trans fer rate corresponding to each value of a parameter. This parameter is not associated with the physical characteris tics of the given unit. As an example, the effective trans fer rate of a magnetic tape unit can be varied by the size of block used. The size of block used affects the total memory requirement. A similar relationship exists with respect to the use of other external storage media, such as disk files, magnetic cards and drums. In general, an increase in the block size used on an external storage device will increase the effective transfer rate of that device at a decreasing rate. However, an increase in block size results in a greater internal storage or memory requirement. Therefore, there exists an optimal relation ship between the effective transfer rate of input-output devices (external storage devices) and the size of the 13 internal storage element (memory), The memory size is made a parameter of the problem. A practical tipper limit to the size of memory exists, because the rate at which input- output transfer rates increase with increasing block size decreases rapidly. The lower limit on memory size is established by the maximum size program which must be exe cuted. Another parameter of the problem which is not a physical characteristic of individual computer units is the number and -type of input-output control units and input- output channels included in the system. The number and type of these devices determine the number of input-output units which can be concurrently transferring information while internal processing takes place. Effective transfer rates are dependent upon memory size and number and types of data transmission paths in addition to the inherent speeds of the input-output units and central processor. Therefore, a model formulated on the basis of finding the set of computer units which satis fies all job time constraints at minimum cost, will provide a solution corresponding to a particular memory size and input-output control and channel configuration. A solution based on a selection of these two parameters may or may not be a global optimum. It may be necessary to try alterna tive sets of these parameters before an optimal solution is lb obtained. However, the number of sets of these parameters which must be tried is usually limited to one, two or three at the most, fhe number of memory size and input-output channel configurations is limited by program size, the num ber of files to be processed in any one run and the rapid increase in cost resulting from increases in memory size and input-output channels. In summary, the model should possess the following characteristics: 1. fhe numerous parameters which are related to the physical and applications characteristics of the computer units should be translated into an effective transfer rate which repre sents the throughput capability of the unit for a given mode of operation. If an attempt is made to build a model on the basis of individ ual parameters, a large number of variables are introduced into the problem; in addition, each manufacturer's equipment must be treated differently, since characteristics which are constant in one system may be subject to varia tion in another system. However, effective transfer rates are common to all systems. 2. If the time of performing a job is treated as a continuous function of effective transfer rates, the choice of computer units "based on the cri terion of choosing the units closest to the continuous variable solution, yields erratic results with respect to obtaining an optimum. The number of choices of each type of unit are too few in number and the values of effective transfer rates are too far apart to give satis factory results. It is necessary to construct the model on the basis of the exact effective transfer rates available in the syste: , being evaluated. This consideration leads to the i formulation of the problem in terms of discrete variables rather than continuous variables. Since the job time constraints and application coefficients are subject to change with time and are also subject to error of estimation and measurement, it would be desirable to have the capability of determining alternative opti mal solutions without resorting to excessive additional computation. A given job time constraint requires the inclu sion of auxiliary time constraints which must be satisfied in order for the job time con straint to hold. Thus, it is desirable to have the capability of easily expressing a complex 16 •timing relationship among computer units. 5. It is possible that a given computer system is infeasible for a given set of job constraints. It is desirable to have the capability in the model of rapidly indicating this condition. Mathematical Formulation The notation used in the development of the model is as follows: A series of one or more computer runs, per formed in sequence, and which comprise a com plete application or a subset thereof. 2. Run r A distinct phase of a job. 3- Mode of Operation, m A particular mode of operation, such as file maintenance or sorting. • Unit u A unit or component of the computer system. 5. Dominant Unit, u* A unit of the computer system which determines run time. 6. Technological Coefficient, A^j The amount of data to be processed or memory references to be made by unit u, run r, job j. Aj^tpj corresponds to data processed or memory references made by a dominant unit. 7. Reciprocal of Effective Transfer Rate, The effective transfer rate is the throughput 17 capability of unit u for mode of operation m. The effective transfer rate is determined from a consideration of all factors which affect processing speed for a particular type of unit and mode of operation. In order to develop a linear system of inequalities which has the unit of time on the right hand side, the reciprocal of the effective transfer rate is employed. 8. Model Index, i An index representing the models available for unit u. 9. Particular Values of Reciprocal of Effective Transfer Rate, X,3Tnj Since there are only certain possible values of corresponding to actual models i of unit u, Xnm-f represents those physically feasible values of the reciprocal of effective transfer rate. 10. Computer Unit Cost, Cai Monthly rental or capital recovery cost in dollars of unit u, model i. 11. Bun Time, Trj Time duration of performing run r, job j. 12. Job Time Constraint, Tq Time in which job j must be completed. 13. Hours in ffirst Shift. H l1 *. Extra Shift Bate, du Percentage of equivalent first shift hourly rate charged for second and third shift usage for unit u. 15. Humber of Jobs. J 16. Humber of Buns per Job. B. 18 17. Humber of Units, IT 18. Humber of Models per Unit. I The following are -the steps in the mathematical formulation of the model: 1. Run Time Run time equals the summation of the product of dominant unit technological coefficients and the reciprocal of effective transfer rate (u * u* when domination is present), I T (1) Tjh — £ -Au* rj u=u* j r=l(l)Rj j=l(l)J J u=l 2* Job Constraints Jobs must be performed within stated intervals^ Tj, T2> " - - - Tj, Job time is the summation of run times occurring in the given job. R R U (2) £ Tp-j - £ £ A^tp-j ®j > ^“U.1 j r=l r=l u=l j=l(l)J 3. Auxiliary Constraints Units which do not determine run time (non dominant) must perform assigned processing tasks in a time less than or equal to the processing time of the dominant unit(s). U r (3) r=l(l)R, j=l(l)J *+. Hon-negativity Constraint M X^ - 0, u = 1 (1) I T 5. Objective Function It is desired to minimize the total equipment rental or capital recovery cost per time period. For a single shift operation, rental is independent of the amount of usage. The formu lation of the multiple stift problem is dis cussed on page 37. U (5) Min Z = £ Cu u=l Discrete Variable Formulation Since there is a finite number of values of X ^ corresponding to values of Cu, where Cu is independent of the mode of operation, m, the problem is to find a subset X*1Tn|, where i represents the model index corresponding to unit u, which will satisfy the constraints (2) and (3) and will minimize Z = !£ C,,-;« This combinatorial problem can u=l be cast into the format of discrete variable problem by 5,6 making a transformation of variables. 7 'George B. Dantzig, linear Programming and Exten- sions (Princeton. Hew Jersey: Princeton University Press. 19537, p, 5^0. ^Charles R. Carr and Charles W, Howe, Quantitative Decision Procedures in Management and Economics (Hew York: McGraw-Hill, 1964), pp. 228-29. (6) (7) (8) (9) (10) (11) 20 • Transform Xumi I ^um = ^umi \l1’ ^ui = ^ ^or ^umi = ^umi I and = 0 for X ^ * X * ^ . S V l “ x» 1=1 11 =1(1)11. Xnm-i is equal to the actual value of corres ponding to model i of unit u and X*1Tni. is equal to the optimal value of X,im^ » • Transformed Statement of the Problem U I Minimize 2 = 2 ^ Gni u=L i=l U1 U1 Subj ect to: R U I ^ ^u'rj -^u'mi ^ fi ” u - u* j r=l u=l i=l j = 1(1)J r=l(l)R j=l(l)J I 2 %i± = 1 9 V = 1 (1) U i=l t^ = (0,1), u = 1 (1) U, i = 1 (1) I A further simplification of equations (8) and (9) can (12) (13) ( 1*0 (15) ( 16) (17) 21 be performed as follows: E U I U I 2 ^ur j ^umi ^ui“ " " ^uij ^ui r=l u=l i=l u=1 i=l where represents the accumulation of pro cessing times over all runs R and for various modes of operation m for each unit u and model i operating in a given job. The final form of the equations follows: - - Minimize Z = 2 Z °ui V l ’ where cui is in- u=l i=l dollars and t^ is a pure number, 0 or 1. Subject to: U I Z Z V u V i s V u * u ’> 5=10 )j u=l i=l j=l(l)J I 'E t = i, u = 1(1 )u i=l U1 ^ * 0,1, u = 1 (1)U, i = 1 (1)1 Linear Integer Programming Solution The above problem in (13) to (16) can be solved by linear programming techniques. However, an integer solution 22 is required because ty^ = C0,1), whereas, in general, 0 " Vi. “ !• The method of Gomory^ can he used to find an optimal integer solution in a finite number of steps after an initial non-integer optimal solution has been found. Additional constraints are introduced, one at a time, such that the reduced feasible set excludes the original non integer solution, but does not exclude any integer solu tions of the original set. The additional constraints are formed from the final tableau at each stage of solution in the following manner: (18) Z (yui) a f . or 2 ^ (yul) - S . f , U1 ° U2 J where fu^ is the fractional part of the coeffi cient of a non-basic variable yu^ in the column ui, row j, after that coefficient has been reduced by the largest integer value; f0- j is the fractional value of the reduced basic variable associated with row j of the tableau; S is a new slack variable. Since 0 - t ^ * 1, ty^ s fQ^; there is no reduction in required in order to make it fractional. Gtomory has shown that (18) will be satisfied by all non-negative ?R. E. Gomory, "Integer Solutions to linear Pro grams," Robert 1. Graves and Philip Wolfe (eds.), Recent Advances in Mathematical Programming (New York: McGraw- Hill, 1963), pp. 269-302. 23 integer solutions of the original problem, but will not be satisfied by the original non-integer solution (the non- basic yu^ = 0 in the original solution). The most effi- cient additional constraint is one in which - max, fui since this corresponds to a maximum value of yu-p a pre viously non-basic variable which now becomes basic. The maximum value of yui reduces the area of the non-integer solution set by the largest amount in the original feasible • £ . set. However, for a given row j, may not be maximum ■^ui for all columns ui. In terms of the primary constraints (1* 0 , yui is the slack in the job time constraints which is introduced into the solution as the slower of two dominant computer units t^^ 0 and the faster unit ^*(2+1) In terms of the auxiliary constraints (15), yu^ is the slack introduced as the slower of two non-dominant computer units t ^ -: * 0 and the faster unit t^Ci+l) -* 1, or the slower of two dominant computer units 1 and the faster -unit ^ 0. The largest cut is obtained by f * finding the maximum — for those constraints which cur- fui rently have zero slack, according to the relationships given above for primary constraints (l*f) and auxiliary con straints (15). If the -t^i, which are in the basis of the optimal integer solution, are in the basis (at a non-zero level) of the non-integer optimal solution, each application of the Gomory method will drive the unwanted one of the pair of tu.i out of the basis and will drive the wanted t ^ to the value 1, The number of additional constraints and iterations corresponds to the number of pairs of fractional tui existing in the non-integer optimal solution* However, if some or all the t ^ which comprise the optimal integer solution are not present in the optimal non-integer solu tion, the number of additional iterations required is unpredictable. This difficulty arises because for each optimal t ^ , t*^, not in the optimal non-integer basis, the initial application of Gomory*s method will drive a tu^ to the value 1 which does not correspond to t*u^; many additional iterations may be required before t*U£ = 1. In general, computational difficulties have been reported in the literature for even modest size prob lems. Tlie pro^em stems from the difficulty in choosing the order of adding constraints in a fashion which will result in convergence in a reasonable number of itera- f ■ tions. The problem is caused by the expression not ________ xui________ ®Saul Gass, "Integer Linear Programming," Franz Alt and Morris Rubinoff (eds.T, Advances in Computers (Hew Yoafe: Academic Press, i960), pp. 302-309* ^R, E. Gomory and C. S. Wade, Integer Programming 2 , 7090 (PKIP 92 and PK IPM2) (Yorktown Heights, New York: IBM Business Theory Group, September, 1961). -*-®G. Hadley, Non-linear and Dynamic Programming (Reading, Mass.: Addison-Wesley, 19o1 +), pp. 251-265. 25 being maximum for every column ui of a given row j • In practice, the selection of row j, is sometimes made on the ^ * basis of the maximum average value of "ui Because of the lengthy computations which are some times required in order to achieve an optimal integer solu tion by the Gomory method, alternative methods of solution have been developed, as described in the next two sections. A Parametric Method of Solution A method for obtaining an approximate solution to the problem in (13) to (17) is to perform parametric linear programming on each of the 30b time constraints simultane ously. The procedure for this method is to decrement on a continuous basis each of the job time constraints until one or more solutions are obtained which are as close to integer solutions as possible. This method corresponds to the reduction of the feasibility set by simultaneously moving the hyperplanes corresponding to the primaiy con straints parallel to themselves at equal rates in order to approximate the points in space where optimal integer solu tions exist. All constraints are reduced by an equal per centage. This percentage is sufficiently small, so that each possible solution can be identified. The criterion for selecting the optimal solution is the followings 1. For each solution obtained, choose the maximum tj^i, t ^ f, for each group of computer units • 26 2. Sum the tu^, from each computer unit group. 3. Choose those solutions (all t^t =1) corres ponding to U (19) 2 \ii* = ^ equals a maximum for a given u=l group of tu^f and (20) (U - F) ^ e, where e is an arbitrarily small positive number. A formal statement of the method follows. The job time constraints, T ^, are made a linear function of a param eter 9. (21) T.j s T^q + c< ^ Q, j = 1(1)J, where T^0 is the original constraint and (22) <* j - -k.j T^0, where k^ » k2 = k^. Equation (22) expresses the fact that the original time constraints are reduced at equal rates, kj, as © is increased. The sum, F, of the maximum t^t in each computer unit group will reach a maximum, for a given group of tgi*, as 9 increases; one of the maximums may be at 9 = 0, The iniif > representing the faster unit of the pair of t ^ if two tg^ are in the basis, or the next faster unit if tg^i = 1, increase as 9 increases from zero. The tgj^ cannot get smaller, since all job time constraints are decreasing. U At some 0 = ©j, v* = P is a maximum for a given group u=l 27 of t ^ »• As 0 is increased from the value 0^, a t ^ * will decrease from the upper hound of 1 and a new will enter the basis. A new set of t^^ will increase as 0 increases, until a new maximum is attained at 02 for the new group of t^t. This process is continued until 9 s 0 max, at which point 0 can no longer be increased and feasibility maintained. The quality of the solution at 0i, 02, ♦ 0 max U can be measured by the size of U - T*, since tu^f = U, corresponds to an integer solution. A positive e can be chosen in order to select only those solutions of quality (U - 3»K e. The optimal integer solution, corresponding to the original set of constraints Tj0, will not necessarily exist at 0^ (all t^i = l). The objective of the method is to identify one or more optimal integer solutions as a function of 0. It is desirable that the first of these solutions be equal to the optimal solution of the original problem. However, if this does not occur, the series of optimal solutions, obtained as a function of job completion time, still provides valuable sensitivity data. The optimal solution i3 normally quite sensitive to changes in job time constraints. An optimal solution to the original problem may be valid for only a small range of one or more T-j. In addition, job 28 time constraints are not inviolate; they are constantly under pressure to be decreased due to competition and a desire to provide better service to the customer* There fore, a series of cost versus job completion time alterna tives must be evaluated. An assumption of this method is that the job time constraints will decrease in the future at equal rates. This assumption was made for purposes of illustration. If the individual jobs have different priorities, the constraints would be decreased at unequal rates• Discrete Programming Solution The two foregoing methods are based on linear pro gramming techniques which are efficient for solving prob lems involving continuous functions, but loses some of its efficiency when applied to discrete functions as a result of the integer solution requirement. A method is needed which is designed to operate in terms of discrete values. One approach is to start with the maximum cost solution, if feasible, and to move in discrete steps in the direction of cost reduction until no further reduction in cost is possi ble without violating feasibility. It is convenient to start with the maximum cost solution for two reasons: 1. An immediate determination is made as to whether the problem is infeasible. 2. If the maximum cost solution is feasible, a 29 starting point for the application of the algo rithm is readily obtained. It is desired to establish a criterion for making an improvement in the objective function at each discrete step which will lead to the minimization of the objective func tion. Ideally, it is desired to obtain a maximum increment in cost reduction accompanied by a minimum increase in job time for each job. Another way of stating the ideal cri terion is that a discrete move should be made which achieves maximum cost reduction accompanied by a minimum step towards infeasibility. In general, there will not exist a = CUj L “ ®u(i-l) i© maximum and also a AU^ij = * ^'(i-Dj which is minimum for all values of j. However, the rate at which Z is reduced with respect to increases in job time can be maximized at each step. In measuring the degree to which job time is increased with cost reduction, the absolute amount ^■^u'ij cannot toe used, since the significance of is dependent upon the amount of time remaining before infeasibility is reached. This time is equal to V I (23) a t , = I, - 2 S Vil» 3 = 1 <1> a 3 3 u=l i=l 3 In order for feasibility to be maintained, in the primaiy constraints at each move, (Sf) s i> i - i (i) a 30 The most critical constraint is the one for which (21 *) is maximum. It is desired to maximize the rate of cost reduc tion with respect to the rate of increase in time of the most critical constraint. The criterion for choosing u, (i - 1) is then Equation (25) is equivalent to moving in the direc tion of the maximum rate of cost reduction which is con sistent with maintaining feasibility. The concept is analogous to the gradient vector of a function, which points in the direction of maximum local increase (rate of change), but modified to take into account the constraints 11 of the problem. The operation of the algorithm is as follows: 1. Start with the maximum cost solution, if feasible. 2. Compute AC^i s C ^ - for each u, where i > 1. 3. Compute = l n ^ _ V(i-l)jl for eaoh u and j of the primary constraints. U I Compute ATi = T. - ^ for each i. J J u—1 i=l J (25) 1 ^Thomas 1. Saaty and Joseph Bram, Nonlinear Mathe matics (New York: McGraw-Hill, 196*+), pp. 70-75• 31 5. Compute — for each u and j. a t • 3 s ^ - D j i * ■ ? i n 6. Select i±-±. « L - 1 = max. For those u in ASj- / \ JY f • j which — 3 ^ 1, for one or more j, mark a Tj 7 0 7 infeasible and eliminate from further consider ation. 7. For those AD^,^ which cause an infeasibility in an auxiliary constraint, mark temporarily infeasible. It is possible that this infeasi bility condition will subsequently be removed, when a change has been made in a dominant unit. 8. Choose u, (i-1) for which, - ■. a max. - * ( 4 3 0 9. Remove any temporary infeasibilities which are dependent upon Step 8 for their removal. 10. Repeat Steps 2 to 9 until further discrete moves are either infeasible or i = 1. It is possible to construct situations in which choosing those moves which lead to the largest cost reduc tion in each step, independent of the rate of closure on the critical constraint, will lead to an optimal solution when the above algorithm will not. These are situations in which there are only a few steps required in order to obtain an optimal solution. However, the algorithm has produced an optimal solution in all of the several problems to which it has been applied thus far* The algorithm is presented as an approximate method based on the common sense approach of making the most efficient "long run" move at each step* The Dual Problem Ttom equations (13) to (17), the primal can be put into the following form: II I (26) Minimize Subject tot U I (27) u s u'j j =1 (1) m 0, 3=1(l)m I (29) 2 *ui 2 1 i=l I (30) “2* tui - - i, j = 1 (1) m» i=l (3D = (0, 1), u = 1 (1) U, u = 1 (1) I The corresponding dual problem is the following: m m" (32) Maximize Z * - T. P, + 'T P», 3=1 3 3 3=1 3 Subject to: (33) -Z ViJ h * 2 Vij + p"3 - °ai. 0 J —1 33 u = 1 (1) U, or i = 1 (1) I (3*0 “ ^7 Ai*ij + ” Gui* J=X J S1 u = 1 (1) U, or i = 1 (1) I (35) P-j 2 0, P'.. - 0, F*1. . (unrestricted in sign) where P. is the dual price of the primary time constraints, P*j is the dual price of the auxiliary time constraints and P"j is the dual price of the equipment constraints, m is the number of primary time constraints, m* is the number of auxiliary time constraints and m" is the number of equip ment constraints. The dual prices are not restricted to integer values unless all and Tj are integer. The P''j s 0 in order that Z be positive and that the left hand side of (33) and (3*+) equal for ty^ = 1. The primal problem is concerned with the minimiza tion of computer equipment cost. The dual problem is con cerned with the maximization of the value of the resources time and equipment. The value of the resource time is given by the dual prices Pi and P%, in units of dollars J j per unit time, corresponding to the primary and auxiliary 30b time constraints, respectively. The P^ is the con trollable price, since it is related to the amount of time ' ' 3^ used in the primary constraints. The is dependent upon the primary constraints. Its value is not of direct con cern, since it is the price of using a resource at zero level and does not enter into the determination of the objective function. The equipment constraint represents a fixed resource in the sense that one of each type of com puter units must be used. The P"^, in units of dollars, are not subject 1 d direct control. The value of the objec tive function (32) is maximized by minimizing the value of Pj. In ordinary linear programming, the value of P-j would represent the cost increase which would result from the use of one less unit of the resource Tj • However, as 12 pointed out by Gomory and Baumol, the dual prices in an integer linear programming are equal to and not ^ ^ui ., as the case in ordinary linear programming. A dis- ^ Tj crete step in time, must be made, corresponding to that required for an integer solution. The AT*. also 4 corresponds to the change required for a new basis. The expression P^ &T*^ represents the increase in cost con tributed by a decrease, ATj, in the time required to per form job j. Any decrease in job j time AT-j s 4 Tfj will result in the full cost increase, P^4T*^. The total increase in cost corresponding to all jobs is given by 12 R. E. Gomory and W. J* Baumol, "Integer Program ming and Pricing." Econometrica. 2XVIII. No. 3 (Julv. I960), 531. ------------- 35 (36) AZ = 2 ^AT* 3=1 J 3 13 As also mentioned by Gomory and Baumol, the dual prices in an integer linear program will be peculiar in several respects. The prices will vary with the choice of the additional constraints which are required in order to ; obtain an integer solution. Prices will be imputed not to : scare resources of the original problem alone, but to the additional constraints as well. Some resources, which were not fully utilized in the original problem, may be fully utilized in the integer solution and possess a dual price greater than zero• The reverse situation may occur with other resources. In general, there will be slack in all constraints in the optimal integer solution of (13) to (17) where, in the original problem, there would have been one or more zero slack constraints. This would imply that all Pj = 0 and the prices originally imputed to the scarce T^ have been imputed to the additional constraints in the integer solution. This problem is resolved by considering the slack, which has been generated as a part of the integer solution, as absorbing part of the resource T^, in order to satisfy the integer solution conditions. This procedure is implemented by treating the generated slacks of the 13Ibid.. p. 528. 36 previous no-slack time constraints as non-slack solution variables in the integer solution. These ''slacks’ 1 become a part of the artificial constraints in order to achieve an integer solution. When viewed in this manner, a constraint which is binding in the non-integer solution is still bind ing in the integer solution. The same solution would be obtained if the right hand side of the constraints in ques tion were reduced by the amounts of these ’ ’slacks" prior to obtaining the optimal integer solution. The above treat ment of the problem allows dual prices, Pj > 0, to be generated for constraints which would be binding in the integer solution, if it were not for the integer solution condition. These dual prices permit the change in cost to be calculated from (36) when moving from one optimal solu tion to another. The > 0 also indicate which jobs cause the increase in cost, when less of the resource time is used. The use of dual prices permits control to be exer cised over the cost of jobs. Any reduction in the adjusted time requirement of job j (adjusted to account for the "slack" required in an integer solution) ^ Ti - AT*.. will " J result in a cost increase of P^AT'^. The time requirement of job j should not be reduced by an amount AT^ unless a cost reduction in clerical activity at least equal to PjAT'^j can be achieved. The time constraint of jobs for which Pj = 0 can be reduced by an amount equal to the slack 37 existing in those jobs, without incurring additional cost. In the optimal solution, the net value of each com puter unit used in the optimal solution must be equal to the cost of that unit. If its value were less, it would not pay to use it. If its value were greater, the solution would not be optimum; some other lower priced unit should be used. The net value of the computer units not used in the optimal solution must be less than the cost of the unit; otherwise, it would pay to use such unit. These con ditions are expressed in equations (33) and (3*f). A nega tive value is imputed to the time resource of dominant con straints; cost is increased when less of the resource is used. A positive value is imputed to the equipment resource; cost would be decreased if the fixed restriction of having to use the unit in question could be removed. The net value is equal to the difference in value of the two resources. Multi-shift Formulation The objective function in (17) must be modified to account for overtime rental charges, if it is possible for one or more units to be used more than one shift. The usage of each unit is determined separately and an overtime charge begins when each unit*s usage exceeds the allowable first shift time. The modification to the objective func tion (17) need not be made in those situations where 38 (37) 2 ®'j^j S ^ where Tf^ is the time adjusted for j —I ll+ estimated chargeable non-productive time, f^ is the frequency of processing job j per month and H is the allowable hours per month in the first shift. When (37) does not hold, it is necessary to deter mine for each unit u and i, whether the following relation ship exists. J R (38) 2 ^ Ayjj 2^^ ~ H u=l(l)U, islCDl where the 3“1 r=l left side of (38) has been adjusted for chargeable non-productive time. That is, it is determined in advance, for each unit and model, whether overtime charges will accrue if the given model is used in the optimal solution. If (38) does hold for a given unit and model, the cost coefficient for that u and i is modified as follows: (39) cui + Aurj Xumi " where ^ Is the percentage of the first shift hourly rent charged for extra shift time on unit u. In the above formulation it is assumed that, if usage is not continuous in a given run, the usage can be Chargeable non-productive time consists of all time lost due to errors on the part of the user: applica tion program error, operator error or input error. 39 metered for each unit. The use of meters allows the user to he charged for active time only. There may be a great difference in the time that a given unit must be available in a run and the active time of that unit. Computer manu facturers will permit recording of charge time on an actual usage basis, with the exception of the central processor; the central processor is charged with the entire run time* At the present time, overtime charges are not sig nificant because the overtime rate *u for major computer systems has been recently reduced from **0 per cent to 10 15 per cent for all computer units comprising the system. Multi-programming Multi-programming is that process whereby two or more programs are run concurrently. The programs may con sist of runs from different jobs or runs from the same job. In the latter case, the runs must not be sequence dependent. Multi-programming may reduce equipment operating costs by making better utilization of the central processor and input-output units. By making use of equipment which would otherwise be idle, multi-programming may result in more jobs being completed in a given interval of time than would be the case in single program operation. A greater number of jobs per unit time is important where the workload is 1^IBM 360 Price list h- 0 expanding rapidly, and additional or upgraded equipment would be required in order to handle the increased work load. The advantage of multi-programming is provided by the ability to run one or more jobs in a time less than the sum of the individual job times on a single operation basis. Another advantage of multi-programming stems from the fact that the most expensive unit in a computer system, the central processor, is idle during a large part of a single program operation, except in those programs which involve a large amount of computing and a small amount of input-output. Idle time is caused by the great speed dis parity between the central processor and the input-output units. The central processor is idle while it waits for input-output operations to be completed. In multi programming, the processor can be executing instructions in one program while an input-output function is being per formed in another program. However, if not used with discretion, multi programming may result in a more expensive operation than would be the case -under single program operation. Although, in general, the use of multi-programming will result in less total computer time, it may require a faster and more costly configuration in order to maintain the individual job time constraints. This situation is illustrated in the following examples 1. The constraint time of Job 1 and Job 2 is 3 hours and 2 hours, respectively. 2. The time required under single program opera tion for Joh 1 and Job 2 is 3 hours and 2 hours, respectively. The total computer time is 5 hours. 3. With multi-programming, Joh 1 is completed in if hours and Joh 2 is completed in 3 hours. The total computer time is if hours, since the two jobs are run concurrently. if. However, the two job constraints have been violated, and a faster system would be required under multi-programming in order to maintain feasibility. The above situation could be caused by a selection of jobs to be multi-programmed which, due to the nature of the hardware requirements of the two jobs, causes delays in processing which would not occur if the jobs were run separately. Ideally, multi-programmed jobs should not compete constantly for access to input-output channels and memory. Jobs should be run concurrently which do not depend on the completion of an operation in one program before an operation in another program can begin. The model in (13) to (17) can be used to analyze the effect on the optimal solution of a multi-programmed operation versus a single program operation. In the multi-programming case, modifications and additions to constraint equations (1*+) and (15) are made in order to account for the additional timing relationships which are operative. CHAPTER II APPLICATION OP THE HARDWARE MODEL General An application of the model described in Chapter I is presented below. Prior to each application of this model, the following parameters must be determined: 1, Configuration in term3 of types and number of computer units to be employed. 2. Pile sizes, card input volumes, print volumes, estimated number of memory references and record sizes. 3• Memory size. if. Number of input-output channels. 5. Reciprocal of effective transfer rates for applicable modes of operation. These values are calculated from the formulas in the section on Input Data for Hardware Model, beginning on page 75, Each application of the method will provide a local optimal solution for given values of memory size, block sizes, and number of input-output channels. The above parameters may be constrained by the nature of the ^3 application to fixed values or they may be variable. The minimum segment of memory will be determined by the maximum program size. Additional memory capacity is added accord ing to the requirements of input-output and work areas. The block size can be varied in order to determine which combination of block sizes and total memory size results in the lowest cost system. Block sizes above 2,500 characters produce little improvement in tape transfer rate. This upper limit on block size plus maximum program size pro vides the upper limit on memory size, and maximum memory size provides the lower limit. These considerations usually limit the number of memory size possibilities to one, two or three at the most. The optimal number of input-output channels and input-output control units required is con strained by the input-output characteristics of the problem and the cost of increased simultaneity. These parameters can be varied in the application of the model. Usually, no more than two input-output channel possibilities need be considered in a given application^) • The following sections are presented in order to indicate how a problem would be formulated and solved using the method described in Chapter I. Figure 1 is a flow chart of three jobs under consideration. It is desired to select a system which will meet the three job time con straints at minimum cost. This model can be applied to J O B 1 Run 1 Input Transactions Card to Tape Transactions J O B 1 R un 2 Sorb Sorted Transaction C P 68 M M em ory Cycles CP Tape Limited 2 W ay Balanced erge CR 20 K Cards T U 3 M C h ar. B-2500 Double Space ^ l&jpT* J O B 1 Run 1 * Sorted Reports W ork Tapes J O B 1 Run 5 Print Reports C P 63 M M em ory Cycles C P Tape Limited 2 W ay Balanced M erge Output y p 6 M Char. -------> Double Space 132 W ork Tapes 21 Tapes Input J O B 3 R un 1 Collate J O B Collated File 333 M M em ory Cycles .(2-yay Merge) C har, 25m W ork Tapes Fig. 1 .—Application Flow Chart 1 Run 2 Sort W ork Tapes 1 Run 5 Print 63 M M em ory Cycles Sorted Transactions J O B 1 Run 3 F ile Maintenance Reports C P Tape Limited 2 W ay Balanced Merge----- 2,999 M M em ory Cycles T U 3 M C h a rJ B=2500 T2 700 M Char* 2500 \B=2500/ Old Master N ew Master Output Input J O B 2 Run 1 Confutation , P 6 M Char. ------- » Double Space 132 Output ) 1216 M M em ory Cycles P li M Char. DoubleSpace 132 p p j LEGEND Collated F ile J O B 3 Run 2 Print 28u M M em ory Cycles Output P 8.32 M Char. )ouble Space 132 C R Card Reader C P Central Processor P Printer T1 T « q ? e Unit 1 T2 Tape Unit 2 T3 Tape Unit 3 T b Tape Unit h K Thousands M Millions pp Print Positions B Block Size in Characters 1 .—Application Flow Chart h6 each manufacturers system under consideration. The steps in setting up the problem are the follow ings 1. Establish mnemonic designations for the various types of computer units and models, corres ponding to the solution variable t^. This is done in order to identify readily the types of computer units and models, 2. From the flow chart in Figure 1, determine the form of the primary and auxiliary time con straints . 3. Calculate the technological coefficients of the problem from the application specifications. (Input Data for Hardware Model.) k» Insert the technological coefficients in the technological coefficient matrix, Table 1. There is an entry in the matrix corresponding to the use of each computer unit for each mode of operation, 5. Calculate the effective transfer rates for each unit and mode of operation. (Input Data for Hardware Model.) 6. Transform the technological coefficient matrix into the processing time matrix, Table 2, by multiplying the technological coefficients in TABLE 1 TECHNOLOGICAL COEFFIC i Row Name Card R ead er CR* M o d e o f Op e : FM ® S o r t. FM ® B*2500 B— 2500 B=100 FM* S ort, B -2500 B -2 # )0 b TAPE UNIT 1 TAPE UNIT 2 T lS T lb T lS T26 T2“ JOB l c 20 5*3 5*3 RUN 11A C - 2 0 2 RUN U B C -2 0 RUN 11C C l ro o RUN 31A C 2 700 RUN 3 2 B c RUN 5 lA C RUN £1B JOB 2d RUN 12A d I* RUN 12Bd JOB 3 e 38*7 3 8 .7 RUN 13Ae -3 8 * 7 - 3 8 .7 RUN 23Ae RUN 23Be ^ F i l e m ain ten an ce* “B lo c k s i z e i n c h a r a c te r s * dJOB 1 t e c h n o lo g i c a l c o e f f i c i e n t s . “ JOB 2 t e c h n o lo g i c a l c o e f f i c i e n t s . ® JOB 3 t e c h n o lo g ic a l c o e f f i c i e n t s . Thousands o f c h a r a c te r s r e a d b y c a r d r e a d e r . ^ M illio n s o f c h a r a c te r s p r o c e s s e d i n f i l e m a in te n a n c e. “M illio n s o f c h a r a c te r s s o r t e d . ^ M illio n s o f c h a r a c te r s p r in t e d . J M illio n s o f p r o c e s s o r memory r e fe r e n c e s * TABLE 1 ' X? 'EC HNOLOGIGAL COEFFICIENT MATRIX Mode of Operation Central Processor cpJ EM® Sort B-2500b B-2^00b FM3- Sort B -2500b B— 2500 EKa Sort B-2500 B— 2500 TAPE UNIT 2 TAPE UNIT 3 TAPE UNIT h Printer T26 T2“ T3B T3L Tk* TAh 5 .3 700 720 5 .3 -7 2 0 -7 2 0 1 .5 5 .3 - 1 .5 - 1 .5 1 .5 6 .2 - 6 68 2999 63 u 1216 -1 2 1 6 -1216 3 8 .7 -38.7 3 8 .7 - 3 8 .7 38.7 -38.7 3 1 .5 8 .3 2 - 8 .3 2 - 8 .3 2 527 281* n te n a n c e TABLE 2 P R O C E S S IN G - T IK E M A T R IX (65,000 Characters of M em ory, 2 Input-Output Channels R ow N am e Card Readers Tape Unit 1 Tape Unit 2 Tape Unit 3 C R 1 C R 2 T il T12 T13 T21 T22 T23 T31 T32 T33 C O S T l a 260 320 578 728 1028 578 728 1028. 578 728 1028 *J0B l b, R U N UA? R U N U $ R U N llC? *R U N 314? R U N 31R? R U N s i r R U N SlB 2000 iSoo -2000 -1S00 -2000 -iSoo -2000 -iSoo 167 83 56 79 lilt 32 79 14; 32 136 83 56 2714*0 15260 10920 28397 35779 11288 -28221; -15696 -11232 -28221; -15696 -11232 JO B 2C #R U N 12AC *R U N 12BC 92k 52k 368 tfJO B 3d - R U N 13Ad R U N 23Ad R U N 23Bd 1518 814; 6oU -1518 -8UU - 601; 1518 814; 601; -1518 -814; -601; 1518 814; 601; -1518 -814; -601; * C R e * Tle * T2e * T3e * Tl;e * P ® * C P e 1 1 1 1 1 1 1 1 1 1 1 a0bjactive function. bjO B 1 time constraints. °J0B 2 time constraints. “J O B 3 time constraints. ^Equipment constraints. ^Seconds. ^Constraints marked with an asterisk are required constraints. TABLE 2 O C E S S IN G T IM E M A T R IX 2 Input-Output Charnels and 1 Tape Controller) Tape linit 3 Tape Unit 1 ; Printers -J — ...j — ■ - ...'i.n i'a -i ■ ! 1 i n 1 m ix Central Processors T IM E T31 T32 T33 TU Tli2 T U 3 PI P2 CPI C P2 C P3 578 728 1028 578 728 1028 1375 1575 10l5O 57U O 9750 28397 15779 11288 226 ll|6 80 U716 2652 -28800^ s 0 158 89 * 0 102 85 31; ■ & 0 -28221; -15696 -11232 - 59 - 33 - 2it * 0 -28221; -15696 -11232 - 59 - 33 - 2U bk99 37 h9 1500 s 0 59 33 2k -2652 * 0 -1;716 -2652 95 79 32 - 0 1821; 1520 608 S 3600f -1821; -1520 -608 « 0 3110; 1768 -1821; -1520 -608 £ 0 1518 810; 601; 1518 810; 601; 6510 3678 S 7200* -1518 -810; -601; -1518 -810; -60 I; 791 6 59 261; * O 1235 687 1;92 -65Uo -3678 - 0 -6510 -3678 U26 355 ll;2 m 0 1 1 1 m 1 m 1 m 1 1 1 1 * X 1 1 8 1 1 1 1 m 1 **9 Table 1 by the reciprocal of effective transfer rates in Tables 9» 11, 12, and 13 in the sec tion on Input Data for Hardware Model and form the sum of the elements applicable to each model• 7. Insert computer unit rental rates from the sec tion on Input Data for Hardware Model in the objective function, C 0 S T 1, Table 2. 8* Insert the right hand side time constraints from the section on Input Data for Hardware Model in Table 2* 9* Solve for t ^ (0, 1) by the methods described in Chapter Is integer programming, parametric programming and discrete programming. Unit and Model Designations Unit u Unit Mnemonic Model i Unit-Model Mnemonic Card Reader 1 CR 1 CR1 2 CR2 Tape Unit 1 2 T1 1 Til 2 T12 3 T13 Tape Unit 2 3 T2 1 T21 2 T22 3 T23 Tape Unit 3 *+ T3 1 T31 2 T32 3 T33 Tape Unit 5 T*f 1 T*fl 2 T*f2 3 T*+3 50 Unit u Unit Mnemonic Model i Unit-Model Mnemonic Printer 6 P 1 PI 2 P2 Processor 7 CP 1 CPI 2 CP2 3 CP3 Constraints 1. Job 1 Job 1 contains the time factors which determine JOB 1 time: HUN 11 - Card read*time. RUN 21 - The sum of tape merge time. RUN 31 - New Master tape time + report tape time. RUN Vl - The sum of tape merge time. RUN % - Print time. 2. RUN 11A Tape output time 5 Card read time. 3. RUN 11B Print time - Card read time. b, RUN 11C Compute time - Card read time. 5. RUN 31A Transaction tape time + old master tape time £ New master tape time + report tape time. 6. RUN 31B Compute time - New master tape time + report tape time. 51 7. RUN 51A Report tape time - Print time. 8. RUN 51B Compute time - Print time. 9. JOB 2 Compute time determines JOB 2 time. 10. RUN 12A Input tape time - Compute time. 11. RUN 11B Print time - Compute time. 12. JOB 3 JOB 3 contains the time factors which deter mine JOB 3 times RUN 13 - The sum of tape merge time. RUN 23 - Print time. 13. RUN 13A Compute time - The sum of tape merge time. 1^. RUN 23A Input tape time - Print time. 15. RUN 23B Compute time - Print time. It should be noted that only the constraints marked with an asterisk in Table 2 are required as input to the linear program, since the remaining constraints will be satisfied by any selection of the applicable units. The above constraints are based on the use of two input-output channels for tape operations, time-shared card read and print operations and a computer system capable of concurrent read, write and compute. Thus, in a given run, card reading, printing, two tape operations and computing may be time shared. It is assumed that the sort runs, ROT 21 and ROT 1*1 are tape limited throughout. Problem formulation 1. Integer Programming Method The solution obtained by employing the Gomory Method of integer programming is shown in Table 3* There are two computer unit groups, T1 and T2, in which frac tional values exist in the first optimal solution. There is zero slack existing in JOB 3 and ROT 31A in Solution 1. Therefore, slack can be introduced into these constraints by the process of Til or T21 becoming zero. Slack cannot be increased by either T12 or T22 becoming zero. The optimal tableau of Solution 1 is examined in Rows T12 and T22 and Columns JOB 3 and ROT 31A (names of slacks in Row JOB 3 Sow ROT 31A). By using equation (18), two possible cuts are formulated: .0011*9 JOB 3 “ .182, Row Til .00823 ROT 31A - .03*+» Row T21 Since Row Til contains the larger cut, the following addi tional constraint is added to the original problem: .0015 JOB 3 ~ 0 T 11 = .18, where C T 11 is a new slack TABLE 3 INTEGER PROGRAMMING METHOD p R I M A L S C L U T I 3ard Readers® Tape Unit l a Tape Unit 2a Tape Unit 3a Tape Unit Solution C R 1 C R 2 T il T12 T13 T21 T22 T23 T31 T32 T33 Till T l*2 l d 1.00 0 .182 .818 0 .031* .966 0 0 1.00 0 0 1.00 2d 1,00 0 0 1.00 0 .035 .965 0 0 1.00 0 0 1.00 3® 1,00 0 0 1.00 0 0 1.00 0 0 1.00 0 0 1.00 i* f 1,00 0 0 1.00 0 0 1.00 0 0 1.00 0 0 0 5f 1.00 0 0 1.00 0 0 0 1.00 0 1.00 0 0 0 6f 1.00 0 0 0 1,00 0 0 1.00 0 1.00 0 0 0 7f 1.00 0 0 0 1.00 0 0 1,00 0 0 1.00 0 0 * D U A L S O I U T I 0 JOB 1 JOB 2 Solution Job Time Constraint (Min.) Slack Time (Min.) Job Timeb (Min.) Dual Solution Job Time Constraint (Min.) Slack Time (Min.) Job Til (Min, l d 1 *80.0 13U-G 31*6.0 0 60.0 29.6 30,1 2d 1*80,0 13lw3 31*5.7 0 60.0 29.6 30.1 3e 1*80.0 13iu3 31*5.7 0 60.0 ■i m aa mm 29.6 30.1 I** l*6l*,0 119.1* 31*1*.6 0 58.0 27.6 30.1 5f 10*8.0 103.8 3UU.2 0 56.0 25.6 30.1 6f 1*32.0 88.3 31*3.7 0 51*.o 23.6 30.1 7f 1*16.0 11*7.1 268.9 0 52.0 21.6 30.1 f"l" indicates selection } "0" indicates non-selection. “Time required to complete job. °Value of objective function. “Infeasible solu tion s. ®Flrst fea sib le solu tion obtained by integer programming. ^€asik le solutions obtained by parametric lin ear programming (se n s itiv ity analys .e c r itic a l job. - TABLE 3 SGER PROGRAMMING METHOD 53 ii . .S O L U T I O N S Lt 3a Tape Unit l;a Printers a Central Processors6 JO B 1° JO B 2b JO B 3b Monthly T33 Tlil Tli2 Tli3 PI P2 CPI C P2 CP3 T IM E T IM E (Min.) (Min.) T IM E (Min.) Rental 0 0 1.00 0 0 1.00 1.00 0 0 3U6.0 30.U 120.0 $ 9,165 0 0 loOO 0 0 1.00 1.00 0 0 315.7 30.1; 118.0 9,192 0 0 1.00 0 0 1.00 1.00 0 0 315.7 30.lt 117.6 9,197 0 0 0 loOO 0 1,00 1.00 0 0 3kk»6 30.lt 113.6 9,k97 0 0 0 1.00 0 1.00 1.00 0 0 310;. 2 30.lt 109.6 9,797 0 0 0 1.00 0 1.00 1.00 0 0 3k3.7 30.lt 105.6 10,097 1.00 0 0 1.00 0 1.00 1.00 0 0 268.9 30.it 101.6 10,397 S O I U T I O N S JOB 2 JOB 3g Slack Time (Min.) Job Tim e*3 Dual (Min.) Solution Job Time Constraint (Min.) Slack Time Job Tim e*3 (Min.) (Min.) Dual Solution (Dollars/Min.) 29.6 30.li 0 120.0 0 120.0 $13.35 29.6 30.li 0 120.0 2.0 118.0 13.35 29.6 30.1; 0 120,0 2.1; 117.6 75.00 i t i 1 CM 1 t 30.1; 0 116.0 2.k 113.6 75.00 25.6 30.li 0 112.0 2.1; 109.6 75.00 23.6 30.li 0 108.0 2.1; 105.6 75.00 21.6 30.li 0 10k ,0 2.1; 101.6 75.00 ing (se n sitiv ity an alysis). variable. In Solution 2, Til has become zero and T12 is equal to one. Two minutes of slack have been introduced into the JOB 3 constraint (JOB 3 = 2 minutes) and C T 11 = Q Constraint EOT 31A now contains slack. Constraint C T 11 is the only constraint containing zero slack. Therefore, the following cut is formulated from Row T21 and Column C T 11 of the optimal tableau of Solution 2: .0001 0 T 11 2 .035 or .0001 C T 11 - CT 21 = .035 An optimal integer solution is achieved in Solution 3 by T21 being driven to zero and the slack in JOB 3 becoming equal to 2.b minutes. Additional solutions are obtained by performing parametric linear programming on the three job constraints, \ f P960I JOB 1, JOB 2 and JOB 3 5 with a change vector V = “120 in ( 72^-0 minutes, equal to a 3*33 P©r cent change in each constraint. These solutions are numbered *f, 5, 6 and 7 in Table 3* JOB 1 and JOB 2 contain slack for each solution obtained. The dual price, Ej? is zero for these two con straints in all solutions. These two jobs are not binding in either the non-integer or integer solutions. JOB 3 is binding in the non-integer solution. Slack is generated in order to satisfy the integer solution conditions. The same solution would be obtained if the slack were subtracted from the constraint prior to solution. The slack would be zero and the job time identical to the situation where slack exists. Therefore, JOB 3? is binding and possesses a dual price greater than zero. The dual prices multiplied by the difference in the job time of two successive solu tions equals the difference in the value of the objective function for the two solutions ($75 per minute x 4- minutes= $300). JOB 3 is the job which is responsible for the increase in cost in each solution, as the constraints are reduced at equal rates. The same set of solutions is obtained if the JOB 3 constraint alone is reduced at the rate used when all three jobs are reduced simultaneously. Therefore, if it assumed that all job time constraints will be reduced by approximately equal rates in the future (equal job priorities), JOB 3 will govern the increase in cost at the rate of $300 per month per k minute decrease in JOB 3 time. The reduction in JOB 1 and JOB 2 time con straints will have no effect on the increase in cost. JOB 3 is the control job, and increases in computer system performance (changes in optimal configuration) will have to be justified by clerical cost reductions at least equal to $300 per month for each minute reduction in JOB 3 comple tion time. If it is anticipated that jobs will be reduced at unequal rates in the future, the appropriate parametric linear programming analysis can be performed. 2. Parametric Programming Method A second method of solution is to employ parametric linear programming on the three job time constraints with a -960 in seconds multiplied by the param- -120 -2*f0 change vector V = eter 9. The assumption is that the three jobs will decrease by approximately equal rates in the future and it is desired to identify a series of optimal solutions as a function of 0. The series of solutions obtained by this method, and the corresponding values of 9, are shown in Table Opti- U mal solutions are chosen on the basis of 21 t ui» = 3? u-1 being a maximum for a given group of t ui, and (U-F) < e, where e is an arbitrarily small positive number* In this problem, all solutions chosen by the above rule are integer solutions* In general, this would not be the case* The value of 0 is increased until the problem becomes infeasible. The same value of the objective func tion is obtained for each of the optimal integer solutions as in the ease of integer programming. In some cases, a different optimal configuration is obtained, because alternative optimal solutions exist. The change in the objective function from one solution to the next is given m r by A Z s &9 ^ V j Pj = 1 96°) (0) + (-120) (0) + (-2^0) (1.25)j = -$300. It is noted that, although Solu tion 6 in the two methods yields identical values of the objective function, Solution 6 of the parametric method requires 83 minutes less time for completing JOB 1. 3* Discrete Programming Method The method of discrete programming is shown in TABLE i t PARAMETRIC PROGRAMMING P R I M A L S 0 L U T lard Readers1 5 Tape Unit l b Tape Unit 2b Tape Unit 3b Tape Unit SolutioE e a C R 1 C R 2 T il T12 T13 T 2 1 T22 T23 T31 T32 T33 Tltl T R 2 l f 0 1.00 0 .182 .818 0 .03lt .966 0 0 1,00 0 0 1.00 2f .31 1.00 0 0 1,00 0 .033 .963 0 0 1,00 0 0 1,00 3* .61 1.00 0 0 1,00 0 0 1.00 0 0 1,00 0 0 1,00 lth 1.61 1.00 0 0 1.00 0 0 0 1,00 0 1.00 0 0 m a 1.00 3h 2.61 1.00 0 0 0 1.00 0 0 1,00 0 1,00 0 0 1.00 6* 3.61 1.00 0 0 0 1.00 0 0 1.00 0 0 1,00 0 1,00 ?h It,61 1.00 0 0 0 1.00 0 0 1.00 0 0 1.00 0 0 D U A L S () L U T I JOB 1 JOB 2 Solution e a Job Tim e Constraint (Min.) Slack Tim e (Min.) Job T im e* * (Min.) Dual Solution Job Time Constraint Slack Time J (Min.) (Min.) 1^ 0 it 80.0 13it.O 3lt6.0 0 60.0 29.6 2f .31 lt71.8 126.1 3it5.7 0 39.0 28.6 38 .61 lt70.3 12it.6 3it3.7 0 38.8 28.lt kh 1.61 It3h.3 109.0 3lt3.3 0 36.8 26.lt 3h 2.61 It38.3 93.3 3ltlt*8 0 3lt.8 2lt.lt 6b 3.61 U22.3 132.3 270.0 0 32.8 22.lt 7h I t .61 U06.3 137.lt 268.9 0 30.8 20.lt Value of change vector parameter* "I" indicates selection; "0" indicates non-selection. ®Sum of the maximum value of solution variable in each computer unit category* aTime required to complete job. ®Value of objective function. ^Infeasible solutions. ®Pirst feasible solution. “Additional feasible solutions (sensitivity analysis). TABLE k PARAMETRIC PROGRAMMING METHOD 57 R I M A L S 0 L U T I 0 N S Tape Unit 3b Tape Unit ltb Printers1 * Central Processors^ J O B l d J O B 2d J O B 3d T IM E T IM E T IM E [Min.) (Min.) (Min.) M onthly® Rental T31 T32 T33 T U I T U 2 tU3 PI P2 C P I C P 2 C P 3 F° 0 1,00 0 0 1.00 0 0 1.00 11.00 0 0 6.79 3lt6.0 30.lt 120.0 $9,165 : 0 1,00 0 0 1,00 0 0 1.00 1.00 0 0 6.97 3lt5.7 30.lt 118.0 9,192 i 0 1,00 0 0 1,00 0 0 1.00 1.00 0 0 7.00 3U5.7 30.U 117,6 9,197 0 1,00 0 0 1,00 0 0 1,00 1.00 0 0 7.00 3lt5.3 30.lt 113,6 9,lt97 0 1,00 0 0 1,00 0 0 1,00 1,00 0 0 7.00 3ltit.8 30.lt 109.6 9,797 0 0 1,00 0 1,00 1.00 0 1.00 1.00 0 0 7.00 270.0 30.lt 105,6 10,097 0 0 1,00 0 0 0 1.00 1.00 0 0 7.00 268.9 30.lt 101,6 10,397 d u a l S 3 L U T [ 0 N S JOB 2 JOB 3 Job Tim e Constraint Slack Tim e (Min,) (Min,) Job Tim ed Dual (Min.) Solution Job Tim e Constraint (Min,) Slack Tim e Job Tim ed Dual Solution (Min,) (Min,) (Dollars/Min,) 60,0 29,6 30.lt 0 120,0 0 120.0 $13.36 $9*0 28.6 30.lt 0 118,0 0 118,0 13.36 58,8 28.lt 30.lt 0 117.6 0 117.6 75.00 56.8 26.it 30.lt 0 113.6 0 113.6 75.00 5lt.8 2lt.lt 30.lt 0 109.6 0 109.6 75.00 52.8 22.U 30.lt 0 105.6 0 105.6 75.00 • CO 20.lt 30.U 0 101.6 0 101,6 75.00 ter unit category. Table 5, The highest cost feasible solution is used to start the process in Step 1. The entries in Table 5 are obtained from the processing time matrix in Table 2. Since any selection of computer units will satisfy the JOB 2 con straint, only the JOB 1 and JOB 3 constraints are analyzed* An entry marked IF(T) means that it is temporarily infeasi ble (violates an auxiliary constraint) and may subsequently become feasible* An entry marked IF is permanently infeasi ble (violates a primary constraint) and is removed from further consideration* The entries marked j represent the critical constraint of the given unit and the entries marked k represent the move made in the given step* Analysis of Solutions From an examination of the series of solutions pre sented in Table 3 and Table ^ (where it is assumed that any future reduction in job constraints will occur at equal rates), the following conclusions can be stated: 1. Adequate slack exists in JOB 1 and JOB 2 of Solution 3. A decrease in the JOB 1 constraint of up to 13^ minutes or a decrease in JOB 2 constraint of up to 29 minutes will not require a change in the optimal configuration. 2. The JOB 3 completion time is so close to the JOB 3 constraint in Solution 3* that it may be necessary to consider an alternative solution. TABLE 5 DISCBETE PROGRAMMING METHOD " (I)* ........... (2)b (3)c T5P™' " 1 W " (8>* (9)i fV.il A T i , A ! l ! d ^ C\ii ui A C^i ADa,il Z1TX ^ ' 1 3 * t3 m x ( ^ ) Step 1 CR2 60 5oo 13,168 .03783 0 1,106 0 l,587k T13 300 27 13,168 .00205 2tf0 1,106 .2173 1,382 T23 300 27 13,168 .00205 2*+0 1,106 .217 IK I) *33 300 if,if 91 13,168 .3**l3 2*+0 1,106 .217. 880 TH-3 300 66 13,168 .00501 2*+0 1,106 .2173 1,382 P2 200 2,06*+ 13,168 .158 2,862 1,106 2.588 IP CH - - — — - - - - Solutions CR2 ^ CR1 Step 2 T13 T23 *33 Ti+3 Solutic 300 300 300 300 jnt T13 -» a 27 27 -•-s 212 12,663 12,668 12,668 12,668 .00213 .00213 .3553 .00521 2*+0 2M-0 2^0 2^0 1,106 1,106 1,106 1,106 .2173 .217 .217. .2173 l,382k iK t ) 81+5 1,382 Step 3 T12 T23 *33 T^-3 Solutic 150 300 300 300 >n: Th- 3 ■* a 8*f 27 "'"S 2M-2 12,61+1 12,61+1 12,61+1 12,61+1 .00665 .00211+ .355 .00522 67*+ 2*+0 2^-0 2^+0 866 866 866 866 .7883 .277 .277. .2773 193 IF(T) 81+5. 1,083 VJl vo TABLE 5 (Continued) (l)a (2)b (3)c 0O d (5)e (6)f (7)S <8)& (9)1 aCui aDfci'il i3 fflflY ui a°ui A D ^ n a ’ V at3 h ) Step 4 T12 T23 233 T4-2 Solut: 150 300 300 i5o Lons T33 84- 27 b M i &*■ 2.32 _ . 12.575 12.575 12.575 12.575 .00669 .00215 .3 57 .00669 67*+ 2b0 2k0 67*f 626 626 626 626 1.076 •383. .3833 1.076 IS* IP(2) 8W IP Step 5 T32 I T23 1 Solutions 150 1 2,528 300 I 27 T23 T22 8,08*f 8,08*f 1.550 .0033^ 67M- 2**0 386 386 1.7^6. .622^ IP . 1*82* Cumulalfcive Job Time Step (Seconds) (Seconds) 6,09^ 6,09** 6 331 * 6^57** 6,8li* 7,0?* Complete Solutions 3^632 16,132 16,225 20,716 20,7^3 CB1, T12, T22, T32, T^-2, P2, CPI; Z = $9,197 per month 1 2 I 5 f-Computer unit and model. "Increment of cost reduction. ^Increment of JOB 1 time increase. dTime remaining Before infeasibility reached in JOB 1. TABUS 5 (Continued) ®Rate at which infeasibility approached in JOB 1, * Increment of JOB 3 time increase. STime remaining before infeasibility reached in JOB 3* “Rate at which infeasibility approached in JOB 3. iRatio of increment in cost reduction to maximum rate of infeasibility. ^Maximum of Col. (5) and Col. (8). “-Maximum of Col. (9) “JOB 1 time constraint (28,800) - Col. (**•). JOB 3 time constraint (7,200) - Col. (7). O ' - H An error in estimating the data processing workload for JOB 3» which would cause the JOB 3 completion time to he less than 118 minutes, would require a different optimal configuration. If a solution other than Solution 3 is contem plated, the optimal configuration should he determined on the hasis of saving at least $300 per month in clerical expense for each k minute reduction in JOB 3 completion time. If a constraint of less than 102 minutes is imposed on JOB 3> the system under considera tion will he infeasible. The choice of printer and central processor units is independent of the constraints. The tape unit models are the only units which must he changed, when changing optimal configura tions . An experienced computer analyst could prohahly obtain the optimal solution to this problem without using the model, because the number of time constraints is not large. However, if the number of time constraints is doubled, the problem becomes unwieldy for solution by trial and error methods. Model Validation The conclusions concerning the solution obtained apply only to the selection of the optimal configuration for a given system and the minimization of hardware costs per time period. The time period is one in which the tech nological coefficients do not change significantly. The complete solution involves the consideration of software selection and the analysis of costs, both initial costs and period costs, developed in Part II. The multi-stage analy sis in Part II takes into consideration the fact that operating costs will vary from period to period due to changes in the application parameters. The effect of purchase on the optimal configuration per time period would be negligible, if any, because the ratio of purchase price to rental for each computer unit is relatively constant. Of course, the question of purchase versus rental is important when considered over several time periods. This analysis is developed in Part II. It is a simple matter to validate the fact that Solution 3 is optimum. JOB 3, the binding job, cannot be performed in the required time if the slower printer PI is used with the highest speed tapes T13, T23> T33, and T^3 (Table 3)* Also, it is impossible to perform JOB 3 within the required time by using printer P2 and any slower tape unit than those in Solution 3« Memory size is constrained to 65,000 characters because *K),000 characters are required 6b in JOB 2 just for the program and operating system* The next lowest memory size available is 32,000 characters. The use of additional input-output channels in this system would require a central processor costing an additional $5,000 per month. Fewer input-output channels (0 or l) would result in an infeasible solution due to zero or insignificant input-output overlap. Multiprogramming and Efficient Use of Re source 3 The optimal solution provides useful information about the utilization of each computer unit and the poten tial for increasing the use of surplus resources via multi programming. In general, it is advantageous to incorporate multiprogramming into jobs where the required equipment is available and slack exists. If multiprogramming is used in non-slack jobs, it may require a higher cost system in order to maintain the original job constraints. The time used by each computer unit for each run is shown in Table 6 for Solution 3> Tables 3 and *f. In the case of the central processor, the time shown is the run time and not the program execution time. These times are accumulated for the purpose of determining overtime charges. The accumulation of use time in this manner is based on the assumption that meters are used to record productive time. It is the practice to charge the central processor for the entire run time, rather than the program execution time, since the central processor must be in the system at all times in order to control the operation of the system. The total usage in minutes per month is shown for each unit. The time for each unit is increased by a factor of 15 per cent to account for non-productive chargeable time. Over time charges begin when usage exceeds 200 hours or 12,000 minutes per month. As additional jobs are evaluated, the economics of single program, with possible use of overtime, versus multiprogramming is considered. In general, assum ing the availability of required equipment and memory, multiprogramming should be used until all slack has been exhausted. Once slack has been exhausted, the decision is not clear-cut. The problem must be resolved for single program operation and then for multiprogram operation and a comparison made of the resulting costs. An example of how the original optimal solution can be used to determine the practicality of multiprogramming is presented below. JOB *+ is a daily card to tape conversion operation with the following parameters: 1. 50 x 10^ cards £ 2. 5 x 10 tape characters 3. 165 x 10^ memory cycles *+. 9 >300 character memory requirement Prom an examination of Table 6, it is seen that a TABLE 6 COMPUTER UNIT UTILIZATION (Minutes) JOB COMPUTER UNIT RUNb TIME Joh 1 (Daily) CR1 T12 T22 T32 T*f2 P2 CPI Run 11 - Card to Tape 3lfa 1 3b 3b 39 Run 21 - Sort Trans. la la la ia b 5 Run 31 " Rile Maint. 1 255 263a la 26b 303 Run *fl - Sort Reports la la la ia b 5 Run 5l ~ Print l b5a b5 52 Total 3^ b 257 265 b 79 351 bck x 1.15 (Non-productive time allowance] 39 5 296 305 5 91 ifrOM- x 21 days 819 105 6,216 6,^05 105 1,911 S,b8b 3,b8b aUnits which determine run time. ^Non-productive allowance added. Ov TABLE 6 (Continued) JOB' COMPUTER UNIT HUNb TIME Joh 2 (Daily) CB1 T12 T22 T32 Tk2 P2 CPI Bun 12 - Computation 9 30 31a 36 x 1,15 (Non-productive time allowance) x 21 days 10 210 35 735 36 756 756 Joh 3 (Monthly) Bun 13 - Collate i5a l5a 15a I5a 60 69 Bun 23 - Print 12 62a 62 71 Total 15 15 15 27 62 122 1*K> x 1.15 (Non-productive time allowance) 17 17 17 31 71 l*+0 Total Minutes per Month 819 332 6,233 6,^22 136 2,717 9,380 9,380 aUnits which determine run time. ^Non-productive allowance added. ON card reader and tape unit are available on a daily basis during Run 51 of JOB 1 and JOB 2. Twenty-four thousand characters of memory are also available for other programs in each of these runs. It is assumed that the input for JOB b becomes available at l*f.80 hours and mast be completed by l6,30 hours, A schedule for the first three jobs is shown in Table 7. JOB 2 is scheduled between 1^.73 and 15*33 hours. If JOB *+ is run after JOB 2, as a single pro gram, there will be ,97 hours (16.30 - 15.33) available in which to complete JOB b» This time is equivalent to .81 * hours (.97 t 1,15) or 3*02*+ seconds of productive computer time. Card read time will be dominant in JOB b* Tape time and central processor time will easily be performed within card read time. The time required per 1,000 cards is 100 seconds for CR1 and 60 seconds for CR2. A change in the optimal configuration would be required if JOB b were ran as a separate job, since CR2 must be used in order to com plete JOB b within the required time. CR2 will complete JOB *+ in 3,000 seconds (60 x 50). The change from CR1 to CR2 will increase rental by $80 per month. The remainder of the optimal configuration will not change, since JOB 3 is the binding constraint and a card reader is not used in JOB 3. If JOB b is run as a multiprogrammed job with JOB 2, the following situation will prevail: TABLE 7 DAILY COMPUTER SCHEDULE Joh Hour Minutes JOB 3 Run 13 05.67 - 06.82 69a Run 23 06.62 - 08.00 7la JOB 1 Run 11 8.00 - 8.65 39 Run 21 8.65 - 8.73 5 Run 31 8.73 - 13.80 303 Run hi 13.80 - 13.88 5 Run 51 13.88 - 1^.73 52 JOB 2 Run 12 1^.73 - 15.33 36 aGnce per month. 70 An input tape, the central processor and printer will he operative in JOB 2. A card reader, the central processor and output tape will he operative in JOB Since there are two input-output chan nels, the tape operations will he overlapped, There is a time-shared channel to which the card reader and printer are attached. Thus, card read ing and printing will he time shared. The central processor is the dominant unit of JOB 2 and the card reader is the dominant unit of JOB *f. The memory cycles required in JOB *f will add to those of JOB 2 and will cause JOB 2 to run slower than it would as a single program. In a multiprogrammed mode, JOB if will he completed in 5,000 seconds (100 x 50) hy using CR1 or .81* - hours pro ductive time. This time is equivalent to .97 hours total time. JOB if will he completed at l*f.80 + .97 = 15.77 hours. JOB 2 time will be increased hy 165 z 1,5 * 2*f8 seconds (1.5 seconds per 10^ memory cycles for CPI) or 5 minutes of productive time. Since there are 29 minutes of slack existing in JOB 2, there will he no change required in the optimal solution. Thus, multiprogramming should he used. If JOB 2 had been a zero slack joh, a faster and higher cost configuration would have been required in order to satisfy the JOB 2 constraint. This cost might have exceeded 71 the additional rental of $80 required in order to run JOB !f as a single job. This multiprogramming example illustrated the prin ciple of minimizing cost by utilizing surplus resources to the extent permitted by the constraints of the problem. Summary and Oonclusions 1. The Hardware Model has been developed in order to select the optimal hardware configuration per time period for a given computer system. 2. The model can be applied to select the optimal configuration for batch processing applications in each of several time periods for every com puter system under consideration. 3. The method of integer programming provides an exact mathematical solution to a given problem formulation. However, the method of integer programming is not computationally efficient. if. Parametric linear programming is used in order to provide an approximate but computationally efficient method of solution. 5. The discrete programming method is also an approximate method, but has the advantage of operating in discrete values. Also, it involves relatively simple calculations. 6. The dual solution is useful for indicating which, jobs are controlling i.e., contribute to cost increases as job time constraints are reduced. 7. With a slight modification, the model can be applied to situations where overtime charges occur. However, the low overtime charges cur rently in effect exert a minor influence on the optimal configuration, 8. An important application of the model is its use in determining the effect of multi-program ming on the optimal solution, A frequently cited objective of multi-programming is the minimization of the time required to process a given batch of programs.1 This concept of optimization may be in conflict with the objec tive of satisfying the individual job time con straints at minimum cost. 9. A limitation of the model occurs in those cases where it is difficult to determine which com puter units will be dominant and which will be non-dominant in a given run. This difficulty is caused by the complex timing relationships which exist among computer units in a system in i E. F. Codd, "Multiprogramming," Advances in Computers. Vol. 3 (New Yorks Academic Press, 1962), pp. 78- which multiple input-output and internal pro cessing operations can take place concurrently. This problem is aggravated by the use of multi programming, where concurrent operations may take place among two or more independent pro grams. In many cases, the speed relationships and processing volumes are such that it is obvious which units are dominant. In other cases, this relationship will not be obvious and some trial and error procedures in the con struction of constraints will be necessary. Although the model does not completely elimi nate trial and error procedures, it does make the problem of computer selection manageable. INPUT DATA POR HARDWARE MODEL INPUT DATA POE HARDWARE MODEL Assumed Technological Coefficients. 1* Rim 1. Job 1 Card to Tape Conversion with, Message Printing Read 20*000 cards A m s 20 x 10 cards Write 20,000 (100 character tape items - 2.500 charac ter block) £ Aon = 2 x 10 tape characters (output tape) Print 2*000 (100 character messages - 132 character print line) A6ll ~ *2 x print characters (double space) Execute 100 instructions per card (average) 100 x 20^000 x 30 characters per instruction = 6 60 x 10 characters Data Movement Card Reader to Input Area = 2x10 Input Area to Work Area = 2x10 Data is representative of applications with which the author has been involved* The source of this data can not be revealed because of the proprietary nature of the data. 75 6 Work Area to Output Area or Printer = 2x10 Output Area to Tape = 2x10 68x10^ characters A1711 = 68x10^ processor memory references (characters) Program Requirement Input Areas 2 x 100 Print Areas 2 x 200 Output Areas 2 x 2,500 Work Area Instructions 1,000 Instructions (8 characters per instruction) Operating System Run 2. Job 1 Tape Sort In-put Transactions (Balanced 2-Way Merge) 65,000 character memoiy -10,000 characters required for sort program -16.000 characters required for operating system 39>000 characters for data Sort 20,000 (100 Character Items - 2,500 character "block) Expected String length in Pre-Sort = 2 x Memoiy Area Available for Data 2 x (39,000 r 100) = = 780 items per string 20,000 t 780 = 26 strings 200 bOO * 5,000 100 B 8,000 = 16.000 29,700 2-Way Balanced Merge Requires 2** < 26 *« 2^ 5-merge passes + 1 pass for pre-sort = 6 tape passes 6 tape passes t tapes =1*5 passes per tape 1.5 passes per tape x 2.10^ characters per pass = 6 3*0 x 10 characters per tape (2,500 character block) Ag2l = 3 x 10 tape characters per tape (work-tape) A^2l - 3 x 10 tape characters per tape (work tape) Alf2i = 3 x 106 tape characters per tape (work tape) A521 = 3 x 10° tape characters per tape (work tape) Processor Activity: Assume pre-sort is tape limited. If this is not the case, the difference in time between compute limited and tape limited pre-sort is negligible. Run 3* Job 1 Pile Maintenance Read 20.000 (100 character tape items - 2.500 character block) c . Ap^i a 2 x 10° tape characters (transaction tape) Read 280.000 (250 character tace item - 2*500 character block) 6 A331 = 700 x 10 tape characters (old master) Write 288.000 (250 character tape items - 2.500 charac ter block) Atfr3i = 720 x 106 tape characters (new master tape) Write 3.000 (500 character tape items - 2.500 character block) L At^-j s 1.5 x 10 tape characters (report tape) 78 Execute 250 instructions per transaction (average) 250 x 20,000 x 30 characters per instruction Data Movement Transaction Tape to Trans action Input Area Transaction Input Area to Transaction Work Area Transaction Work Area to Master Work Area Old Master Tape to Old Master Input Area Old Master Input Area to Master Work Area Master Work Area to Hew Master Output Area Hew Master Output Area to Hew Master Tape Master Work Area to Report Area Report Area to Beport Tape = 150.0 x 10^ characters 2.0 x 10 2.0 x 10 2.0 x 10' = 700.0 x 10c * 700.0 x 10' = 720.0 x 10' = 720.0 x 10' ,6 1.5 x 106 = 1.5 x 106 2,999.0 x 106 characters A731 = 2,999 x 10*> processor memory references (characters) Program Requirement (Transaction Input Areas 2 x 2,500 = 5,000 characters Old Master Input Areas 2 x 2,500 = 5>000 New Master Output Areas 2 x 2,500 = 5,000 Report Output Areas 2 x 2,500 = 5,000 Transaction Work Area = 100 Master Work Area = 250 2,500 Instructions x 8 characters per Instruction = 20,000 Operating System = 16.000 56,350 characters Run *f. Job 1 Tape Sort Report Pile Sort 3.000 t 500 Character Items (1.5x10^ characters) 2 x (39,000 * 500) = 156 items per string 3,000 * 156 = 20 strings 2 Way Balanced Merge Requires 2^ - 20 - 2^ 5 merge passes + 1 pass for pre-sort = 6 tape passes 6 tape passes r tapes =1.5 passes per tape 1.5 passes per tape x 1,5 x 10^ characters per pass = 2.3 x 10^ characters per tape (2,500 character block) k2bi = 2.3 x 106 tape characters (work tape) A.3^1 ■ 2.3 x 106 tape characters (work tape) / L A*+*fl - 2.3 x 10° tape characters (work tape) / • A^i - 2.3 x 10 tape characters (work tape) Processor Activity! Assume pre-sort is tape limited. Run 5. Joh 1 Print Reports Read 3,000 (500 character tape items - 2.500 character block) A^^i = 1.5 x 106 tape characters (input tape) Print 3.000 (2.000 reports - 132 character print line) A651 = 6 x 10^ print characters (double space) Execute 600 instructions per report (average) 600 x 3»000 x 30 characters c per instruction = 5^,0 x 10 characters Pat a Movement Report Tape to Report Input Area * 1.5 x 10 Report Input Area to Output Area = 1,5 x 10^ Output Area to Printer = 6.0 x 10^ 63.0 x 10^ characters 6 A75i = 63 x 10° processor memory references (characters) Program Requirement Report Input Areas 2 x 2,500 = 5>000 characters Report Output Areas 2 x 2,000 » *+,000 2,000 Instructions x 8 Charac ters per Instruction = 16,000 Operating System = 16.000 *+1,000 characters Run 1. Job 2 Make Calculations on Input Tape Data and Print Results Read *+0.000 (100 character items - 100 character block) A212 = *+ x iO tape characters (input tape) Print M-0.000 (100 character resorts - 132 character srint line) A^i2 = * * ■ x 10^ print characters (double space) Execute 1.000 Instructions ser Insut Item 1,000 x *+0,000 x 30 memory references per instruction = 1,200 x 10^ characters Data Movement Input Tape to Input Area *f x 106 Input Area to Work Area *+ x 106 Work Area to Output Area b x 106 Output Area to Printer b x 106 1,216 x 10^ characters ■^712 - Ij2l6 x 106 processor memory references (characters) Program Requirement Input Areas 2 x 100 = 200 characters Work Area 100 Print Areas 2 x 200 *+00 3,000 Instructions x 8 Charac ters per Instruction = 2*+, 000 Operating System - 16.000 1*0,700 characters Run 1. Joh 3 Tape Collate (2-Way Merge) Collate 21 Tapes Each Containing 3*000 (500 character items) (Overlapped Reading and Writing: Collate Time - Total of Pile Input Time) O C \ J 1.5 = 30.0 x 10^ characters 10 x 3.0 a 30.0 x 106 * * X 6.0 = 2^.0 x 106 1.5 + 12.0 2 : 13.5 x 106 13.5 + 12.0 = 25.5 x 106 25.5 + 6.0 a 31.5 x 106 l5**.5 x 10^ characters / ■ s 15**.5 x 10 f ** tapes = 38*7 x 10 characters per tape (2,500 character block) ^213 = 38.7 x 10 tape characters (work tape) ■A-313 = 38.7 x 106 tape characters (work tape) ^•13 = 38,7 x 10^ tape characters (work tape) * ■A-513 - 38*7 x 10° tape characters (work tape) Execute 33 Instructions per Original Input (63.000 total input) 33 x 63*000 x 30 memory 62.5 x 10^ characters references per instruction 83 Data Movement Input Tapes to Input Area 15**.5 x 10 Input Area to Output Area 15**.5 x 10^ Output Area to Output Tapes 15**.5 x 10^ 526.0 x 10^ characters Z L ^713 ~ z 10 processor memory references (characters) Program Requirement 2 Input Areas 2 x 2 x 2,500 = 10,000 characters Output Area 2 x 2,500 = 5,000 500 Instructions x 8 Characters per Instruction * = **,000 Operating System - 16.000 35,000 characters 8. Run 2. Job 3 Print Reports Read 63.000 (500 character tape items - 2.500 character block:) a523 s 31»5 x 106 tape characters (input tape) Print 63.000 (132 character messages) •^623 = 8.32 x 106 print characters (double space) Execute 100 Instructions per Input Item 100 x 63>000 x 30 Characters per Instruction = 189.0 x 10^ characters Data Movement Input Tape to Input Area = 31*5 x 10^ characters 8b Input Area to Work Area Work Area to Output Area = 31.5 x 106 = 31.5 x 106 283.5 x 10^ characters A723 - 28^ x 106 processor memory references (characters) Program Requirement Input Areas 2 x 2,500 Work Area Output Areas 2 x 200 1,000 Instructions x 8 Char acters per Instruction Operating System 5.000 characters 500 1*00 8.000 16.000 29,900 characters Reciprocal of Effective Transfer Rate, 1. Magnetic faoes a. Pile Maintenance Mode Sf = forward speed Sr = rewind speed t = reel change time L = reel length = fraction of data which 2 is alphabetic (inches per second) (inches per second) (seconds) (feet) 2 Application specification. 85 I' d = fraction of data which n . is decimal^ = alphabetic density (characters per inch) dD = decimal density (characters per inch) g = inter-block gap (inches) B = block size^ (characters) n^ = number of items^ njj = number of items per block^ ^ = reciprocal of effective transfer rate (seconds £ per 10° characters) 1 - Time Required to Transfer Bata R Amount of Bata Transferred Time Required to Transfer Bata = Humber of Reels x Time per Reel taber of Heels „ keel Inches Required = (Humber of Blocks) ^ + Gap Inches \ Block j = number of blocks nB Bata Inches _ BI^ BFp Slock + dD Inches Required = + |®B + gj 3 Application specification* / 86 a± /BFa BFp \ Number of Reels « X2t ng l“5I + dD + s/ Time per Reel « Forward Time + Rewind Time + Tape Change Time 12 L 12 i Time per Reel = a + -§— + t n-; f BFa . BFt) Total Time = 12 ImB (“cK “*5 + 8 12 I 12 1 SJ" + — s; +t| - 5i2 Pa . si [1 + 1 + Jl_ ~ “B I3! ^ Bj I3? Sf 12 L Amount of Rata = n-B 1 (^A s\/l 1 t\ ( . K ( d Atd D * I J ( s ;+s rn21) "1 0 sSiSss^. ters h. Sort Mode With a read backward feature the time for re-winds is eliminated. Also, there is no time lost due to reel change. 1 flk *D s W 1\ 6 _ - .-a + — + — — |x 10 seconds per million I dA dB B/ \ Sfj characters 2. Printers k Nc = total number of characters to be printed % * number of horizontal print positions** Sp = speed of printing one line and spacing one line (lines per minute) Application specifications. IBM System 360 Specifications. P Z8 to ro h Model i 37.5 75.0 112.5 Forward Speed, sf (Inches per Second) -F u> ro to r t ) h O' o CO Rewind Speed, sr (Inches per Second) • • • On On On Inter-Block Gap, 8 (Inches) 00 CO 00 O O O O O O Alpha Density, dA (Characters per Inch) 1,600 1,600 1,600 Decimal Density, dD (Characters per Inch) ro ro ro X « 4 -r -r -r o o o o o o Reel Length, 1 (feet) 03 ' -d t e J a H h r | H O !» f - 3 M O f c ? 03 O h c j j - 3 H3 H O E, H 1 - 9 03 P 00 88 TABLE 9 RECIPROCAL OP EPPECTIVE TRANSFER RATE OP MAGNETIC TAPE UNITS Model i Mode of Opera tion, m < 3 . o o o a>ft •H |Q •P C f l * - 0 C t f $S ftP 1 Fraction ofa Decimal Data, Pp i o 0 ) 0 1 ( 5 5 o -P > r J h r - t o W O < D OO O {> 0 ) O *H 0 ) OXO ft o © *h © d h © o 3 p h < o «h ?h 3d) a> « WEhH ft+> 1 Pile Maintenance 2,500 Character Block .50 .50 39.2 1 Pile Maintenance 100 Character Block .50 .50 231 1 Sorting 2,500 Character Block .50 .50 31.^ 2 Pile Maintenance 2,500 Character Block .50 .50 21.8 2 Pile Maintenance 100 Character Block .50 .50 131 2 Sorting 2,500 Character Block .5o .50 15.7 3 Pile Maintenance 2,500 Character Block .50 .50 15.6 3 Pile Maintenance 100 Character Block .50 .50 91.9 3 Sorting 2,500 Character Block .50 • 5o 10.5 aPart of the application specifications. ^Tape change time, t = 60 seconds. 89 Wg = number of space lines minus one Ny < = number of vertical lines per inch Ss = spacing speed (inches per second) 1 _ Time Required to Print Data R “ Amount of Data to be Printed Time Required to Print Data = Print Time + Spacing Time Humber of Print Lines Print Time Humber of Print Lines Print Time Spacing Time Number of Space Lines Number of. Inches Spaced Spacing Time Total Time _ Nc Printing Speed % 60 Nc W T R seconds Humber of Inches Spaced Spacing Speed Nc Ns " % . Nc Ns * seconds Nh Nv 's % i r T s . seconds 1 6. 10' K = % 1. s , Nc v°s * , n «y «g seconds per million characters T - Job Time Constraints, Tj Job 3 1 8 hours 2 1 hour 3 2 hours (28,800 seconds) ( 3*600 seconds) ( 7,200 seconds) ^Application specifications. TABLE 10 SPECIPICATIONS OP PRINTERS0 1 Model i Print Positions Print Speed, Sp (Lines per Minute) Vertical Spacing, Nv (Lines per Inch) Spacing Speed, So (Inches per Second) 1 132 600 8 33 75 over 8 lines spaced 2 132 1,100 8 33 75 over 8 lines spaced aIBM System 360 Specifications. vo o TABLE 11 RECIPROCAL OP EFFECTIVE TRANSFER RATE OF PRINTERS Model 1 Mode of Operation, m Number of Horizontal Print Positions, % Number of Space Lines - 1, Na Reciprocal of Effec tive Transfer Rate, £ Xnm-,- (Seconds per 10 Characters) 1 Double Space 132 1 786 2 Double Space 132 1 1*2 vO H 92 TABLE 12 RECIPROCAL OP EFFECTIVE TRANSFER RATE OP CARD READERS8 Reciprocal of Effective Model Transfer Rate, X 1 (Seconds per 1,000 cards) 1 100 2 60 aIBM System 360 specifications. TABLE 13 PROCESSOR MEMORY CYCLE8, Memory Cycle Time Per Model Character (Seconds per i 10b Memory References) 1 1,50 2 1.25 3 .50 aIEM System 3&0 Specifications. 93 TABLE 14 RENTAL RATES9, Unit, u Model i Rental, Card Reader 1 $ 260 per month i i i t 2 320 Tapes 1 578 includes pro-rata share of tape controller i t 2 728 " n 3 1,028 " Printer (132 print positions) 1 1,375 including control unit n 2 1,575 " Processor (65 K) 1 4,450 including 2 input- output channels and miscellaneous features n 2 5,7“ f 0 t t 3 9,750 aIBM System 360 Price List. PART II HARDWARE - SOFTWARE MODEL CHAPTER III DEVELOPMENT OP THE HARDWARE-SOFTWAKE MODEL General Approach The optimal selection of a computer system must he made on the basis of evaluating a combination of hardware, software and financial factors in a growth environment. These three elements are interdependent. The Hardware Model of Part I is used to find the optimal hardware con figuration in a given system during the time in which the technological coefficients and job time constraints remain constant. This optimal solution will need to be modified in subsequent time periods due to the following growth factors: 1. A compound annual rate of growth occurs in the applications due to the growth in input volumes, file sizes and output volumes. This growth is the result of serving a growing body of data processing customers. 2. A second type of compound annual rate of growth occurs due to the expansion and addition of types of services* 95 96 The first type of growth is reflected in the model as an increase in the speed of computer units necessary in order to process a larger amount of data within the origi nal time constraints. The second -type of growth is reflected as an increase in the size of programs necessary to accommodate expanded and additional services. This latter growth may require larger memory sizes in subsequent time periods. It is assumed that the job time constraints remain constant while the two types of growth take place. This need not be the case. If changes in the job time con straints are anticipated, alternative optimal solutions ean be evaluated as described in Part I. In addition to the affect of application growth on the original optimal solution, modifications to the origi nal optimal solution in subsequent time periods may be necessary due to the choice of programming language. The object program execution times and program sizes used in Chapter II were based on the use of an assembly or machine oriented programming language. If a compiler language is used, there will be, in general, an increase in object pro gram execution time and an increase in object program memory requirement. In general, it will also take longer to compile a set of programs than it does to assemble the same set of programs. These inefficiencies of the compiler 97 language are incorporated in the model by using average ratios of compiler language to assembly language object program execution time and object program memory require ment. Assembly time and compile time constraints must be satisfied. If these constraints cannot be satisfied by the original optimal hardware configuration, the original opti mal solution must be modified by using faster computer units, more memory or both. The advantages of using a compiler language are the lower costs of program preparation, program maintenance* program debugging and program conversion. The compiler language also produces superior program documentation.'1 ’ This last advantage is difficult to measure in dollars but may be a deciding factor when there is little difference in the costs of an assembly language system and a compiler language system. Replacement is always involved in a computer selec tion for business applications, whether there is an exist ing computer or not. If the computer selection study is being performed for the original installation of a com puter, the computer may replace either part of a clerical or a tabulating system. According to George Terborgh the ■^Arthur J • Whitmore, MOOBOL at Westinghouse,n Datamation, (April, 1962), 31“32. replacement process can be thought of as occurring in two 2 parts. The first part is the "challenge round" or the selection of the best present challenger. This is the sub ject of Part II* In order to select the best present challenger, the best hardware-software combination and method of paying for the computer must be determined for each computer manufacturer under consideration. In Part H, the selection of the best challenger is performed for one manufacturer's system only. The complete selection process would involve a determination of the best hardware-software alternative for each manufacturer's system and a selection of the best of these finalists to challenge the defender. Future challengers must be considered when determining the best present challenger. Consideration of future challeng ers is necessary because of obsolescence which present challengers develop with respect to future challengers. However, with the exception of conversion costs, there will be no difference in the cost of present computer systems which is related to the installation of future challengers. The same best future challenger may replace any present challenger. This replacement may occur in the same year for any present challenger. Differences in cost of conver sion of present challengers may arise due to differences in 2&eorge Terborgh, Dynamic Equipment Policy (Hew York: McGraw-Hill, 1 9 * 9 p. 5*+-. 99 equipment and program compatibility between present and future challengers. Since the analysis of present challeng ers in this example is limited to a single manufacturer, there will be no difference in cost of conversion due to possible incompatibility of present and future equipment. However, there is a difference in the cost of future pro gram conversion due to the choice of programming language for the present challenger. This cost is included in the analysis of present challengers. The second part of the replacement process is the challenge itself which pits the best present challenger against the defender, making allowance for future challeng ers which may be superior to the present challenger. The challenge part of the replacement process is covered in Part HI. In Part II, it is shown how to determine the best hardware-software combination for a single manufacturer's system, using alternatives of rental and a combination of purchase and rental. Other possibilities, such as leasing from a leasing company and rental with an option to pur chase are not covered. It is not the purpose of Part II to test exhaustively all possible methods of paying for the use of the computer, but rather, to indicate how the model is used to evaluate the hardware-software alternatives. The best hardware-software alternative for a given 100 system is determined by finding the alternative with the lowest equivalent uniform annual cost. There is no physi cal life of a computer; its life is determined solely by technological obsolescence. Annual costs are calculated for a number of years sufficient to determine the minimum annual cost for each alternative. A sensitivity analysis is performed on the optimal hardware-software solution for two types of variables: technical and financial. The technical variables are the relative efficiencies of the compiler language with respect to the assembly language. The financial variables are the interest rate, disposal value, purchase versus rental of additional equipment, method of depreciation and tax rate. Only those costs which affect the choice of a hardware-software system are considered in Part II. Some costs affect the decision to replace an existing system, but have no effect on the choice of the present challenger. The costs of system design and file conversion, associated with replacing an existing system, have no effect on the present challenger selection* These costs are contrasted with the cost of program conversion, which affects both the decision to replace and the choice of the present challenger. Mathematical Notation The following mathematical notation is used in the 101 development of the Hardware-Software Model, Za: total cost of program preparation using an assembly language. Ceps annual cost per programmer of efficiency e. Cems computer cost incurred during program prepara tion phase by programmer of efficiency e, ]Jeas number of programmers of efficiency e required for assembly language programming during program preparation or program main tenance • Npa* number of assembly language instructions which must be written during program prepara tion phase. Tpt time allotted for program preparation. Rea* average production rate of programmer of efficiency e using an assembly language (instructions per day). Ta: computer time required for assembling pro grams per assembly run (minutes), Tdaf computer time required for debugging in assembly language per debugging run (minutes), total computer time allotted per assembly or compilation and debugging run (minutes). Sa: average required assembly speed (instructions per minute). total cost of program preparation using a compiler language, number of programmers of efficiency e required for compiler language programming during pro gram preparation or program maintenance, estimated ratio of program size for compiler language to program size for assembly langu age (Pm 2 1), estimated production rate of programmers of efficiency e using a compiler language (instructions per day). computer time required for compiling programs per compilation run (minutes), computer time required for debugging in com piler language per debugging run (minutes), average number of computer instructions per compiler language statement, estimated required compiling speed (state ments per minute). estimated compound annual rate at which pro gram size increases. value of technological coefficients in year n. estimated compound annual rate at which application workload increases. 103 estimated ratio of object program execution time using a compiler langu age to program execution time using an assembly language (I^ - 1)* Pn * program size in year n (instructions), eat relative efficiency of programmer using assembly language, ec: relative efficiency of program using compiler language, K: estimated ratio of number of program maintenance instructions to number of original program instructions, is interest rate. (crf-i$-n)s capital recovery factor for n years at interest = [i (1 + i)n] * [(1 + i)n - 1]. (pwf *-i$-n)i single payment present worth factor for n years at interest a (1 + i)‘ -n . salvage value in year n. ps computer purchase price. Program Preparation Programmers of various efficiencies e should be allocated in order to satisfy the program preparation time constraint Tp and to minimize the total cost of programming 10*f in a given language during the program preparation period. The cost per programmer during the program prepara tion period includes the costs of writing, translating and debugging programs. Computer costs are incurred for trans lating programs using an assembler or compiler and for com puter time consumed in debugging. Assembly Language 0*0) Minimize Za » ^ (Cep + Cem) Nea, subject to 0*2) Uea - 0 and integer Equation (*+0) expresses the total cost function as the sum of programmer cost, Cep and computer cost Cem (including the cost of computer operators) multiplied by the number of required programmers Nea in each efficiency classification. In equation (*+1) the time of writing IFpa instructions must be less than or equal to the time allotted for program preparation Tp. In addition to writing the programs within a time period Tp, two other constraints must be satisfied with respect to the machine time used for program assembly and debugging. ft3> Ta + Taa 5 *t 10? The sum of the program assembly time and the pro gram debugging time must be less than or equal to the total time allotted for program testing in a given test run. This restriction is imposed in order to keep the turnaround time within an acceptable limit. ( i f ! *) ®pa 5 T&j g a I f pa sa V Equation (M+) expresses the minimum rate at which instructions must be translated in order to maintain the total time constraint (*+3). I £ this constraint cannot be satisfied with the optimal configuration it may be possible to satisfy it by adding memory and/or using faster computer units. Compiler Language 0*5) Minimize Zc * (if6) ^Pa ^m 5 Ip^ec ^ec (**7) Ne(J - 0 and integer In equation (*f6) the number of equivalent assembly language instructions are obtained by multiplying Npa by the average ratio of compiler instructions to assembler instructions Em. (W> T0 + Tdo - Tt S (C + Cem) Neo, subject to e tp 106 BmS^ ~c' "u - (1 +9) 2|aJa - Tc, Sc - Sga -fe The equivalent assembly language instructions which must be compiled is Npa Pm* However, it is desired to determine the minimum required rate of compilation Sc in terms of compiler statements and not equivalent assembly language instructions. Therefore, Npa Pm is divided by % which is the average of number of assembly instructions per compiler statement. It may be necessary to modify the optimal equipment configuration in order to satisfy the compile rate constraint. Compile speed is becoming competitive with, and in some instances is exceeding, assembly speed. The increase in compiler speed is due primarily to two factors: 1. Some compilers are now syntax-directed, in which the translation from source to target language occurs on the basis of the syntactical and semantic description of the source language 3 ^ in terms of the target language. ’ In this context, the target language can be object code ^Robert Steven Ledley, Programming and Utilizing Digital Computers (New York: McGraw-Hill, 1962), pp. 295- J & T * ^D. Morris, "The Use of Syntactic Analysis in Com pilers," Peter Wegner (ed.), Introduction to System Pro gramming (London: Academic Press, 19640, pp. 249-255. 107 or another automatic programming language. This type of translator identifies the syntac tical elements of the source language and makes the appropriate substitution of target language according to the relationship between the syn tactical elements of the source language and the associated semantics of the target language. This method is contrasted with the method used in some compilers of substituting relatively large segments of generalized coding for the source language statements. 2. The new compilers translate from source langu age to machine code, without going through an intermediate assembly language. Assembly languages have been used in some compilers in order to achieve rapid implementation of the compiler at the expense of compilation time. This method aided the compiler writers, since an existing assembly language could be used for writing and debugging the compiler. Program maintenance will be required after programs become operational in order to incorporate modifications Program Maintenance 108 and improvements which later prove desirable. An increase in the services provided by the applications will affect the rate at which program size increases with time. It is assumed that the average annual rate at which program size increases is equal to Gm. An assumption must also be made regarding the level of programming activity £ which is required for maintenance purposes as compared to the origi nal programming effort. This level of programming effort is reflected in the number of instructions which must be written per year in order to maintain the system. Assembly language The number of instructions which must be written in year n for an assembly language system is given by the following expressions (50) K Npa (1 + Gm)n where K is the estimated ratio of number of program maintenance instructions to the number of original program instructions and Hpa is the number of original program instructions. This number of instructions must be written and debugged each year (250 working days). Minimize (H-0) sub ject to (51) K V (1 * sm>n~1 5 250 J>]^ea ®ea e 109 (52) Ta + Ida * It (53) K % a (1 + sf f l )n~1 £ T s £ K gpa (1 + Gm)n~1 xa Compiler language Minimize (*f5) subject to (9f) ^ V pm (1 + sa2n 1 s 25o. E Kec e (55) T0 + Tdc S It (56) 6 T , W so i *0 It may turn out that the computer time required for program maintenance is negligible compared to the computer time required for program preparation. Equations (52) and (53) or (55) and (56) may be of no consequence. Object Program Execution lime The optimal computer configuration will change for a given system as the workload increases for existing applications* The increase in workload of existing appli cations is reflected as an increase in technological coefficients in the model. Additional applications will be 110 reflected as new constraints in the model. The analysis of additional applications is identical to that of applications already discussed, For the purpose of developing the model, it will he assumed that JOB 1, JOB 2 and JOB 3 constitute the major applications, No purpose would he served hy con sidering additional applications. It will he assumed that all technological coeffi cients increase at a compound yearly rate of Ga. The size of the technological coefficients in year n is given hy: (57) % = (1 + Ga) n"1 Assembly Language An increase in processing time will occur which is directly proportional to the increase in technological coefficients. The time used hy unit u, model i, job j, in year n is the following: <58> JWjn ■ Duijl (1 ♦ Ga> n_1 Compiler Language The use of a compiler language will result in an increase in object program execution time and an increase in the size of program as compared to the use of an assem bly language. The average ratio of program execution times is given by - 1. The expected increase in program execu tion time can he incorporated in the model as follows: (59) Baijn = ^uijl ^ + An analysis is performed on the binding job (s) in order to determine what effect (58) or (59) will have on the optimal configuration and equipment cost. Object Program Memory Requirements The growth in program size Gm may also cause a change in the optimal configuration, if program size plus data requirements exceed memory capacity for one or more programs. If memory size is exceeded in a slack job, it may be possible to reduce block sizes and still be within the job time constraints. If memory size is exceeded in a zero slack constraint, additional memory or faster computer units will be required in order to maintain feasibility. Assembly Language The program size in year n is given by the follow ing equations (60) Pn = Pi (1 + &m) n“1 Compiler Language (61) Pn « = Pi (1 + Cfo) 11-1 Pm The compiler language program size in each year is greater than the assembly language program size by the fac tor 1^, the average ratio of compiler instructions to assembly instructions. 112 Program Conversion A program conversion cost will be incurred at the time of replacement of a computer system. The conversion cost under consideration is the one which will occur at some future replacement date, when a new challenger is acquired, as the result of using a given programming langu age in the present challenger to be selected. The cost of reprogramming under consideration is restricted to that part of programming related to the use of a programming language. The cost of systems design associated with installing the new computer is virtually independent of the programming language used. Assembly Language If an assembly language is used in both the present and future challenger systems, the expected cost of reprogramming is equal to the original assembly language program preparation cost. It is assumed that the new challenger is not program compatible with the present challenger. It is required that the converted programs be at least as efficient as the old programs. This restric tion excludes the use of simulators and conversion programs designed to simulate the operation of the old computer on the new computer or to automatically translate programs from one computer to another. Translators can never pro duce a program as efficient as the original program due to 113 differences in hardware characteristics, data dependency and the difficulty of translating some of the subtleties employed by programmers in order to produce efficient pro- 6 grams. The assumption of dissimilar present challenger and new challenger computers and the restriction on con verted program efficiency lead to the conclusion that the use of an assembly language in old and new computers will result in a repetition of program preparation costs. How ever, if a compiler language is used to program the new computer, the program conversion cost will be approximately equal to the cost of program preparation for using a com piler language on the old computer. When appropriate, the sensitivity of the programming language selection to the cost of program conversion can be tested. Compiler Language The cost of reprogramming if COBOL is used on both the old and new computers is about one-half of the original 7 COBOL programming cost because of the following reason: The three major divisions of a COBOL program are the Environment Division, Data Division and Procedure Division. The first two divisions are machine dependent and the third Ascher Opler. "Automatic Program Translation," Datamation. (May, 1963), M-5-MJ. ^Stanley M. Naftaly, "Correcting Obfuscation by Ordained Linguists," Datamation. (May, 1963)» 2^-28. ll*f 8,9 division is machine independent. The Environment Division describes the computer on which the program is to be compiled and the computer on which the object program is to be executed. The time spent in rewriting this portion is insignificant. The Data Division describes the charac teristics of the files to be processed and must be rewritten in order to make the converted programs efficient with respect to the data handling characteristics of the new computer. The Procedure Division contains the action statements of the compiler. These statements, written in English language narrative form, were designed to be machine independent. Thus, the major reprogramming task in COBOL consists of rewriting the Data Division which consti tutes approximately 50 per cent of the original program preparation effort. When appropriate, the sensitivity of the programming language selection to the cost of repro gramming in compiler language can be tested. D Howard Bromberg, "COBOL and Compatibility," Datamation. (February, 1961), 30-3^. ^Robert M. Gordon, "COBOL and Compatibility," Datamation. (July, 19&3), CHAPTER IV APPLICATION OP THE HARDWARE-SOFTWARE MODEL The application of the model for selecting the optimal hardware-software system takes place in the follow ing steps! 1. The total cost of program preparation, Za and ZG, is minimized subject to the constraints of program preparation time Tp and computer assembly or compilation and debugging time T^.. The optimal number of programmers, N|a and Ng0, required to minimize Za and Zc, respectively, is determined. In addition, the assembly rate Sa and compile rate S0 required to satisfy the constraint T-fc are determined. In some computer systems a larger configuration is required for compilation than for assembly. This is not the case in the system under consideration. 2. Step 1 is repeated for program maintenance. 3. The effect of the growth in application volume Ga on the optimal assembly language equipment configuration is determined by examining the 115 effect of Ga on the critical job(s). A series of optimal equipment configurations is obtained for each year for a period of eight years. The optimal equipment cost per time period is obtained in terms of equipment rental. The optimal configuration in each time period in terms of purchase price is the same as that for rental because the purchase to rental ratios for the various computer units are approxi mately equal. The combination of Ga and the ratio of object program execution times F^ is applied to the critical job(s) in order to obtain a series of optimal equipment configura tions for the use of a compiler language in each of the eight year periods. The growth in program size G r m is applied to the size of critical program(s) in assembly langu age in order to determine whether additional memory capacity is required during the time span under consideration. The combination of ®m and the ratio of object program size Fm is applied to the critical program(s) in compiler language. Equivalent uniform annual costs are calculated for various optimal equipment configuration- programming language alternatives, under conditions of rental and a combination of pur chase and rental. The selection of the optimal hardware-software system is made on the basis of minimum equivalent uniform annual cost. Programming Language Characteristics Several studies have been performed on determining the efficiency of a compiler language versus that of an assembly language for several technical characteristics. The results of these studies are tabulated in Table 15. The values which are used in the analysis to follow are shown in the last column. The example used in the applica tion of the model is representative of a production oriented data processing facility, i.e., one in which there is a large amount of object program execution time in relation to program compilation or assembly time and program debug ging time. This type of operation is also characterised by the repeated running of the same programs as opposed to the many "one shot programs" of the research oriented data processing facility. In the production type data proces sing facility, the efficiency and cost of the hardware is of greater importance relative to the efficiency and cost of programming than is the case in a research computation facility. Therefore, the results obtained from the appli cation of the hardware-software model cannot be generalized. TABLE 15 EAT 10 OP COMPILER TO ASSEMBLER CHARACTERISTICS Characteristic COBOL/ Assembly8- POL/MOL*3 GECOM/GAP0 Value Used in Analysis Ratio of Object Program Size, Pm 1.10-1.15 1.16-2.63 1.03-1.05 1.10 Ratio of Object Program Execution Time, 1.18-1.25 .96-5.18 1.10-1.67 1.10 Ratio of Program Translation Time 1.16-1.20 1.18 Ratio of Computer Debugging Time .73- .78 .75 Ratio of Program Preparation Time .53- .56 .*+5 Ratio of Reprogramming Time .10- .1+0 .50 aStanley M. Naftaly, "Correcting Obfuscations by Ordained Linguists," Datama tion . (May, 1963% 2^-28. bo Several Procedure Oriented Languages vs. Machine Oriented Languages, Christopher J. Shaw, "More Instructions • . . Less Work," Datamation. (June, 196*0, 3*+-35. cGeneral Electric Co., Compiler vs. Assembler. Jerry Schwalb, "Compiling in English," Datamation. (July, 1963)» 28-30. 118 119 The results are applicable to a specific type of operating environment • Economic Analysis The following assumptions and conditions pertain to the economic analysis of the next sections: 1. The personal property tax and cost of insurance for purchase and the use tax on rental are ignored. 2. A combined effective federal and state income tax rate of 50 per cent is assumed. 3. It is permissible for a lessee to receive an investment tax credit for eligible property if the lessee adopts the useful life used by the lessor and the lessor waives his right to the tax credit.*^ It is assumed that the lessor does not waive his tax credit. if. It is assumed that disposal value is equal to book value in the case of purchase, so that there is no gain or loss on disposal. 5. The salvage value is equal to the estimated disposal value. There is no inferior use con templated for the asset within the organization when the asset is replaced. iOtT. S. Master Tax Guide (Commerce Clearing House. Inc., 1965), pp. 20-21. 120 The first analysis is for an assembly language rental system versus a compiler language rental system. The annual disbursements for the rental systems are shown in Table 16. The disbursements for program preparation, program maintenance, equipment rental and program conver sion correspond to the values developed in the Input Data for Hardware-Software Model Section, p. 177. Since the economic life of a computer system is determined solely by technical obsolescence, the economic life of a system is dependent upon the availability of superior replacements. The choice of alternatives may depend on the year in which replacement occurs. The mini mum cost point represents the economic life only in the case where there is a series of replacements with identical investment and operating costs. Since this is not the case with computer systems, replacement may occur before or after the minimum cost point, depending upon the charac teristics of future challengers. However, the number of years corresponding to minimum cost is the best estimate of life which can be made at the time of the study. Each alternative is compared on the basis of its most favorable life, as seen at the time of the study. The sensitivity of the choice of present challenger to assumed life can be determined by graphing the equivalent uniform annual cost curves over a number of years. 121 TABLE 16 ANNUAL DISBURSEMENTS8 , RENTAL SYSTEM Year n Disbursement Assembly Language (Thousands) Compiler Language (Thousands) 0 Program Preparation $36.1 $ 22.0 1 Program Maintenance 10.0 10.0 Rental llO.lf 121.2 2 Program Maintenance 10.0 10.0 Rental 11*+.0 121.2 3 Program Maintenance 10.0 10.0 Rental 117.6 12^.8 b Program Maintenance 11.3 10.0 Rental 117.6 1^2.5 5 Program Maintenance 11.3 10.0 Rental 121.2 1^2.5 6 Program Maintenance 11.3 10.0 Rental 12^.8 1^2.5 7 Program Maintenance 11.3 10.0 Rental lb2.5 1^2.5 8 Program Maintenance 11.3 10.0 Rental 1^2.5 177.1 Year of Replacement Program Conversion 22.0 11.0 aS*rom Input Data for Hardware-Software Model, 122 From Table 17, Table 18 and Figure 2, it is seen that the assembly language system is superior on a rental basis. The next analysis is performed in order to deter mine what effect purchase has on the selection of the opti mal hardware-software system. Assembly and compiler langu ages are compared on the basis of purchasing equipment which will not be replaced during the eight year period and renting equipment which is required to modify the original configuration. In the Sensitivity Analysis Section, the effect of also purchasing the equipment required to modify the original configuration is determined. The schedule of purchase and rental for the two language systems is shown in Table 19. The schedule of maintenance costs is shown in Table 20. Annual disbursements for the purchase-rental system are shown in Table 21. The same costs of program preparation, program maintenance and program conversion are used for purchase-rental as were used for the rental sys tems. Salvage and depreciable values are shown in Table 22. The schedule of salvage values is based on predicted dis posal values. Disposal value cannot be predicted with pre cision in a field marked by rapid technological progress. However, a decline in value of about 10 per cent per year seems reasonable based on the fact that it would have required a system costing $1,000,000 in i960 to perform TABLE 17 AFTER-TAX EQUIVALENT UNIFORM ANNUAL OOSTS, ASSEMBLY LANGUAGE RENTAL SYSTEM (Interest Rate = 5 Per Cent) (1) Year n n ) ra -p / —i a ) c d oq c \ j g <D«g III m • h rfet A ■P U Q*~s 0 2 ro T3 - % 3 © £ r a o o > A jh ^ COo Cumulative Present Worth (Thousands) (5) d Present Worth o f Program Conversion (Thousands) <6) e Total Cumulative Present Worth (Thousands) (7)* Equivalent Uni form Annual Cost (Thousands) 0 $ 36.1 $ 36.1 1 120 .b 11^.6 $150.7 $ 20.0 $171.6 $180.2 2 12^.0 112.5 263.2 20.0 283.2 l?2.b 3 127.6 110.2 373 A 19.0 392A lM f.O b 135.2 111.3 b 8b. 7 18.1 502.8 lM-1.8 5 137.0 107.4 592.1 17.2 609.3 lbQ.7^ 6 138.8 103.5 695.6 l6.b 712.0 lbo.3s 7 153.8 109 .*f 805.0 15.6 820.6 lb2.Q 8 165.0 111.7 916.7 l*f.9 931.6 lMf.^ ^Disbursements of Table 16 plus difference in income tax disbursements. J?[Col. (2)] (pwf * - % - n). ^Cumulative total of Col. (3). d22(pwf* - - n). «Col. flt) + Col. (5). £[Col. (6)] (erf - 5% -n). sMinimum cost. TABLE 18 APTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS, COMPILER LANGUAGE RENTAL SYSTEM (Interest Rate * (1) Year n Di sbur semen t sa during Year (Thousands) Present Worth (Thousands ) 0O o Cumulative Present Worth (Thousands) d Present Worth o f Program Conversion (Thousands) (6) e Total Cumulative Present Worth (Thousands) I (7)f Equivalent Uni form Annual Cost (Thousands) 0 $ 29.1 $ 29.1 1 131.3 125.0 &l5*f.l $ 10.5 $l6*f.6 $172.8 2 133.1 120.7 27^.8 10,0 28*f.8 153.2„ 3 136.7 118.1 392.9 9.5 *f02.*f 1**7.7« If 152.5 125.5 518.** 9.1 527.5 l*f8.8 5 152.5 119.6 638.0 8.6 6*f6.6 l*f9.** 6 152.5 113.8 751.8 8.2 760.0 1**9.7 7 152.5 112.8 86*f.6 7.8 872.*f 150.9 8 187.1 126.7 991.3 7.V 998.7 15**.8 ^Disbursements of Table 16 plus difference in income tax disbursements. "[Col. (2)] (pwf’ - 5% - n). ^Cumulative total of Col, (3). 11 (pwf"* - 5% - n). ®Col. (k) + Col, (5). f[Col. (6)] (erf - 5$ - n), ^Minimum cost. 15: TABLE 19 HJRCHASE-RENTAL SCHEDULE Assembly Language Year Item Annual XL Disbursements3- 0 Purchase Card Reader Purchase Tape Controller Purchase Printer and Controller $ 15,500 ^6,700 73,000 Purchase Central Processor (65*0 .207,3.50 $3^2,550 Rent Tape Units $b95 1 at each 23,760 2 Rent 3 Tape Units at $b95 each 17,820 Rent 1 Tape Unit at $795 9. Sto 27,360 $^95 ll,880 3 Rent 2 Tape Units at each Rent 2 Tape Units at $795 each 19,080 30,960 b Same as Year 3 $1*95 5,9to 30,960 5 Rent 1 Tape Unit at each Rent 3 Tape Units at $795 each 28,620 3^,560 6 Rent b Tape Units at $795 each 38,160 38,160 7 Rent b Tape Units at $795 each Rent 1 Tape Controller at $930 each 11,160 Rent 1 Tape Switching Unit at $55o each 6,600 55,920 8 Same as Year 7 55,920 f —1 to o> TABES.19 (Continued) Compiler Language Tear Item Annual n Disbursementsa 0 Purchase Card Reader Purchase Tape Controller $ 15,500 *f6,700 Purchase 3 Tape Units at $38,350 each 115,050 Purchase Printer and Controller 73,000 Purchase Central Processor (65K) 207 alQ b9 5 $*f57,600 1 Rent 1 Tape Unit at $ 5,9^0 2 Same as Year 1 5,9^0 3 Rent 1 Tape Unit at $ 795 9,5^0 9,5^0 b Rent 1 Tape Unit at $ 795 Rent 1 Tape Controller at $ 930 11,160 Rent 1 Tape Switching Unit at $ 550 6.600 27,300 5 Same as Year ^ 27,300 6 Same as Year *f 27,300 7 Same as Year b 27,300 8 Same as Year b 27,300 Plus Rent Central Processor (131K) at $ 7,3^0 88.080 $115,380 aIBM System 360 Price List. 128 TABLE 20 COMPUTER MAINTENANCE COST SCHBLULE Monthly Maintenance Cost as a Percentage of Monthly Rental0 , Computer Unit Years of Service 1-3 3-6 6-9 Central Processor 3.5% 3.8% b.3% Card Reader 8.2 11.2 l5.b Printer and Control 2*f.2 26.9 29.9 Magnetic Tapes and Control 11.6 12.1 13 A Annual Maintenance Costs Year n Assembly Language (Thousands) Compiler Language (Thousands) 1 $8,0 $11.3 2 8.0 11.3 3 8.0 11.3 b 8.8 12.3 5 8.8 12.3 6 8.8 12.3 7 9.9 13.8 8 9.9 13.8 aComputed from an average of monthly maintenance cost to monthly rental for several computer systems. ^Computed from data in Table 19 and Table 20. 129 TABLE 21 ANNUAL DISBURSEMENTS- -PURCHASE-RENTAL SYSTEM Assembly Compiler Year Disbursement Language Language n (Thousands) (Thousands) 0 Computer Purchase3 , , $3^2.6 $*f57.6 Program Preparation® 36.1 $378.7 22.0 $V79.6 1 Computer Rentala 23.8 6.0 Computer Maintenance Program Maintenance1 5 8.0 10.0 *fl.8 11.3 10.0 27.3 2 Computer Rental3 27 .*f 6.0 Computer Maintenance® Program Maintenance® 8.0 10.0 *f5.*f 11.3 10.0 27.3 3 Computer Rental0 , 31.0 9.6 Computer Maintenance® Program Maintenance® 8,0 10.0 ^9.0 11.3 10T0 30.9 if Computer Rental0 , 31.0 27.3 Computer Maintenance0 Program Maintenance® 8.8 12.3 11.3 10.0 51.1 if9.6 5 Computer Rental0, 3*+.6 27.3 Computer Maintenance® Program Maintenance® 8.8 12.3 11.3 5*+.7 10.0 *f9.6 6 Computer Rental0 , 38.2 27.3 Computer Maintenance® Program Maintenance® 8.8 11.3 12.3 10.0 *f9.6 56.0 58.3 7 Computer Rental0 , 27.3 Computer Maintenance® Program Maintenance® 9.9 ll*3 13.8 10.0 51.1 56.0 77.2 8 Computer Rental0 , 115J+ Computer Maintenance® Program Maintenance® 9.9 11.3 13.8 10.0 77.2 139.2 Year of Program Conversion1 3 22.0 11.0 Replacement aProm Table 19• ®From Table 16. eFrom Table 20. TABLE 22 STRAIGHT LIKE REDUCTION IN SALVAGE VALUE SALVAGE VALUE DEPRECIABLE VALUE Assembly Language First Cost = $3^2,600 (1) Years Retained (2) Fraction ofa First Cost (3) * Salvage0 Value (Thousands) (h) Col. (2) less 0.10° Reduced^ Salvage Value (Thousands) (6) e Depreciable Value (Thousands) 1 .90 $308.3 * $ 3^.3 2 .80 27^.1 - - 68.5 3 .70 239.8 .60 $205.6 137.0 k .60 205.6 .50 171.3 171.3 5 .50 171.3 .k0 137.0 205.6 6 .*f0 137.0 .30 102.8 239.8 7 .30 102.8 .20 68.5 27k.1 8 .20 68.5 .10 3^.3 308.3 H U> O TABLE 22 (Continued) Compiler Language First Cost = $5-57,600 (1) Years Retained (2) Fraction of8 . First Cost (3) b Salvage Value (Thousands) (5) Ool. <2)_ less 0.10 (5) d Reduced Salvage Value (Thousands) (6) p Depreciable6 Value (Thousands) 1 .90 $511.8 — — $ 55.8 2 .80 366.1 - - 91.5 3 .70 320.3 .60 $275.6 205.0 5 .60 275.6 .50 228.8 228.8 5 .50 228.8 .50 183.0 275.6 6 .50 183.0 .30 137.3 320.3 7 .30 137.3 .20 91.5 366.1 8 .20 91.5 .10 55.8 511.8 aFraction of original cost remaining at the end of the year shown. v Actual salvage value. cFraction of original cost reduced by .10 in order to account for the reduction of salvage value by 10$ of first cost permitted by I.U.S. regulations if asset is retained three or more years. U. S. Master Tax Guide (Hew York: Commerce Clearing House, Inc., November, 1965), p. 397. j Reduction in salvage by 10 per cent of first cost if asset kept three years or more. eFirst cost - Col. (5)* l - * ' 132 JOB 1, JOB 2, and JOB 3, whereas the same jobs can be per formed in 1965, within the original time constraints, at a cost of $500,000. A price of $500,000 is the upper limit at which the system of I960 could be sold for today. A decline in salvage value at an average rate of 10 per cent per year also corresponds to the Internal Revenue Service guideline life of ten y e a r s A deterrent to purchase, in the past, has been the lack of an organized used computer market. Today, a used computer market is rapidly develop- 12 ing around the services provided by computer brokers. In Table 22 it is assumed that the salvage value has been adjusted for disposal costs. The depreciation and investment tax credit sched ules for the purchase-rental systems are shown in Table 23. Annual costs are calculated in Table 2k, From Table 2k and Figure 3 it is seen that the two systems have approximately equal minimum equivalent uniform annual costs. When the alternatives are this close on a cost basis, the selection is based on irreducible factors. The major irreducible factor is the superior documentation produced by a program written in English language over that written in a machine oriented language. Therefore, the compiler language system - * - - * • 11. S. Master Tax Guide (Commerce Clearing House, Inc., 1965), p. *+04. -^George H. Heilbom, "The Used Computer Market," Datamation. (July, 1965), *+8-53. CO w P P § 4 g U > tv> H O _ t o ONpG • • • O • O - P " 0 0 1 —1 On HI H to "O O ro -P'on ro VJT.C 0 • • • <0 ONVJX lo to o-roj-r * v 3 1 —1 ONfO • • • • -TOOl— 'On Hi jJUJ - r - r • • • O'LoCo a FOND • • toOO -r -T'foiorovji • • • • • NOtOtOO on H ro o vr • o H t o ONO -rooro • • • rotovji 03 NO *o H eei u»ro (ovji co-<3 ro - < ] * • * • W W O O N H i NO H • OS - r - r VjrvNA * • 00 Oo OOM-O • * • ooNnH h & P < D K H W 8 S‘gI O m to Onto -T to ONpO • • • O O f - J On H j m t o -r t o ■m -r toroNJi 00 ( V ) -o • • • t o O ON N -r NJl co 4# - 0^1 • • GH Year n Di sbur sement sa (Thousands) Depreciation3’ 0 (Thousands ) Investment Tax Credit (Thousands) Additional Income Tax (Thousands) t o C O P ( 3 o' M f c H P P fa £ P fa P Di shursement s (Thousands) Depreciation (Thousands) Investment Tax Credit (Thousands) Additional Income Tax (Thousands ) o o P o p> H P H in P P fa P P fa P &1 1 O 02 P G g •tfpl i i Id ^ t t-3 vnw h re { » > & to t o TABLE 23 (Continued) Year n A s s e m b 1 y L a n g u a g e Comp i 1 e r L a n g u age Disbursements3 , (Thousands) b G Depreciation ’ (Thousands) Investment Tax Credit (Thousands) Additional Income Tax (Thousands) Disbursements8 , (Thousands) Depreciation*1,c (Thousands) j Investment Tax Credit (Thousands) Additional Income Tax (Thousands) Replace 0 $342.6f $>♦57.6* after 36.1 22.0 $7.1 4 Years 1 * *1.8 $ 68.5 $8.0 $7.0 27.3 I 91.5 $10.7 2 45.4 51 .4 27.3 68.6 .5 3 49.0 34.3 30.9 45.8 3.3 4 73.1 17.1 60.6 22.9 3.4 171.3" 228.8 Replace 0 342.6f 457.6f after 36.1 22.0 7.1 5 Years 1 41.8 68.5 8.0 6.9 27.3 91.b 10.7 2 45.4 54.9 .2 27.3 73.3 3 49.0 ^ 1.1 30.9 5*+.9 2.2 4 5i.l 27.3 3.9 49.6 36.5 5 76.7 13.8 60.6 18.4 5.8 205.6 274.5 TABLE 23 (Continued) Replace after 6 Years 0 1 2 i 5 6 A s s e m b l y L a n g u a g e o s co a © < D *d © r o c o c o II C O E H •rlv^ & $31*2.61' 36.1 i f l * 8 k9.0 51.1 5^.7 80,3 o # 1 O C O >H*d s s •H C O o d C D O fc,d ft£H o r *d co +> +a *d d-n 0>t3 a r o -P u coo ro I 02 d a M E H £H r o co h M'd as ro ' den O 0 3 •p © d -p a o -dOEc $ 68.5 $16.0 .7 .5 5.0 1.3 239. C o m p i l e r L a n g u a g e ro C O d ® © *d a d © ro co co pp co eh ■ r l '- ' P O # s ^d^ O C O • h »d -p ■ ro • H 0 3 o d © o d,d ft EH o ■d r o -P - p 'd 83 § flgg coo o r o ,d IIs $^57.6f 22.0 27.3 27.3 3°.? ^9.6 *f9.6 60.6 & 91.6 76.2 60,9 ^5.8 30 15.if 320.3 $21 .if © © H « d g ! § O C Q •H © d •p a o ■rl O d 'd O E H 3fl~ 7.1 l.if 7.9 Replace after 7 Years 1 2 3 if 5 6 7 3if2.6f 36.1 i f l . 8 if5A if 9.0 51.1 5if.7 58.3 99.2 68.5 58.7 *f9.1 39.2 29.3 19.5 16,0 if57.6f 22.0 7.1 9.7 27.3 91.5 21 .if •8 27.3 78.3 30.9 65.5 .9 5.9 if 9.6 52.if 2.i f if9.6 39.2 if9.6 26.0 1.1 62.1 13.2 366.1 16.9 U3 vn TABLE 23 (Continued) a h c d © A s s e m b 1 y L a n g u a g e : Comp i 1 e r L a n g u age cd 09 +3 y \ ri a> a>'d S 3 a ) © m ra ,5 x i 0 1 E H • H s _ / P I Depreciation^,C (Thousands) investment Ta x Credit (Thousands) a > m ■3 si s6*! ■H f f l j J 4 » a o tH o , eJ 'd O E H ' C f d'~' < t ! P l o d m + s d © © » d § 3 0 9 0 9 s § ,Q - d 0 9 E H C l O 0 9 * H < d SS • H 0 9 O J j © O H i £ | P t E H Investment T a x Credit (Thousands) Additional Income Tax (Thousands) 1 Replace 0 $3^2.6f $i+57.6f after 36.1 $ 1.0 22.0 $ 7.1 8 Years 1 M.8 $ 68 A $2b.O 12.1+ 27.3 $ 91.b $32.1 2 *+5.*+ 59.8 1.0 27.3 79.9 3 *+9.0 51.5 30.9 68.8 . b b 51.1 1+2.9 6.1+ 1+9.6 57.2 5 5M-.7 3^.2 3.2 !+9.6 1+5.7 6 58.3 25.6 1+9.6 3*+.2 7 77.2 17.3 51.1 23.1 10.2 8 99.2 8.6 27.0 150.2 11.5 308.3 1+11. o aFrom. Table 21. From Table 22. cDepreciation is computed on the basis of the sum of the year's digits method, except for years 1 and 2 where I.E.S. regulations require the straight line H method, TABLE 23 (Continued) ^Investment tax credit which is allowed by I.E.S. regulations for the first year that the asset is placed in service. k- - years held <6 7 ^ of 1/3 of basis. 6 s years held <8 7% of 2/3 of basis. 8 s years held 7% of 3/3 of basis. It is assumed that the tax liability is such that the amount of credit shown does not exceed the allowable limits. (1965 U.S. Master Tax Guide [New York: Commerce Clearing House, Inc.], p. * +18.) eDifferences in income tax disbursements between the two alternatives which will occur due to differences in investment tax credit, normal disbursements and depreciation. ■^Capitalized. 137 138 TABLE 2b AFTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS PURCHASE-RENTAIi SYSTEM (Interest Rate = 5%) > r Year n Assembly Language Compiler Language Cl) « Disbursements (Thousands) (2) " h Present Worth0 (Thousands) (3) a Disbursements (Thousands) (*+) Present Worth0 (Thousands) Replace after 1 Year 0 1 $378.7 $378.7 63.8 60.7 $*+86.7 - $1+86.7 *+5.3 *+3.1 Totals *+h2.5 *+39.*+ Less Present Worth of Salvage Value* 1 239.5 Net Present Worth Equivalent Uniform Annual Cost® 153.2 532.0 529.8 392?0 137.8 l*+*+.7 Replace after 2 Years 0 1 2 378.7 378.7 *+1.8 39.8 67.*+ 61.1 !+86.7 *+86.7 28.8 27.*+ *+7.1 *+2.7 Totals h87.9 *+79.6 Less Present Worth of Salvage Value* 1 2h8.6 Net Present Worth 231.0 Equivalent Uniform Annual Coste 12*+. 3 562.6 556.8 332.1 22*+. 7 120.9 139 TABLE 2b (Continued) Assembly Language Compiler Language Tear n Di sbur sement sa (Thousands) (2) h Present Worth (Thousands) a < n • P - ' - N S 3 m < D »d ^ g j c i ro a ) r o 'W' t Q O Q M EH ( = J c Present Worth (Thousands) Replace 0 $378.7 $378.7 $1+86.7 $1+86.7 after 1 51.6 1+9.1 27.3 26.0 3 Tears 2 1+7.7 1+3.3 27.3 21+.8 3 71.0 61.3 50.8 1+3.9 Totals 5^9*0 532.1+ Less Present Worth of Salvage Value d 207.2 Net Present Worth 325.2 Equivalent Uniform Annual Coste 119.3 592.1 581.1+ 276.7 301+.7 111.8 Replace 0 378.7 378.7 1+86.7 1 +86.7 after 1 1+8.8 i+6.5 27.3 26.0 1+ Tears 2 i+5.!+ 1 +1.2 27.8 25.2 3 1+9.0 1+2.3 31+.2 29.5 b 73.1 60.1 6i +.0 52.7 Totals 595.0 568.8 Less Present Worth of Salvage Valued 169.2 Net Present Worth 399.6 Equivalent Unifoim Annual Coste 112.7 61+0 iO 620.1 226.0 39^.1 111.1 I*f0 TABLE 2k (Continued) Assembly Language Compiler Language Year n 0 3 < = 3 / - V . © r d h a £ 0 3 0 3 1 1 W E H ^ o C O O J S r n 3 4 3 £ t o O 4 ) , 3 0 3 0 3 • P ' - ' d ® 0 3 nJ maS 4 > 0 5 C O O Q 1 1 0 3 E - * • H ' - ' fl W c Present Worth (Thousands) Replace 0 $378.7 $378.7 $W6.7 $**86.7 after 1 ^8.7 46.4 27.3 26.0 5 Years 2 ^5.6 27.3 2*f.8 3 ^9.0 ^2.3 33.1 28.6 k 55.0 ^5.3 *+9.6 ^0.8 5 76.7 60.1 66,4 52.1 Totals 653*7 Less Present Worth of Salvage Value* 1 Net Present Worth Equivalent Uniform Annual Coste 61^.2 1 3 ^ ,3d *+79.9 U0.98 690 .k 659.0 179^ ^79.6 110.6 Replace 0 378.7 378.7 ' k86.7 M-86.7 after 1 51.5 ^9*0 27.3 26.0 6 Years 2 ^•5.9 kl.6 27.3 2*f.8 3 ^9*0 ^2.3 32.3 27.9 k 56.1 ^6.2 ^9.6 if0.8 5 56.0 ^3*9 ^9.6 38.9 6 80.3 59.9 68.5 51.1 Totals 717.5 Less Present Worth of Salvage Value* 1 Net Present Worth Equivalent Uniform Annual Cost 661.5 102.2d 559.V 102.2f 7^1.3 696.2 136.5 559.7 110.3f TABLE 24 (Continued) l4l Assembly Language Compiler Language a a (1) a Disbursements (Thousands) (2) . Present Worth (Thousands) < 6 0 1 •p.'-N d ro /"-V 0 r o a d ' © 5 $ { £ > 0 3 S§ men • h w ft j-sc 8 CO • P 0 3 r f g < D O 0 1 , C j < D Replace 0 $378.7 $378.7 $486.7 ®>t86.7 after 1 51.5 *+9.0 27.3 26.0 7 Years 2 *+6.2 41.9 27.3 2M-.8 3 49.0 42.3 31.8 27.5 4 57.0 46.9 49.6 *t0.8 5 57.1 44.8 49.6 38.9 6 58.3 43.5 50.7 37.8 7 99.2 70.5 79.0 56.2 Totals 797*0 Less Present Worth. of Salvage Valuea Net Present Worth Equivalent Uniform Annual Coste 717.6 111.5 802.0 738.7 64i.i 110.9 Replace 0 379.7 379.7 486.7 486.7 after 1 54.2 51.6 27.3 26.0 8 Years 2 2 46.4 49.0 57.5 42.1 42.3 47.3 27.3 , 31-? 49.6 24.8 27.0 40.8 - 5 57.9 45.4 49.6 38.9 6 58.3 43.5 49.6 37.0 7 77.2 54.9 61.3 43.6 8 126.2 85.4 150.2 101.7 Totals 906.4 Less Present Worth of Salvage Value ^ Net Present Worth Equivalent Uniform Annual Coste 792.2 46.4 7453 115.6 932.9 826.5 61.9 764.6 118.5 aThe sum of disbursements and additional income tax from Table 23. b[Col. (X)] (pw£> - % - n). TABLE 2M- (Continued) c[Col. (3)] (pwf* - 5% “ n). ^■(Salvage value of Table 22) (pwf* - 5% - n). e(Iet present worth) (erf - 5% ~ n). f Minimum cost. rrH C r E V g b B tw C H V S ia C r E V B b W J M J . b V b E B C O ’ C 39' S O X S O D I A I S 30M 3 b E B I M C H ’ J 30 X S O O D I A I S I O M 3 ' b B I M i - E D I H r V S ’V O H C T E V B b t J I H X l E C H M I C V r b V b E B H O ’ J O J3 is selected in the case of purchase-rental. Since the equivalent uniform annual cost curves are flat and approxi mately equal over a range of several years, the choice of alternatives is sensitive to the assumed values of and Larger values of F^ and/or F^ would result in the selection of the assembly language system. The next analysis is to compare the winners of the previous analysis. An assembly language rental system is compared to a compiler language purchase-rental system in Table 25 and Table 26. From Table 26 and Figure it is seen that the compiler language system has a clear advan tage over the assembly language system for an after tax interest rate of 5 per cent and for F^. * 1.10 and Fm = 1.10. The effect on the solution of varying the technical and financial variables is determined in the next section. Sensitivity Analysis Program Preparation Let it be assumed that t^ie cost of programming in compiler language is equal to the cost of programming in assembly language ($36,100). This is the upper limit on program preparation cost for the compiler, since it is unlikely that the use of a compiler language will result in a greater program preparation cost than the use of an assembly language. From Table 25, disbursements for to co p o p 4 4 0 C D C D U )fO H O H H H -rtvxfooj VO - T O ON • • « • O O -T h •r -vl -r •P' 1 0 rurovji H^j-vjjo-'j * « * • • \OOjOJ O on HI U>ONO -Toofu • • • rooooi O J H o v -r -o • • coro p I? a s u p cd p 4 4 0 at ( 0 ro m h o H H -Troco ■ T O ON • • • O f H ■ F ' u»io (ovn OOMrO'vl • • • • ( jJ O J O ON H -r -r vjvoi • • 0000 uxo O O -’-v] • * • 0 * s3 H H _ . C D c+ H S C O P ^ 5 ( D H O 4& H -ru> ro on • • • T H -r corovn ooio- vj • • • O JO O N H 0 3= -T vn 00 to O n- n ] • • ro h Year n Disbursements during Year (Thousands) a Additional Income Tax (Thousands) Disbursements during Year (Thousands) Depreciation^ (Thousands) Investment® Tax Credit (Thousands ) Additional Income Tax (Thousands) o o a H* H C D 4 JE % C D !> CO CO W WCOCQ tr!«H 1 - 3 W ro vn TABLE 25 (Continued) Year n Assembly Language Compiler Language - i Li sbur sement sa during Year (Thousands) Additional Income Tax (Thousands) Lisbursements during Year (Thousands) Lepreciation (Thousands) Investment Tax Credit (Thousands) Additional1 3 Income Tax (Thousands) Replace $'t57.6f after 0 $ 36.1 22.0 $ 7.1 * t Years 1 120 A $9.9 27.3 $91.5 $10.7 2 12*+.0 27.3 68.6 l.if 3 127.6 30.9 ^5.8 25.5 b 150.9 60.6 22.9 33.7 Replace V57.6f after 0 36.1 22.0 7.1 5 Years 1 120 A 9.9 27.3 91.^ 10.7 2 12^.0 27.3 73.3 11.7 3 127.6 30.9 5^.9 20.9 b 128.9 ^9.6 36.5 21 5 l5*f.5 60.6 18 .*f 37.8 •r Os 3]ABLE 25 (Continued) S I S h Assembly Language Compiler Language Disbursements during Year (Thousands) Additional Income Tax (Thousands) Disbursements during Year (Thousands) Depreciation^ (Thousands) Investment Tax Credit (Thousands) Additional Income Tax (Thousands) Replace $1 +57.6f after 0 $ 36.1 22.0 $ 7.1 6 Years 1 120.1 * $20.6 27.3 $91.6 $21.1 * 2 12**.0 27.3 76.2 10.3 3 127.6 30.9 60.9 17.9 i+ 128.9 ^9.6 1*5.8 16.8 5 132.5 1+9.6 30.1* 26.3 6 158.1 60.6 15.1* 1 *1.1 Replace i*57.6f after 0 36.1 22.0 . 7.1 7 Years 1 120.1 * 31.3 27.3 91.5 32.1 2 12*+. 0 27.3 78.3 9.2 3 127.6 30.9 65.5 15.6 I f 128.9 1*9.6 52.1+ 13.5 5 132.5 1*9.6 39.2 21.9 6 136.1 1*9.6 26.0 30.3 7 175.8 62.1 13.2 50.3 £ "0 TABLE 25 (Continued) Year n Assembly Language Compiler Language Disbursements during Year (Thousands) Additional Income Tax (Thousands) Disbursements during Year (Thousands) iDepreciation^ (Thousands) i Investment6 Tax Credit (Thousands) Additional Income Tax (Thousands) Replace $*f57.6* after 0 $ 36.1 22.0 $ 7.1 8 Years 1 120.^ $31*2 27.3 $91 .*f $32.1 2 12M-.0 27.3 79.9 8.*f 3 127*6 30.9 68.8 l*f.O If 128.9 ^9.6 57.2 11.1 5 132.5 i f 9.6 b5.7 18.6 6 136.1 M-9.6 3^.2 26.2 7 153.8 51.1 23.1 39.8 8 175.8 150.2 11.5 7.1 ^rom Table 16. ^Differences in income tax disbursements due to differences in disbursements, depreciation and investment tax credit. cErom Table 21. Straight line depreciation for first two years and sum of the years digits depreciation thereafter, using depreciable value of Table 22. eE'rom footnote "d" Table 23. ^Capitalized. 00 TABLE 26 AFTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS ASSEMBLY LANGUAGE-RENTAL vs. COMPILER LANGUAGE-PURCHASE-RENTAL (Interest Rate = 5%) (1) Year n Assembly Language Compiler Language C tf rQ © .d ■P /-v •P '" '' d d m d m Q ) (O .rcJ j '-n O >d C M g © d m 3S 5 f f l ' —'TO m m -pm 2 Ft 0 © 0 m ,d tQ d E H (D E H •H p w d'—' P rd (L o s 0 m -P +3X-S d d ro d m © to 0 J- SL 0 ) d u-Nts d m m -pro d d d 2 B d Q © 0 xt 'H xi ro .d r o d EH ©EH •h d ' - ' d 1 - - ’ P > d Ph Replace after 1 Year 0 1 $ 36.1 ^ 36.1 lh2.*f 135.6 $h86.7 $^86.7 6h.5 61 .h Totals 178.5 171.7 Less Present Worth of Salvage Value Net Present Worth 171.7 Equivalent Uniform Annual Cost® 180.2 551.2 5^8.1 3g2.o i56.i 163.9 Replace after 2 Years 0 1 2 36.1 36.1 120.^ llh.6 lMf.O 130.6 h86.7 h86,7 51.0 ^8.6 68.3 59.0 Totals 300.5 281.3 Less Present Worth of Salvage Value4 Net Present Worth 281.3 Equivalent Uniform Annual Coste l5l.3 606.0 59^.3 332.1 262.2 1^1.1 i5o TABLE 26 (Continued) Assembly Language Compiler Language (1) Year n (2) Li sbursement a during Year (Thousands) (3) b Present Worth (Thousands) (4) Disbursements during Year (Thousands) (5) . Present Worth (Thousands) Replace 0 $ 36.1 $ 36.1 $486.7 $486.7 after 1 125.1 119.1 27.3 26.0 3 Years 2 124.0 112.5 41.5 37.6 3 149.6 129 *3 78.7 68.0 lotals 1 434.8 Dess Present Worth of Salvage Value** 397.0 mm 634.2 618.3 276.7 Jet Present Worth Equivalent Uniform Annual Coste 397.0 145.7 341.6 125.4 Replace 0 36.1 36.1 486.7 486.7 after 1 130.3 124.0 27.3 26.0 4 Years 2 124.0 112.5 28.7 26.0 3 127.6 110.2 56.4 48.7 4 150.9 124.2 94.3 77.6 Totals 568.9 Less Present Worth of Salvage Value ^ Jet Present Worth Equivalent Uniform Annual Coste 507.0 507.0 143.0 693.4 665.0 226.0 439.0 123.8f TABLE 26 (Continued) Assembly Language Compiler Language H a (2) Disbursements during Year (Thousands) +3/— \ U t o 0 fd 5 ro 2 3 w -P C O d d © 0 0 1 si 0 ) EH PL, (L° a Disbursements during Year (Thousands) _ (5) 0 Present Worth (Thousands) Replace 0 $ 36.1 $ 36.1 $486.7 $486.7 after 1 130.3 124.0 27.3 26.0 5 Years 2 12if.O 112,5 39.0 35.4 127.6 110.2 51.8 44.8 128.9 106.1 71.0 58 .if 5 154.5 121.1 98 .if 77.1 Totals 701.4 Less Present Worth of Salvage Value Wet Present Worth Equivalent Uniform. Annual Coste 610.0 610.0 lif0.9f 774.2 728 .if 179 .if 349.0 126.8 Replace 0 36.1 36.1 i+86.7 if86.7 after 1 lifl.O I3if.2 27.3 26.0 6 Years 2 12if.O 112.5 37.3 33.8 3 127.6 110.2 48.8 42.2 i f 128.9 106.1 66 .if 54.6 5 132.5 103.9 75.9 59.5 6 158.1 117.9 101.7 75.9 Totals 8if8.2 Less Present Worth of Salvage Valued- Net Present Worth Equivalent Uniform Annual Coste 720.9 720.9 lif2.0 8ifif.l 778.7 136*5 642.2 126.5 TABLE 26 (Continued) 152 Assembly Language Compiler Language Year n a l 0 3 ■ p H t o CM C D < B nd 0 © g C O w m 1 5 a joreh •H pi'-' M ^ (3) . Present Worth (Thousands) (l f) « Disbursements during Year ( Thousands ) (5) 0 Present Worth (Thousands) Replace 0 $ 36.1 $ 36.1 $if86.7 $if86.7 after 1 151.7 lifif.if 27.3 26.0 7 Years 2 12if.O 112.5 36.5 33.1 3 127.6 110.2 if6.5 if 0.2 if 128.9 106.1 63.1 51.9 5 132.5 103.9 71.5 56.1 6 136.1 101.5 79.9 59.3 7 175.8 125.0 112.if 79.9 Totals 1*012.7 Less Present Worth of Salvage Valuea Net Present Worth Equivalent Uniform Annual Coste 839.7 923.9* 833.2 m m 839.7 l*f5.3 97.6 735.6 127.3 Replace 0 36.1 36.1 if 86.7 if86.7 after 1 151.6 lifif.3 27.3 26.0 8 Years 2 l-2if.O 112.5 35.7 32.i f 3 127.6 110.2 Mf.9 38.8 if 128.9 106.1 60.7 50.0 5 132.5 103.9 68.2 53.5 6 136.1 101.5 75.8 56.5 7 153.8 109 A 90.9 6i f .6 8 175.8 119.0 157-3 106.5 Totals 1,166 .if Less Present Worth of Salvage Value* 1 STet Present Worth Equivalent Uniform Annual Coste 968.0 968.0 lif6.2 l,Oif7.5 915.0 61.9 853.1 132.2 aSum of disbursements and additional income tax from Table 25. TABLE 26 (Continued) 153 *[Col. (2)] (pwf - 5% - n). °[Col. (if)] (pwf* - 5% ” n). ^(Salvage value from Table 22) (pwf* - 5% ~ n) • e(Net present worth) (erf - 5% “ n). f Minimum cost. C r k v t J b H J U l bVbEB C O ' C 3 8 5 0 X 5 0 D t A 1 2 l Q W a bEB I W C H ' I S O X S O O D I A I S I O M S * b t f 1 M J . E D I M n a V O H C T E V B b B l H A l E C H H I C V r b V b E B H O * 1 0 1 2 155 program preparation and additional income tax are equal to $29,100 for the compiler language in Year 0, The differ ence in disbursements ($36,100 - $29,100) in Year 0, when multiplied by the appropriate capital recovery factor and added to the compiler language annual costs, will not cause the annual cost in compiler language to be equal to or greater than the annual cost in assembly language at any time in the eight year period. Therefore, within expected limits of the ratio of compiler language to assembly langu age programming cost, the choice of the two alternatives is not sensitive to differences in cost of program preparation* Program Maintenance and Oomputer Maintenance The costs of program maintenance and computer main tenance are fairly predictable. Neither of these costs are significant with respect to the choice of hardware-software alternatives in this illustration. Object Program Execution Time Tabulated on the following page are the values of which must not be exceeded in the year shown in order for the use of a compiler language not to cause an infeasi bility in the hardware system under consideration. These values of are calculated from the data presented in the Object Program Execution Time Section for JOB 3 (Input Bata for Hardware-Software Model). They are based on the use of 156 the fastest tape units available in the system and the use of two tape controllers. Maximum Year Value of F - f c 1 1.36 2 1.32 3 1.28 k 1.25 5 1.22 6 1.18 7 1,15 8 1.12 In order to test the sensitivity of the program- ming language-hardware system selection to changes in the value of Ft, an equal to 1.20 will be used. The origi nal value of Ft was equal to 1.10. In order to remain feasible in Year 6, it is necessary to add a fifth tape unit in Year 6 in order to perform a way merge instead of a 3-way merge in the collate of JOB 3* This procedure will result in a time reduction of approximately 25 per cent in JOB 3 and will permit feasibility to be maintained in all years under consideration. The additional rental, beginning in Year 6, for the fifth tape unit is $9,5^0 per year or $ * ■ * • , 770 after tax cost. This amount when converted to annual cost values at an interest rate of 5 per cent for Years 6, 7, and 8, is too small to have an effect on the choice of alternatives. Therefore, within reasonable limits of the ratio of compiler language object program execution time to assembly language object program 1 5 7 execution time (1,00 to 1.20), the choice of alternatives is insensitive to object program execution speed. Ob.iect Program Memory Requirements It will be assumed that Pm is equal to 1.20 rather than 1.10 for the compiler language system, as was pre viously the case. Prom the data presented in the Object Program Memory Requirements Section (Input Data for Hard- ware-Software Model), it is determined that the program size in Run 31 cannot exceed 28.6K, the difference between the 652 memory capacity and the data and operating system memory requirements. Therefore, with Gm = .05 and Pm = 1.20, memory capacity will be exceeded in Year 5 k [(1.05) (1.20) (20K) = 29.22], The additional rent for a central processor with a larger memory capacity is $88,080 per year. An additional after tax cost of R*f.l (thousands) will be incurred in Years 5, 6, 7 and 8. The calculation of the revised annual cost for the compiler language is shown in Table 27. Program Conversion The cost of program conversion when a compiler language is used for the new challenger was estimated as 50 per cent of the compiler language program preparation cost of the present challenger. The upper limit on the cost of conversion would be 100 per cent of the original 158 TABLE 27 AFTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS COMPILER LANGUAGE-PURCHASE-RENTAL (Fm = 1.2Q,a Interest Rate = 5$) (1) rear n rO X t 4» h 0 3 OOtJ '-'SO E j co i —) c 3 4a « • p i d ) E H f n c wv —' P4 o s i - p ^ 0 f n m 0 ) o o •Vi > :s o 5 CO*H H C O - P + = • » 0 3 C O 33; 3 r - j 03- O I,®, 001 n a Additional Equivalent Uni form Annual Cost (Thousands) (5) Revisede Equiva lent Uniform Annual Cost (Thousands ) 5 $3^.7 $ 3^.7 $ 8.0 $13^.8 6 33.0 67.7 13.3 139.8 7 31 99.1 17.1 lifif.if 8 29.9 129.0 20.0 152.2 aEstimated ratio of memory size using compiler language to memory size using assembly language. bM*.l (pwf1 - 5$ - n). 0 Cumulative sum of Col. (2). d[Col. (3)] (erf - - n). eCol. (^) plus equivalent uniform annual cost for compiler language system in Table 26. NOTE: From a comparison of the above data and the data in Table 26, it is seen that the compiler language system is still the better alternative. 159 programming cost. An additional $11,000 spent on program conversion would result in an after-tax disbursement of $5,500 in the last year of life of the present challenger. This amount, when converted to an annual cost figure, will not affect the selection of alternatives. Interest Eate The equivalent uniform annual costs of the two systems using an after-tax minimum attractive rate of return (interest rate) of 10 per cent is calculated in Table 28 and shown in Figure 5* It is seen that the assem bly language system is now superior to the compiler langu age system over almost all the time span under considera tion. This result is due primarily to the economics of purchase versus rental rather than to the use of a particu lar programming language. As shown in Figure 6, the capital recovery cost for the purchased system is higher for the 10 per cent interest rate than it is for the 5 per cent interest rate. The higher capital recovery cost causes the total equivalent uniform annual cost to be higher for each year of assumed life. The equivalent uni form annual costs of disbursements are approximately equal for the two interest rates. The initial outlay for the assembly language system is small compared to that for the compiler language system. Therefore, the capital recovery cost has negligible effect on the assembly language ro W j o © © cf H P © JB s * s 0 3 ( 0 H H C D f3 O cf P H 0 9 Co O O • vr\ H vrv ro CO I t ro V W H O H H -Froco -F O O N • • • O -FH H H H OU) OOnOOn • • • N O -TH L f e i t e j H J& iQ 0 0 (0 5 fj cf H > 0 0 PH- T O p < 4 O r ) 01 © < 2 < j © oP f f i id m O d-PW ® CO cf © P Cf Cl cf ©a «;<4 Hops: H>4 H O H (y ® cf & PP* H 00 ro i i * f3 O cf P H G O 00 vox H O N vn * N JX O N o ON o -r ONvnCo 00H0N • • • U> O-Nj VA W H ro Lo cn On 00 O 00 N J l - n3 ro N O • • • • - f H -r N Jl F* NA-FCO ON ON ON • • • -r-r^a H H CO "J OO H • • H O lo V Ul ¥ n j t . co fieplace after 1 Year H O <1 > Year n Sfr H ■Fw ro on • • -T H (2) Disbursements during Year (Thousands) = 1 8 = H ro co NO ON • • FH (3) b Present Worth (Thousands) -F QN 00 -F On • * V O X ' S ] W a Di sbursement s during Year (Thousands) -F V J \ 0 0 00 On • • ONNJ (5) c Present Worth (Thousands) > m 0 3 © g - a 0 1 H* H © H S P © cp £p <1 02 1 -3 H r a g 1 *! J Bo O £4 0 8 e«if3 • f3 B ro 00 O N o 161 TABLE 28 (Continued) Assembly Language Compiler Language A H -9 JH (2) Li sbursement sa during Tear (Thousands) (3) ^ Present Worth (Thousands) (1° a L i sbursements during Tear (Thousands) C5) . Present Worth (Thousands) Eeplace 0 $ 36.1 $ 36.1 W 6 . 7 $*♦86.7 after 1 125.1 113.7 27.3 a*f .8 3 Tears 2 12*f.0 102.h ifl.5 3*f.3 3 1^9.6 112.3 78.7 59.1 Totals **3^.8 36if.5 Less Present Worth of Salvage Valued- Net Present Worth - Equivalent Uniform Annual Coste 1^6.5 63^.2 6o*t.9 2*f0.5 36k.h 1^6.5 Beplaee 0 36.1 36.1 if 86.7 ^ 86.7 after 1 130.3 118 .h 27.3 2if.8 if Tears 2 120 .if 102 .b 28.7 23.7 , 3 127.6 95.8 56 A if2.if k 150.9 103.1 9^.3 6if A Totals 568*9 *+55.8 Less Present Worth of Salvage Value4 “ Net Present Worth Equivalent Uniform Annual Cost® 1^3*6 693.^ 6if2.0 I87.6 W J * lif3.1 162 TABLE 28 (Continued) r - \ d a £ Assembly Language Compiler Language <8 ,Q 09 Si 4a 4a d U m u ro ^ <» cij s 0 *d cm S a> d c o s d ® {H ffl w 00 ro ro 4> 09 1 11 SI r O f l si m x j 09 M e n < 1 ) EH •H P ' ' - ' h ' - ' (lf) a Disbursements during Tear (Thousands) (5) „ Present Worth (Thousands) Beplaee after 5 Tears 0 1 2 t $ 36.1 $ 36.1 130.3 118.1 * 12^.0 102A 127.6 95.8 128.9 88.0 154-.5 95.9 $1*86.7 $**86.7 27.3 2V.8 39.0 32.2 51.8 38.9 71.0 1*8.5 98.1* 61-;l Totals 701A 536.6 Less Present Worth. of Salvage Value® Wet Present Worth Equivalent Uniform - innual Cost® 1^1.7 77l*.2 692.2 ll*2.1 550.1 ll*5.2 Beplaee after 6 Tears 0 1 2 ■2 5 6 36.1 36.1 1^1.0 128.2 12^f .0 102.1 * 127.6 95.8 128.9 88.0 132.5 81.7 158.1 89.3 1*86.7 1*86.7 27.3 2M-.8 37.3 30.8 1 *8.8 36.6 66 A i*5.i* 75.9 1*7.1 101.7 57.5 Eotals 81*8.2 621.5 Less Present Worth of Salvage Value® Wet Present Worth Equivalent Uniform Annual Cost® 1^2.9 81 *1*.1 728.9 103.1* 625.5 li*3.9 163 TABLE 28 (Continued) Assembly language Compiler language Year a } m 4» P l M a ? a J • x s ✓ — > . g j) r i CM 3 w 09 0 3 Hi P»*CJ (3) b Present Worth (Thousands) a Disbursements during Year (Thousands) h 09 O w + 3 m S 3 p l 0 ) o 0 9 ,3 C DE -J H W PM Beplaee 0 $ 36.1 $ 36.1 $486.7 $486.7 after 1 151.7 137.9 27.3 24.8 7 Years 2 124.0 102.4 36.5 30.1 3 127.6 95.8 46.5 34.9 4 128.9 88.0 63.1 *+3.1 5 132.5 82.3 71.5 . 44.4 6 136.1 76.9 79.9 45.1 7 175.8 90.2 112.4 57.7 Totals 1,012.7 less Present Worth, of Salvage Value3 709.6 923.9 766.8 70.4 Net Present Worth Equivalent Uniform Annual Cost® 145.5 696.4 142.8f Beplaee 0 36.1 36.1 486.7 ^ 86.7 after 1 151.6 137.8 27.3 24.8 8 Years 2 t 124.0 127.6 128.9 102.4 95.8 88.0 35.7 44.9 60.7 29.5 33.7 41.5 5 132.5 82.3 68.2 42.4 6 136.1 76.9 75.8 42.8 7 153.8 78.9 90.0 46.6 8 175.8 82.1 157.3 73.5 Totals 1,166.4 less Present Worth of Salvage Value3 Net Present Worth Equivalent Uniform Annual Cost® 780.3 145.9 1,o47.5 821. ^ 42.7 mJk 145.6 aFrom Table 26, b[Col. (2)] (pwff - 10% - n). TABLE 28 (Continued) c[Col. (if)] (pwff - 10$ - n). ^(Salvage value from Table 22) (pwf' - 10$ - n). e(Net present worth) (erf - 10$ - n). ■^Minimum cost. as - -1 - 4 - H ~h i - r C r e V H b B I M J . b V f e E M C O ’ C 3 8 ' S O X S O D I A I 2 I O H 2 b E B 1 H C H ' 1 2 0 X S O O D l A 1 2 1 0 1 4 2 ' O H V g b B IH lC H V K ia b B l M l E D IVl r v e ’V * O H C T E V B b B l H X l E C H M l C V T b V b E U M O ' 1 0 1 2 ator .on flasAR j a o i h h o b t TutnqHABJD m o ,A,a.u m » aaTMiyt'i ZTaA H ^iTM iaqflA ajli ,enoi2ivia oos x oar .homi na^ 3Moi2ivia os x os .eeo .00 TnrR«iHA3jp 11 1 167 equivalent uniform annual costa. A comparison of Figure ^ and Figure 5 reveals that the equivalent uniform annual costs of the assembly language system are approximately equal for interest rate of 5 per cent and 10 per cent, respectively. Disposal Value The superiority of the compiler language purchase- rental system over the assembly language rental system is to some extent dependent on the resale value of the pur chased system. In order to test the sensitivity of the solution to estimated salvage value, a salvage value 10 per cent lower for each year than the value used in Table 26 is used in Table 29. The calculation of equivalent uniform annual cost is made for a life of five years. With this modification the equivalent uniform annual cost for the compiler language system is still less than that of the assembly language system for an assumed life of five years. Purchase of Additional Equipment If the equipment which is required in Tears 3 and b under the compiler language plan (Table 19) is purchased rather than rented and the system is replaced after five years, a slight improvement in annual cost is obtained. The equivalent uniform annual cost of this system for a five year life is $12*+,000 (calculations not shown) versus TABLE 29 AFTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS-REBUCED SALVAGE VALUE ASSEMBLY LANGUAGE-RENTAL vs. COMPILER LANGUAGE-PURCHASE-RENTAL (Interest Rate = 5$) H e place i n 5 Years Assembly Language -Rental Compiler Language-Purchase Rental Disbursements8 , Additional^ Disbursements8 In-vestment8 Additional0 Year during Year Income Tax during Year Depreciation0 Tax Credit Income Tax n (Thousands) (Thousands) (Thousands) (Thousands) (Thousands) (Thousands) $*57.6d 0 $ 36.1 22.0 $ 7.1 1 120 . * $17.5 27.3 $106.7 $10.7 2 12*.0 27.3 85.5 5*6 3 127.6 30.9 6*.l 16.3 k■ 128.9 *9.6 *2.6 18.* 5 ■ 15*.5 60.6 21.5 36.2 320.* H o\ oo TABLE 29 (Continued) Assembly Lanffuase-Hental Compiler Lansuajsre-Purchase Rental (1) < • 2) e Disbursements \b) Disbursements® W Tear during Tear Present Worth1 during Tear Present Worth^ n (Thousands) (Thousands) (Thousands) (Thousands) 0 $ 36.1 $36.1 $*f86.7 $*f86.7 1 137.5 130.9 27.3 26.0 2 12>*.0 112.5 32.9 29.8 3 127.6 110.2 **7.2 1*0,8 b 128.9 106.1 68.0 56.0 . i ____________ 15^.5 _ _ 121.1 96.8 .. 2. . Totals Less Present Worth of Salvage Value® Net Present Worth Equivalent Uniform Annual Cost11 616.9 616.9 1^2.5 715.2 571.7 132.1 aProm Tahle 25. ^Differences in income tax dishursements due to differences in disbursements, depreciation and investment tax credit, ®Sum of the years digits depreciation of $320,^00. ^Capitalized. ®Sum of disbursements and additional income tax from Table 29. f[Col. (2)1 (pwf1 - % - n). fCCol. (*f)j (pwf* - % - n). Salvage value ($183-000)] (pwf* - 5% “ 5). ••■(Net present worth) (erf - 5^ ~ 5). H O' vo 170 $126,800 for the purchase rental system (Table 26), The method of paying for equipment modifications and additions does not exert a great influence on annual cost because of the relatively small costs of modifications and additions in relation to the original purchase price. Method of Depreciation Equivalent uniform annual costs for an assumed life of five years for the assembly language-rental system and compiler language-purchase-rental system, using straight line, sum of the year’s digits and double declining balance methods of depreciation are shown in Table 30. The assem bly language system equivalent uniform annual costs change with method of depreciation because of the impact of income tax payments relative to the computer language system. It is seen that the choice of alternatives is not sensitive to the method of depreciation for an assumed life of five years. The straight line method of depreciation is not at a disadvantage relative to the rapid write-off methods because disbursements increase with the age of the equipment. Income Tax Rate If it is assumed that the federal income tax rate will continue to decrease over the next several years, the 171 TABLE 30 COMPARISON OP METHODS OP DEPRECIATION (Pive Year Life, Interest Rate = 5%) Method Assembly Language- Rental (Thousands) Compiler language Purchase-Rent al (Thousands) Straight Line ' $L*t0.6 $12^.0 Sum of the Year's Digits Xb0.9 126.8 Double Declining Balance 151.0 135.6 aBased on data in Table 25* 172 margin by which the purchased compiler language system is superior to the rented assembly language will increase. An examination of Table 25 reveals that the disbursements for income taxes are higher for the compiler language system in every year except Tear 1. Therefore, a decrease in the income tax rate will lower the equivalent uniform annual cost of the compiler language system relative to the equivalent uniform annual cost of the assembly language system. If it is assumed that the income tax rate increases at a rate of 2 per cent per year for the next five years, the equivalent uniform annual cost of the compiler language system increases from $126,800 (Table 26) to $132,300 for a five year life. This cost is still lower than the assem bly language system equivalent uniform annua.l cost of $1^0,900 for a five year life (Table 26). Thus, the choice of alternatives is not sensitive to either an increase or decrease in income tax rate. Analysis of Solutions The following conclusions can be stated as a result of the solutions obtained in the hardware-software analysis: 1. An assembly language system is favored in the case of rental for even small values of Fm and (Figure 2). This superiority increases for larger values of Fm and F - j . . The choice of alternatives is insensitive to the interest rate, either before or after taxes, because of the absence of a large initial investment. There is little difference in minimum equiva lent uniform annual cost between language systems in the case of purchase-rental for small values of P^ and Pm (Pigure 3)* The assembly language system would be superior for larger values of P^. and Pm- The choice of alternatives is not sensitive to the interest rate because both alternatives have approxi mately the same pattern of disbursements. In the case of an assembly language rental system versus a compiler language purchase- rental system, the choice of alternatives is more sensitive to the interest rate than it is to changes in P^ or Pm (Pigure b and Pigure 5)« The compiler language purchase-rental system is favored at low interest rates and the assem bly language system is favored at high interest rates. The minimum attractive rate of return (inter est rate) on alternative investments may be the single most important factor in the selec tion of an optimal hardware-software system. This factor may he more important than the technical characteristics of the programming language. At low rates of interest the oppor tunity cost of the initial investment in the compiler language purehase-rental system is not large in comparison to the advantage to he obtained from lower annual disbursements over the assembly language rental system. At higher interest rates the opportunity cost of the initial investment is too high in relation to the lower annual disbursements and an assembly language rental system is favored. Summary and Conclusions The Hardware-Software model has been developed in order to provide a single vehicle for the selection of a minimum cost computer system which satisfies the constraints imposed by applications which grow with respect to time and the constraints imposed by the use of a given programming language. The use of the model permits the best present challenger to he selected from the list of available present challengers. This best present challenger and the best future challengers are used to contend the defender in order -to determine if and when replacement of the present system is appropriate. The model can he used to study the effect on the choice of alternatives of a variety of technical and financial variables* IKPtJT DATA EOR HARDWARE-SOFTWARE MODEL 177 INPUT DATA POE HARDWARE- SOFTWARE MODEL Program Preparation Program Requirements' JOB 1 Run 1 1,000 instructions Bun 2 - (supplied "by manufacturer) Run 3 2,500 Bun h - (supplied by manufacturer) Run 5 2,000 JOB 2 Run 1 500 Run 2 1.000 %>a = 10,000 instructions Npc = Npa Fm = 10,000 x 1*10 — 11,000 instructions Assembly Language The average computer time required per programmer for assembly and debugging during the program preparation o period = .007 (2,000) - lM- hours per year. Fifty-two •^From program requirements, Input Data for Hard ware Model. 2 ^Assumed ratio of assembly plus debugging time to total programming time. Jeriy Schwalb, "Compiling in English," Datamation. (July, 1963)» 28-30. 178 TABLE 31 ESTIMATED PROGRAMMER PRODUCTION RATES Assembly Language Compiler Language Programmer Efficiency ea Estimated Produc tion Rate3 , (Instructions/Day) ^ea Programmer Efficiency ec 1 1 O C Q g O a *H » C J 0)+> ® +» O • h rt rap^ W -P w 1.00 19.0 1.00 37.6 .80 15.2 .90 33.8 •60 1 1.1+ .70 26.3 aEstimate based on data from Christopher J. Shaw, "More Instructions— Less Work," Datamation, I June. 196*+), 3*+-35. TABLE 32 ESTIMATED ANNUAL PROGRAMMER COSTS Programmer Efficiency (Assembly Language) Programmer Efficiency (Compiler Language) Estimated Annual Programmer Cost 1.00 1.00 $15,000 .80 .90 11,250 •60 .70 i 10,000 179 per cent of the computer time is consumed in assembly and *+8 per cent is consumed in debugging.3 She prime shift computer cost per hour is $*+6 based on the optimal configuration cost of $9,1^7 (Solution 3, Sable 3) plus $10 per hour for two operators. The total computer cost per programmer during the program preparation period is given in Table 3*K Minimize Za * 15,627 \ m0a> + 12,03*+ ST.ga + 11 »°lf7 ^ a (from eq. *+0), subject to: --------- .......... . ...... ..... - 250 working days N1.0a + 1^-2 K.Sa + *.6a <froH1 eq.^1) ^•0 N1#0a + N#8a + 11A U#6a * 1+0 H.0* - fl » “tea ■ a « ^ a ■ i - za - 135,112 Computer Debugging Time U*3a = 2 ; 2 (6.7) = 1 3 hours ®?6a - 1 t 1 (9.0) - _^0 " 2 2. 1+ » It is assumed that debugging will take place during the last two months of the one year program preparation period and that all programs will be reassembled and debugged on a daily basis during this period. ' ^^*1+2 - 32 minutes per debugging run 3Ibid.. p. 30. 180 TABLE 33 ANNUAL COMPUTER TIME REQUIRE]) POR ASSEMBLY LANGUAGE PROGRAMMING Programmer Efficiency ea Assembly Time per Year per Programmer (Hours) Debugging Time Per Year per Programmer (Hours) Total Computer Time per Year per Programmer (Hours) 1.00 5.8a 5.1fb 11.2 .80 7.3° 6.7d l*f .0 .60 9.7® 9.0f 18.7 a(.80) (Ilf) (.52) c(l*f)(.52) e(1.33)(lif)(.52) f(1.33)(l^)(A8) TABLE 3k ANNUAL COST OP ASSEMBLY LANGUAGE PROGRAMMING Programmer Efficiency ea Programmer Costa cep Computer Cost® em Total Cost ®ep + ^em 1.00 $15,000 $ 627 $15,627 .80 11,250 78k 12,03^ •60 10,000 1,0k? 11,0^7 aProm Table 32. b(Computer time from Table 33) ($56 per hr.). Assembly Time 2 (7.3) = 1^.6 hours 1 (9.7) = 9*2 " W i 3 " . . Ci^. ^ .I-C^S) = 35 minutes -per assembly run It is assumed that a sixty minute limitation T - j . has been established as a total time for assembly and debugging Since the actual time is 35 + 32 = 67 minutes, the first solution is not feasible, The revised solution is J'f.Oa = 2 . H?8a » 2 . - ° , Z % = 3.6^102. Revised Computer debugging Time 3 (6.7) = 20.1 hours (2Q.l^, (60j. _ 29 minutes per debugging run Revised Assembly Time 3 (7.3) = 21.9 hours , (. ^l- t . 9 . ^ (60l = 31 minutes per assembly run Total time per run = 31 + 29 = 60 minutes Assembly Time Constraint iPa. QQQ 31 bb), sa 4 ^Qi. QQA _ 323 instructions a per minute It should be possible to achieve this minimum assembly rate with the optimal configuration (Solution 3> Table 3). 182 Compiler Language The average ratio of compilation time to assembly I f . time is assumed to be equal to 1.18. The average ratio of computer debugging time when using a compiler language to computer debugging time when using an assembly language is 5 assumed to be equal to .75. These ratios are applied to the previously developed assembly and debugging times language in order to develop the corresponding times for the compiler language. Again, using a prime shift rate of $56 per hour (including operators) the total computer cost per program mer is shown in Table 36. Minimize Zc = 15,610 % #oc + 12,012 H#9c + 11,019 ^#7C (from eq. *+5) subject to: -------- HAaQSQ - 250 (from eq. *f6) 37.6 N1#0e + 33.8 2 T#9e + 26.3 H.7c 37.6 N1#0c + 33.8 F.9c + 26.3 ^ bb K£.0c = 0 » ®*9c - 0 » Nt7c * ^ » ZS s $22,038 Computer Debugging Time Required H*7c s 2 ; 2 (6.8) ■ 13.6 hours per year = 20 minutes per debugging run ^Sehwalb, Datamation. (July, 1963)j 28-30. ^Ibid.. p. 30. 183 TABLE 35 ANNUAL COMPUTER TIME REQUIRED FOR COMPILER LANGUAGE PROGRAMMING Programmer Efficiency ec Compile Time8 , per Year per Programmer (Hours) Debugging Time1 3 per Year per Programmer (Hours ) Total Computer Time per Year per Programmer (Hours ) 1.00 6.8 *+.1 10.9 .90 8.6 5.0 13-6 .70 11A 6.8 18.2 a(l.18)(Assembly Time of Table 33)• k( .75)(Debugging Time of Table 33). TABLE 36 ANNUAL COST OF COMPILER LANGUAGE PROGRAMMING Programmer Efficiency ec Programmer Cost8 , °ep 1 . Computer Cost °em Total Cost ^ep + cem 1.00 $15,000 $ 610 $15,610 .90 11,250 762 12,012 .70 10,000 1,019 11,019 aFrom Table 32. ^Computer Time from Table 35) ($56 per hour). l8*f Compilation Time Required = 16 minutes per oompilation run Total time per run = 16 + 20 = 36 minutes Compilation Time Constraint It is assumed that there is an average of two machine language instructions per one compiler language statement. Npa f 2 s 5,500 equivalent compiler statements ■ Si?QS s 16 ^9), Sc - - 3*& statements per minute c There should be no difficulty in obtaining this minimum compilation speed with a COBOL compiler for the optimal equipment solution (Solution 3, Table 3)* Program Maintenance Assembly Language, Gm = .05 •25 &q»poo> £ 25o ( P r o r a ea 51) 19.0 N1-0a + 15.2 N>8a + 11A U,6a 250 (Pr0m e4> 51)- n-1 19.0 K1-0a + 15.2 M>8a + 11A IT.6a 2 10 (1.05) A uAuthor’s experience on tests conducted on COBOL, University of Southern California Computer Sciences Labora tory, February, 196*f. 7 'Average ratio obtained from tests of several com piler and assembly languages. Shaw, Datamation. (June, 19<&), 3^-35. TABLE 37 ANNUAL PROGRAM MAINTENANCE COST ASSEMBLY LANGUAGE <G* - .05) Year n Programmer8 , Efficiency ea Optimal Number8 of Programmers N* aea Programmer Cost1 3 Gep 1 .60 1 $10,000 2 .60 1 10,000 3 .60 1 10,000 b .80 1 11,250 5 .80 1 11,250 6 .80 1 11,250 7 .80 1 11,250 8 .80 . ____ ... - 1 11,250 aComputed from equation 5l. ^From Table 32. 186 TABLE 38 COMPUTER TIME REQUIRED POE PROGRAM MAINTENANCE ASSEMBLY LANGUAGE (K = .25) Year n Programmer8, Efficiency ea Optimal Number8 of Programmers 4 a Assembly- Time (Min./Week) Debugging Time (Min./Week) 1 .60 1 ll£ 10d 2 .60 1 n £ 10d 3 .60 1 llb 10d b .80 1 gc 8® 5 .80 1 8® 6 .80 1 8® 8e 7 .80 1 8 8® 8 .80 1 8C 8® aProm Table 37. *>(9.7) (60) t 52. °(7.3) (60) f 52. d(9.0) (60) r 52. e(6.7) (60) r 52. 187 The only program maintenance cost is the cost of personnel, since the computer time required for assembly and debugging amounts to approximately twenty minutes per week* This small amount of time will not require overtime in any year. There will be no additional computer rental required for assembly and debugging for the purpose of program maintenance* The constraints on total assembly and debugging time (52) and assembly speed (53) are of no con sequence • Compiler Language G r m = .05, c 1,10 | 7 . * • - 37'6 H1.0o + 33.8 H>9e + 26.3 H,7c a 11 (1.05)n_1 The computer time required for compilation and debugging will not require the use of overtime in any year. The only program maintenance cost is the cost of personnel. There will be no additional computer rental required for compilation and debugging for the purpose of program maintenance. The constraints on total compilation and debugging time (55) and compile speed (56) are of no con sequence. 188 TABLE 39 ANNUAL PROGRAM MAINTENANCE COST COMPILER LANGUAGE (G* = *05, Fm = 1.10) Tear n Programmer8 , Efficiency ec Optimal Number8 of Programmers N* ec Programmer1 3 Cost ep 1 .70 1 $10,000 2 .70 1 10,000 3 .70 1 10,000 if .70 1 10,000 5 .70 1 10,000 6 .70 1 10,000 7 .70 1 10,000 8 .70 1 10,000 aComputed from equation 51 *. ^From Table 32. 189 TABLE ^-0 COMPUTER TIME REQUIRED FOR PROGRAM MAINTENANCE COMPILER LANGUAGE (K = .25, * 1.10) Year n Programmer8 , Efficiency ec Optimal Number8 of Programmers N* ec Compile1 3 Time (Min./Week) Debugging0 Time (Min./Week) 1 .70 1 13 8 2 .70 1 13 8 3 .70 1 13 8 k- .70 1 13 8 5 .70 1 13 8 6 .70 1 13 8 7 .70 1 13 8 8 .70 1 13 8 aFrom Table 39* b(ll.W (60) T 52. °C 6.8) (60) f 52. 190 TABLE >fl OBJECT PBOGBAM EXECUTION TIME ASSEMBLY LANGUAGE, Ga = .03 (JOB 3 Constraint = 7,200 Seconds) Year 1 Tape Time = 8Mfa x 1.00 x k- = 3,376 (eq. 58) Printer Time = 3,678s x 1.00 x 1 = 3.678 (eq. 58) JOB 3 Time = 7,05*f seconds, Z=$9.197 Configuration! CB1, T12, T22, T32, T**2, P2, CPI and one tape controller. Year 2 Tape Time = 8Mfa x 1.03 x 3 = 2,608 Tape Time a s 60^-a x 1.03 x 1 = 622 Printer Time = 3,678s x 1.03 x 1 = 3.788 JOB 3 Time = 7,018 seconds, Z=$9.^97 Configuration! CB1, T12, T22, T33.** T^2, P2, CPI and one tape controller. Year 3 Tape Time = 8M+a x 1.06 x 2 = 1,789 Tape Time = 60*+a x 1.06 x 2 = 1,280 Printer Time = 3,678a x 1.06 x 1 = 3.899 JOB 3 Time = 6,968 seconds, Za$9.797 Configuration: CHI, T12, T23**, T33, T*+2, P2, CPI and one tape controller. 191 TABLE ifi (Continued) Tear Tape Time * = 8U4a x 1.09 x 2 = l,8*+0 Tape Time * 60^a x 1.09 x 2 = 1,317 Printer Time = 3>678a x 1.09 x 1 = *f>009 JOB 3 Time = 7,116 seconds, Z*$9*797 Configurations CB1, T12, T23^» T33, T*f2, P2, CPI and one tape controller. Tear 5 Tape Time « SM*8, x 1.12 x 1 = 9^5 Tape Time = 60*fa x 1.12 x 3 = 2,029 Printer Time = 3»678a x 1.12 x 1 = * k !19 JOB 3 Time s = 7,093 seconds, Z=$10,097 Configuration! CEL, T 1 3 T23, T33? T^2, P2, CPI and one tape controller. Tear 6 Tape Time a 6Ch& x 1.15 x b = 2,778 Printer Time * 3j678a x 1.15 x 1 = *».230 JOB 3 Time = 7,008 seconds, Z=$10,397 Configuration! CEL, T13, T23, T33> T ^ 3 ^2, CPI and one tape controller. TABLE h i (Continued) 192 Year 7 Tape Time = 60*fa x 1*19 x ^ = 2,875 Printer Time = 3*678a x 1.19 x 1 = **-.377 JOB 3 Time = 7,252 seconds Year 7 (Eevised)c Tape Time * 60»*a x 1*19 x h x .67 * 1,927 Printer Time = 3,678a x 1.19 x 1 = *<-.377 JOB 3 Time * = 6,30*f seconds, Z * $11,877 Configuration: CKL, T13, T23» ^33 > ^3, P2, CPI and two taoe controller and tape switching units Year 8 Tape Time = 601 +a x 1.23 x h x .67 Printer Time a 3,678a x 1.23 x 1 JOB 3 Time Configuration: Ho change. aProm Table 2. ^Change in configuration. cAdd a second tape controller and tape switching unit at an additional rental of $1,^80 per month (IBM System 360 Price List). This equipment will permit a 3“way merge, with overlapped reading and writing, to be performed in the collate in JOB 3 rather than the present 2-way merge. A reduction in JOB 3 time of approximately one-third will be achieved with no reduction in the other job times. * 1,985 = iti-528 = 6,^93 seconds, Z = $11,877 193 TABLE 1+2 OBJECT PROGRAM EXECUTION TIME COMPILER LANGUAGE, Ga = .03, P-t = !*10 (JOB 3 Constraint ■ 7200 seconds) Year 1 Tape Time « 8*fA+a x 1.00 x 1.10 x 1 — 928 (eq[. 59) Tape Time = 60*+a x 1.00 x 1.10 x 3 = 1,993 (e<i. 59) Printer Time = 3,678a x 1.00 x 1.10 x 1 = if,0*+6 (ea. 59) JOB 3 Time = 6,987 seconds, z = $10,097 Configuration s CB1, T12,b T23,b T33,1 ) 1^2, P2, CPI and one tape controller. Year 2 Tape Time = 8Mfa x 1.03 x 1.10 x 1 = 95*f Tape Time = 60*fa x 1.03 x 1.10 x 3 = 2,OU-9 Printer Time * 3T678a x 1.03 x 1.10 x 1 = *+.156 JOB 3 Time = 7,159 seconds Z = $10,097 Configuration: No change. Year 3 Tape Time - 6oM-a x 1.06 x 1.10 x *+ » 2,828 Printer Time = 3,678a x 1.06 x 1.10 x 1 » *+,303 TABLE ^2 (Continued) 19*f Tear 3 (Continued) JOB 3 Time = 7,131 seconds Z = $10.397 Configuration: CS1, T13, T23, T33, Tif3,b £2, CPI and one tape controller. Tear if Tape Time = 60ifa x 1.09 x 1.10 x if = 2,900 Printer Time = 3*678a x 1.09 x 1.10 x 1 = if .ifjif JOB 3 Time = 7,31*+ seconds Add a second tape controller and tape switching unit at an additional rental of $1 ,if80 per month, as in the case of the assembly language p. 192. Tear if (Revised) Tape Time » 60ifa x 1.09 x 1.10 x if x .67 « 1,9^3 Printer Time = 3»678a x 1.09 x 1.10 x 1 = if.iflif JOB 3 Time « 6,357 sec. 2 = $11-877 Configuration: CR1, T13, T23, T33, Tif3, P2, CPI and two tare controller and tane switching units TABLE 2 (Continued) 195 Year 5 Tape Time - 60**a x 1.12 x 1.10 x x .67 = 1,992 Printer Time = 3,678a x 1.12 x 1.10 x 1 = * f , ; ? 2 N - JOB 3 Time 6,516 see. Z - $11,877 Configuration t No change. Year 6 Tape Time ® 60^a x 1.15 x 1.10 x x .67 = 2,052 Printer Time = 3,6?8a x 1.15 x l.io x 1 = !ti£Zk JOB 3 Time - 6,723 sec. Z = $11,877 Configuration: No change. Year 7 Tape Time = 6Ch& x 1.18 x 1.10 x *+ x .67 = 2 , l l *f Printer Time = 3,678a x 1.18 x 1.10 x 1 - fct&l JOB 3 Time = 6,925 sec. Z = $11,877 196 TABLE b2 (Continued) Year 8 Tape Time = 60^a x 1,.22 x 1.10 x x .67 = 2,177 Printer Time = 3,678a x 1,.22 x 1.10 x 1 JOB 3 Time = 7,132 sec. Z = $11,877 Configuration: Ho change aFrom Table 2. ^Change in configuration. Object Program Memory Requirements 197 Assembly Language. G r m « .05 The greatest demand on memory occurs in Run 31* Bata 20,350 characters Program 20,000 Operating System 16.000 56,350 characters Pg = P^ (1.05)7 = 20,000 (1*^-1) s 28,200 characters are required in the program in Year 8 (from eq.. 60). A total of 6*t,550 characters is required in Year 8. The memory capacity of 65,000 characters is not exceeded. Compiler Language, = .05, Pm = 1.10 ?8 - l*i (1.05)7 (1.10) = 20,000 (1.55) * 31,000 characters are required in the program in Year 8 (from eq. 6l). A total of 67,350 characters is required in Year 8. The memory capacity of 65,000 characters is exceeded by 2,350 characters. However, the block size in the master file in Run 31 can be reduced from a 2,500 character to a 1,500 character block size. This change results in a reduction in memory requirement of 1,000 characters for each of four input-output areas or *t,000 characters total. The reduction in block size increases the reciprocal of effective transfer rate from 15.6 to 17.7 seconds per million characters in the file maintenance mode for the 198 o Model 3 tape unit. This results in a time increase of 13*5 per cent in the tape collating phase of JOB 3* JOB 1 and JOB 3 are related jobs. Therefore, the block size used in the master file must be the same in both jobs. This increase in tape processing time causes the JOB 3 time to be increased to 7,^26 seconds (see p. **8) or a violation of the job time constraint. Therefore, the memory capacity must be increased in order to satisfy the job time con straint. The memory capacity may be increased to 131*000 characters by using a different central processor unit. 9 The additional rent is $2,890 per month. This amount increases the rent in Year 8 to $1^,337 per month. ^Calculated from tape unit specifications Table 8. ^IBM System 360 Brice List. PART III REPLACEMENT ANALYSIS CHAPTER V DEVELOPMENT OP REPLACEMENT ANALYSIS Introduction The final requirement of a computer selection model is that it provide the ability to determine at what point in time it is economical to replace the existing system (defender) with a better system (challenger). The model which has been described thus far permits the identifica tion of the best presently available challenger. The chal lengers which are presently available are computer systems which have been announced at the time of making the com puter selection study. The specifications of these systems usually exist in sufficient detail in order to make the kind of technical and economic analysis presented in the previous chapters. However, if the replacement analysis is limited to a comparison of the defender and the best present challenger, a premature replacement may result. The correct replacement policy is one which makes an allow ance for superior future challengers. According to Terborgh,'3 ' the determination of when to replace should not ■^George Terborgh, Dynamic Equipment Policy (New York: McGraw-Hill, 19^+9), p. 55. 200 201 toe made toy comparing the defender with the present chal lenger over the anticipated life of the present challenger. The choice is not between replacing the defender now or retaining it for the life of the present challenger. Rather, it is a choice between taking the present chal lenger now or retaining the defender, not for the life of the present challenger but, until a better challenger comes along. The question arises as to how far into the future to carry the analysis of future challengers. One of Terborgh*s two major assumptions is that the adverse mini mum (the time-adjusted average of capital recovery cost and O operating inferiority ) of the present challenger will toe repeated in future challengers. It follows from this assumption that future challengers will repeat the first cost, salvage value and operating inferiority of the present challenger. Hence, all challengers have equal lives. The adverse minimum of the entire series, consist ing of the present challenger and its successors, is equal to the adverse minimum of the present challenger alone. Also, since the successors to the present challenger are also the successors to the defender, the difference in adverse minimums of the series headed by the present ^The difference in operating cost between the defender and the best available challenger at a given point in time. ^Terborgh, o p . cit.. p. 6M-. 202 challenger and the series headed by the defender is equal to the difference in adverse minimums of the present chal lenger and the defender. The problem is reduced to one of calculating the time-adjusted sum of capital recovery cost and operating inferiority for the defender and present challenger. A replacement is made if the adverse minimum of the defender is greater than that of the present chal lenger. It is pointed out by Terborgh that the defender’s adverse minimum will usually occur in the next year of operation. This is always the case when the defender has zero salvage value and increasing operating costs. With zero salvage value, there is no capital recovery cost. The only component of annual cost is the increasing operating cost. In the case where salvage value is present, the decline in capital recovery cost is normally less rapid than the rise in operating cost, resulting in a rising equivalent uniform annual cost curve. If this is the case, the replacement analysis is further simplified by adding the defender's next year operating inferiority to its next year capital recovery cost and comparing this sum with the challenger’s adverse minimum. The second major assumption of lynamic Equipment ^Ibid.. p. 120. 203 5 6 Policy is that the inferiority gradient is constant. The operating inferiority consists of a constant first year operating cost gap plus a uniform gradient. In his later work, Business Investment Policy. Terhorgh develops the MAPI Urgency Bating, which is the after-tax rate of return on net investment of next year's 7 « operating advantage. This model is also based on the assumption that all future challengers will have the same first costs, salvage values and operating inferiority as the present challenger. A constant, decreasing or increas ing inferiority gradient may be used in this model. The assumption that future challengers will have the same adverse minimum as the present challenger cannot be used in the computer selection model. Prom its incep tion, the computer field has been characterized by rapid technological progress. This progress has resulted in dramatic reductions in the cost/performance ratios of vari ous technical characteristics• Several of these ratios are tabulated in Table M-3 and are plotted in Pigures 7, 8, 9> and 10. The history of computer development suggests that significant cost reductions can be expected in the future ^The rate at which operating inferiority is accu mulated. £ Terborgh, o p . cit.. p. 65. ^George Terborgh, Business Investment Policy (Washington, D. C.: Machinery and Allied Products Insti tute, 1958). TABLE 1*3 T E C H N O L O G I C A L I M P R O V E M E N T O F S C O M P U T E R C H A R A C T E R I S T I C S Model IB M 650 IB M 70511 ibm 705m I B M 7070 IB M 11 * 0 1 H-800 Date o f F irst D elivery 12-51+ 12-56 6-59 6-60 9-60 12-60 M emory S ize Used1 3 (Thousands o f Characters) 10 l+ o 1 + 0 50 1 + 32 Storage Cycles per Second (Thousands) 2.1 (av e.) 58.8 125 833 87.0 1,333 Central Processor Rental® $3,200 $16,650 $17,500 $11,800 $2,680 $10,150 D ollars per 1,000 Storage Cycles per Second per 1,000 Characters I52.li 7.08 3.50 .283 7.70 .117 Tape Transfer Rate (EC) 7.5d 15 62.5 62.5 31.3d 6 1 + Tape Unit Rental® $ 1 , 0 3 8 $ 1,775 $ 2 , 1 5 0 $ 2 , 1 7 5 $ 900 $ 1 , 1 + 0 0 D ollars per K C 138.1* 118.1* 3 1 + .1 + 3 1 + .8 28.8 21.8 Cards Read Speed (Cards per Minute) 200 250 800 800 650 Card Reader Rental0 $ 550 $ 2,1*00 $1,890 $ 550 $ 1,1+25 D ollars per Cards per Minute 2.75 9.60 2.36 .69 2.19 2 0 1 ; TABLE 1*3 j O G I C A L im provem ent of several I C M P U T E R C H A R A C T E R IS T IC S a IB M 1 1 * 0 1 H-800 J IB M 7080 I B M 7 0 7 1 * B-1 * 0 0 H-1800 B-200 IB M 360 (Mod. 30) 9-60 12-60 9-61 12-61 12-61 9-63 1-61* 9-65 1 * 32 80 50 32 65 65 65 87*0 1,333 500 1,250 26 ? 1*,000 5oo 666 $2,680 $10,150 $33,500 $11,800 $5,825 $16,050 $ 1 * ,1 * 6 5 lit, 050 7.70 .117 co v»\ CO . .189 .682 .0617 .137 .0936 31.3d 61 62.5 l*8 e 121 * 83 90 $ 900 $ 1,1 * 0 0 $ ■ 1,725 VO 8 $ 1,675 $ 750 $1,118 28.8 21.8 26.6 18.8 13.5 9.0 1 * 12.1 * 800 650 800 650 800 1,000 $ #0 $ 1,1*25 $1,890 $ 325 $ 560 $ 320 .69 2.19 2.36 .50 .70 .32 TABLE 1*3 (Continued) Model IB M 650 IB M 705II ibm 705111 IB M 7070 IB M 11*01 B-600 Date o f F ir st D elivery 12-51* 12-56 6-59 6 — 6 o 9-60 12-60 P rin ter Speed (Lines per Minute) 150 1,000 600 600 900 P rin ter Rental® $ h00 $ t,5 o o $2,115 $775 $3*000 D ollars per Line per Minute 2.67 lu 5o 3.53 1 .2 9 3.33 ^Computed from data presented in Martin H. Weik, A Third Survey o f Domestic E lectron ic Di (Aberdeen Proving Ground, Maryland: March, 1961). Charles W . Adams, ‘ ’Computer C h a ra cteristics," ^Memory s iz e used in c a lc u la tio n s. This memory s iz e does not n ecessa rily correspond to m cA ll r en ta l fig u res are in d o lla rs per month. ^FLfty per cent o f tra n sfer ra te used where there is no concurrent read, w rite and computi ® Seventy-five per cent o f tra n sfer ra te used where there i s concurrent reading and w ritinj 205 T A B L E 1 (3 (Continued) I B M 11(01 B-800 IB M 7080 I B M 707l( Moo 3-1800 3.200 IB M 360 (Mod. 30) 9-60 12-60 9-61 12-61 12-61 9-63 1-61( 9=65 600 900 600 900 950 1,100 1775 13,000 $2,115 $2,3U0 $1 , 3 1 * 0 $1,575 1.29 3*33 3.53 2,60 1*1 ( 1 1.1(3 Domestic E lectronic D ig ita l Computing Systems, B a llis tic Research Laboratories Report No* 1115 iuter C haracteristics," Datamation, (November, 1963), 31-38 and Datamation, (July, 1961*.), 1(0-1}1 isarily correspond to m axim um memory siz e available* read, w rite and compute. ent reading and w riting but not concurrent computing. H-1800 :uum 'pejrforiaance * I - U - U I tax H r E j r ; t £ L - x r p c CLEARPfl J NT PAPER CO. NO. C315. 20 DIVISIONS PER INCH < I ZO DIVISIONS \ BY 3 3-INCH CYCLES RfcTIO RULING POINTED I N U WHUriUTOI * ii StOi O N -IVOiMi-fL^J. iNiada-/3"0 f.O • V ST. N; SSiriiyd ' ONJ-Jflfcl OI-LVi: S313JL3 HONI ■ C E A3 fsworsi/vio OS i > kOKI tfSU SNOISIAIG O? -SLSTO ON '03 HddVrJ i.i-ilMdaV3'T:> 210 for given speed and storage requirements* This means that the lives of the present challenger and future challengers will be unequal. Obsolescence of computer systems is caused by both first cost and operating cost inferiority. Therefore, total obsolescence cannot be measured by an inferiority gradient. In order to obtain a correct replacement decision in a situation of future price decreases, it may be neces-. sary to consider more than one year of additional defender operation. If Challenger 1 is better than one more year of defender operation followed by Challenger 2, Challenger 1 may not be better than two more years of defender opera tion followed by Challenger 3* It is also necessary to put the defender and challenger on an equivalent basis, when the time con straints of the challenger are less than those of the defender. The additional cost which would be incurred in order for the defender to process within the time con straints established for the challenger should be added to the cost of defender operation. Given the price and rental decreases of future challengers, the operating costs of future challengers will be a function of year of purchase or the year in which rental begins. The operating costs will also be a function of the age of the installation. Operating costs which are 211 a function of the age of the installation are computer maintenance costs and the rental of additional equipments required in order to accommodate expanded applications. It is the latter cost which causes the cost of operation to increase at an increasing rate with respect to the age of the installation. However, because of the exponentially decreasing purchase price and rental of future challengers, the operating costs will start from a first year operating cost which is an exponential function of the year of pur chase or the first year of rental. Therefore, the operat ing inferiority function is not a constant plus a gradient; it is a variable plus a gradient. Other replacement models employ the methods of finite differences (discrete functions) or the calculus (continuous functions) in order to determine the minimum value of the present worth or equivalent uniform annual cost function of an indefinite series of identical 0 ^ 10 1 machines. ’ * 9 The life for which the present worth ®C. West Churchman, Russell 1. Ackoff and E. Leonard Amoff, Introduction to Operations Research (Hew York: John Wiley and Sons, Inc., 1957), pp. 4&1-4-91. ^Maurice Sasieni, Arthur Yaspan and Lawrence Friedman, Operations Research— Methods and Problems (Hew York: John Wiley and Sons, Inc., 19^9), pp. 102-108. 10Herbert E. Schweyer, Analytic Models for Mana gerial and Engineering Economics (Hew York: Reinhold~Pub lishing Company, 19w), pp* 308-336. ■^William T. Morris, Engineering Economy (Homewood, Illinois: Richard D. Irwin, Inc., I960), pp. 155-187. 212 or annual cost is minimum is the optimal life. Since all machines are identical with respect to first cost, salvage value and operating costs, the lives of all challengers are equal. Hence, an optimal life can he computed. This approach has no relevance in a situation where challenger lives are unequal. Because the characteristics of the computer replacement problem differ significantly with respect to the assumptions used in the foregoing models, none of these models is used in the computer replacement analysis. Instead, a method of analysis suitable to the computer replacement problem is developed in the sections to follow. Development of the Replacement Criterion In order for replacement to take place, the series of machines consisting of the present challenger and its successors should be superior to the series of machines consisting of the defender and its successors. The series headed by the present challenger is superior if the minimum equivalent uniform annual cost of this series is less than the minimum equivalent uniform annual cost of the series headed by the defender. The challengers which will be available as replacements for the present challenger and the successor to the defender, and all subsequent challeng ers, can be eliminated from consideration. This elimination is possible because the identical challenger can be used to replace either the present challenger or the defenderfs successor. The replacement problem is then one of deter mining whether the present challenger’s minimum equivalent uniform annual cost is less than that of the defender and its successor. If this is the case, the present challenger replaces the defender. Otherwise, the defender remains in service at least one more year and the same analysis is per formed next year. An alternative way of viewing the prob lem is that it is necessary to determine whether there is a better challenger worth waiting for. In order for replace ment to occur, the present challenger must prove itself superior to every combination of additional defender ser vice, followed by its successor, which can occur within the assumed life of the present challenger. If the present challenger is designated as Challenger 1 and challengers appearing in subsequent years are designated as Challenger 2, Challenger 3j st cetera, then Challenger 1 must be better than one more year of the Defender followed by Challenger 2, two more years of the Defender followed by Challenger 3> et cetera. These tests will be designated respectively as Test 1, Test 2, et cetera. A great number of comparisons do not have to be made. If Challenger 1 fails Test 1, replacement is deferred at least one year. If Challenger 1 passes Test 1, Test 2 is performed. If Challenger 1 fails Test 2, replacement is deferred at least one year. If Challenger 1 increases its advantage in terms of minimum equivalent uniform annual cost between Test 1 and Test 2, it indicates that additional Defender service will increase cost faster than Challenger 2 can decrease cost. The Defender should he replaced by Challenger 1. If Challenger 1 loses ground in Test 2, the tests are con tinued. Eventually the difference in minimum costs will become and remain constant due to the large discounting of distance cost differences. The maximum number of tests is governed by the life at which the present challenger is physically incapable of processing the required workload within the required time constraints. If Challenger 1 fails any of the tests up to this point, replacement does not occur. If Challenger 1 passes all tests, replacement does occur. The logic of the replacement analysis is shown in Figure 11 and Figure 12. It is extremely difficult to estimate the lives of the present challenger and the defenderfs successor in an environment of rapid technological growth. Therefore, in the absence of precise service life information, the best estimate of life at the time of the replacement study is that life corresponds to the number of years where the equivalent uniform annual cost is minimum. This does not necessarily mean that replacement will occur at these points. Replacement may occur before or after this point, depending upon the characteristics of future challengers. V\V> Test 1 Test 2 Test n Test K- 1 Year D (1) 2 Years J L (.2). n Years D (n) (n'-l) Years C (2) n” Years Cl (n*-2) Years 0 (3. ) nH Years Cl (n*-n) Years C (n+1) I n1 1 Years K Tears LEGEND 2>(n): n more years of defender operation C(n+1): Challenger which suc ceeds defender in year n+1 Cli n ’ n Cl { (NHL) Years (n*-N+l) Years 3> (H-l) C (JO f ---------------< N: Present challenger Number of years corres ponding to minimum equivalent uniform annual cost of defender followed by its suc cessor (variable) Number of years corres ponding to minimum equivalent uniform annual cost of present challenger (variable) Estimated physical life of present challenger Cl Eig. 11.— Emplacement Analysis Time Chart VJ\ 216 1 n Cl*[D(n),C(n+l)] n+l-> n [IKxi),CCn.+l)] Cl-[D(n) ,C(n+l) ] Stop, Don't Replace >iCl:\CD(n),C(n+l)]-Cl^[D(n-l),C(n)]-Cl ^ Stop, Replace i [D(n) ,C(n+l) ]-Cl£[D(n-l) ,C(n) ]-Cl n=N Stop, Replace IEGEND Cl: Minimum equivalent uniform annual cost of present challenger. The value of 01 will change from test to teBt. [D(n),C(n+l)]: Minimum equivalent uniform annual cost of defender for n more years followed by challenger installed in year n+1. N: Estimated physical life of present challenger. Pig. 12.--Replacement Analysis Plow Chart 217 The minimum cost point comparison does permit a comparison on the basis of the most favorable life for each alterna tive, as seen at the time of the replacement study. By graphing the equivalent uniform annual cost curves over a number of years, it is possible to determine the sensitivity of the replacement decision to the assumed lives of the present challenger and the defender’s succes sor. Outline of Replacement Analysis The replacement analysis is performed as follows: A. Purchased Challengers. Using Compiler Language and a . 5 Per Cent Interest Rate 1. The best present challenger at an interest rate of 5 per cent is compared with an assumed defender and its best successors. In Chapter 17, it was shown that the best available present challenger at an interest rate of 5 per cent, and for a given set of equipment specifications, is a purchased compiler language system. In practice, the analysis of Chapter IV would be performed for all sets of equip ment specifications. For the purpose of the replacement analysis it will be assumed that the purchased compiler language system of Chapter IV is the best available present challenger. This chal lenger is designated Challenger 1. It-is assumed that the best challengers which follow Challenger 1 (Challenger 2, Challenger 3* et cetera) are also compiler language systems. There is no reason to believe that the best future challengers at the stated interest rate, will differ from the present challenger as to type of programming language and method of acquisition. The first cost and operating costs of future chal lengers will decrease. The rate of decrease is based on past technological progress and an esti mate of the rate at which future technological progress will be made. It is assumed that price and rental decreases will be proportional to future technological growth. Since the assumed rate at which cost will decrease has an important bearing on the replacement decision, the sensitivity of the replacement decision to cost decreases is deter mined by making the replacement analysis for two assumed rates of future technological improvementi 50 per cent and 100 per cent of the past technolog ical improvement rates. A purchased defender is assumed, rather than a rented defender, because a replacement study is of greater interest when the capital recovery cost of the defender*s salvage value must be considered. It is assumed that the defender time constraints are equal to the challenger time constraints, 5. When a decrease in future prices based on 50 per cent of past technological improvement is used (9 per cent decrease in price per year), an 11 per cent compound annual decrease in salvage value is used for the defender and challengers. This rate of decrease is based on a weighted average of the rate of decrease in salvage value of several hundred computers. When a decline in future prices based on 100 per cent of past technological improvement is used (19 per cent decrease in price per year), a 23 per cent compound annual decrease in salvage value is used for the challengers, based on the assumption that salvage values will decrease at a rate commensurate with the rate of decrease in prices. Rented Challengers. Using Assembly language and a 10 Per Cent Interest Rate 1. In Chapter IV, the best present challenger at an interest rate of 10 per cent, and for a given set of equipment specifications, was shown to be a rented assembly language system. This challenger is designated Challenger 1R. 2. It is assumed that the best challengers which 220 follow Challenger 1R (Challenger 2R, Challenger 3&j et cetera) are also rented assembly language sys tems . 3# A rate of decrease in future costs is forecast, based on future technological progress at 50 per cent and 100 per cent of past technological improve ment rates. b, The purchased defender which is used in the pur chased compiler language replacement analysis is used in the rental assembly language replacement analysis. Forecasts of Price and Rental of Future Challengers Four rental/performance ratios which pertain to the applications under consideration are used to forecast the price and rental of future challengers. A ratio of cost to performance is used because both cost and performance change rapidly with respect to time. It is not possible to obtain cost as a function of time for a given value of per formance characteristic. The cost/performance ratios are used as the assumed rates of decrease in future prices and rentals of individual computer components because the per formance characteristic used in computing the cost/perform ance ratios is constant for a given computer component. The rate of decrease used for rental systems can also be 221 used for purchased systems because the purchase/rental ratios of the systems used in the eost/performance ratio calculations are approximately constant, The cost/perform ance ratios are tabulated in Table b$ and plotted in Figures 7, 8, 9 and 10. A least squares fit of the exponen tially decreasing cost/performance ratio curves has been made, as indicated in Figures 7, 8, 9 and 10, in order to determine the average compound annual rates at which the cost/performance ratios have decreased. These rates are tabulated in Table kb* Future challengers will require an initial comple ment of equipment equal to that which would be required in future years by the present challenger. That is, the initial complement of equipment for Challenger 3 is equal to that required by Challenger 1 in its third year of operation. The compound rates of decrease of the four eost/performance ratios are used to forecast the compound rates of price and rental decrease for each of the four computer components: central processor, magnetic tape unit, printer and card reader. It should be noted that the cost/performance ratios are applied to a configuration which requires higher performance equipment in future years in order to accommodate a larger workload. This means that the rate of decrease in price or rental will be less than the rate of decrease in the cost/performance ratio for 222 TABLE kb COMPOUND ANNUAL HATES OP DECEEASE IN COST/PERFORMANCE RATIOS FORECASTED COMPOUND RATES OP PRICE AND RENTAL DECREASE Unit 100 Per Cent of Co st/Performance Rate 50 Per Cent of Co st/Performance Rate Central Processor *+8 per centa per centa Magnetic Tape Unit 2k per cent*5 12 per cent** Card Reader 20 per cent0 10 per cent0 Printer 8 per cent^ *+ per centd Forecast Rate of Price Decrease, Rp 19 per cente 9 per cente Rate of Rental Decrease, Ry 22 per cent 10 per cent1 aRate of decrease in monthly rental dollars per 1.000 storage cycle per second per 1,000 characters of memory. Determined form least squares fit of data in Figure 7. fcRate of decrease in monthly rental dollars per 1.000 characters per second tape transfer rate. Determined from least squares fit of data in Figure 8. cRate of decrease in monthly rental dollars per cards per minute. Determined from least squares fit of data in Figure 9. ^Rate of decrease in monthly rental dollars per lines per minute. Determined from least squares fit of data in Figure 10. eForecast based on data in Table 19 and Table kb, ■^Forecast based on data in Table 16 and Table . those computer components which must be changed in future years in order to obtain higher performance. An eight year price forecast is made on the basis of the initial comple ment of equipment required for future challengers for the purchased compiler language system of Table 19? P* 126, and is shown in Figure 13. The forecast is made on the basis of 100 per cent of the historical rates of decrease of the cost/performance ratios and at 50 per cent of these rates, A. least squares fit of the exponentially decreasing price curves has been made in order to forecast the average com pound rates of price decline. The forecasted compound rates of price decrease are shown in Table Mf, The corres ponding rental forecasts, based on the equipment required for future challengers over an eight year period for the rented assembly language system of Table 16, p. 121, are shown in Figure l^f, The forecasted compound rates of rental decrease are shown in Table M+, Forecasts of Operating Costs of Future Challengers The operating costs of a purchased system consist of computer maintenance, program maintenance and rental for additional equipment which may be required as applications grow in size. The program maintenance cost for the pur chased compiler language system (Table 21, p. 129) is constant at $10,000 per year. The sum of the computer ■ 5 I ti 20 DIVISIONS PES INCH. 150 X 200 DIVISIONS. PRINTED I N U.S.A. ON CLaARPRINT TECHNICAL PAPER NO. IOIS gLEAHPRiNf C h a r t s * o N r > 0 0 u tr c. c. 2 cr a' j L) I Csi*f3 d »* H M Srs d p "jf o ■ t s nHnmuMmn^Suni - I t ;0 Ef nr 'E maintenance cost and additional rental of equipment increases at an increasing rate (Table 21, p. 129). A least squares fit of an exponentially increasing function is made to the sum of the computer maintenance cost and additional rental of Table 21, p. 129* ®his least squares fit produces the function = 10 + 15 (1.27)^, in thou sands of dollars, where ; j is an index of the year of opera tion (3 = 0 is the first year). However, this function pertains to the operating costs of the present challenger only. The operating costs of future challengers will increase at an increasing rate from an initial operating cost which is lower than that of the preceding challenger. Since computer maintenance costs and additional rental costs are proportional to the price of the system, these costs will be a function of the year of purchase n(n = 0 for the present challenger) and the index j of the year of operation. If Hp * 1 is the compound annual rate at which price decreases, the operating cost forecasting equation for purchased compiler systems is the followings (62) Cn. j = 10 + 15 (&p)n (1*27)^ (thousands of dollars) The operating costs of a rental system consist of rental costs and program maintenance costs. The constant component of program maintenance cost is $10,000 per year (Table 16, p. 121). Rental costs for a given system increase with time due to the requirement for higher per formance equipment as applications grow in size. This cost increases at an increasing rate. A least squares fit of an exponentially increasing function is made to the rental costs and the non-constant component of program maintenance cost for the assembly language rental system of Table 16, p. 121. The least squares fit produces the function Cj = 10 + 108 (l.O^f)3, in thousands of dollars, where j is an index of the years of operation (j s 0 is the first year of operation). This equation pertains to the operating costs of the present challenger only. The operating costs of future challengers will increase at an increasing rate from an initial operating cost which is lower than that of the preceding challenger. If Bp A 1 is the compound annual rate at which rental decreases, n is the first year of operation (n = 0 for the present challenger) and j is an index of the year of operation, the operating cost fore casting equation for rented assembly language systems is the following: (63) Cnj t s 10 + 108 (Kp)11 (1.0*0** (thousands of dollars). forecast of Rate of Decrease of Salvage Value"of Defender. Present Challenger and Future Challenger There is little information available on the pre cise manner in which the salvage values of computer systems have decreased. One reason for this situation is that until recently there was no active used computer market in which prices were quoted. A strong used computer market did not develop for vacuum tube machines because of high 12 maintenance costs. There is not a long history of depreciation rates for solid state computers, since these machines were not introduced until 1959* Sales of used computers are normally negotiated on an individual basis. Used prices are seldom published* In a competitive situa tion, the sales price may be confidential information. These are situations in which the manufacturer buys a com petitive system from a user, for resale to a broker, in order to sell the user a new system. In other cases, the manufacturer makes a trade-in allowance on his earlier model in order to sell a new model. One survey of approximate used computer prices was conducted by Computers and Automation Magazine and is pre sented in Table *0.^ Weighted averages of compound and constant rates of decrease of salvage value have been cal culated from this data. The values are 11 per cent and 8 per cent per year, respectively. "^George H. Heilbom, "The Used Computer Market— 196^: A Broker*s View," Computers and Automation. (July, 196* 0 , 22, 26. ■^Ueil Macdonald, "The Used Computer Market," Computers and Automation. (November, 19o2), Mf-M-5. TABLE 15 decline in s a l v a g e v a l u e o f s e v er a l c o m p u t e r s y s t e m s ( 1) C om p u ter ( 2) Year of® First Delivery (3) b Salvage Value0 (Per Cent of First Cost) (U b Year of Salvage Value Determination (5) C o m p o u n d Annual Rate of Decline in Salvage Value (Per Cent) ( 6) Constant Annual Rate of Decline in Salvage Value (Per Cent) IBM 650 1951| IBM 70i| 1955 IBM 7090 I960 Burroughs 205 1 9 5 1 * Alwac III E 1955 IPG 30 1956 Univac I 1951 Univac File 1958 50 30 55 30 55 35 25 5o 1962 1965 1965 1962 1962 1962 1962 1962 8 11 11 Hi 8 16 12 16 C om p u ter Weighted average of co m p o u n d annual rates of decline in salvage value “ 21,573 7 l)99k * 11 per cent. Weighted average of constant annual rates of decline in salvage value ■ 15,715 7 l,99l| n 8 per cent. 6 7 9 9 6 11 7 13 ®Charles W . A d a m s, "(Com puter Characteristics) O n ce Again," Datamation, (November, 1963), 31* C e il Macdonald, " T h e Used C om puter Market," C om puters and Automation, (November, 1962), Ui-li5* Calculated on the assumption that decline in salvage value takes place at a co m p o u n d annual rate. "M onthly C om puter Census," C om puters and Automation, (November, 1962), 39-l|0 and (January, 1961i), 36-37 229 TABLE 1(5 E N S A L V A G E V A L U E O F S E V E R A L C O M P U T E R S Y S T E M S & a C om pound Annual Rate o f Decline in Salvage Value (Per Cent) (6) Constant Annual Rate o f D ecline in Salvage Value (Per Cent) (7) a N um ber o f In sta lla tio n s in Year o f Salvage Value Determination (8) (5) x (7) (9) (6) x (7) 8 6 997 7,976 5,982 11 7 61 671 r— CM -= t 11 9 270 2,970 2,130 m 9 92 1,288 828 8 6 32 256 192 16 11 1*00 6,1*00 li,it00 12 7 65 780 Ii55 16 13 77 1,232 1,001 age value Totals 1,99k a , 573 15,715 age value )atamation, (November, 1963), 31* rtoraation, (November, 1962), U;-Ii5* ;akes place at a compound annual rate* ier, 1962), 39— U O and (January, 1961;), 36-37* 230 The salvage value information of Table *4-5 provides a measure of salvage value for one year only. It is not known what the values were in intervening years. A good assumption is that the decline is exponential rather than linear. As was seen in the previous section, eost/perform- ance ratios have declined exponentially. It is reasonable to assume that the salvage value function has the same shape as the new computer price function. It is also reasonable to assume that the value of a computer will decrease faster in the early years of life, after newer models have been delivered, and decrease slower in the later years of life as the program library is built up and as the number of programmers and operators trained in the use of the system increases. The availability of programs and trained personnel can be as important in determining the price of a used computer as the hardware itself. A constant rate of salvage value decline of 10 per cent per year was used (Table 22, p. 130) in comparing the purchased compiler language present challenger with the rented assembly language present challenger at an after tax interest rate of 5 per cent (Table 26, p. 1*4-9) and at an after tax interest rate of 10 per cent (Table 28, p. 160). The constant 10 per cent rate of decrease does not differ significantly from the compound 11 per cent rate of decrease during the first four years of operation, as 231 illustrated below. Salvage Value as a Per- Salvage Value as a Per- Years of centage of Pirst Cost centage of Pirst Cost doerstion ^or a Constant Sate of for a Compound Sate of Decrease of 10 Per Cent Decrease of 11 Per Cent Per Year Per Year 1 90 2 80 3 70 60 5 50 6 ifO 7 30 8 20 The capital recovery cost of the purchased system is reduced in the later years, if the compound rate of decline in salvage value is assumed. However, the effect of the difference in salvage value in later years is less than that indicated in the above table because of the effect of discounting the differences in salvage values in later years. There is an additional compensating factor due to the larger depreciation deductions which can be taken in later years if the constant rate of decline in salvage value is assumed. The equivalent uniform annual costs of the purchased compiler language system and the rented assembly language system are computed, using an 11 per cent compound rate of salvage value decrease for a six year life. These costs and the original costs, obtained by using a constant 10 per cent rate of salvage 89 79 70 62 55 ^9 Mi- 39 value decrease, are shown in Table *+6. Thus the choice of salvage value function is not a decisive factor in the selection of the present challenger. The 11 per cent compound rate of salvage value decrease will be used in the replacement analysis for the defender, present challenger and future challengers because this function is consistent with the forecasted exponential decrease in price of future challengers. 233 TABLE ^6 EQUIVALENT UNIFORM ANNUAL COSTS CONSTANT RATE vs. COMPOUND RATE OP SALVAGE VALUE DECREASE (SIX YEAR LIRE) Interest Rate = 5 Per Cent Rented Assembly Language System Purchased Com piler Language System Constant 10 Per Cent Annual Salvage Value Decline $l1 +2,000a $127,000a Compound 11 Per Cent Annual Salvage Value Decline 1^+1,000 123,000 Interest Rate = 10 Per Cent Constant 10 Per Cent Annual Salvage Value Decline 1^3,000^ l*4*,000b Compound 11 Per Cent Annual Salvage Value Decline 1^-1,000 11*1,000 aProm Table 26. kprom Table 28. CHAPTER YI APPLICATION OP REPLACEMENT ANALYSIS After-Tax Replacement Analysis: Purchased Compiler Language Challengers. Interest Rate = Per Cent. Income Tax Rate = 5>Q Per Cent The coats of the Defender and four purchased com piler language challengers are shown in Table *+7. Chal lenger 1 is the purchased compiler language challenger of Table 21, p. 129. Challenger 1 is available for purchase in Year 0. Challenger 2 is available for purchase in Year 1 at a price 9 per cent less than that of Challenger 1. Challenger 3 is available for purchase in Year 2 at a price 9 per cent less than that of Challenger 2. Challenger 2A is available for purchase in Year 1 at a price 19 per cent less than that of Challenger 1. Time is measured with respect to the purchase of Challenger 1, i.e., Year 0. In all of the analysis to follow, the number of years n is measured from Year 0 and is equal to the assumed life of Challenger 1. A one year conversion period is required in order to prepare for the installation of any of the challengers. Therefore, the operating cost of the Defender in Year 0 is not relevant. 23*f TABLE k-7 SCHEDULE OP COSTS OP DEFENDER AND POUR PURCHASED CHALLENGERS Conversion Costs® (Thousands) Operating Costs (Thousands) Challenger Defender Challenger Challenger Year 2A $67 $27 $67 $379 $22 139 forecasting equations C (n,j) = 10 + l5(RpP (l,27)*J> where n is the year of purchase and j is an index of years after purchase \ j = 0 corresponds to the first year of operation); Bp s t ,91 for a rate of purchase price reduction based on a 50 per cent rate of past technological progress. Rp = .81 for a rate of purchase price reduction based on a 100 per cent rate of past technological progress. ^Computed from C j) = 10 + 15 (.9l)~lf (1.27)^. ro UJ TABIiE b ? (Continued) cFrom Table 21. ^Computed from C (1 ^ = 10 + 1? (.91) (1.27)^. eComputed from C (1>3) *10 + 15 (.81) (1.27)^. ■^Computed from C (2,3) = 10 + 15 (.91)2 (1.27)**. ^Consists of $22,000 in program conversion costs from Table 21 plus an estimated $+5,000 in systems and file conversion costs. k^rom Table 21. Computed from *+58 (.91) , where n is the year of purchase. 3Computed from ^58 (#89)n, where n is the year of purchase. & ON 237 This cost would he incurred whether Challenger 1 is installed or not. The decision to install Challenger 1 is made at the beginning of Year 0 because, if Challenger 1 is to be installed, conversion must take place in Year 0 and Challenger 1 must be purchased in Year 0. The first year of Challenger 1 operation is in Year 1. Challenger 1 vs. Defender (One More Year) and Challenger 2 (Test 1) The estimated remaining salvage value of the Defender is shown in Table * * - 8. The estimated salvage values of Challenger 1 and Challenger 2 are shown in Table ^9 and Table 50, respectively. In all of the replace ment analysis to follow, it is assumed that book value is equal to disposal value, so that there is no gain or loss on disposal. The capital recovery costs for n years of operation for Challenger 1, the Defender and Challenger 2 are shown in Table 51, Table 52, and Table 53, respectively. The capital recovery cost of the loss in salvage value for one more year of Defender operation is computed for n years in Table 52. This capital recovery cost is added to the capital recovery cost of Challenger 2 in Table 53 in order to obtain the total capital recovery cost of one more year of Defender operation followed by the installation of Challenger 2. The depreciation and investment tax credit schedule 238 TABUS kQ DEFENDER SALVAGE VALUE SCHEDULE Pirst Cost = $668,000 (1) Years Retained (2) Salvage Valuea (Thousands) (3) * Reduced0 Salvage Value (Thousands) Of) Depreciable Value (Thousands) b $*♦•20 $353 $315 5 37b 307 361 6 333 266 *f02 aComputed from 668 C.89)n where n is the number of years the system is retained. ^Reduction in salvage value by 10 per cent of first cost if asset is retained three or more years. cFirst cost minus Col. (3). 239 TABLE 4-9 CHALLENGER X SALVAGE VALUE SCHEDULE First Cost a $4-58,000 (1) Years Retained (2) Salvage Yaluea (Thousands) (3) t Reduced Salvage Yalue (Thousands) (1+) c Depreciable Value (Thousands) 1 $1+08 $4-08 $ 50 2 363 363 95 3 323 277 181 4- 287 24-1 217 5 256 210 24-8 ft 228 182 176 7 203 157 301 8 181 135 323 aComputed from 4-58 (.89)n , where n is the number of years the system is retained. ^Reduction in salvage value by 10 per cent of first cost if asset is retained three or more years. cFirst cost minus Col. (3)» 2^0 TABLE 50 CHALLENGES 2 SALVAGE VALUE SCHEDULE Eirst Cost » $*+17,000 (1) Years Retained (2) Salvage Value3, (Thousands) (3) Salvage Value ^ (Thousands) 00 c Depreciable Value (Thousands) 1 $371 $371 $ be 2 330 330 87 3 29b 252 165 b 262 220 197 5 233 191 226 6 207 165 252 7 18b lb2 275 aCompute& from *+17 («89)n > where n is the mLmher of years system is retained. Reduction in salvage value by 10 per cent of first cost if system is retained three or more years, cEirst cost minus Col. (3). 2kl TABLE 51 CHALLENGER 1 CAPITAL RECOVERY COST Pirst Cost = $L f58,OOOa (Interest Rate = 5 Per Cent) Year n Salvage Value ( Thousands) Capital Recovery Cost0 (Thousands) 0 $>*58 1 *f08 $73 2 363 69 3 323 66 b 287 62 5 256 60 6 228 56 7 203 9* 8 181 52 aProm Table 21* kprom Table 1 +9* c(P-Le1) (erf - 5# - n) + *051*^ 2k2 TABLE 52 DEFENDER CAPITAL RECOVERY COSTS (Interest Rate « 5 Per Cent) Defender Kept One More Year Year n Salvage3 Value (Thousands) Capital Recovery Cost (Thousands) 0 $^20 1 37^ $67 2 3^ 3 23 18 5 15 6 13 7 11 8 10 Defender Kept Two More Years Year n Salvage3, Value (Thousands) Capital Recovery Cost (Thousands) 0 20 1 37^ 2 333 $63 3 **3 C 33 5 27 6 23 7 20 8 18 aFrom Table *+8 • fe[L0 - (pwf* - 5% “ 1)] [erf - 5% - n]. °[L0 - L2 (pwft - 5% - 2)] [erf - 5% - n]. 2*4-3 TABUS 53 DEFENDER-CHALLENGER 2 CAPITAL RECOVERY COST First Cost = $*H7,OOOa (Interest Rate » 5 Fer Cent) Challenger 2 Defender** Year n v Salvage Value (Thousands) Capital Recovery Cost (Thousands) Capital Recovery Cost (Thousands) Total Capital Recovery Cost (Thousands) 0 1 $**17 $ 0 $67 $67 2 371 33 3^ 67 3 330 *fl 23 6*f b 29b bb 18 62 5 262 bb 15 59 6 233 bb 13 57 7 207 b3 11 5*f 8 l8*f b2 10 52 aErom Table *f?. ^From Table 50. C£P (pwf* - 5% - 1) L^ (pwff - 5% - n](crf- 5$-n) dFrom Table 52. 2¥f is shown in Table 51 * for a period of eight years. This is the number of years for which cost data was determined for Challenger 1 in Chapter IV. It is also the number of years required in order to obtain the minimum equivalent uniform annual costs of the two alternatives. These costs are shown in Table 55 and plotted in Figure 15* The costs are plotted against the number of years n of assumed Challenger 1 life. Any value of n is also equal to one more year of Defender operation followed by (n - 1) years of Challenger 2 operation. In Figure 15, Challenger 1 is superior to the Defender and Challenger 2 for all assumed lives of Challenger 1 and Challenger 2. The minimum cost of Challenger 1 is less than the minimum cost of Defender - Challenger 2, regardless of what lives are assumed for Challenger 1 and Challenger 2. This result means that Challenger 1 has passed Test 1 and has qualified to compete in Test 2 against two more years of Defender operation followed b.v the installa tion of Challenger 3. Challenger 1 vs. Defender (Two More Years) and Challenger 3. (Test 2) The salvage value schedule of Challenger 3 is shown in Table 56. The capital recovery cost of Challenger 3 is calculated in Table 57* The total capital recovery cost of two more years of Defender operation followed by the TABLE Bh D E P R E C IA T IO N A N D I N V E S T M E N T T A X C R E C H A L L E N G E R 1 v s . D E F E N D E R (O N E M O R E T E A R ) A N D (Income Tax Rate = f> 0 Per Cent) C h a l l e n g e r 1 Disbursements* Investment* 5 A dditional0 Tear during Tear D epreciation Tax C redit Income Tax n (Thousands) (Thousands) (Thousands) (Thousands) Replace Challenger 1 0 |6 7 After 1 Tear 1 27 !5oc $1 * 1 Replace Challenger 1 0 67 A fter 2 Teal's 1 27 U8 e 12 2 27 1 *8 ® Replace Challenger 1 0 67 After 3 Tears 1 27 91f 20 2 27 60f 3 31 30f 5 Replace Challenger 1 0 67 A lter U Tears 1 27 87 J 111 11 2 27 65* 8 3 31 h k 50 22 f Replace Challenger 1 0 67 • f t After B Tears 1 27 83 11 13 2 27 66 ! 15 3 31 50^ 3 h 50 33f 5 50 l 7f Replace Challenger 1 0 67 A fter 6 Tears 1 27 79" 22 k 2 27 66 ^ 13 3 31 52! 2 k 5o 39! 5 5o 26 ! 6 5o l 3f 2 1 4 5 FABLE 514 3 IN V E S T M E N T T A X C R E D IT (O N E M O R E T EA R ) A N D C H A L L E N G E R 2 Hate = 50 Per Cent) Disbursements during Year (Thousands) A d d ition al Income Tax (Thousands) Investm ent Tax C redit (Thousands) A dditional Income Tax (Thousands) m m 1 3 1 4 13U 11 13U TABLE 5k (Continued) C h a 1 1 e n g e r 1 Disbursements8 - Investment^ Additional0 Year during Year Depreciation Tax Credit Income Tax n (Thousands) (Thousands) (Thousands) (Thousands) Replace Challenger 1 0 $67 After 7 Years 1 27 175* $22 ¥> 2 27 6itJ 23 3 31 I t 50 k£ 5 50 32* 6 50 21 * 7 51 l l f 3 Rqjlace Challenger 1 0 67 x » After 8 Years 1 27 72* 33 2 27 63* 22 3 31 5 1 t* 1 I t 50 50 36* 6 5o 27* 7 51 18i it 8 139 9 aF rom Table i* 7 » bF rom Table 23, C . Difference in income tax disbursement due to differences in disbursements, depreciation and L Computed according to Internal Revenue Service regulations given in the footnotes of Table 23 S tra ig h t lin e depreciation o f Challenger 1. f ■ " S u m of the years d ig its depreciation of Challenger 1, % u m of the years d ig its depreciation on remaining l i f e of Defender, ^Straight lin e depreciation of Challenger 2, ■ “ ■ S u m of the years d ig its depreciation of Challenger 2. TABLE (Continued) 216 e r 1 De fe nd e r and Challe nge r 2 ivestment^ Additional0 Disbursements® Investment* 1 Additional0 ax Credit Income Tax during Year Depreciation Tax Credit Income Tax Thousands) (Thousands) (Thousands) (Thousands) (Thousands) (Thousands) $ 3 1 4 $22 $6 m k $2 1 * ? 23 2k 72? $20 27 60 ? 1 32 I4 8 ? 2 38 *1 li ue 2h; ’ 1 3 56 121 3 1 4 33 13U 2 1 4 ? 3 22 2 1 4 69? 20 1 27 59? 32 7 38 39? 5 29 l k 56 20 * - 68 101 35 bursements, depreciation and investment tax cred it, a in the footnotes o f Table 23. TABLE 55 AFTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS CHALLENGER 1 vs. DEFENDER (ONE MORE YEAR) AND CHALLENGER 2 (Interest Rate « 5 3?er Cent) Challenger 1 Defender-Challenger 2 C M O J m P rf Hq U) (0 {Q £ < 1 > ' C J ' M R r d '-— ,Q 05 05 m ' C J m m h g h hh Eh t f 0 5 "S ■ ©>H ' 0 5 U 0 5 •H ^ R ' c J C o 0 5 0 5 p j E " l o R P * H tt\ v^P rfR 0 5 0 5 S'S 0 1 - p S h 2 0 1 05 05 nS 0 5 I SH Replace Challenger 1 after 1 Year Totals 0 1 $67 68 $67 65 Equivalent Uniform Annual Cost of , Disbursements 139 Capital Recovery Cost 73 Total Equivalent Uniform Annual Cost 212 135 132 f 35- lit 168 l 3k 128 162 170J 67s 237 Replace Challenger 1 after 2 Years 0 1 2 Totals 67 69 2Z. T63 67 66 g* T W Equivalent Uniform Annual Cost of = Disbursements 8V Capital Recovery Cost 69€ Total Equivalent Uniform Annual Cost 153 13^ _27 U T 3*+ 128 2>f TB3T 100- 67s 167 TABLE 55 (Continued) Challenger 1 Defender-Challenger 2 |H « a 09 -p , d H ' ” ' < D C O 0 0 £ 3 Q) - r f <M O M d v-* © a U JjOW 3 & iQ >H O © U 43 -H f5B P n j '- ' H EH a 81 +> d Ho a ) r o © s a <u<a - ©Jh d ■ » © r a Hi © H ^ •h 9^ P nd'-' © 5 2 o © ra fn U\ jd d V —< - +jp 0 3 d © r a ©«h d TOP O © A ? ■ » <HEH P O'-- Replace ChkLlenger 1 after 3 Years. 0 1 2 JL k7 27 J L $67 b5 2k -JL. Totals 177 16 7 Equivalent Uniform Annual Cost of ^ Disbursements 6l Capital Recovery Cost -66e Total Equivalent Uniform Annual Cost 127 $3^ I3fc J £ L 229 %3k 128 31 -21 216 79 &*& 1*4-3 Replace Challenger 1 after k Years 0 1 2 I 67 3? 35 35 50 67 36 32 30 *fl Totals 225 206 Equivalent Uniform Annual Cost of Disbursements 58 Capital Recovery Cost 62e Total Equivalent Uniform Annual Cost 120 3}* 27 38 257 3k 128 22 23 31 238 67 129 TABLE 55 (Continued) 2*f9 Challenger 1 Defender-Challenger 2 (1) ?ear n ■ P Ctf r O d BQ r d 4) • p + » a d ?H ^ < D r o r o 0 10 « o ( M S ( D ' O r O ^ s i s S . 2 g ,£» « H O 0 3 R O 03 d , d 0 ) , d • H d E H h < H E H H ' d ' w C 4 o s ^ 03 d © cd © . d - g © > d ® H d © © g | 9 to f t j j • H 3 & n ' d w (5) 0 Present Worth 3 f Disbursements (Thousands) Replace 0 « 6 7 $ 6 7 $ 3^ $ 3^ Challenger 1 1 I f O 3§ 13^ 128 after 5 Tears 2 b2 38 2if 22 3 31 * 29 27 23 K 50 M-l 38 31 5 50 39 ^3 3k Totals 283 252 300 272 Equivalent Uniform Annual Cost of Disbursements 58 63 Capital Recovery- . rt» Cost 60e 59S Total Equivalent Uniform Annual Cost 118 122 Replace 0 67 6 7 3^ 3 o Challenger 1 1 31 39 13^ 128 after 6 Tears 2 36 2^- 22 3 33 29 27 23 H - 50 h i 38 31 5 50 39 k 2 33 6 50 37 ± 7 35 Totals 321 279 3b6 306 Equivalent Uniform Annual Cost of • 1 Disbursements 55d 60f Capital Recovery A . np Cost 56 57g Total Equivalent Uniform Annual Cost 111 117 TABLE 55 (Continued) 250 Challenger 1 Defender-Challenger 2 n c t f <D JH a J 03 d ^ © a s r a c v j g < u '■'£*§ H tcOra |.g§ r a H rd • r f 3 £H R •3 R , d xl r a ^ o rn^s ®/-N ra ra H'd •p , d © ra ra _ * h d raR o a> xt d <H EH R ow a ) ra -p d ^ ra aJ ra J- S.ra'd w ratn d r a c o l?l X I T 1 o r a dxl •H den ( —l fO-w" u s~\ o U~\ )s U d § ra ra ra ■d •px d © ra-H raR ra R O v r a g xs Replace Challenger 1 after 7 Years 0 $ 67 $ 67 $ 34 $■ 34 1 33 31 134 128 2 50 45 2*f 22 3 31 27 28 24 4 50 4i 34 28 5 50 39 42 33 6 1° 37 47 35 7 54 .....3.8, - 56 bo Totals 385 325 Equivalent Uniform Annual Cost of Disbursements Capital Recovery- Cost Total Equivalent Uniform Annual Cost 56c no1 399 f 60 114* Replace Challenger 1 after 8 Years 0 67 67 i 27 26 2 49 kb 3 32 28 b 50 bl 5 50 39 6 50 37 7 55 3? 8 -139 94 Totals 519 415 Equivalent Uniform Annual Cost of - Disbursements 64 Capital Recovery Cost 52e Total Equivalent Uniform Annual Cost _________116 34 3^ 137 130 2b 22 27 23 39 32 43 34 b7 36 56 40 103. __ , 70 - Tio 421 65f U2. TABLE 55 (Continued) 251 aThe sum of disbursements and additional income tax from Table b[Col. (2)] (Pwf* - 5$ - n). c[Col. OO] (Pwf - 5$ - n). d(Sum of Col, 3) (erf - 5f° “ n). eProm Table 5l. f(Sum of Col. 5) (erf - 5$ - n). &Prom Table 53* ■^Minimum cost. "Iln'liii LkSSumod i., TTrnyjT1 . 1 253 TABLE 56 CHALLENGER 3 SALVAGE VALUE SCHEDULE Eirst Cost = $379>000 (1) Years Retained (2) Salvage Valuea (Thousands) (3) b Reduced Salvage Value (Thousands) M 0 Depreciable Value (Thousands) 1 $337 $337 $ b2 2 300 300 79 3 267 229 i5o h 238 200 179 5 212 17*+ 205 6 189 151 228 aComputed from 379 (•89)n > where n is the number of years system is retained. v "Reduction in salvage value by 10 per cent of first cost if system is retained three or more years. cEirst cost minus Col. (3). 25^- TABLE 57 DEFENDER- CHALLENGER 3 CAPITAL RECOVERY COST Pirst Cost = $379,OOOa (Interest Rate = 5 Per Cent) Challenger 3 Defender --------------- Year n Salvage** Value (Thousands) Capital Recovery Cost (Thousands) Capital** Recovery Cost (Thousands) Total Capi tal® Recov ery Cost (Thousands) 0 1 2 9 379 $ 0 $63 $63 3 337 19 **■3 62 b 300 27 33 60 5 267 31 27 58 6 238 33 23 56 7 212 33 20 53 8 189 33 18 51 aProm Table *+7. ^Prom Table 56* C[P (pvf* - 5% - 2) - Lq (pwf - If - n)](orf-5^-n) ^Prom Table $2* Challenger 3 capital recovery cost plus Defender capital recovery cost. installation of Challenger 3 is determined by adding the capital recovery cost of two more years of Defender opera tion from Table 52 to the capital recovery cost of Challenger 3 in Table 57. The depreciation and investment tax credit schedule is shown in Table 58. Equivalent uni form annual costs are calculated in Table 59 and plotted in Eigure 16. Challenger 1 is again superior to the Defender and its successor for all assumed lives of Challenger 1 and Challenger 3. The minimum cost of Challenger 1 is less than the minimum cost of Defender-Challenger 3, regardless of what lives are assumed for Challenger 1 and Challenger 3* A comparison of Eigure 15 and Figure 16, reveals that Challenger 1 has increased its margin over the Defender and its successor in Test 2, This means that the cost of con tinuing the Defender in operation rises at a faster rate than the reduction in cost which can be achieved by defer ring replacement and waiting for an improved challenger. The Defender should be replaced by Challenger 1. An Analysis of the Replacement Decision— Purchased Compiler language Challengers In order to determine why replacement is indicated in the above analysis, the component parts of the equiva lent uniform annual costs of the Defender, Challenger 1, Challenger 2 and Challenger 3 are analyzed* Defender capital recovery costs are plotted in TABLE 58 DEPRECIATION AND INVESTMENT TAX CHALLENGER 1 vs. DEFENDER (TWO MORE YEARS) (Income Tax Rate » 50 Per Cent Year n C ha ll e ng e r 1 Disbursements® during Year (Thousands) Depreciation (Thousands) Investment Tax Credit (Thousands) A dditional0 Income Tax (Thousands) Replace Challenger 1 0 $67 a fter 2 Years 1 27 $35 2 27 1 * 8 ® 1 * 7 Replace Challenger 1 0 67 after 3 Years 1 27 91* 2 27 60f la 3 31 30 2 Replace Challenger 1 0 67 a fter k Years 1 27 87* $11 2 27 65* 38 3 31 1 * 3 * I t 50 22* Replace Challenger 1 0 67 a fter 5 Years 1 27 83? n 2 27 66* 3 31 50* k 5o 33* 5 5o 17* Replace Challenger 1 0 67 • f * a fter 6 Years l 27 79* 22 2 27 66* 38 3 31 52* k5 k 50 39? 5 5o 26* 6 5o 13* 256 ; TABLE 58 J N D IN V E S T M E N T T A X C R ED IT R ( T W O M O R E YEARS) A N D C H A L L E N G E R 3 Rate * 50 Per Cent) D e f e n d e r a n d C h a l l e n g e r 3 A d d itio n a l0 Income Tax (Thousands) Disbursements® Investm ent1 3 A d d ition al0 during Year D ep reciation Tax C red it Income Tax (Thousands) (Thousands) (Thousands) (Thousands) $15 kl # 67 $38^ Ut9 19s 1 <1~ 2 * 67 3 8 f 7 lit9 19g 22 bZ 38 3lt 67 38g 16 1h9 1 9g 22 itO “ 6 26 U0h 3 & 67 38g lit lii9 IS? 22 75* 26 5Cr- it 30 25* 6 38 k5 67 38g 23 2h 9 19g 22 72. $9 26 5 l£ 5 30 36* 5 35 18* 5 TABLE 58 (Continued) fear n C h a l l e n g e r 1 Disbursements® during Year (Thousands) Depreciation (Thousands) Investment Tax Credit (Thousands) A dditional0 Income Tax (Thousands) Replace Ghallenger 1 0 |6 7 a fte r 7 Years 1 27 *75; C M C M # 3 = 2 27 139 3 31 11 1 * 50 5 50 32; 6 5o 21* 7 51 l l f Replace Challenger 1 0 67 a fter 8 Years 1 27 72* 33 2 27 6 3? 39 3 31 19 h 50 1 * 5 * 5 50 6 50 27^ 7 51 18* 8 139 9 aFrom Table 1*7. bFrom Table 23. d iffe r e n c e in income tax disbursement due to differences in disbursements* depreciation and Computed according to Intern al Revenue Service regulations given in the footn otes o f Table eS traigh t lin e depreciation o f Challenger 1 . f S u m o f the years d ig its depreciation o f C hallenger 1, eS u m o f the years d ig its depreciation on remaining l i f e o f Defender. ^Straight lin e depreciation o f Challenger 2. •^ S u m o f the years d ig its depreciation o f Challenger 2. TABLE $8 (Continued) 257 D e f e n d e r a nd C h a l l e n g e r 3 3 Additional0 Disbursementsa Investment Additional0 Income Tax during Year Depreciation Tax Credit Income Tax I (Thousands) (Thousands) (Thousands) (Thousands) (Thousands) £ r m $ 67 138® 20 $39 1 1 * 9 19, 1 11 22 68* $ 9 26 6 30 1 * 1* 6 35 27* 5 1 * 2 li** 3 3 1 * 67 38g 30 39 1 1 * 9 19? 19 22 65* 18 26 5 tf 8 30 1 * 3 * 7 35 33* 5 1 * 2 22 * 3 51 11* 1 * 3 jn ts, dep reciation and investm ent ta x credit* le fo otn otes o f Table 23* 258 TABLE 59 AFTER-TAX EQUIVALENT UNIFORM ANNUAL COSTS CHALLENGER 1 vs. DEFENDER (TWO MORE YEARS) AND CHALLENGER 3 (Interest Rate = 5 Per Cent) Challenger 1 Defender-Challenger 3 rH d o j i* (2) ' Disbursements during Year (Thousands) (3) * Present Worth of Disbursements (Thousands) & cs 0 h +3 +a g d * Q ) CO I D *'■'*00301 j- a o>nd H'd '-'ffllH ti v-* fj d 0 0 S -p^i S W) 03 pj 00 0 1 ? C|j (Drl f j *H O 00 R O ro d'd © si •h den R 'OV p - | O'*-' Replace Challenger 1 after 2 Years 0 1 2 $ 67 $ 67 b2 bo 7b 67 $ 3^ $ 3b 67 6^ ... 1^9 135. Totals 183 lib Equivalent Uniform Annual Cost of d Disbursements 9*+ Capital Recovery Cost 69e Total Equivalent Uniform Annual Cost 163 250 233 125f 63« 188 Replace Challenger 1 after 3 Years 0 1 2 3 67 67 27 26 68 62 33 29 3^- 3b 7b 70 1^9 135 22 19 Totals 195 181* - Equivalent Uniform Annual Cost of d Disbursements 68 Capital Recovery Cost .66 Total Equivalent Uniform Annual Cost 13^ 279 258 95f 62^ 157 TABLE 59 (Continued) 259 Challenger 1 Defender-Challenger 3 0 3 4 3 r , a ra & oj +=> -p a d k < D ' ~ ' < a m +5 d 0 3 o j t, 6 H © ^ ''-■.© C i J ® ^ O 01 0 3 © «} ro /-n o ra « CM S ®-d ro3s -j- a .©•d U'n J s fn » cj ^ + 5 | § v-' © M d ^ p d m 3 +5,q 3 ~ g !Hs d s d bp 0 3 dm® D f i © - H d , 0 -H O 0 3 R O CD M © 41 •h d e h ,£>.3 O ra d x l •H p £H d 0 3 03 © • H d ©R O © 41 0 £ < P & H PH O''-" tH R Ph Ow R *d w Replace 0 $ 67 $ 67 $ 3^ $ 3^ Challenger 1 1 27 26 83 79 after b Years 2 65 59 1^9 135 3 31 27 28 2b b 50 *n 29 2b Totals 2^0 220 Equivalent Uniform Annual Cost of _ Disbursements 62 Capital Recovery- Cost 62 Total Equivalent Uniform Annual Cost 12M- 323 296 83f 6og l*+3 Replace 0 67 67 3b 3b Challenger 1 1 27 26 81 77 after 5 Years 2 65 59 m-9 135 3 39 3*+ 22 19 b 50 bi 30 25 5 50 . 39 36 28 Totals 298 266 Equivalent Uniform Annual Cost of d Disbursements 6l Capital Recovery Cost 60 Total Equivalent Uniform Annual Cost 121 352 318 73f 58^ 131 TABLE 59 (Continued) 260 Challenger 1 Defender-Challenger 3 H a 8 $ * I ri u 8-^ <d a m 0 ra r a ^ S O'd ^ H ^ CM <t)H [j (V) £ S3 w G Q g ' - ' + 9 ^ 3 ( j f - t bora g r a r a 3 Pj 2 <d*h si iO“ n 0 rap 0 C Q H p © P •H ?EH ? - l <H 6H P rcJv-/ p O ' — 00 Disbursements8, during Year (Thousands) (5) Present Worth o f Disbursements (Thousands) Replace 0 $ 67 $ 67 $ 3^ $ 3*f Challenger 1 1 27 26 9° 86 after 6 Years 2 I 65 59 b6 *+0 50 bl l*+9 22 31 135 26 5 50 39 35 27 6 50 37 bo __.. 30 Totals 355 309 Equivalent Uniform Annual Cost of * Disbursements 61 Capital Recovery Cost ?6e Total Equivalent Uniform Annual Cost 117 bOl 357 70f 56® 126 Replace 0 67 67 3^f 3*f Challenger 1 1 27 26 87 83 after 7 Years 2 66 60 0+9 135 3 k2 36 22 19 k 50 bl 32 26 5 50 39 36 28 6 5° 37 1+0 30 7 51 _ 36 b5 32 Totals *H)3 3**2 Equivalent Uniform Annual Cost of , Disbursements 59 Capital Recovery Cost 9+e Total Equivalent Uniform Annual , Cost 113n M+5 387 67f 53g 120h TABLE 59 (Continued) 26l Challenger 1 Defender-Challenger 3 (1) fear n o s m -p P S <D C O 0 1 CM S o W a>|H d to S F-i ttO tQ 0 a p f (D'ri O vH 3 EH (3) b Present Worth o f Disbursements (Thousands) a s I Q -p 0 ® a Jr H i ra Ji & isls 1 C5) c Present Worth o f Disbursements (Thousands) Replace 0 $ 67 $ 6 7 $ 3^ $ 3b Challenger 1 1 27 26 97 92 after 8 Tears 2 66 60 1^9 135 3 50 ^3 22 19 b 50 ¥ l 3>+ 28 5 50 39 37 29 6 g° 37 *f0 30 7 51 36 >*5 32 8 139 Sib _ 6b Totals 550 ¥+3 Equivalent Uniform Annual Cost of d Disbursements 69 Capital Recovery Cost 52e Total Equivalent Uniform Annual Cost 121 552 b63 72f 5lg 123 aThe sum of disbursements and additional income tax from Table 58. H goI. (2)] (Pwf - % - n). c[Col. (1+)] (Pwf* - 5% - n). d(Sum of Col. 3) (erf - 5% - n). eErom Table 5l. f(Sum of Col. 5) (erf - 5% - n). ^Erom Table 57. ■^Minimum cost point. c l e a r p r i n t TAPEF? C O KvukS i 11 II n r . r i : Figure 17* The capital recovery costs of Challenger 2 and Challenger 3 are plotted in Figure 18. The decrease in capital recovery cost between Challenger 2 and Challenger 3 is greater than the increase in capital recovery cost between one more year of Defender operation and two more years of Defender operation. Hence, the total capital recovery cost is less when Challenger 3 is used. Total capital recovery costs are plotted in Figure 19* The capital recovery cost, using Challenger 3? is always less than the capital recovery cost of Challenger 1. Hence, it is not an adverse capital recovery cost of the Defender and its successor which leads to replacement. Hext, the equivalent uniform annual cost of disbursements of the Defender, Challenger 2 and Challenger 3 are analyzed. These costs are shown in Table 60 and Table 6l and are plotted in Figure 20 and Figure 21. The increase in cost of disbursements from one to two more years of Defender operation is greater than the reduction in cost of dis bursements from Challenger 2 to Challenger 3* Hence, the total cost of disbursements increases as the life of the Defender is extended. The total cost of disbursements is plotted in Figure 22. The cost of disbursements for Challenger 1 is less than the cost of disbursements of the Defender and its successor. Thus* it is the increasing operating disbursements of the Defender which cause tl 3 20 X 20 DIVISIONS PER INCH. 550 X 200 DIVISIONS. PRINTED IN U.S.A. ON CLEARPRiNT TECHNICAL PAPER NO. 5CHS C l e a r p iT int C h a r t s 2 . 0 DIVISIONS P-R INCH 150 X 200 DIVISIONS. PRINTED I N U.S.A. ON CLEAf’pPlN'T TECHNICAL PAPER NO. >015 C ^ iE A JtP friN r ( C h a r t s ti vt IrtlTF-^ iiiiiL liiiS t TJ M I a 111 liawMdiBi C i . E A R P R t N Y P 4 P K R C O . C 3 8 . 2 0 V 2© O l V I S ' . O M S P E R I N C H 5 5 C X 2 Q 0 D I V I S I O N : P R I N T E D IN U . S . A . U N C LEW, R P R I N T TECHNICAL P A P E R N O , 1 0 5 5 ClI^EECHarI I ildhi-d 267 TABLE 60 EQUIVALENT UNIFORM ANNUAL COST OF DISBURSEMENTS OF DEFENDER (Interest Rate « 5 Per Cent) (1) Year n (2) _ (3) Disbursements Present Worth during Year of Disbursements ( Thousands ) ( Thousands ) 00 Equivalent Uni- form0 Annual Cost (Thousands) Defender Kept One More Year 0 $ 6kb 1 $ 67 $ 67 2 3^ 3 23 h 18 5 15 6 13 7 11 8 10 Defender Kept Two More Years 0 $137d 1 $ 67 2 81 $ 1b 3 50 39 5 32 6 27 7 2b 8 21 aFrom Table 1 +7. *67 Cpw£» - 5% - 1). c[Col. (3)] [erf - 5% - a]. d67 Cpwf' - 5# - 1) + 81 (pwf» - 5% - 2). 268 TABES 6l EQUIVALENT UNIPOEM ANNUAL COST OB DISBURSEMENTS OB CHALLENGER 2 AND CHALLENGER 3 (Interest Rate « 5 Per Cent) ra © N ! pfra f i t " r l g r a R.ra ra ra += ra r a r a o Si m ra - p « 5 J ra '^•Sra'cJ ^-'!3f£i, ra • r a oj ©H ra ro 3 d j ra \r\53ra TJB ^ o o ra *h ra .ra H'H&f 3fcHB o^oe< Q ■ r a * - " ______________ PM o ^ ___________ o f o _______________ H b o - - ^ Challenger 2 1 $67 $6k $ 6k $67 2 2k 22 86 k6 3 27 23 109 ko k 32 26 135 5 38 30 165 38e 6 1 + 6 ?* 199 39 7 56 ko 239 kl 8 68 k6 285 M+ 2 5 6 7 8 Challenger 3 167 $61 $ 6l $33 22 19 80 29^ 26 19 101 28e 30 125 29 35 26 l5l 30 *+2 30 181 31 51 35 216 33 aRrom Table k?, *[Col. (2)] [pwf* - 5% - n]. cCnnrulative sum of Col. (3)* d[Col. 0+)] [erf - 5$ - n]. e Minimum cost point. m im m m ! i : j : ; : f t’ h !' ' I . 1 » at m ii i a: OP-i CU^AFipRINT TECHNICAL PAPER NO. 1015 272 replacement to be required. At the 3qw interest rate of 5 per cent, the capital recovery cost of the 1 naa An Defender salvage value is not as important as the Defender's operating disbursements in determining when replacement should take place. The Effect of a Larger Price Decrease on the Replacement Decision The previous analysis was for a price decrease of 9 per cent between Challenger 1 and Challenger 2. This forecase of price decrease was based on a future rate of technological improvement equal to 50 per cent of the rate of past technological improvement. It is desired to deter mine what effect a price decrease of 19 per cent, based on a future rate of technological improvement equal to 100 per cent of the rate of past technological improvement, will have on the replacement decision* Challenger 2k is used in place of Challenger 2 for this analysis. Since the rate of decrease in salvage value is geared to the rate of decrease in new computer prices, the salvage values of Challenger 1 and Challenger 2A are decreased at a compound annual rate of 23 per cent ( ^ x 11) instead of the 11 per cent rate previously used. It is assumed that the Defender*s salvage value at the end of Year 1 is the same as it was in the previous analysis. This assumption is based on the obser vation that used computer prices do not decrease in 273 relation to new computer prices until after the new com puters have been put into operation*'1 ' Therefore, the capital recovery cost of the Defender will remain the same, as will its cost of disbursements. The costs for the two rates of price decrease are shown in Table 62 for a seven year Challenger 1 life and a six year life for Challenger 2 and Challenger 2A. These lives correspond to the minimum cost points in Mgure 15. The increase in capital recovery cost of Challenger 1 is greater than the increase in capital recovery cost caused by using Challenger 2A instead of Challenger 2. However, there is a decrease in the cost of disbursements of Challenger 1 due to a larger deduction for depreciation in the first year of Challenger 1 life, when a reduced sal vage value is used. The net effect of the additional price decrease, based on the assumed lives of Challenger 1 and Challenger 2A, is to make one more year of Defender opera tion equally attractive to replacing the Defender with Challenger 1• If the 19 per cent price decrease is taken as the umoer limit of possible price decrease* the analysis reveals that replacement should take place for anv rate of price decrease up to that decrease based on 100 -per cent of past technological improvement. ■^George H* Heilborn, "The Used Computer Market," Datamation. (July, 1§65), W-53. 27*f TABLE 62 COMPARISON OP COSTS FOR TV/O RATES OP PRICE DECREASE CHALLENGER 1 (7 YEAR LIRE) vs. DEFENDER (I MORE YEAR) AND CHALLENGER 2 (6 YEAR LIFE) CHALLENGER 2A (6 YEAR LIRE) 9 Per Cent Price Decrease3, 19 Per Cent Price Decrease*3 Challenger 1 (Thousands) Defender Challenger 2 (Thousands) Challenger 1 (Thousands) Defender Challenger 2A (Thousands) Equivalent Uniforn Annual Cost of Disbursements Capital Recovery Cost Total Equivalent Uniform Annual Cost $ 56 $ 60 51 * % $ 51 $ 59 70 63 100 ll*f 121 122 aFrom Table 55. ^Computed on same basis as values in Table and Table 55 but using Challenger 2A instead of Challenger 2 and salvage value decrease of 23 per cent per year for Challenger 1 and Challenger 2A. 275 After-Tax Replacement Analysis? Rented Compiler Language Challengers. Interest Rate - 10 Per Cent. Inoome Tax Rate = 50 Per Cent The costs of the Defender and four rented assembly language challengers are shown in Table 63, Challenger IE is the rented assembly language challenger of Table 16, p. 121. Challenger 1R is available to begin operation in Year 1. Challenger 2R is available to begin operation in Year 2 at a first year rental 10 per cent below the first year rental of Challenger 1R. Challenger 3R is available to begin operation in Year 3 at a first year rental 10 per cent below the first year rental of Challenger 2R. Chal lenger 2RA is available to begin operation in Year 2 at a first year rental 22 per cent below the first year rental of Challenger 1R. A one year conversion period is required prior to the installation of any of the challengers. The cost of conversion is higher than the cost of conversion for purchased compiler language challengers because as assembly language is used. Challenger 1R vs. Defender (One More Year and Challenger 2R (Test lEt) The capital recovery cost of loss in salvage value for n years, for one and two more years of Defender opera tion, are shown in Table 6^. Equivalent uniform annual costs are calculated from the data in Table 65, Table 66 and are plotted in Figure 23. In Figure 23, Challenger 1R 276 TABLE 63 SCHEDULE OF COSTS OF DEFENDER AND FOUR RENTED CHALLENGERS Operating Costs8, (Thousands) Conversion Costs® (Thousands) 1 Defender Challenger Challenger xear n 1RC 2Rd 2RAe 3Kf IE 2 2RA 3R 0 $81 1 $67 $120 $81 $81 2 82 12*+ $107 $ 9*+ $81 3 128 111 97 $ 97 *+ 129 115 100 100 5 132 119 10*+ 10*+ 6 3.36 123 108 108 7 l5*+ 128 112 112 8 15*+ 133 116 116 forecasting equations C(n ■?) = 10 + IO8R3 . (1.0*+)J where n is the first year of rental; j is an index of the years after the first year of rental (j = 0 repre sents the first year of operation); Rr = .90 for a rate of rental reduction based on a 50 per cent rate of past tech nological progress. Rr - .78 for a rate of rental reduc tion based on a 100 per cent rate of past technological progress. ^From Table *+7. cFrom Table 16. ^Computed from C^ = 10 + 108 (.90) (1.0*+)*5. eComputed from C^ = 10 + 108 (.78) (1.0*0*5. ■^Computed from C^-j = 10 + 108 (.90)^ (l.0*f)^. ^Consists of $36,000 in program conversion costs from Table 21 plus an estimated $*+5,000 in systems and file conversion cost. 277 TABLE 6*+ DEFENDER CAPITAL RECOVERY COST (Interest Rate = 10 Per Cent) Defender Kept One More Year Salvage Capital^ Year Value Recovery Cost n (Thousands) (Thousands) 0 $*f20 1 37^ $88 2 b6 3 32 b 25 5 21 6 18 7 16 8 15 Defender Kept Two More Years Year Salvagea Capital n Value Recovery Cost (Thousands) (Thousands) 0 $*+20 1 37b 2 333 $8»+ 3 58 if if6 5 38 6 33 7 30 8 27 aFrom Table **8. b[L0 - Lx (pwf' - 10$ " 1)] (erf - 10$ - n). C[L0 - L2 (pwf1 - 10$ - 2)] (erf - 10$ - n). 278 TABLE 65 DIFFERENCES IN INCOME TAX DISBURSEMENTS CHALLENGER 1R vs. DERENDER (ONE MORE YEAR) AND CHALLENGER 2R (Income Tax Rate = 50 Per Cent) Challenger 1R Defender-Challenger 2R £ a ) a ) Disbursements during Year (Thousands) Additional Income Tax (Thousands) c S o a < u c 3 r a < i > ! = h £ 0 3 5 j 111 & * H O m £•£ P i Q Depreciation (Thousands) Additional Income Tax (Thousands) 0 $ 81 1 $^-2 1 120 $26 $lb8 $2^ 2 12*f 107 9 3 128 111 9 b 129 115 7 5 132 119 7 6 136 123 7 7 l5b 128 13 8 15+ 133 11 aFrom Table 63. ^Difference in income tax disbursement due to differences in disbursements and depreciation, cSum of the years digits depreciation on remaining life of Defender. TABLE AFTER-TAX EQUIVALENT UN CHALLENGER IR vs. DEFENDER (ONE H (Interest Rate * Challenger 1R < d ra -p cm a w < D o o S < D W W ■ < { ) ( 0 to t < d l < D IH ID ID £ -P to T O 0 ) & § » £> O a ) < a ja S - i *H (H CUD'-' •3 • ® Q -H *H H ° 1 9 0 ) u £ <0 1 3 Cum of disbursements and additional income tax from Table 6£. b[Col. (2)] [pvf« - lOJg - n]. Cumulative total of Col. (2). d fcol. (10 ] [erf - 10$ - n} e[Col. (6)] [pwf* - 10$ - n , ] . £Cumulative total of Col. (7). £[Col. (8)] [erf - 10$ - n]. bSTom Table 61*. Cum of Col. (9) and Col. (10). 0 t o vO E J U* 'w ' Q) C O H & ' S N C Q I h W) P 'U M O v — 4 — J — 0 CO NO > 0 VO $=- — j rv> v o c d v o o O v v n r o & crwo po-fc" Ov- ^ j ia VO -o frovw U —J <*■ HHHMH (-»*-* M U> VO V jO U> t * t - v r i V ? C O - J —3 0 3 O Ov — 4 V T V £ : €&s HHlMNWjroo O v co I — * vn. p o o n co £ « « £ ! £ £ ; § I v o \ o v n . v o v n o » v o v o (6) a D isbursem ents d u rin g Year (Thousands) (7 ) P r e se n t Worth® o f D isbursem ents (Thousands) ( 8 ) f C um ulative P re se n t* Worth o f D isb u r se ments (Thousands) (9 ) E q u iv a le n t Uniform6 Annual C ost o f D isbursem ents (Thousands) ( 10) D efender C a p ita l E ecovery C ost (Thousands) ( 11) T o ta l E q u iv a le n t1 Uniform Annual C ost (Thousands) < o < o P p* © I o £3* & I - * M C D £ S tn C D *1 ro W Ov O N o r o -o VO :3e fender i--an ' I i : (jJHal linger G H alle srnnec Year ui^al :a sse TP .g£:isf c H a lle E l seiibl est : 281 is superior to the Defender and Challenger 2R for all assumed lives of Challenger IS and Challenger 2R. Thus. Challenger 1R has passed Test 1R and has qualified to com pete in Test 2R against two more years of Defender operation followed by the installation of Challenger 3R* Challenger 1R vs. Defender (Two More Years) ~ and Challenger 3R (Test 2R) Equivalent uniform annual costs are calculated from the data in Table 67, Table 68 and are plotted in Figure 2^. Again, Challenger 1R is superior to the Defender and its successor for all assumed lives of Challenger 1R and Chal lenger 2R. A comparison of Figure 23 and Figure reveals that Challenger 1R has increased its margin over the Defender and its successor. Therefore. a better challenger is not worth waiting for. The Defender should be replaced b.v Challenger 1R. An Analysis of the Replacement Decision-Rented Assembly language Challengers In order to determine why replacement is indicated in the above analysis, the capital recovery and disburse ments cost of the Defender and the disbursements costs of the challengers are analyzed. The capital recovery cost of the Defender for one more and two more years of operation are plotted in Figure 25. In order to indicate the affect of interest rate on 282 TABLE 67 DIFFERENCES IN INCOME TAX DISBURSEMENTS CHALLENGER IE vs. DEFENDER (TWO MORE YEARS) AND CHALLENGER 3R (Income Tax Rate = 50 Per Cent) Challenger IE Defender-Challenger 2R Year n Disbursements during Year (Thousands) Additional Income Tax (Thousands) Disbursements during Year (Thousands) Depreciation0 (Thousands) Additional Income Tax (Thousands) 0 $ 81 $1+2 1 120 $ 67 $38 8 2 12>+ $29 163 19 3 128 97 16 k 129 100 15 5 132 10*+ l*+ 6 136 108 li+ 7 19* 112 21 8 19* 116 19 aFrom Table 63. Difference in income tax disbursement due to differences in disbursements and depreciation. cSum of the years digits depreciation on remaining life of Defender. TABLE 68 AFTER-TAX EQ UIVALENT UNIFO] C H A L L E N G E R 1R v s . DEFENDER (T W O M O R E ( I n t e r e s t Rate * 6 Pi u < D C h a l l e n g e r 1R C Q CV? ( 3 < 3 (0 XI 0 3 34 H S II ^ rC Eh O •P S-t r ^ t o w Se 0 0 "S. -P £0 to pJ m xi © C Q fn *h (U Q S1 1 0 I E h •H«H I ° - S 4 s C ✓-% ■ L T \ - P 1 0 © W ' f i O E ' g oio 3 g H C Q C D £•3*3 B •H g.O O P C 03 XJ irigs 4§ <"N C vO V C Q C Q 0 2 a > H . JPS •§f JS h -oS. 0 $ 81 $ 81 1 120 109 2 153 126 $316 $182 3 128 96 1*12 1 66 k 129 88 5oo 1J > 8 5 132 82 582 15U 6 136 77 659 152 7 15U 79 738 1 51 8 I51t 72 810 151 Cum o f disbursem ents and a d d itio n a l income ta x from Table 67 . b [Col. (2 )] [pwf* - 10% - n ]. C u m u lative t o t a l o f Col# ( 2 ) . ^ [C ol. (ij.) ] [ c r f - 10$ - n ]. e [C o l. ( 6 ) ] [pwf* - 10$ - n ], f Cumulative t o t a l o f C ol. (7). g[C o l. (8)] g r f - 10% - n ]. bFrom Table 61*. ^•Sum o f C ol. (9 ) and C ol. (1 0 ). * U2 75 163 113 115 118 122 133 135 UWMHHHOn-J vnvo ro o o w iw u v n H ^ O O v On-J —3 OOU» O-P" vo oovo v*>vo vnvnaoro o o w n p - j r - v o ro r o H U l C o O U J i r M ' O H f O ' O O V l H M H H H H H M M M N N V r i i r O B S I - v J - J \ O U ) H -*f)r r o v o v o v o p - v n e » -JOWQD^Oor H H H H H H N v n v n . ex*, o s — a vo ro Vn—3 o vnvn h vn z n $ ( 6) Disburs ements during Tear (Thousands) (7) e Present Worth of Disbursements (Thousands) (6) f Cumulative Present Worth of Disburse ments (Thousands) (9) Equivalent Uniform® Annual Cost of Disbursements (Thousands) ( 10) h Defender Capital Recovery Cost (Thousands) tn < D < V P. © ► * I O P* P H H © P TO a > VO Ov CO Total Equivalent1 Uniform Annual Cost (Thousands) C tEACPaiNTCtfAaTS s,iavH!2iNi«davai| capital recovery cost, the capital recovery costs at an interest rate of 5 per cent from Figure 17 are plotted, in addition to the capital recovery costs at an interest rate of 10 per cent. The equivalent -uniform annual cost of dis bursements of the Defender for one more and two more years of operation are shown in Table 69 and plotted in Figure 26. The cost of Defender disbursement at an interest rate of 5 per cent differs only slightly from the cost of disburse ments at an interest rate of 10 per cent (compare Figure 20 with Figure 26). Therefore, the difference in Defender costs at an interest rate of 10 per cent is due solely to the difference in capital recovery cost of the loss in sal vage value. The cost of disbursements of Challenger 2R and Challenger 3® are shown in Table 70 and plotted in Figure 27. From a comparison of Figure 26 and Figure 27 it is observed that thr decrease in cost of disbursements between Challenger 2R and Challenger 3R is greater than the increase in disbursements between one more year of Defender opera tion and two more years of Defender operation. Therefore, the total cost of disbursements is less when Challenger 3& is used. Total disbursements costs are plotted in Figure 28. The disbursement co3t of the Defender and its Sv. icessor is always less than the disbursement cost of Challenger 1R. Hence, it is not an adverse cost of disbursements of the Defender and its successor which leads to replacement. 287 TABLE 69 EQUIVALENT UNIFORM ANNUAL COST OF DISBURSEMENTS OF DEFENDER (Interest Rate * 10 Per Cent) (l) ■ ' ■ (27... . ■ ..... U ) ....... . CM-) Disbursements Present Worth of Equivalent Uniform Tear during Tear Disbursements Annual Cost n (Thousands) (Thousands) (Thousands) Defender Kept One More Tear 0 $ 61° 1 $67 $ 67 2 35 3 25 5 19 5 16 6 1** 7 13 8 11 Defender Kept Two More Tears 0 $128d 1 $67 $1^1 2 81 7b 3 51 k 1 * 0 5 31* 6 29 7 26 8 2lf aFrom Table 63. b67 (pwf* - 10# - 1). c[Col. (3)] [erf - 10# - n], d67 (pwf1 - 10# - 1) + 81 (pwf - 10S5 - 2). TABLE 70 EQUIVALENT UNIFORM ANNUAL COST OF DISBURSEMENTS OF CHALLENGER 2R AND CHALLENGER 3R (Interest Rate = 10 Per Cent) 289 ( H o j < D IH OJ 0 ) P ri 0 ) O J 0 3 a < » s ^ g i?? r o £«,§ s i s «P O IS p -p H s-'O © ® rnCS S«d ss m . £ £ £ 2 P © Challenger 2R •h 8 P ° ^ p to CO d P'd v_/ d) * ‘ ■a .ft P * ro I fclAliri? 1 $ 81 $7*+ t lb $ 81 2 107 88 162 93 3 111 83 2^5 98 if 115 79 32^f 102 5 119 7tf 398 105 6 123 69 *f67 107 7 128 66 533 109 8 133 62 595 111 Challenger 3H 2 $ 81 $lf7 $ b 7 $ 27 3 97 39 86 35 b 100 32 118 37 5 10*f 27 lb5 38 6 108 25 170 39 7 112 23 193 *f0 8 116 22 215 bO aFrom Tahle 63. ’’[Col. (2)] Cpwf* - 10# - a]. cCumalative sum Df Col. (3). d[0ol. ft)] [erf - 10# - * *1* rnrtti mtx a *1 & C L E A R P f l I N T P A P E R C O C 3 b 2 0 X 2 . O D I V I S I O N ? P E R IN’C.H 1 5 0 X D I V I S I O N S P R I N T E D JN' U . S . A . C N C L E A ’ r t P H I N T T E C H N I C A L P A P E R N O . 1 0 5 5 Cuflcpg^TraiHs 292 Rather, it is the capital recovery cost of the loss in Defender salvage value whieh pauses replacement to he required. At the high interest rate of 10 per cent, the capital recovery cost of the loas in Defender salvage value is more important than the Defender’s operating disburse ments in determining when replacement should take place. The Effect of a Larger Rental Decrease on the Replacement Decision The previous analysis was for a rental decrease of 10 per cent between Challenger 1R and Challenger 2R. This forecast of rental decrease was based on a future rate of technological improvement equal to 50 per cent of past technological improvement. It is desired to determine what effect a rental decrease of 22 per cent, based on a future rate of technological improvement equal to 100 per cent of the rate of past technological improvement, will have on the replacement decision. Challenger 2RA. is used in place of Challenger 2 for this analysis. For the reason cited previously (p. 272), it is assumed that the Defender’s sal vage value at the end of Year 1 is the same as it was in the previous analysis. The costs of Defender operation remain the same. The costs for the two rates of rental decrease are shown in Table 71 for a seven year Challenger 1R life and a six year life for Challenger 2R and Chal lenger 2RA. TABLE 71 COMPARISON OP COSTS FOR TWO RATES OP RENTAL DECREASE CHALLENGER 1R (7 YEAR LIFE) vs. DEFENDER (l MORE YEAR) AND CHALLENGER 2 (6 YEAR LIPE)/CHALLENGER 2RA (6 YEAR LIFE) 10 Per Cent Rental Decrease8, 22 Per Cent Rental Decrease** Challenger 1R (Thousands) Defender Challenger 2R (Thousands) Challenger 1R (Thousands) Defender Challenger 2RA (Thousands) Equivalent Uniform Annual Cost of Di sbursements $i5l $137 $151 $131 Capital Recovery Cost — 16 — 16 Total Equivalent Uniform Annual Cost i5i 153 151 lb? aFrom Table 66. ^Computed on same basis as values in Table 65 and Table 66 but using Challenger 2RA instead of Challenger 2R. ro \o UJ 29^ The decrease in the operating costs achieved by- using Challenger.2RA instead of Challenger 2R is sufficient to overcome the capital recovery eost disadvantage of the Defender. Therefore* when the decrease in rental takes Place at a rate based on 100 -per cent of past technological improvement, replacement should not take place: there is a better challenger worth waiting for. Summary and Conclusions 1. In order to arrive at a correct decision in replacement analysis, future challengers must be evaluated in addition to evaluating present challengers. This is of particular importance in computer replacement studies, where rapid technological progress may make it economical to defer replacement now in favor of replace ment with an improved challenger at a later date. 2. A method of replacement analysis has been developed which permits future technological growth to be incorporated into the replacement analysis in terms of both first cost inferior ity and operating cost inferiority. 3. At low rates of interest, defender operating disbursements are more important than defender capital recovery cost in determining when replacement will he required* At high rates of interest, defender capital recoveiy cost is more important than defender operating dis bursements . SUMMARY AND CONCLUSIONS 1* The Hardware Model is a deterministic model for the selection of computer hardware systems operating in a hatch processing environment. The model can also he applied to existing computer installations, where a modification or expansion of an installation is con templated. 2. The Hardware Model can be applied to either single pro gram or multiprogram hatch processing operations. It cannot be applied to applications in which inputs occur randomly. 3. Where complex timing relationships exist, due to the time-sharing of the central processor and multiple input-output channels, it may he necessary to try several constraint equation formulations before an optimal solution is obtained. *+. The Hardware Model is a linear integer programming model. Constraint equations stipulate the times in which jobs must be performed. The constraint equations are a summation of the product of reciprocal of through put rates and technological coefficients. The solution 296 to the linear programming model can be obtained by the following three methods: a. Gomory method of integer programming. b. Parametric linear programming. c. Discrete programming. In the absence of efficient integer programs, the approximate methods of (b) and (c) are preferred. The dual solution of the linear integer programming model is useful for identifying critical job(s). A critical job is one which will result in a different optimal solution if a reduction in the time constraint of the critical job occurs or if the technological coefficients of the critical job are increased. The Hardware Model is used to determine an optimal hardware configuration for each time period under con sideration. The time period is one in which the job time constraints and technological coefficients remain constant. An optimal solution applies for the duration of a given time period. A one year time period is used in the analysis. It is assumed that the technological coefficients (file sizes, input volumes and output volumes) grow at a compound annual rate. It is also assumed that programs grow in size at a compound annual rate. However, any rate of application growth may be assumed in the model. A new optimal solution is obtained for each year, based on the increase in tech nological coefficients and the increase in memory requirement. The optimal solutions are based on the use of an assembly or machine oriented programming .language. 7. The Hardware-Software Model is used to obtain an opti mal solution for each time period, when the use of compiler languages is considered in addition to the use of assembly languages. The constraint equations are modified in order to reflect the use of a programming language different from a machine oriented language. This is accomplished by using estimated ratios of com piler object program execution time and program size to assembly language object program execution time and program size. 8. The costs of program preparation and program maintenance for alternative programming languages are calculated, based on satisfying a program preparation time con straint and a machine time constraint on assembly/ compilation time plus program debugging time. 9. The best available hardware-software system (present challenger) is chosen on the basis of minimum equiva lent uniform annual cost. 10. Future hardware-software systems (future challengers) must be considered in the replacement analysis. The price and rental of future challengers is forecasted on the basis of historical technological progress. 11. A replacement analysis is made in order to determine whether the existing system (defender) should be replaced by the present challenger. The analysis is made by comparing the present challenger with the defender and its successor (future challenger). In order for replacement to occur, the minimum uniform annual cost of the present challenger must be less than the minimum uniform annual cost of every combina tion of additional defender service, followed by its successor, which can occur within the assumed life of the present challenger. When the defender time con straints are greater than the challenger time con straints, the defender operating costs should be increased to reflect a set of time constraints equal to the challenger time constraints. 12. At low rates of interest, defender operating disburse ments are more important than defender capital recov ery cost in determining when replacement will be required. At high rates of interest, defender capital . recovery cost is more important than defender operat ing disbursements. 13. The replacement analysis used in the model is appli cable to the selection of computers for all applica tions. 300 1*+. The technical and financial variables involved in the selection of a computer system are interdependent. A purchased compiler language system is favored at a low minimum attractive rate of return (interest rate)* A rented assembly language system is favored at a high interest rate. In order to execute production pro grams within the same time constraints as an assembly language system, the compiler language system requires higher speed components and a larger memory than does the assembly language system. However, at a low rate of interest, the capital recovery cost of the pur chased compiler language system is small in comparison to the cost of disbursements of an assembly language system. The low capital recovery cost results in a lower total equivalent uniform annual cost for the purchased compiler language system. At a high rate of interest, a high capital recovery cost results in a higher total equivalent uniform annual cost for the purchased compiler language system. 15. The analysis of qualitative factors is important where the quantitative analysis does not produce a significant difference in alternatives. Sub.iects for Further Research 1. It appears possible to use the model in the product planning and design of new computer systems in addition to the selection of computer systems* The model could be used to simulate the characteristics of a variety of data processing applications* The model would provide the speed, memory capacity, configuration and programming language required in order to minimize the cost of given applications. The resulting dis tributions of component speed and memory size could be used to establish the range of values required for various technical characteristics for given price and rental brackets* A stochastic model could be developed for those appli cations in which random inputs occur. The replacement analysis developed in this dissertation could be used as part of such a model. b i b l i o g r a p h y BIBLIOGRAPHY ACM Sort Svnroosium. Communications of the ACM. Vol. 6. Ho. 5 (May, 19537:--------------------- Amdahl, Gene M. "New Concepts in Computing System Design," Proceedings of the I.R.E.. (May, 1962;, 1073-1077. Arbuckle, Edward C., and Saxberg, Emil J. "Improving Debugging Efficiency," Datamation. (April, 1962), 33-35. Barish, Norman. Economic Analysis for Engineering and Managerial Decision-Making- New York-; McGraw- Hill, 1962. Bromberg. Howard. "COBOL and Compatibility," Datamation. (February, 1961), 30-31 *. ________ . "Compilers Today," Business Automation. (December, 1962), 25, 32. Brooks, Frederick P., and Iverson, Kenneth E. Automatic Data Processing. New York: John Wiley, 1962. Cantrell, Harry N. "Where Are the Compiler Languages Going?", Datamation. (August, 1962), 25-28. Carr, Charles E., and Howe, Charles W. Quantitative Decision Procedures in Management and Economics. New York: McGraw-Hill, 196m-. Chames, A., and Cooper, W. W. Management Models and Industrial Applications of Linear 3-frnffrflinming. New York: Wiley, 1961. 5X1 Churchman, C. West, Ackoff, Bussell L., and Amoff, E. Leonard. Introduction to Operations Eesearch. New York: McGraw-Hill, 195?. Clippinger, Richard E. "Progress in Language Standards," Datamation. (April, 19&3), 26-27. 303 304 "COBOL Is the Language!" Datamation. (January-February, i960), 54-55. "CQDASYL Emphasis Shifting to Development Committee; More on COBOL," Datamation. 70-71. "CODASYL O.K.'s Publication of COBOL - 6l," Datamation. (July, 196l),~40-4l. Codd, E. P. "Multiprogramming," edited by Franz L. Alt and Morris Rubinoff, Advances in Computers, III. New York: Academic Press, 1962. Pp. 78-152. Critchlow, A. J. "Multiprogramming and Multiprocessing," I.E.E.E. Spectrum. (March, 1964), 192-198. Dantzig, George B. Linear Programming and Extensions. Princeton, New Jersey: Princeton University "Press, 1963. Dennis, Jack B. Mathematical Programming and Electrical Networks. New York; Wiley, 1959* Desmonde, W. H. Real Time Data Processing Systems: Intro ductory Concents. Englewood Cliffs: Prentice-Hall, 19&f. Doig, A. G., and Land, A. H. "An Automatic Method of Solv ing Discrete Programming Problems," Econometrics, Vol.^28, No. 3 (I960). Dorfman, Robert, Samuelson, Paul A., and Solow, Robert M. Linear Programming *md Economic Analysis. New York: McGraw-Hill, 1958. Evans, G. W., and Perry, C. L. Programming and Coding for Automatic Digital ComoutersT New York: tec&raw- Hill, 1961. . Pisher, Peter P., and Swindle, George P. Computer Program ming Systems. New Yorks Holt, Rinehart and Winston, 1964. Gass, Saul. "Integer Linear Programming." Edited by Franz L. Alt and Morris. Advances in Computers. New York: Academic Press, i960. II, tp. 302-309. Gomory, R. E. "Integer Solutions to Linear Programs." Edited by Robert L. Graves and Philip Wolfe, Recent Advances in Mathematical Programming. New York: McGraw-Hill, 1963. 305 Gomory, R. E., and Baumol, W. “Integer Programming and Pricing," Econometrics. Vol. 28, No. 3 (I960). Gomory, R. E., and Wade, C. S. Integer Programming 2. 7090 (PK IP92 and PK IPM2). Yorktown Heists, New York: IBM Business Theory Group, 1961. Goodman, Richard (ed.). Annual Review in Automatic Pro gramming. Vol. 2, 1961; Vol. 3> 1963* Gordon, Robert M. "COBOL and Compatibility.” Datamation. (July, 1963), lf7-W. ---------- Grant, Eugene L., and Ire son, Grant W. Principles of Engineering Economy. Fourth edition. New York: The Roland Press Co., i960. Grosch, H. R. J. "Software in Sickness and Health," Datamation. (July, 1961), 32-33* Hadley, G. Nonlinear and Dynamic Programming. Reading, MasaTl Wesley Publishing Co., 196V. Halpem, Mark. "Evolution of the Programming System," Datamation. (July, 196*0, 51-53* Heilbom, George H. "The Used Computer Market," Datamation. (July, 1965), 48, *f9, 5l, 53* Herman, Donald, and Ihrer, Pred C. "The Use of a Computer to Evaluate Computers," American federation of Information Processing Societies Proceedings. "l96*f. Vol. 25. Honeywell Electronic Data Processing. Advanced Linear Pro gramming System. Wellesley Hills~7 Mass.: 1964. International Business Machines. System 360 Specifica tions (various documents), 3^24 Wilshire Blvd., Los Angeles, California, 196*t. ________ • Srstem 360 Price List. 3^2*f Wilshire Blvd., Los Angeles, California, 1965. Jordan, Ray V. "Programmer Talent Can Be Measured," Datamation. (July, 1962), *f9-5l. Joslin, E. 0. "Cost-Value Technique for Evaluation of Com puter Systems," American Federation of ^Information Processing Societies donference !fcrooeedings. 196*fr. Vol. 25. 306 Lambert, Frank P. "What's So Hard about Software?'* Datamation. (June, 1965), 50-52.- Ledley, Robert Steven. Programming and TTtlHzing Digital Computers. Hew Yorks McGraw-Hill, 19627 ”” Lemke, C. E. “The Constrained Gradient Method of Linear Programming," Society of Industrial and Applied Mathematics. Vol. la. 1 (fttareh, l$)bl), 1-1 Llewellyn, Robert. Linear Programming. Hew York: Holt, . Rinehart and Winston, 196*f • Lonergan, William and King, Paul, "Design of the B-5000 System," Datamation. (May, 1961), 28-32. Manne, A. S. Economic Analysis for Business Decisions. New York: McGraw-Hill, 1961. Masse, Pierre. Optimal Investment Decisions. Englewood Cliffs, N. J.: Prentice-Hall, 1962. McCraken, Daniel M. "IBM's New Programming Language," Datamation. (July, 196*0, 3l“36. ________ • "Object Program Efficiency," Datamation. (June, 1962), 32. Morris, William T. Engineering Economy. Homewood, Illinois: Richard D. Irwin, Inc., i960. Naftaly, Stanley M. "Correcting Obfuscations by Ordained Linguists," Datamation. (May, 1963)* 2*f-28. Opler, Ascher. "Measurement of Software Characteristics," Datamation. (July, 196*0. Orchard-Hays, William. "The Evolution of Programming Systems." Proceedings of the I.R.E.. TJanuarv. 1961), 2835295: ------------ Phillips, Charles. "CODASYL'S Phillips Writes about COBOL," Datamation. (May-June, i960), 2*f-25, 3^* Rosen, Saul. "Programming Systems and Languages: A His torical Survey." AFIPS Conference Proceedings. 106U-. Vol. 25, pp. 1-1*:----------- --------- Rosenthal, Solomon. "Analytical Technique for Automatic Data Processing Equipment Acquisition," American Federation of Information Processing Societies Jbmfi&edinga, 19^. ?ol. 2:?. 307 Saaty, Thomas L., and Bram, Joseph, Nonlinear Mathematics. New York: McGraw-Hill, 196*4-. Sasieni, Maurice, Yaspon, Arthur, and Friedman, Lawrence. Operations Research— Methods and Problems. New York: John Wiley and Sons, 1959* Schwalb, Jerry. "Compiling In English." Datamation. (July, 1963), 28-30. Schweyer, Herbert E. Analytical Models for Managerial and Engineering Economics. New York: Reinhold tub- lishing Corporation, 196*f. Shaw, Christopher J. "The Language Proliferation," Datamation. (May, 1962), 3*+-36. Singer, E. "Nonlinear Programming," Datamation. (July, 196* 0. Solo, Myron B. "Selecting Electronic Data Processing Equipment," Datamation. (1958), 28-32. "Some More Pacts about COBOL," Datamation, (March-April, i960), 29. -------- Teichroew, D. Introduction to Management Science (Deter ministic Models), New Yorks John Wiley and Sons, 196*f. Terborgh, George. Dynamic Equipment Policy. New York: McGraw-Hill, 1949. ________ • Business Investment Policy. Washington, D. C.: Machinery and Allied Products Institute, 1958. Thuesen, H. G., and Pabrycky, W, J. Engineering Economy. 3rd edition. Englewood Cliffs, New Jerseys Prentice-Hall, 196*4-. United States Department of Defense. COBOL, Report to Con ference on Data Systems Languages. Washington, D. C.s United States Government "Printing Office, April, I960. Taj da, S, Mathematical Programming- Reading, Mass.: Addison-We sley, 1961. Vazsonyi, Andrew. Scientific Programming in Business and Industry. New York: Wiley, 1958. " Wegner, Peter. Introduction to Systems Prn^ramminp;- London: Academic Press, 1964. Weik, Martin H. Third Survey of Domestic Electronic Data Processing Systems. Aberdeen Proving GroundT Ha^Eand: Ballistic Research Laboratories, June, Whitmore, Arthur J. "COBOL at Westinghouse." Datamation, (April, 1962), 31-32. Wofsey, Marvin M. "Compatibility of COBOL Specifications," Data Processing. (February, 1963), ll. Wolfe, Philip, "Recent Developments in Nonlinear Program ming," Edited by Franz L. Alt and Morris Rubinoff, Advances in Computers. New York: Academic Press, 1962. “ill, pp. 1%-186. Worthington, John H. "The Anatomy, of an Assembly System," Computers and Automation. (January, 196^;, 18-20.

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Creator Schneidewind, Norman Floyd (author)

Core Title Analytic Model For The Design And Selection Of Electronic Digital Computing Systems

Degree Doctor of Business Administration

Degree Program Business Administration

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Advisor Beckwith, Richard E. (committee chair), Fleischer, Gerald A. (committee member), White, C. Michael (committee member)

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