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Queueing loss system with heterogeneous servers and discriminating arrivals

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Content Queueing Loss System with Heterogeneous Servers and Discriminating Arrivals Babak Haji Advisor: Professor Sheldon M. Ross Dissertation Submitted in Partial Fulllment of the Requirements for the Degree of Doctor of Philosophy Daniel J. Epstein Department of Industrial and Systems Engineering University of Southern California May 2015 i COMMITTEE MEMBERS Professor Sheldon M. Ross Professor Maged M. Dessouky Professor Amy R. Ward ii ACKNOWLEDGMENTS This research thesis was conducted in the Daniel J. Epstein Department of Industrial and Systems Engineering at the University of Southern California, under the supervision of Professor Sheldon Ross. I would like to thank my advisor, Professor Sheldon Ross for guiding and supporting me over the years. You have set an example of excellence as a researcher, mentor, instructor, and role model. I would like to thank my thesis committee members for all of their guidance and help through this process. I would also like to thank the amazing people of the Daniel J. Epstein Department of Industrial and Systems Engineering, faculty, sta and fellow students for their kindness and support. Finally, thank you to my family for the encouragement, love, and continued support I have gotten over the past years. Contents iii CONTENTS Contents: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : iii List of Tables : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : v 1. Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1 1.1 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2. A Queueing Loss Model with Heterogeneous Skill Based Servers under Idle Time Ordering Policies : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 General Erlang Random Variables and The Method of Stages . . . . . . . 8 2.4 Stationary Probabilities for No-Memory Rules with General Service Time Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5 Finding The Limiting Probabilities using Markov Chain Monte Carlo Meth- ods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3. On the Adan-Weiss Loss Model having Skill-Based Servers and Longest Idle As- signment Rule : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 16 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.2 Model Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.3 Finding The Limiting Probabilities using Markov Chain Monte Carlo Meth- ods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 4. General Case Optimality, Heuristic Policy : : : : : : : : : : : : : : : : : : : : : 25 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.2 Heuristic Approaches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3 Lower Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 5. General Case, Idle Time Ordering Policies : : : : : : : : : : : : : : : : : : : : : 39 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 5.2 Limiting Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Contents iv 5.2.1 Longest Idle Rule and Shortest Idle Rule . . . . . . . . . . . . . . . 40 5.2.2 Random Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 5.2.3 Random Priority Ordering Rule . . . . . . . . . . . . . . . . . . . . 42 5.3 i = for all i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 5.4 p i =p for all i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 5.5 General Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Bibliography : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 54 List of Tables v LIST OF TABLES 2.1 Estimates for P(0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.1 Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.2 Eciency of Heuristic 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.3 Non-Transitive Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 4.4 Eciency of Heuristic 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.5 Examples Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.6 Eciency of Heuristic 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 5.1 Comparison between LIR, RR, SIR, and RPO . . . . . . . . . . . . . . . 48 5.2 Comparison between ITOP and RPO . . . . . . . . . . . . . . . . . . . . 49 5.3 Comparison between ITOP and RPO . . . . . . . . . . . . . . . . . . . . 50 5.4 Comparison between LIR, RR, SIR, and RPO . . . . . . . . . . . . . . . 53 5.5 Comparison between LIR, RR, SIR, and RPO . . . . . . . . . . . . . . . 53 1. Introduction 1 1. INTRODUCTION 1.1 Model In this research we consider a queueing loss system with n heterogeneous servers having arbitrary service distributions ,G i ,i = 1;:::;n. Customers arrive according to a Poisson process with rate and arrivals are discriminating, in the sense that each arrival has an eligibility vector (X 1 ;:::;X n ) withX i = 1 if serveri is eligible to serve that customer and X i = 0 if ineligible,i = 1;:::;n. The binary vectors of successive arrivals are independent and identically distributed having a specied distribution. An arrival can be assigned to any of the servers that are both currently idle and eligible to server that arrival; if there are no such servers, the arrival is lost. In queueing systems with heterogeneous servers one of the main questions is how to assign arrivals to the idle and eligible, call it "idle eligible", servers in order to minimize the cost structure of the system. One common assignment policy that has been stud- ied extensively in the literature is to assign arrival customers to one of the idle servers at random. One of the earliest works which used this policy was by Gumbel [12] on a queuing model with Poisson arrivals and heterogeneous exponentially distributed servers. He assumed when all the servers are busy the customers will join a queue and wait until they get served. He obtained the closed form expressions for the steady state probabilities of the system and also found the expected length of the waiting line. In queueing loss systems arrival customers are lost when all the servers are busy. Fakinos [8] has studied the equilibrium behavior of an M=G=k loss system with heterogeneous servers under the policy which assigns arrival customers to the idle servers uniformly at random. By uti- lizing the method of supplementary variables, he has given a generalization of the Erlang B-formula and shown that this formula holds true under the assumption of heterogeneous servers with equal mean service times. In another paper, Fakinos [9] considered the same model but this time with balking customers. That is, customers will immediately depart the system upon their arrival with a certain probability, otherwise they will be served by one of the idle servers at random. Once again by assuming equal means for the service time distributions he showed Erlang B-Formula is valid for the new model and proved that, in equilibrium, customer (served, balking or rejected) departures from the system 1. Introduction 2 form a Poisson process. A list policy picks a specic ordering of servers and assign arrival customers to the idle and eligible servers in accordance with this ordering. Cooper and Palakurthi [6] proposed a loss system with Poisson arrivals and heterogeneous servers with general service distri- butions. They showed, using a counter example, that the loss probability of their model with list assignment policy does not depend on the service time distributions only through their means. Gregory and Litton [11] considered a queuing loss model with exponentially distributed service times. In their model arrivals arrive according to a Poisson process and are assigned to idle servers according to a list policy. By dening over ow, if an arriving customer nds a server busy is said to over ow to the next server in the ordering, they found the inter-arrival time distributions of arrival customers at servers and showed that prioritizing servers in a descending order of their service rates minimizes the rate at which arrivals are lost. Furthermore, Yao [25] has proved that the best ordering of idle servers for a GI=M=n system with n heterogeneous servers is also to give the rst priority in the ordering to the fastest server, second priority to the second fastest and so on, by fastest we are referring to the server with highest service rate. To do so, he has represented the arrangement of servers by a permutation vector of their service rates, and develops a partial order on these vectors. He has shown that one can reduce the rate of lost customers and improve system performance by moving in the direction pointed out by this partial order. Hordijk and Koole [13] considered a system with parallel queues where each queue has a server with exponentially distributed service time. They assumed the servers are heterogeneous and the queues may have nite capacities. They supposed that the arrival process is a stochastic process which can depend on the number of cus- tomers in the parallel queues and the assignment policy. They showed that when there is no queue for each server for a large class of cost functions and any time horizon, the policy of assigning customers to the fastest idle server is the optimal assignment decision which minimizes the cost at any time point in stochastic sense. Sobel [20] investigated an M=M=n=n system with heterogeneous servers. He showed that for any isotone cost function, a real-valued functiong on a partially ordered set ( ;) is said to beisotone if h(!)h(! 0 ) for all!! 0 2 , the policy of assigning customer to the fastest idle server minimizes the expected discounted cost and the long-run average cost per unit time. The cost may include incremental costs such as resource expense, costs of lost customers, and the cost per unit time of holding a customer inside the system. Let S n be the time till n th arrival, for a similar model Derman, Lieberman, and Ross [7] showed that, for any arrival process, the list policy which assigns arrivals to the fastest idle server stochastically maximizes the total amount of time in [0;S n ] that customers in the system are less than k, for every k and n. Because their arrival process is general the result in [7] also shows that the total amount of time in [0;t] that customers in the system are less than k, for 1. Introduction 3 every k and t, is stochastically maximized by the fastest idle assignment policy. Ross [16] has considered a queueing loss system with heterogeneous servers having exponential distributions with rates depending on the identity of the servers. In his model, for every arrival there is a vector (X 1 ;:::;X n ) which determines the eligible severs for serving that arrival. He also assumed that the X i are exchangeable. Let I i be the indicator of the event that arrival i gets served. By using a coupling argument, he showed that for any arrival process which is independent of the service times, the vector (I 1 ;:::;I r ), for every r> 0, is stochastically maximized when each arrival is assigned to the idle eligible server with the highest service rate. When servers are humans, the assignment policy may potentially in uence the rates at which servers work. For instance, if arrivals are always assigned to the fastest avail- able server then that server might start working slower to avoid being overloaded. Thus, policies that give special priority to servers, such as assigning arrival to the idle eligible server with the highest service rate, can damage the performance of the system and can be considered as "unfair". From a managerial point of view, the perception of fairness aects employee performance (see [5],[4]). This motivates us to investigate assignment policies that are considered as \fair" and also try to nd the policy within this class that in some sense minimizes the cost structure of the model. For more references on the issue of fairness see [10]. In this work, we study a class of no memory policies which we refer to as \idle time ordering policies". These policies are considered as fair policies in the sense that there is no special priority given to any of the servers. Letting the \idle servers vector" be i 1 ;:::;i k if there are currently k idle servers, with i 1 having been idle the longest, i 2 the second longest, and so on, an idle time ordering policy is one whose assignment decisions are based solely on the number of idle servers and the positions in the idle servers vector of those servers that are eligible to be assigned. Examples of such policies would be to assign to a randomly chosen idle and eligible server, or to assign to idle eligible server that has been idle the longest, or that has been idle the shortest. Call centers provide a strong motivating example of a service organization that cares about the issue of server fairness. In particular, many call centers follow the longest idle policy. We are interested in measuring the performance of policies within this class with respect to their rate of lost customers. In Chapter 2 we assume X 1 ;:::;X n are exchangeable and study the behavior of our system under idle time ordering policies. Utilizing the method of stages for this model we nd the limiting probabilities of the list of idle servers in the form which is only specied up to multiplicative constant and show that no matter what idle time ordering assigning policy has been used the results will be the same. Moreover, we show when there are k servers idle, i 1 ;:::;i k , all possiblek! orderings of the list of idle servers are equally likely. 1. Introduction 4 We also show that the limiting probabilities of the ordered list of idle servers depend on the service distributions only through their means. Again we prove, given the set of busy servers, their remaining service times (as well as their provided service times) are independent with their respective equilibrium service distributions. In practice, as n, the number of servers, gets larger nding the multiplicative constant becomes com- putationally intractable. Therefore, we provide an algorithm using the Gibbs sampler method to simulate our model in order to estimate the limiting probabilities and other desired quantities. It has also recently been brought to our attention that results having some similarity to ours have been obtained by Doroudi and Gopalakrishnan in [10]. The model of [10] is similar but less general than our model in that it assumes exponential service distributions and that all servers are always eligible (that is, that P (X i = 1) = 1 , i = 1;:::;n); it does, however, allow for a queue. They show for their model that the limiting probabilities are the same no matter which idle time ordering policy is employed. In chapter 3, by dropping the exchangeability assumption our model becomes the model in Adan and Weiss [2]. In [2], Using a supplementary variable approach to make their model Markovian, Adan and Weiss derived the limiting distribution for this model, and in doing so showed that the limiting distribution of the ordered list of idle servers depend on the service distributions only through their means. We obtain results of [2] using the method of stages. This method considers the service time at each server as a weighted average of dierent gamma distributions with the same rate. Our approach will be to conjecture the reverse chain, and then show the conjecture is correct by nding, up to a multiplicative constant, the limiting probabilities of the ordered list of idle servers. We also show that given the set of busy servers, the remaining service times (as well as the provided service times) are independent with their respective equilibrium distribu- tions (This result is only implicitly noted in Adan and Weiss [2]). In practice, when n is large nding the limiting probabilities or any other quantity of interest becomes compu- tationally intractable. Therefore, we will provide an algorithm for simulating this model using the Gibbs Sampler method to estimate these desired quantities. In the chapter 4, we assume X 1 ;:::;X n are independent and X i = 1 with probability p i , i = 1;:::;n. We also assume that service time at server i is exponential with rate i . We intend to nd the best stationary policy for assigning customers to idle eligible servers which minimizes the rate of lost customers. A stationary policy, depending on the set of idle servers and the set of idle eligible servers, assigns arrivals to one of the idle eligible servers. Note that list policies can be categorized under this group of policies. For any given policy, we are able to nd the rate of lost customers when n = 2. However, nding the limiting distribution for dierent policies when n> 2 seems computationally intensive and even having the limiting distribution comparing the rates that customers 1. Introduction 5 are lost for dierent policies is a formidable task. Hence, we rst show how some of these stationary policies can be eliminated from our search for the best stationary pol- icy which signicantly reduces the complexity of our problem and allows us to nd the best stationary policy when n = 3. Later, we develop two dierent heuristic approaches for nding an optimal or close to optimal assignment policy. In the rst heuristic, for each pair of servers we consider a system consisting of those 2 servers and nd the best ordering among them. Let i > 2 j show that among servers i and j, the best ordering is (i;j). If the pairwise orderings are transitive, i> 2 j and j > 2 k imply i> 2 k, we use the ordering for our original problem that satises these pairwise transitivity relations, and if transitivity does not hold among all of the pairs, we use the ordering that in some sense violates the transitivity relations less frequently. The second heuristic sorts the servers in the decreasing order of the ratio i =p i . For n = 3 we are able to compare these heuristic with the best stationary policy. For n > 3, we use simulation to nd an estimate of the rate of the lost customers for our heuristics. In this case, we compare our heuristics with the random rule and we also dene some lower bounds to examine their performance. In chapter 5, we again assumeX 1 ;:::;X n are independent andX i = 1 with probabil- ity p i , i = 1;:::;n. We also assume that service time at server i is exponential with rate i . We study the behavior of this system under dierent no memory policies, specically, random rule, longest idle rule, shortest idle rule, and a new rule which we refer to as "random priority ordering" rule. By using random priority ordering policy we assume that the system chooses one of the possible n! permutations at random (before starting to work) and assigns arrival customers to idle eligible servers in accordance with this permutation. We compare the performance of these assignment policies under dierent assumptions on service rates and eligibility vectors. First we let i = , for all i, and we conjecture that rate of lost customer is the smallest under longest idle rule. We give an intuitive explanation for this belief and, when n = 2, we analytically show that the longest idle rule has lower rate of lost customers than the random rule or the shortest idle rule. For n = 3, we also provide some numerical evidence. Later we let p i =p, for all i, and numerically compare the rate of lost customers under the random priority ordering policy with the other three policies (which, as shown in chapter 2, have the same rates of lost customers). The numerical examples show that the random priority policy has lower rate of lost customers than idle time ordering policies. For a special case, where n = 2, 1 2 , and p 1 =p 2 = 1, we show that the number of customers lost using the random priority ordering rule is stochastically smaller than the same for idle time ordering policies. A Queueing Loss Model ... 6 2. A QUEUEING LOSS MODEL WITH HETEROGENEOUS SKILL BASED SERVERS UNDER IDLE TIME ORDERING POLICIES 2.1 Introduction In this chapter, we assume that a random eligibility vector (X 1 ;:::;X n ) is exchangeable. The binary vectors of successive arrivals are independent and identically distributed. Al- though our exchangeability assumption is restrictive, we do allow for a general class of operating policies. Letting the \idle servers vector" be 0 if there are currently no idle servers, or i 1 ;:::;i k if there are currently k idle servers, with i 1 having been idle the longest, i 2 the second longest, and so on, we dene an idle time ordering rule as one whose assignment decisions are based solely on the number of idle servers and the posi- tions in the idle servers vector of those servers that are eligible to be assigned. Examples of such policies would be to assign to a randomly chosen idle and eligible server, or to assign to the idle eligible server that has been idle the longest, or that has been idle the shortest. Our analysis uses the method of stages, which starts by assuming that the service distributions are all general Erlang. Doing so, enables us to analyze the model as a continuous time Markov chain. Using a conjecture concerning the reverse chain enables us to nd, up to a multiplicative constant, the limiting probabilities for this model, which surprisingly are the same no matter which idle time ordering policy is employed. We show that the limiting distribution of the idle servers vector depends on the service dis- tributions only through their means, and that given the set of idle servers (a) all possible idle server vectors are equally likely, and (b) the remaining service times (as well as the amounts of service time each has so far provided) of the busy servers are independent and distributed according to their respective equilibrium service distributions. Application of a continuity argument then establishes these results for arbitrary service distributions. Because the determination of the multiplicative constant by summing all the probabilities is computationally intractable for large n, we show how the Gibbs sampler can be used to simulate a Markov chain whose stationary probabilities enable us to determine desired quantities of interest for our model. In Section 2:2, we introduce the model and provide some further notation. In Section 2:3, we review general Erlang distributions and the method of stages. In Section 2:4 we derive the stationary probabilities, and in Section 2:5 we present the Gibbs sampler simulation approach. A Queueing Loss Model ... 7 2.2 Model Arrivals come to an n server system in accordance with a Poisson process with rate : Each arrival has a vector of binary values (x 1 ;:::;x n ) with the interpretation that serveri is eligible to serve that arrival ifx i = 1 and is ineligible ifx i = 0;i = 1;:::;n: The binary vectors of successive arrivals are assumed to be independent and identically distributed having the distribution of (X 1 ;:::;X n ); where X 1 ;:::;X n are assumed to be exchange- able. That is, there are probabilities j ; P n j=0 j = 1; such that j =P ( n X i=1 X i =j) and, given that P n i=1 X i = j; all n j sets of j servers are equally likely to be the set of servers that are eligible for the job. For instance, if j = n j p j (1p) nj thenX 1 ;:::;X n would be a vector of independent random variables, each equal to 1 with probability p. An arrival that nds all of its eligible servers busy is assumed lost. Otherwise an arrival can be assigned to any one of its eligible servers. The time it takes server i to serve a customer has a general distribution G i ;i = 1;:::;n: Set 0 = 0; and fork> 0; let k =P ( P k i=1 X i > 0); thus, k is the probability that at least one of a specied set of k servers is eligible for the next job. Note that P ( k1 X i=1 X i = 0; X k = 1) = k k1 ; k 1 Each idle time ordering rule results in a set of probabilities P j;k ; j k n, where P j;k is the probability that a job arrival when the idle servers vector is i 1 ;:::;i k will be assigned to worker i j . Note that P k j=1 P j;k = k : Examples (a) If the idle time ordering rule in use is to give an incoming job to the idle eligible server that has been idle the longest, then P j;k = j j1 . (b) If the idle time ordering rule in use is to give an incoming job to the idle eligible server that has been idle the shortest, then P j;k = kj+1 kj . (c) If the idle time ordering rule in use is to give an incoming job to a randomly chosen idle eligible server then P j;k = k =k: A Queueing Loss Model ... 8 2.3 General Erlang Random Variables and The Method of Stages Distribution of a random variable X is called general Erlang, GE(N;), if it can be expressed in the following form X = N X i=1 W i where W i , i = 1; 2; 3;:::, are iid exponential random variables with rate and N is a positive integer valued random variable that is independent of the W i . We can write the distribution of the random variable X as the weighted average of dierent Gamma distributions with the same rate . That is, G X (x) = 1 X j=1 p(j)G Y j (x) where Y j is Gamma(j;), p(j) =P (N =j), and P 1 j=1 p(j) = 1. Let G be the distribution function of a GE(N;) random variable X, and let G e be the equilibrium distribution of G. That is, G e (x) = E(N) Z x 0 (1G(y))dy We then have Lemma 1: G e is the distribution function of a GE(N e ;) random variable, where P (N e =j) =P (Nj)=E(N) Proof: G e is the limiting distribution of the excess of a renewal process having the interarrival distributionG. BecauseG is aGE(N;) distribution, we can interpretG e as the limiting distribution of the time until the next visit to state 1 of a continuous time Markov chain which spends an exponential time with rate in each state, and which when leaving state i goes to state j with probability P i;j where P i;i1 = 1; i> 1 P 1;j =P (N =j); j 1 It is well-known (and quite easy to verify) that the limiting probabilities of the preceding embedded Markov chain are i =P (Ni)=E(N). Because the time spent in each state A Queueing Loss Model ... 9 has the same distribution, i ;i 1, is also the limiting probability distribution for the state of the continuous Markov chain. But if the state of the continuous time chain is i, then the time until the next visit to state 1 is distributed as the sum of i independent exponentials with rate , which proves the lemma. For every non-negative random variable Z we can nd a sequence of general Erlang variables that converges to Z, see for instance Whitt [6]. Because of this, our approach will be to rst assume that all service distributions are general Erlang. Using this we can express service time at each server in our model as a general Erlang random variable. In particular, we can imagine that every time a customer enters server i his service consists ofN i exponential stages each with rate i , whereN i is a random variable independent of the service time at each stage withp i (j) =P (N i =j), and P 1 j=1 p i (j) = 1.The method of approximating any arbitrary distributed random variable by a general Erlang distribution is called the method of stages and has been used to analyze various models with general distributions. For instance, see Schassberger [17, 18, 19] , and Kelly [14]. To show that the results we obtain under the assumption that all service distributions are general Erlang remain valid when the service distributions are arbitrary requires that our quantities of interest are continuous functions of the service distributions. Barbour [3] and Whitt [23] have shown that such a continuity exists. 2.4 Stationary Probabilities for No-Memory Rules with General Service Time Distribution To start the process, we rst suppose that the service distribution of server i is general Erlang GE(N i ; i ), i = 1;:::;n and we will analyze the model as a continuous time Markov chain. In order to analyze our system we will dene the \state vector" as (0; r) with r = (r 1 ;:::;r n ), if there are currently no idle servers and server i has r i remaining exponential stages with rate i in order to complete its service. Or as (i 1 ;:::;i k ; r), if i 1 ;:::;i k are the idle servers, with i 1 having been idle the longest, i 2 the second longest, and so on, and each server i has r i exponential stages to complete. Clearly r i = 0 for all i2fi 1 ;:::;i k g. Proposition 1: For general Erlang service times, GE(N i ; i ) i = 1;:::;n, where N i is a random variable with p i (s) = P (N i = s), all idle time ordering policies have the same stationary probabilities. Namely P (i 1 ;:::;i k ; r) = i k i 1 k 1 k P (0; 1) Y m= 2fi 1 ;:::;i k g P (N m r m ) A Queueing Loss Model ... 10 where P (0; 1) is such that P (0; 1)(1 + X (i 1 ;:::;i k ;r) i k i 1 k 1 k Y m= 2fi 1 ;:::;i k g P (N m r m )) = 1 Proof: Assuming an arbitrary idle time ordering policy, P j;k is the probability that the idle time ordering policy assigns an arrival to server i j when the state is (i 1 ;:::;i k ; r). For states x = (i 1 ;:::;i k :r 1 ;:::;r n ); x = (i 1 ;:::;i k :r 1 ;:::;r j 1;:::;r n ); x + = (i 1 ;:::;i k ;i k+1 :r 1 ;:::;r i k+1 1 ; 0;r i k+1 +1 ;:::;r n ); x = (i 1 ;:::;i j1 ;i j+1 ;:::;i k :r 1 ;:::;r i j 1 ;s;r i j +1 ;:::;r n ); then the innitesimal rates of the resultant continuous time Markov chain are q x;x = j ; for r j > 1 q x;x + = i k+1 ; for r i k+1 = 1 q x;x = P j;k p i j (s) for r i j = 0 We now make the following conjecture about the reverse process. (a) It is a queueing model with n servers all of whom are eligible to serve any arriving customer; (b) The state is x = (i 1 ;:::;i k : r 1 ;:::;r n ) if (i 1 ;:::;i k ) is the current idle state vector; r j is the current stage of server j if that server is busy, and r j = 0 if j is idle. (c) An arrival to server j begins in stage 1; and the time it takes server j to complete a stage is exponential with rate j , j = 1;:::;n. (d) Customer at server j leaves the system upon completion of stage m with probability j (m) = p j (m) P km p j (k) ; otherwise it goes to stagem + 1 with probability j (m) = 1 j (m); (e) The arrival rate of customers when the idle state vector is i 1 ;:::;i k is k , and the arriving customer is assigned to server i k ; (f) If server r becomes idle when the idle state vector is i 1 ;:::;i k then the new order list of idle servers becomes i 1 ;:::;i j1 ;r;i j ;:::;i k with probability P j;k+1 k+1 ; Note that under our conjecture, with x; x ; x + and x as dened in the preceding, the A Queueing Loss Model ... 11 innitesimal rates of the reversed chain would be q x ;x = j (r j 1) j q x + ;x = k+1 q x ;x = i j i j (s) P j;k k because it is clear that when in state (i 1 ;:::;i k ; r) the rates at which the forward and the conjectured reverse processes leave that state are both equal to P i= 2fi 1 ;:::;i k g i + k , it follows from Theorem 1.13 of [15] that the conjecture will be veried if we can nd probabilities P (x) such that P x P (x) = 1 and P (x)q x;x = P (x )q x ;x for r j > 1 P (x)q x;x + = P (x + )q x + ;x for r i k+1 = 1 P (x)q x;x = P (x )q x ;x for r i j = 0 Thus, we must nd probabilities that satisfy P (x) j = P (x ) j (r j 1) j for r j > 1 (2.1) P (x) i k+1 = P (x + ) k+1 for r i k+1 = 1 (2.2) P (x)P j;k p i j (s) = P (x ) i j i j (s) P j;k k (2.3) rewrite the rst and second equations as P (x ) = 1 j (r j 1) P (x) for r j > 1 (2.4) P (x + ) = i k+1 k+1 P (x) (2.5) by rst iterating over (3.5) and then (3.4) we will have, P (x) = i k i 1 k 1 k P (0; 1) Y m= 2fi 1 ;:::;i k g rm1 Y i=1 m (i) using the fact n1 Q i=1 m (i) =P (N m n) we can rewrite the preceding as follow A Queueing Loss Model ... 12 P (i 1 ;:::;i k ; r) = i k i 1 k 1 k P (0; 1) Y m= 2fi 1 ;:::;i k g P (N m r m ) as it is straightforward to verify that the preceding, with P (0; 1) chosen to make the probabilities sum to 1, satisfy the reversibility equations (3.1), (3.2), and (3.3) so the proposition is proven. Theorem 1: Suppose the service distributions G 1 ;:::;G n , and let E[S j ] be the mean of the distribution G j . IfI is the set of idle servers in steady state, then P (I =fi 1 ;:::;i k g) =k! 1 k 1 k E(S i 1 )E(S i k ) P (0) where P (0) is the probability that all servers are busy. Furthermore, given thatI = fi 1 ;:::;i k g , (a) all k! possible ordering of idle servers are equally likely; (b) the remaining service times of the busy servers are independent, and are distributed according the to their respective equilibrium service distributions; (c) the amount of service time already provided on their current customers by the busy servers are independent, and are distributed according to their respective equilibrium ser- vice distributions. Proof: To begin, suppose that the service distribution of server i is general Erlang GE(N i ; i ), i = 1;:::;n. Let P (i 1 ;:::;i k ) be the steady state probability that i 1 ;:::;i k is the idle servers vector. Using proposition 1 and summing P (i 1 ;:::;i k ; r) over all the consistent vectors r (that is all r such that r j = 0;j2fi 1 ;:::;i k g) yields P (i 1 ;:::;i k ) = i k i 1 k 1 k P (0; 1) X r Y m= 2fi 1 ;:::;i k g P (N m r m ) Now assuming that the set of the busy servers is b 1 :::;b nk , we can write P (i 1 ;:::;i k ) = i k i 1 k 1 k P (0; 1) X r b 1 ::: X r b nk P (N b 1 r b 1 ):::P (N b nk r b nk ) = i k i 1 k 1 k P (0; 1) Y m= 2fi 1 ;:::;i k g E(N m ) (2.6) A Queueing Loss Model ... 13 Let P (0) be the probability of all servers are busy. Proposition 1 gives, P (0) = X r P (0; r) = X r P (0; 1) n Y i=1 P (N i r i ) =P (0; 1) n Y i=1 E(N i ) Hence, by using the fact that the mean service time at server i is E(S i ) = E(N i ) i , i = 1;:::;n, we can rewrite the limiting probabilities of the idle servers vector (3.6) as follow P (i 1 ;:::;i k ) = 1 k 1 k E(S i 1 )E(S i k ) P (0) which yields that P (I =fi 1 ;:::;i k g) =k! 1 k 1 k E(S i 1 )E(S i k ) P (0) as well as showing that, given thatI =fi 1 ;:::;i k g; allk! possible ordering of idle servers are equally likely. Moreover, it follows from proposition 1 and (3.6) that P (i 1 ;:::;i k ; r) P (i 1 ;:::;i k ) = Y m= 2fi 1 ;:::;i k g P (N m r m ) E(n m ) which, using Lemma 1, proves that conditional on the set of busy server their remaining service times are independent, and are distributed according to their respective equilib- rium service distributions. In addition, because the reverse chain has the same stationary probabilities as does the forward chain, and as the interpretation ofr i for the reverse chain is that serveri is currently at stager i , part (c) also follows. Hence, the theorem is proven when all service distributions are general Erlang type. Because any service distribution is the limit of a sequence of general Erlang distributions, the approach of Barbour [1] can be used to establish necessary continuity. Thus, the theorem is proven for arbitrary service distributions. 2.5 Finding The Limiting Probabilities using Markov Chain Monte Carlo Methods In practice, when the number of servers is large, it is almost impossible to nd the mul- tiplicative constant by directly adding all the limiting probabilities up to one. Even if it A Queueing Loss Model ... 14 were known, the derivations of other quantities of interest, such as average waiting time in the system, rate at which customers are lost, etc., remain computationally intractable. However, these quantities can be determined by using the Gibbs sampler Markov chain Monte Carlo method to generate a Markov chain whose limiting distribution is the sta- tionary distribution of the set of idle servers. That is, interpreting Y i as the indicator of whether serveri is idle, we want to generate a Markov chain whose stationary distribution is p(x 1 :::;x n ) = P (Y i =x i ;i = 1:::;n) = C k! k 1 k Q n i=1 E[S i ] ; x 1 = 0; 1; k = n X i=1 x i When the current state of the Markov chain is x = (x 1 ;:::;x n ), the Gibbs sampler method chooses a coordinate that is equally likely to be any of 1;:::;n. If coordinate j is chosen then the next state will be (x1;:::;x j1 ; 0;x j+1 ;:::;x n ) with probability = p(x1;:::;x j1 ; 0;x j+1 ;:::;x n ) p(x1;:::;x j1 ; 0;x j+1 ;:::;x n ) +p(x1;:::;x j1 ; 1;x j+1 ;:::;x n ) = p(y) PfX i =x i ;i6=jg wherer = P i6=j , or it will be (x1;:::;x j1 ; 1;x j+1 ;:::;x n ) with probability 1. The sta- tionary distribution of the successive values of a Markov chain generated by the preceding is the limiting distribution of the set of idle servers. Consequently, we can approximate the steady state probability there are exactly k idle servers, call it P (k), by the propor- tion of states (x 1 ;:::;x n ) such that P n i=1 x i =k. We can then use our estimates of P (k), k = 0;:::;n, to estimate P n i=1 P (k) k , equal to the proportion of arrivals that enter the system. Similarly we can estimate the proportion of arrivals that are served by server i by letting (i :k) be the proportion of states of the chain for which x i = 1, P n i=1 x i =k and then using the estimate P n i=1 (i :k) k =k where the preceding used that conditional on the set of idle servers all orderings are equally likely, and so, given that an arrival is eligible, each server in this set is equally likely to be the one used. Example In Table 2.1 we compare the eciency of the Gibbs sampler method in nding the multiplicative constant with a discrete event Monte Carlo simulation. We simulated a model with 5 servers with deterministic service times, D i = 1=i,i = 1;:::; 5. Arrival process is a Poisson process with rate = 15. We have also assumed that the component of the eligibility vectors are independent Bernoulli random variables with mean 1=2. The A Queueing Loss Model ... 15 results are based on 500 runs where each run has a length equal to 3 seconds. The per run estimator in the discrete event simulation is the proportion of the run simulated time that all servers are busy; the per run estimator in the Gibbs sampler approach is the proportion of states having all of its components equal to 0. The actual value of P (0) was analytically determined and, for each run, we computed A, the absolute value of the dierence between the run estimator andP (0). Table 2.1 lists both the sample mean and the sample variance of the observed A values for the two methods. The estimators of P (0) are the average of the estimators over all 500 runs. Tab. 2.1: Estimates for P(0) { Estimates of Gibbs sampler Raw simulation P(0) 0.14779 0.14440 E(A) 0.02813 .03210 Var(A) 0.0004325 0.0008913 Exact Value of P (0) = 0:15149 On the Adan-Weiss Loss Model ... 16 3. ON THE ADAN-WEISS LOSS MODEL HAVING SKILL-BASED SERVERS AND LONGEST IDLE ASSIGNMENT RULE 3.1 Introduction In this chapter we drop the exchangeability assumption and study our model under the longest idle assignment policy. Adan and Weiss [2] considered the same model and used the method of supplementary variables to obtain the limiting probabilities of the ordered list of idle servers. They also showed that the limiting probabilities of the ordered list of idle servers depend on the service distributions only through their means. Here we obtain results of [2] using the method of stages. This method considers the service time at each server as a weighted average of dierent gamma distributions with the same rate. Our approach will be to conjecture the reverse chain, and then show the conjecture is correct by nding, up to a multiplicative constant, the limiting probabilities of the ordered list of idle servers. We also show that given the set of busy servers, the remaining service times are independent with their respective equilibrium distributions (This result is only implicitly noted in Adan and Weiss [2]). In practice, when n is large nding the limit- ing probabilities or any other quantity of interest becomes computationally intractable. Therefore, we will provide an algorithm for simulating this model using the Gibbs Sampler method to estimate these desired quantities. Servers that can only serve certain arrivals are refereed to as "skill based servers". Other papers concerned with skill based servers are Whitt [24], Adan et al. [1], Ross [16], Talreja et al. [21], and Visschers et al.[22]. 3.2 Model Analysis Dene the "idle server vector" as 0 if there are currently no idle servers, or as i 1 ;:::;i k if there are currentlyk servers idle, wherei 1 has been idle the longest,i 2 the second longest, and so on. Let P j:i 1 ;:::;i k be the probability that an arrival will be assigned to server i j given that the idle server vector is i 1 ;:::;i k . That is, On the Adan-Weiss Loss Model ... 17 P j:i 1 ;:::;i k =P ( j1 X m=1 X im = 0; X i j = 1): Let 0 = 0, and for k > 0 let i 1 ;:::;i k = P ( P k m=1 X im > 0); hence, i 1 ;:::;i k is the prob- ability that an arrival nding the idle server vector i 1 ;:::;i k will be served. Note that P k j=1 P j:i 1 ;:::;i k = i 1 ;:::;i k and P j:i 1 ;:::;i k = i 1 ;:::;i j i 1 ;:::;i j1 : Example If we assume thatX 1 ;:::;X n are independent andX i = 1 with probabilityp i ,i = 1;:::;n, then i 1 ;:::;i k = 1 Q k m=1 (1p im ) and P j:i 1 ;:::;i k =p i j Q j1 m=1 (1p im ). Recall that a random variable X is called general Erlang, GE(N;), if X = N X i=1 W i where W i , i = 1; 2; 3;:::, are iid exponential random variables with rate and N is a positive integer valued random variable that is independent of the W i . Let G e be the equilibrium distribution of X. From Lemma 1 in chapter 2, we know that G e is the dis- tribution function of a GE(N e ;) random variable, where P (N e =j) =P (Nj)=E(N): (3.1) In our model, we suppose that the service distribution of server i is general Erlang GE(N i ; i ), i = 1;:::;n, and we analyze our model as a continuous time Markov chain. To do so, we dene the "state vector" as (0; r) with r = (r 1 ;:::;r n ), if there are currently no idle servers and server i has r i remaining exponential stages with rate i in order to complete its service; or as (i 1 ;:::;i k ; r), if i 1 ;:::;i k is the current idle server vector and each server i has r i exponential stages to complete, where r i = 0 for all i2fi 1 ;:::;i k g. Proposition 1: For general Erlang service times, GE(N i ; i ), i = 1;:::;n, where N i is a random variable with p i (j) = P (N i = j), the stationary probability of the state vector (i 1 ;:::;i k ; r) has the following form P (i 1 ;:::;i k ; r) = i k i 1 k i 1 i 1 ;:::;i k P (0; 1) Y m= 2fi 1 ;:::;i k g P (N m r m ); On the Adan-Weiss Loss Model ... 18 where P (0; 1) is such that P (0; 1)(1 + X (i 1 ;:::;i k ;r) i k i 1 k i 1 i 1 ;:::;i k Y m= 2fi 1 ;:::;i k g P (N m r m )) = 1: Proof: Using longest idle rule, P j:i 1 ;:::;i k is the probability that the idle time ordering policy assigns an arrival to server i j when the state is (i 1 ;:::;i k ; r). For states x = (i 1 ;:::;i k :r 1 ;:::;r n ); x = (i 1 ;:::;i k :r 1 ;:::;r j 1;:::;r n ); x + = (i 1 ;:::;i k ;i k+1 :r 1 ;:::;r i k+1 1 ; 0;r i k+1 +1 ;:::;r n ); x = (i 1 ;:::;i j1 ;i j+1 ;:::;i k :r 1 ;:::;r i j 1 ;s;r i j +1 ;:::;r n ); the innitesimal rates of the resultant continuous time Markov chain are q x;x = j for r j > 1; q x;x + = i k+1 for r i k+1 = 1; q x;x = P j:i 1 ;:::;i k p i j (s): We now make the following conjecture about the reverse process. (a) It is a queueing model with n servers all of whom are eligible to serve any arriving customer. (b) The state is x = (i 1 ;:::;i k :r 1 ;:::;r n ) if (i 1 ;:::;i k ) is the current idle server vector; r j is the current stage of server j if that server is busy, and r j = 0 if j is idle. (c) An arrival to server j begins in stage 1; and the time it takes server j to complete a stage is exponential with rate j , j = 1;:::;n. (d) Customer at server j leaves the system upon completion of stage m with probability j (m) = p j (m) P km p j (k) ; otherwise it goes to stagem + 1 with probability j (m) = 1 j (m). (e) The arrival rate of customers when the order list of idle servers isi 1 ;:::;i k is i 1 ;:::;i k , and the arriving customer is assigned to server i k . (f) If serverr becomes idle when the idle server vector isi 1 ;:::;i k then the new idle server vector becomes i 1 ;:::;i j1 ;r;i j ;:::;i k with probability P j:i 1 ;:::;i j1 ;r;i j ;:::;i k i 1 i 1 ;:::;i k i 1 i 1 ;:::;i j1 i 1 ;:::;i j1 ;r i 1 ;:::;i j1 ;r;i j+1 i 1 ;:::;i j1 ;r;i j+1 ;:::;i k : On the Adan-Weiss Loss Model ... 19 With x; x ; x + and x as dened in the preceding, the innitesimal rates of the reversed chain under our conjecture are q x ;x = j (r j 1) j ; q x + ;x = i 1 ;:::;i k+1 ; q x ;x = i j i j (s) P j:i 1 ;:::;i k i 1 i 1 ;:::;i j1 i 1 ;:::;i j1 ;i j+1 i 1 ;:::;i j1 ;i j+1 ;:::;i k i 1 i 1 ;:::;i k : In order to verify our conjecture we rst need to show that when in state (i 1 ;:::;i k ; r) the rates at which the forward and the reverse process leave that state are equal. Lemma 1: The rates at which the forward and the reverse process leave the state (i 1 ;:::;i k ; r) are equal; i.e. X x k 6=x q x;x k = X x k 6=x q x;x k: Proof: Using the innitesimal rates of the forward and the reverse process the preceding equality can be written as k X j=1 P j:i 1 ;:::;i k + X i j = 2fi 1 ;:::;i k g i j = i 1 ;:::;i k + X i j = 2fi 1 ;:::;i k g k+1 X m=1 P m:x m i 1 i 1 ;:::;i k i 1 i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k i j ; where x m = (i 1 ;:::;i m1 ;i j ;i m ;:::i k ) andP m:x m =P m:i 1 ;:::;i m1 ;i j ;im;:::i k . Since P k j=1 P j:i 1 ;:::;i k = i 1 ;:::;i k , to show that the preceding equality holds it suces to show that k+1 X m=1 P m;x m i 1 i 1 ;:::;i k i 1 ; i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k = 1: (3.2) This will be proven by rst showing that for 1nk, n X m=1 P m;x m i 1 ; i 1 ;:::;i k i 1 i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k = i 1 ;:::;in i 1 ;:::;i k i j ;i 1 ;:::;in i j ;i 1 ;:::;i k : (3.3) On the Adan-Weiss Loss Model ... 20 We will prove (3.3) by induction. It is clear that for n = 1 this is true. Now assuming that (3.3) is true for n we will show that it is also true for n + 1. Namely, n+1 X m=1 P m;x m i 1 i 1 ;:::;i k i 1 i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k = n X m=1 P m;x m i 1 i 1 ;:::;i k i 1 i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k + P n+1;x n+1 i 1 i 1 ;:::;i k i 1 i 1 ;:::;in i 1 ;:::;in;i j i 1 ;:::;in;i j ;i n+1 ;:::;i k = i 1 ;:::;in i 1 ;:::;i k i j ;i 1 ;:::;in i j ;i 1 ;:::;i k + P n+1;x n+1 i 1 ;:::;i n+1 i 1 ;:::;i k i 1 ;:::;in;i j i 1 ;:::;in;i j ;i n+1 ;:::;i k = i 1 ;:::;i n+1 i 1 ;:::;i k i j ;i 1 ;:::;i n+1 i j ;i 1 ;:::;i k ; the last equality holds because P n+1;x n+1 + i 1 ;:::;in = i j ;i 1 ;:::;in . Using (3.3) it is easy to show that (3.2) holds. That is, k+1 X m=1 P m;x m i 1 i 1 ;:::;i k i 1 i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k = k X m=1 P m;x m i 1 i 1 ;:::;i k i 1 i 1 ;:::;i m1 i 1 ;:::;i m1 ;i j i 1 ;:::;i m1 ;i j ;im;:::;i k + P k+1;x k+1 i 1 i 1 ;:::;i k i 1 i 1 ;:::;i k i 1 ;:::;i k ;i j = i 1 ;:::;i k i j ;i 1 ;:::;i k + P k+1;x k+1 i 1 ;:::;i k ;i j = 1: Therefore, Lemma 1 is proven. Now that we showed the rates which the forward and the reverse process leave state (i 1 ;:::;i k ; r) are equal our conjecture will be veried if we can nd probabilities P (x), P x P (x) = 1, such that P (x)q x;x = P (x )q x ;x for r j > 1; P (x)q x;x + = P (x + )q x + ;x for r i k+1 = 1; P (x)q x;x = P (x )q x ;x : On the Adan-Weiss Loss Model ... 21 thus, we must nd probabilities that satisfy P (x) j =P (x ) j (r j 1) j for r j > 1, (3.4) P (x) i k+1 =P (x + ) i 1 ;:::;i k+1 for r i k+1 = 1, (3.5) P (x)P j;k p i j (s) =P (x ) i j i j (s) P j:i 1 ;:::;i k i 1 i 1 ;:::;i j1 i 1 ;:::;i j1 ;i j+1 i 1 ;:::;i j1 ;i j+1 ;:::;i k i 1 i 1 ;:::;i k : (3.6) Rewriting the rst and the second equations we have, P (x ) = 1 j (r j 1) P (x) for r j > 1, (3.7) P (x + ) = i k+1 i 1 ;:::;i k+1 P (x): (3.8) By rst iterating over (3.8) and then (3.7) we have, P (x) = i k i 1 k i 1 i 1 ;:::;i k P (0; 1) Y m= 2fi 1 ;:::;i k g rm1 Y i=1 m (i): Using that n1 Q i=1 m (i) =P (N m n), the preceding gives P (x) = i k i 1 k i 1 i 1 ;:::;i k P (0; 1) Y m= 2fi 1 ;:::;i k g P (N m r m ): As it is straightforward to verify that the preceding, withP (0; 1) chosen to make the prob- abilities sum to 1, satisfy the reversibility equations (3.4), (3.5), and (3.6), the proposition is proven. Theorem 1: Suppose service distributions are G 1 ;:::;G n , and let E[S j ] be the mean of the distribution G j . If i 1 ;:::;i k is the idle server vector in steady state, then P (i 1 ;:::;i k ) = 1 k i 1 i 1 ;:::;i k E(S i 1 )E(S i k ) P (0); where P (0) is the probability that all the servers are busy. Furthermore, given that idle server vector is i 1 ;:::;i k , (a) the limiting probabilities of the idle server vector i 1 ;:::;i k depend on the service dis- tributions only trough their means; On the Adan-Weiss Loss Model ... 22 (b) the remaining service times of the busy servers are independent and are distributed according to their respective equilibrium service distributions; (c) the amount of service time already provided on their current customers by the busy servers are independent and are distributed according to their respective equilibrium ser- vice distributions. Proof: To begin, suppose that the service distribution of server i is general Erlang GE(N i ; i ), i = 1;:::;n. Let P (i 1 ;:::;i k ) be the steady state probability that i 1 ;:::;i k is the idle server vector. Using Proposition 1 and summing P (i 1 ;:::;i k ; r) over all the consistent vectors r (that is all r such that r j = 0;j2fi 1 ;:::;i k g) yields P (i 1 ;:::;i k ) = X r i k i 1 k i 1 i 1 ;:::;i k P (0; 1) Y m= 2fi 1 ;:::;i k g P (N m r m ) = i k i 1 k i 1 i 1 ;:::;i k P (0; 1) X r Y m= 2fi 1 ;:::;i k g P (N m r m ): Now if we let the set of the busy servers be b 1 ;:::;b nk which is the compliment of the set of idle servers i 1 ;:::;i k we can write P (i 1 ;:::;i k ) = i k i 1 k i 1 i 1 ;:::;i k P (0; 1) X r b 1 ::: X r b nk P (N b 1 r b 1 ):::P (N b nk r b nk ) = i k i 1 k i 1 i 1 ;:::;i k P (0; 1) Y m= 2fi 1 ;:::;i k g E(N m ): (3.9) Let P (0) be the probability that all the servers are busy. Clearly, P (0) = X r P (0; r) = X r P (0; 1) n Y i=1 P (N i r i ) =P (0; 1) n Y i=1 E(N i ): Hence, by using the fact that the mean service time at server i is E(S i ) = E(N i ) i , i = 1;:::;n, we can rewrite the limiting probabilities of the ordered list of idle servers (3.9) as follow P (i 1 ;:::;i k ) = 1 k i 1 i 1 ;:::;i k E(S i 1 )E(S i k ) P (0); On the Adan-Weiss Loss Model ... 23 which shows that, conditional on the idle server vectori 1 ;:::;i k , the limiting probabilities depend on the service distributions only through their means. Moreover, it follows from Proposition 1 and (3.9) that P (i 1 ;:::;i k ; r) P (i 1 ;:::;i k ) = Y m= 2fi 1 ;:::;i k g P (N m r m ) E(n m ) ; which, using Lemma 1 in chapter 2, proves that conditional on the set of busy servers, their remaining service times are independent and are distributed according to their re- spective equilibrium service distributions. In addition, because the reverse chain has the same stationary probabilities as does the forward chain and as the interpretation of r i for the reverse chain is that server i is currently at stage r i , part (c) also follows. Hence, the theorem is proven when all service distributions are general Erlang type. Because any service distribution is the limit of a sequence of general Erlang distributions, see for instance Whitt [23], the approach of Barbour [3] can be used to establish necessary continuity arguments for extending the results for general Erlang service distributions to any arbitrary service distributions. Thus, the theorem is proven for arbitrary service distributions. 3.3 Finding The Limiting Probabilities using Markov Chain Monte Carlo Methods In practice, when the number of servers is large, the determination of the constant P (0) is computationally intractable. Indeed, even if it were known, the derivation of other quantities of interest, such as average waiting time in the system, rate at which customers are lost, etc., remain computationally intractable. However, these quantities can be de- termined by using the Gibbs sampler Markov chain Monte Carlo method to generate a Markov chain whose limiting distribution is the stationary distribution of the ordered list of the set of idle servers. That is, by letting Y i to be equal to m if server i is the m th server on the idle server vector and 0 if it is busy, we want to generate a Markov chain whose stationary distribution is p(x 1 ;:::;x n ) = P (Y i =x i ;i = 1;:::;n) = 8 > > < > > : C 1 k i 1 i 1 ;:::;i k k Q j=1 E(S i j ) ; x im =m;mk and x im = 0;m>k 0; otherwise ; On the Adan-Weiss Loss Model ... 24 When the current state of the Markov chain is x = (x 1 ;:::;x n ), the Gibbs sampler method chooses a coordinate that is equally likely to be any of 1;:::;n. If coordinate j is chosen then with r = max i6=j fx i g if 0<x j <r the next state is x, otherwise the next state is (x 1 ;:::;x j1 ; 0;x j+1 ;:::;x n ) with probability = p(x 1 ;:::;x j1 ; 0;x j+1 ;:::;x n ) p(x 1 ;:::;x j1 ; 0;x j+1 ;:::;x n ) +p(x 1 ;:::;x j1 ;r + 1;x j+1 ;:::;x n ) = i 1 ;:::;ir;j E(S j ) 1 + i 1 ;:::;ir;j E(S j ) ; or it will be (x 1 ;:::;x j1 ;r + 1;x j+1 ;:::;x n ) with probability 1. The stationary distribution of the successive values of a Markov chain generated by the preceding is the limiting distribution of the ordered list of the set of idle servers. Thus, for instance, we can approximate the steady state probability there are exactlyk idle servers, call it P (k), by the proportion of states (x 1 ;:::;x n ) such that P n i=1 x i =k(k + 1)=2. We can also approximate the steady state probability that the ordered list of the set of idle servers isi 1 ;:::;i k , call itP (i 1 ;:::;i k ), by the proportion of states (x 1 ;:::;x n ) such that P k j=1 x i j =k(k + 1)=2 and max i (x i ) =k. We can then use our estimates of P (i 1 ;:::;i k ), k = 0;:::;n, to estimate P n k=1 P (i 1 ;:::;i k ) P (i 1 ;:::;i k ) i 1 ;:::;i k , equal to the proportion of arrivals that enter the system. General Case Optimality, Heuristic Policy 25 4. GENERAL CASE OPTIMALITY, HEURISTIC POLICY 4.1 Introduction In this chapter, we assume that X 1 ;:::;X n are independent and P (X i = 1) = p i , i = 1;:::;n. We also assume that the service time at server i is exponential with rate i , i = 1;:::;n. Upon an arrival, call server i \idle eligible" if it is both idle and eligible. In this chapter, our objective is to nd the stationary policy that minimizes the rate at which customers are lost. Stationary policies are policies that upon each arrival, depending on the set of idle servers and the set of idle eligible servers, assign that arrival to one of the idle eligible servers. Hence, a policy of this type has a rule for assigning arrivals for every combination of the set of idle servers and the set of idle eligible servers. For instance, consider a model with 3 servers, a policy of this type can give priority to server 1 when the set of idle eligible server isf1; 2; 3g, while it gives priority to server 2 when the set of idle servers isf1; 2; 3g and the eligible servers are 1 and 2, and so on. For each set of idle eligible servers of sizei, we havei dierent choice of actions. Also for each set of idle servers of size k we have k i possible sets of idle eligible servers of size i and the number of sets of idle servers of size k is n k . Thus, total number of stationary policies = n Y k=1 k Y i=1 i ( k i ) ! ( n k ) As n increases the number of stationary policies grows rapidly. However, for nding the best stationary policy we do not need to consider all these policies. Let S be the set of idle servers and E be the set of idle eligible servers upon an arrival. Since the decisions of how arrivals should be assigned to servers have to be made at arrival epochs we only draw our attention to these points. Dene (S; E) as the state at each arrival epoch. Upon an arrival, for j2 E 0 E, if it is optimal to assign an arrival to server j when the state is (S; E) it is also optimal to assign that arrival to server j whenever the state is (S; E 0 ). This is true since the state at the next arrival epoch depends solely on S , the set of idle General Case Optimality, Heuristic Policy 26 servers immediately after the decision is made for an arrival. For instance, whenn = 3, if upon an arrival all three servers are idle eligible and the decision is to assign the arrival to server 1, then if all are idle but only 1 and 2 are eligible again the arrival should be assigned to server 1. Thus, for nding the optimal stationary policy we only need to focus on the possible sets of idle servers after arrival epochs and the possible orderings of all these sets. Therefore, for every k n, we should consider all the possible orderings of the sets of idle servers of size k, N k !, where N k = n k : Thus, the number of stationary policies that should be considered in our search for the best stationary policy, N, is N = n Y k=1 N k !: For instance, for n = 3, there exist 192 stationary policies where for nding the best stationary policy we only need to consider 36 policies in our analysis. This signicantly reduces the complexity of our search for the best stationary policy. However, even for the case n = 3, we still need to consider 36 dierent policies which shows that direct analytical comparison of these policies is computationally intensive. In special cases, the number of policies that need to be considered can be further decreased. Before analyzing these cases we rst need to dene a preference relation between two servers and following that we provide a useful lemma. Say that server i is preferred to server j, i p j, if p i p j and i j . Let I m be the indicator of whether the m th arrival enters service or not and also let N r = P r m=1 I m . Lemma 1: For an arbitrary arrival process, the optimal policy never rejects an arrival if there is at least one idle eligible server upon that arrival. Proof: Upon the initial arrival, consider two dierent scenarios one of which, scenario 2, assigns that arrival to server j and the other scenario, scenario 1, rejects it. By using a simple coupling argument we show that, for every r > 0, N r is stochastically greater in scenario 2. Couple the arrival processes in both scenarios so that arrivals occur at identical times and also couple the eligibility vectors in both models so that they are the same for the second arrival (the arrival right after the initial arrival). Let 1 be the policy that is employed in scenario 1 and let B i be the initial set of busy servers in scenario i, clearly B 1 B 2 . Couple the initial service time of the busy server i, for all i2 B 1 , in both models so that their service times are identical. By induction on r we show that there exists a policy 2 for scenario 2 such that E 2 [N r jB 2 ]E 1 [N r jB 1 ]; General Case Optimality, Heuristic Policy 27 where E [XjB] denotes the expected value of X given that policy is employed and B is the initial set of busy servers. The statement is clearly true for r = 1. So suppose that this is true forr and we show that this is true forr +1 by dening a policy 2 for scenario 2 in the following manner. For each arrival, whoever policy 1 assigns to that arrival let 2 make the same assignment with the exception that if 1 assigns to server j and server j is still busy in scenario 2, let 2 assign that arrival to one of the idle eligible servers in scenario 2 (if there is no idle eligible server in scenario 2 that arrival is lost). Let A(B i ) be the set of busy servers after the second arrival in scenario i, by using policy 2 one of the following cases happens: Case 1: A(B 2 ) A(B 1 ) : This is true when either server j in scenario 2 has completed service before the second arrival or when there is no idle eligible server for the second arrival except server j in scenario 1 and this server is still busy in scenario 2. Since the remaining service time of server j in scenario 2 is exponential with rate j , the situation is the same as both systems have started with the same set of busy servers and it is easy to see that our induction hypothesis is true. Case 2: A(B 2 ) A(B 1 ) : This is true when server j has not yet completed service in scenario 2 before the second arrival and server j is not the only idle eligible server in scenario 1. Again using the fact that the remaining service time of serverj is exponential with rate j , we can see that after the second arrival, the problem is the same as it was started withA(B 1 ) andA(B 2 ) as its initial sets of busy servers. Hence, we can apply the induction hypothesis to complete our proof. The following theorem helps us exclude policies that assign arrivals to server j, when both servers i and j are idle eligible and i p j, from our search for the best stationary policy. Theorem 1 (Preference Rule) Upon each arrival, if servers i and j are idle eligible with i p j, then the optimal policy never assigns that arrival to server j. To prove Theorem 1 we use the same technique that has been used in Ross [16]. We rst dene a relation between certain pairs of subsets. Let S 1 andS 2 be subsets off1;:::;ng, say that S 2 dominates S 1 if either S 2 S 1 ; or there are distinct integers i;j;i 1 ;i 2 ;:::;i k , i p j, such that S 1 =fj;i 1 ;i 2 ;:::;i k g and S 2 =fi;i 1 ;i 2 ;:::;i k g. Now consider the following lemma which has a key role in the proof of Theorem 1. Lemma 2 Let S 1 and S 2 be two dierent sets of busy servers. If S 2 dominates S 1 then General Case Optimality, Heuristic Policy 28 for r 1 sup E [N r jS 1 ] sup E [N r jS 2 ]: where E [XjS] denotes the expected value of X given that policy is employed and S is the initial set of busy servers. Proof: Assuming that S 2 dominates S 1 , we show by induction on r that for any policy 1 there is a policy 2 such that E 1 [ r X m=1 I m jS 1 ]E 2 [ r X m=1 I m jS 2 ]: For the rest of the proof we closely follow the steps of Lemma 2 in [16]. To begin, con- sider two scenarios, scenario one having S 1 as the initial set of busy servers and scenario 2 having S 2 , and let 1 be a policy that is to be used in scenario one. Couple the arrival streams, the eligibility vectors, and the initial service times in both scenarios in the same manner that is introduced in Lemma 2 [16]. It is easy to show that the lemma follows when r = 1. So assume that it is true for r and let A(S i ) be the set of busy servers right before the rst arrival in scenario i. To continue our proof consider two dierent cases: Case 1 : A(S 2 )A(S 1 ). Case 2 : A(S 1 ) =fj;i 1 ;i 2 ;:::;i k g and A(S 2 ) =fi;i 1 ;i 2 ;:::;i k g where i p j. In case 1, couple the eligibility vectors in both scenarios for the rst arrival so that they are identical. For case 2, letU 1 ;:::;U n ben independent uniform [0; 1] random variables, and set X 1 k = IfU k p k g; k = 1;:::;n X 2 k = IfU k p k g; k6=i;j X 2 i = IfU j p i g X 2 j = IfU i p j g where X i 1 ;:::;X i n is equal to the eligibility vector for the rst arrival in scenario i. Note that coupling the eligibility vectors in this manner would guarantee that serverj is eligible General Case Optimality, Heuristic Policy 29 in scenario two whenever serveri is eligible in scenario one. In both cases, whoever policy 1 assigns to the rst arrival in scenario one should be the one that 2 assigns to the rst arrival in scenario two, with the exception that in case 2 if server i is assigned to the rst arrival in scenario one then 2 assigns server j to that arrival. In both cases, if the rst arrival is rejected by 1 then let 2 reject that arrival as well. Following the same steps as in Lemma 2 [16] we have E 1 [N r jS 1 ]E 2 [N r jS 2 ]: Now using Lemma 1 we have 2 0 for scenario 2, which does not reject arrivals if there exist at least one idle eligible server, that works at least as good as 2 . Thus, E 1 [N r jS 1 ]E 2 0[N r jS 2 ]; which completes the proof of Lemma 2 and establishes the result. We are now ready to prove theorem 1. Proof of Theorem 1 : Consider any given policy and assume that upon an arrival, serversi andj are idle eligible withi p j. If violates the preference rule and assigns the arrival to server j, then the system will enter a state that is dominated by the state that system would have been entered if the arrival was assigned to server i. By using Lemma 2, it follows that there is a policy that does not violate the preference rule, whenever the preference relation between servers exists, for the initial period that is as good as in the sense ofE[N r ]. Continuing this argument for each subsequent period shows that following this rule will result in higher E[N r ] than if it was not employed. As a result the policy which violates the preference rule cannot be the optimal policy, therefore, our theorem is proven. Remark : Theorem 1 implies that when 1 2 ::: n and p 1 p 2 :::p n the priority list policy (1;:::;n) stochastically maximizes N r . This result is also mentioned in [16], where the author mistakenly stated that the list policy (1;:::;n) stochastically maximizes the vector (I 1 ;:::;I r ). Recall thatI m is the indicator of the event that arrival m is served and N r = P r m=1 I m . For any given stationary policy we can nd the proportion of lost costumers when n = 2. However, nding the limiting distribution for dierent policies when n> 2 seems General Case Optimality, Heuristic Policy 30 computationally intensive, and as we mentioned earlier, even with the limiting distri- bution comparing the rates that customers are lost for dierent stationary policies is a formidable task. Hence, we will develop heuristic approaches for nding an acceptable stationary policy. In the next section we introduce two dierent heuristics one of which is more computationally involved than the other. We also consider some numerical ex- amples to examine the performance of these heuristics when n = 3. In section 4.3, we dene some lower bounds for our model and numerically compare the heuristics and the random rule using simulation for n> 3. 4.2 Heuristic Approaches We introduce two dierent heuristics each of which provides a permutation ordering policy for assigning arrivals to the idle eligible servers. A permutation ordering policy assigns arrivals to the idle eligible servers in accordance with a specic permutation order of the servers. let (i 1 ;:::i n ) be a permutation ordering policy when the instructions is to give an arrival to the idle eligible server that appears earlier on the list. Say server i beats j, written i > 2 j, if (i;j) is the best ordering in a model which consists of only these two servers. Our rst heuristic which we refer to as \Heuristic 1" nds the best ordering for all n 2 possible pairs and then checks if these pairwise orderings are transitive. That is, for servers i, j, and k i> 2 j;j > 2 k =)i> 2 k and then uses the permutation ordering that follows these pairwise transitivity relations. In case we do not have full transitivity, choose the permutation ordering that violates the transitivity relations among the ordered pairs least frequently. Specically, we choose the ordering (i 1 ;:::;i n ) that maximizes n1 X k=1 X j>k Ifi k > 2 i j g . If the heuristic does not yield in a unique policy, we pick one at random. From Theorem 1, we know that whenever server i is preferred to server j we should give priority to server i. However, we do not always have this preference property among each pair of servers. One reasonable approach to address this problem is to give priority to the server that has higher i p i . This is the idea of \Heuristic 2", which is the list policy General Case Optimality, Heuristic Policy 31 that sorts the servers in decreasing order of the ratio i p i . We compare both heuristics with the random rule (randomly assigns to one of the idle eligible servers) to measure their performance in terms of the rate of lost customers. When n = 3, for each numerical example we nd the exact rate of lost customers for all stationary policies by using Matlab to solve the balance equations, enabling us to nd the best stationary policy. Thus, we can compare the performance of the heuristics with the best stationary policy. In Table 4.1 and Table 4.2, we consider 20 dierent numerical examples and compare the rate of lost customers under both of the heuristic policies, the random policy, and the best stationary policy. Tab. 4.1: Numerical Examples Case ( 1 ; 2 ; 3 ;p 1 ;p 2 ;p 3 ) Case ( 1 ; 2 ; 3 ;p 1 ;p 2 ;p 3 ) 1 9.7 (7.2, 0.5, 9.7, 0.65, 0.77, 0.66) 11 7.1 (1.5, 2.6, 2.8, 0.82, 0.71, 0.51) 2 7.5 (9.4, 5.6, 8.8, 0.29, 0.23, 0.17) 12 8.2 (2.9, 1.7, 4.4, 0.15, 0.04, 0.8) 3 2.2 (3, 9.8, 5.3, 0.48, 0.39, 0.91) 13 8.3 (7.8, 3.4, 0.2, 0.81, 0.49, 0.24) 4 7.8 (4.3, 9, 0.4, 0.07, 0.96, 1) 14 3.5 (7.1, 7.7, 6.7, 0.57, 0.94, 0.44) 5 4.7 (5, 8.9, 2.5, 0.64, 0.39, 0.33) 15 4 (1, 1.8, 2, 0.2, 0.55, 0.6) 6 2.1 (7.4, 2.9, 5.8, 0.5, 0.51, 0.94) 16 8.3 (2.8, 5.9, 1.6, 0.89, 0.13, 0.45) 7 8.6 (7.8, 9.2, 8.1, 0.93, 0.32, 0.42) 17 0.1 (3.1, 0.1, 4.9, 0.66, 0.07, 0.6) 8 0.2 (6.8, 2.3, 5.1, 0.5, 0.51, 0.94) 18 6.8 (3.3, 5.6, 9.4, 0.65, 0.51, 0.51) 9 8.8 (3.4, 1.5, 6, 0.88, 0.18, 0.39) 19 2 (1.9, 9.3, 2.3, 0.49, 0.13, 0.33) 10 6.5 (3.1, 0.6, 9.5, 0.77, 0.26, 0.81) 20 4.3 (7, 5.5, 5.2, 0.28, 0.09, 0.38) For the numerical examples ; 1 ; 2 , and 3 are chosen at random according to a uniform [0; 10] random variable. p 1 ;p 2 , andp 3 are generated according to a uniform [0; 1]. Table 4.1 shows the input parameters for all of the numerical examples. In Table 4.2, for each of the cases in Table 4.1 we calculate the rate of lost customers for all the aforemen- tioned policies. In some of these cases, Theorem 1 enables us to eliminate some of the 36 stationary policies from our search for the best stationary policy. In all of the examples, Heuristic 1 performs better or at least as good as Heuristic 2, and in 90 percent of the cases results in the best policy. Even in the rest of the cases, where Heuristic 1 is not optimal, the rate of lost customers for Heuristic 1 is close to the best policy. General Case Optimality, Heuristic Policy 32 Tab. 4.2: Eciency of Heuristic 1 Case Best Policy Heuristic 1 Heuristic 2 Random 1 0.310149 0.310149 * 0.327077 2 0.521638 0.521638 * 0.524135 3 0.083175 0.083175 * 0.104626 4 0.387368 0.387368 * 0.405830 5 0.268324 0.268324 0.272185 0.283628 6 0.052894 0.052894 0.063903 0.070979 7 0.206835 0.206835 * 0.236525 8 0.139781 0.139781 0.141682 0.140736 9 0.444485 0.444485 * 0.461690 10 0.240494 0.240522 * 0.260913 11 0.421056 0.421056 * 0.433767 12 0.561939 0.561939 * 0.570611 13 0.386431 0.386431 * 0.390348 14 0.062873 0.062873 * 0.082477 15 0.439334 0.439334 0.439710 0.443277 16 0.537110 0.537110 * 0.550194 17 0.131040 0.131041 0.131072 0.131905 18 0.233770 0.233770 * 0.255437 19 0.407089 0.407089 * 0.418072 20 0.478001 0.478001 * 0.482912 This table compares the rate of lost customers of the best stationary policy, heuristic 1, heuristic 2, and random policy. When both heuristics result in the same policy we use "*". It worth mentioning that where all those examples turn out to be transitive, transi- tivity need not always hold. Table 4.3 shows an example were transitivity does not hold. For this example, the best ordering among servers 1 and 2 is (1,2), for servers 1 and 3 is (1,3), but the best ordering among servers 2 and 3 is (3,2) which shows that transitivity does not necessarily hold in all cases. Yet again, we can see that the rate of lost customers for Heuristic 1 is close to the best policy and better than Heuristic 2. 4.3 Lower Bounds As we mentioned before, for n > 3, the number of stationary policies that need to be considered grows rapidly. This growth makes the brute-force search for nding the best stationary policy a formidable task. This forces us to nd other ways to examine the per- formance of our heuristic approaches. One way is to compare them with each other and the random rule as we did for the casen = 3. Another way is to dene some lower bounds and use them to examine the performance of the heuristics. Lower bound models can be General Case Optimality, Heuristic Policy 33 Tab. 4.3: Non-Transitive Example Server 1 2 3 i 1 2 3 p i 0.125 0.22463851 0.3054560568436 Server i vs j Ordering i,j Ordering j,i Best Ordering 1 vs 2 0.707355541 0.707355540 21 1 vs 3 0.639934010188197 0.639934010188199 13 2 vs 3 0.572662498 0.572661043 32 Best Policy Heuristic 1 Heuristic 2 Random 0.505837 0.505846 0.505858 0.505852 In this example = 1. For each possible ordering of servers i and j we have calculated the rate of lost customers. obtained by letting p i = max 1in (p i ) for i = 1;:::;n, or by letting i = max 1in ( i ) for i = 1;:::;n. The advantage of these boundary models is that their optimal assignment policies are known. Ross [16] showed that, for r > 0 , when i = , for all i, assigning arrivals to the idle eligible server with the lowest p i and when p i =p, for all i, assigning customers to the idle eligible server with highest i , stochastically maximizes N r . Recall that I m is the indicator of the event that arrival m is served and N r = P r m=1 I m . It is worth mentioning that these lower bounds are usually good (useful) when either service rates are close to each other, or p i 's are close to each other. In the following theorems we show that the preceding models are indeed the lower bounds for our original model. Theorem 2 compares two dierent models both with n servers having i , i = 1;:::;n, as their service rates, but in one model servers are more likely to be eligible to serve arriving customers. let (X k 1 ;:::;X k n ) be the eligibility vector in model k, p i = P (X 1 i = 1), and q i = P (X 2 i = 1) where q i p i for all i. We show that for any r > 0, N r is stochastically greater in model 2 than the same quantity in model 1. This theorem enables us to nd a lower bound for each numerical example by letting p i = max 1in (p i ) for i = 1;:::;n. Theorem 2 : Consider the following two models with n servers where each server i, i = 1;:::;n, in both models has service rate i : Model 1 with P (X 1 i = 1) =p i , i = 1;:::;n. Model 2 with P (X 2 i = 1) =q i , i = 1;:::;n. where q i p i for all i. For any r> 0 we have sup E 1 [N r ] sup E 2 [N r ]; General Case Optimality, Heuristic Policy 34 where E i [X] denotes the expected value of X given that policy is used in model i. Proof : Let B be the initial set of busy servers in both models. Couple the arrival processes in both models so that they are at identical times. Also couple the initial service time of server i, i2 B, in both models so that they are equal. For each arrival, couple the eligibility vectors in both models so that if server m is eligible in model 1, it is also eligible in model 2. That is, for U 1 ;:::;U n independent uniform [0; 1] random variables set X 1 k = IfU i p i g; for all i X 2 k = IfU i q i g; for all i Now dene a policy 2 that exactly follows the same assignments made by policy 1 in model 1. That is, every time that policy 1 in the rst model assigns a customers to a server, let policy 2 make the same assignment and couple the service times of the newly busy servers so that they are identical in both models. Every time that there is no idle eligible server in model 1 then let 2 reject that arrival even if there exist an idle eligible server in model 2. Now by using Lemma 1 we have E 1 1 [N r ] =E 2 2 [N r ]E 2 0 2 [N r ] In the next theorem we compare two dierent models where one has service rates 1 ;:::; n and the other has 1 ;:::; n as its service rates. In both modelsP (X i = 1) =p i and i i for all i. We show, by using a coupling argument, that for any number of arrivals, r, the vector (I 1 ;:::;I r ) for the model with larger service rates is stochastically greater than the same vector for the model with smaller service rates. Using this theorem we can nd a lower bound for numerical examples by letting i = max 1in ( i ). Theorem 3 : Consider two models with n servers one having service rates 1 ;:::; n , model 1, and the other having 1 ;:::; n , model 2, as its service rates, where i i . In both models P (X i = 1) =p i , i = 1;:::;n. For r> 0 we have sup E 1 [h(I 1 ;:::;I r )] sup E 2 [h(I 1 ;:::;I r )]: Note thath is a non decreasing function of its coordinates andE i [X] denotes the expected value of X given that policy is used in model i.. General Case Optimality, Heuristic Policy 35 Proof : We start our poof by showing that for every policy 1 in model 1 there exists a policy 2 that works at least as good with respect to the criterion E[h(I 1 ;:::;I r )]. To do so, rst couple the arrival processes in both models so that customers arrive at identical times. Couple the eligibility vectors of the i th , i = 1;:::r, arrival in the two models so that they are identical. Again note that both models start with the same set of busy servers,B. Using Lemma 1 in [16], couple the initial service time of the busy serverj, for allj2B, in both models so that they are either identical or the service time in model 1 is larger by a random amount that is independent and exponential with rate j . Coupling the service times in this manner will guarantee that the set of idle servers seen by the rst arrival in model 1 is a subset of the set of idle servers seen by the same arrival in model 2. Now, if policy 1 assigns the rst arrival to server i in model 1, then let 2 make the same assignment and couple the service times of the newly busy servers in the preceding fashion. If there is no idle eligible server for the initial arrival in model 1 and there is at least one idle eligible server in model 2 then let 2 assign the arrival to any of the idle eligible servers. Use the same method as was used for the initial service times to couple the service time of the newly busy server in model 2 with the service time of the same server in model 1. This is possible since we have coupled the service times such that if the service time of server j in model 1 exceeds the service time of server j in model 2, the excess time is an exponential random variable with rate j . Therefore, if server j in model 1 has not yet completed service before the second arrival, because of the memory less property, the remaining service time after the arrival is also exponential with rate j . By letting 2 follow the same pattern for the subsequent arrivals, it is easy to see E 1 1 [h(I 1 ;:::;I r )]E 2 2 [h(I 1 ;:::;I r )]: Let i be the optimal policy in model i. Recall that in both models the optimal policy assigns arrivals to the idle eligible server with the highest service rate. Thus, using the preceding train of reasoning we can show that there exists a policy in model 2 that is at least as good as 1 , and clearly is at most as good as 2 . Therefore, our theorem is proven. Now using theorems 1,2, and 3 we introduce the following procedure for nding the lower bounds: Pick the server that has the highest i and check if it has the lowest p i , if it does, pick the second highest i and check if it has the second lowest p i ; continue doing so till you see otherwise. Dene the lower bound models by either setting the service rates of the remaining General Case Optimality, Heuristic Policy 36 servers (including the server that the preceding procedure stopped at) equal to highest remaining service rate or setting the p i 's of all the remaining servers equal to the highest remaining p i . The optimal policy for the rst lower bound model is to assign arrivals to the eligible server with the lowest p i and for the latter case is to assign to the eligible server with the highest i . The rates of lost customers of the preceding models under their optimal policies are the lower bounds for the rate of lost customers of our original model. In Table 4.4 we see the lower bounds for the numerical examples in Table 4.2. Again we use Matlab to calculate the rate of lost customers under the optimal policy for these boundary models. Note that in cases where the server with highest i has the lowest p i , the server with second highest i has the second lowest p i , and so on; the optimal policy is known (see the Remark) and there is no need for lower bounds. In Table 4.4 in front of these cases we have used the term \N=A". In some cases, the rate of lost customers under the heuristics are close to the lower bounds which is an indication of their acceptable performance. This agrees with our previous ndings for the case n = 3, where we are able to compare our heuristics with the best stationary policy. As it can be seen, these boundary models give better lower bounds when the heterogeneity among servers is low. As n increases nding the limiting distribution by solving balance equations becomes computationally intensive. Therefore, for n > 3, we use simulation to nd an estimate of the rate of lost customers under the heuristic policies and the random policy. The estimates are based on simulation of r = 10000 renewal cycles. A renewal cycle consists of a busy period, where at least one of the servers is busy, and an idle period, where the system is empty. If we look at the model under each of these assignment policies, we see that every time the system becomes empty everything starts all over again. Therefore, we can say that renewals occur every time that our system becomes empty. For estimating the quantities of interest, we have used the regenerative approach. Using this approach at each run i, we record a vector (L i ;A i ), where L i is the number of lost customers and A i is the total number of arrivals in cycle i. We use r X i=1 L i , r X i=1 A i as our estimate for the rate of lost customers. General Case Optimality, Heuristic Policy 37 Tab. 4.4: Eciency of Heuristic 1 Case Heuristic 1 Heuristic 2 max( i ) max(p i ) 1 0.310149 * 0.136090 0.285258 2 0.521638 * 0.510669 0.441674 3 0.083175 * 0.075630 0.039364 4 0.387368 * 0.145526 0.197633 5 0.268324 0.272185 0.214163 0.145213 6 0.052894 0.063903 0.046633 0.022183 7 0.206835 * N=A N=A 8 0.139781 0.141682 0.139082 0.042068 9 0.444485 * 0.329565 0.296103 10 0.240522 * 0.123492 0.212653 11 0.421056 * N=A N=A 12 0.561939 * 0.549451 0.347224 13 0.386431 * 0.239812 0.336349 14 0.062873 * 0.059860 0.015946 15 0.439334 0.439710 0.406620 0.377478 16 0.537110 * 0.489848 0.513184 17 0.131041 0.131072 0.129911 0.046756 18 0.233770 * N=A N=A 19 0.407089 * N=A N=A 20 0.478001 * 0.466851 0.320247 This table shows the rate of lost customers for the boundary solutions for 20 dierent numerical examples. When both heuristics result in the same policy we use "*". For examples in Table 4.5, is generated according to a uniform [0; 10] random vari- able. i , i = 1;:::; 5, is generated according to uniform [0; 5] and p i , i = 1;:::; 5, according to a uniform [0; 1]. In Table 4.6 for both of the heuristics, the random rule, and the lower bounds we see the estimate of the rate of lost customers. Like the case where n = 3, we see that Heuristic 1 outperforms Heuristic 2. In 50 percent of the cases heuris- tics 1 and 2 are the same and in the rest of the cases, Heuristic 1 has a smaller estimate for the rate of lost customers than Heuristic 2. In all of the numerical examples, both of the heuristics perform better than the random rule where in some cases this improvement is up to 50 percent. Thus, using the heuristics rather than random rule may potentially result in substantial improvements. Furthermore, the numerical results indicate that the dierence between the estimate of the rate of lost customers by heuristics 1 and 2 is not signicant. Hence, sometimes using Heuristic 1 instead of Heuristic 2 might not worth the extra computational cost. General Case Optimality, Heuristic Policy 38 Tab. 4.5: Examples Inputs Case 1 Server 1 2 3 4 5 Case 6 Server 1 2 3 4 5 i 1.6 3 1.1 3.5 1.4 i 1.5 0.4 3.9 2.2 1 = 7:3 p i 0.53 0.01 0.9 0.61 0.47 = 2:6 p i 0.83 0.39 0.27 0.81 0.39 Case 2 Server 1 2 3 4 5 Case 7 Server 1 2 3 4 5 i 3.1 1.5 0.3 4 4.5 i 0.2 2 3.3 3.3 3.5 = 8:2 p i 0.05 0.35 0.73 0.49 0.15 = 6:4 p i 0.48 0.54 0.01 0.7 0.61 Case 3 Server 1 2 3 4 5 Case 8 Server 1 2 3 4 5 i 4.9 4 2.7 3.2 1.2 i 3.5 3 2.9 4.7 4 = 8:3 p i 0.25 0.46 0.18 0.66 0.56 = 1:7 p i 0.96 0.5 0.19 0.65 0.12 Case 4 Server 1 2 3 4 5 Case 9 Server 1 2 3 4 5 i 3.2 1.6 2.3 4 0.5 i 2.2 0.1 1.4 0.9 3.8 = 2:5 p i 0.94 0.48 0.76 0.78 0.76 = 3:8 p i 0.32 0.47 0.02 0.6 0.45 Case 5 Server 1 2 3 4 5 Case 10 Server 1 2 3 4 5 i 2.5 2.7 1.4 0.5 1.1 i 3.1 3.8 2 4.6 1.5 = 3:9 p i 0.03 0.73 0.02 0.5 0.75 = 2:8 p i 0.05 0.47 0.81 0.33 0.42 Tab. 4.6: Eciency of Heuristic 1 Case Heuristic 1 Heuristic 2 Random max( i ) max(p i ) 1 0.367610 0.369093 0.388303 0.166453 0.152548 2 0.487938 * 0.506363 0.250353 0.166850 3 0.267637 * 0.301528 0.161655 0.125213 4 0.009471 0.012143 0.022520 0.002078 0.002938 5 0.362044 * 0.381882 0.189527 0.086013 6 0.097364 0.100377 0.128028 0.026101 0.055419 7 0.300049 * 0.311995 0.163812 0.137855 8 0.021908 0.022320 0.040042 0.017576 0.000277 9 0.360511 * 0.387069 0.153266 0.151727 10 0.121210 0.127624 0.149472 0.073742 0.004735 When both heuristics result in the same policy we use "*". General Case, Idle Time Ordering Policies 39 5. GENERAL CASE, IDLE TIME ORDERING POLICIES 5.1 Introduction Again in this chapter, we assume that X 1 ;:::;X n are independent with P (X i = 1) =p i , i = 1;:::;n and the service times are exponentially distributed with rate i . We study the behavior of this system, for n = 2 and 3, under dierent idle time ordering policies, specically, the random rule, the longest idle rule, and the shortest idle rule, and a new rule which we refer to as the \random priority ordering" rule. This new policy randomly chooses one of the n! possible permutations of the servers and assigns arrivals to the idle eligible servers in accordance with this permutation. We compare the performance of these assignment policies under dierent assumptions on service rates and eligibility vectors. In section 4.2, by analyzing our model as a continuous time Markov chain we derive its limiting distribution under each of the four aforementioned assignment policies when n = 2. In section 4.3, we let i = , for all i, and we conjecture that rate of lost customer is smallest under longest idle rule. We give an intuitive explanation for this belief and, when n = 2, we analytically show that the longest idle rule has lower rate of lost customers than the random rule or the shortest idle rule. For n = 3, we also provide some numerical evidence. In section 4.4, we let p i = p, for all i, and nu- merically compare the rate of lost customers under the random priority ordering policy with the other three policies (which, as shown in chapter 2, have the same rates of lost customers). Finally, a numerical comparison of the rate of lost customers for the general case, i 6= j andp i 6=p j , under all of the four dierent policies is presented in section 4.5. 5.2 Limiting Distribution When n = 2, the general case under each of the assignment policies can be analyzed as a continuous time Markov chain. By solving the balance equations we nd the limiting distribution of our model under each policy. General Case, Idle Time Ordering Policies 40 5.2.1 Longest Idle Rule and Shortest Idle Rule Dene (i 1 ;i 2 ) as the state of the resulting Markov chain under the longest and the shortest idle rules when both servers are idle, with i 1 having been idle the longest time. Or (i 1 ) if i 1 is the only idle server and (0) if there is no idle server. Let P j:i 1 i 2 , j = 1; 2, be the probability that an arrival will be assigned to serveri j given that the state is (i 1 ;i 2 ). Then for Longest idle rule P 1:i 1 i 2 =p i 1 , and P 2:i 1 i 2 =p i 2 (1p i 1 ): Shortest idle rule P 1:i 1 i 2 =p i 1 (1p i 2 ), and P 2:i 1 i 2 =p i 2 : Let =P (X 1 +X 2 > 0) = 1 Q 2 m=1 (1p im ); be the probability that at least one of the servers is eligible to serve the next arrival when both servers are idle. Also let P (1; 2), P (2; 1), P (2), P (1) and P (0) be the stationary probabilities when the states are (1; 2), (2; 1), (2), (1), and (0), respectively. For the longest and the shortest idle policies we can write the balance equations as follows State Rate at which the process leaves = Rate at which it enters 1; 2 P (1; 2) = 2 P (1) 2; 1 P (2; 1) = 1 P (2) 2 ( 1 +p 2 )P (2) = 1 P (0) +P 1:12 P (1; 2) +P 2:21 P (2; 1) 1 ( 2 +p 1 )P (1) = 2 P (0) +P 2:12 P (1; 2) +P 1:21 P (2; 1) 0 ( 1 + 2 )P (0) =p 1 P (1) +p 2 P (2) Note that P (1; 2) +P (2; 1) +P (2) +P (1) +P (0) = 1. The balance equations yield the following stationary probabilities For the longest idle rule P (1; 2) = 1 2 2 p 1 P (0); P (2; 1) = 1 2 2 p 2 P (0); P (2) = 2 p 2 P (0); P (1) = 1 p 1 P (0): General Case, Idle Time Ordering Policies 41 For the shortest idle rule P (1; 2) = 1 2 2 p 1 + ( 1 + 2 )(1p 2 ) + 1 (1p 1 ) + 2 (1p 2 ) P (0); P (2; 1) = 1 2 2 p 2 + ( 1 + 2 )(1p 1 ) + 1 (1p 1 ) + 2 (1p 2 ) P (0); P (1) = 1 p 1 + ( 1 + 2 )(1p 1 ) + 1 (1p 1 ) + 2 (1p 2 ) P (0); P (2) = 2 p 2 + ( 1 + 2 )(1p 2 ) + 1 (1p 1 ) + 2 (1p 2 ) P (0): The limiting distribution under the longest idle rule can also be found from the results in chapter 3. For both of these policies we can nd R, the rate of lost customers, as follows R = " (1p 1 )(1p 2 ) P (12) +P (21) + (1p 2 )P (2) + (1p 1 )P (1) +P (0) # : 5.2.2 Random Rule For the random rule, let the state of the resulting Markov chain be the set of idle servers. Then we have State Rate at which the process leaves = Rate at which it enters 1; 2 ( 1 + 2 )P (1; 2) = 2 P (1) + 1 P (2) 2 ( 1 +p 2 )P (2) = 2 P (0) +P 1:12 P (1; 2) +P 2:21 P (2; 1) 1 ( 2 +p 1 )P (1) = 1 P (0) +P 2:12 P (1; 2) +P 1:21 P (2; 1) 0 ( 1 + 2 )P (0) =p 1 P (1) +p 2 P (2) where 1 = 1 2 [p 1 +p 1 (1p 2 )] and 2 = 1 2 [p 2 +p 2 (1p 1 )]. By solving these balance equations we can nd the following stationary probabilities for this model P (1; 2) = 1 2 2 (p 1 +p 2 ) + 1 + 2 p 1 p 2 ( 1 + 2 ) + 2 p 1 1 + 1 p 2 2 P (0); P (2) = 2 (p 2 + 1 ) 1 + 2 p 1 + 1 1 p 1 p 2 ( 1 + 2 ) + 2 p 1 1 + 1 p 2 2 P (0); P (1) = 1 (p 1 + 2 ) 2 + 1 p 2 + 2 2 p 1 p 2 ( 1 + 2 ) + 2 p 1 1 + 1 p 2 2 P (0): General Case, Idle Time Ordering Policies 42 Note that P (0) is such that P (1; 2) +P (2) +P (1) +P (0) = 1. Let R RR be the rate of lost customers under random rule, then we have R RR = h (1p 1 )(1p 2 )P (1; 2) + (1p 2 )P (2) + (1p 1 )P (1) +P (0) i : 5.2.3 Random Priority Ordering Rule As we mentioned earlier, using the random priority ordering rule the model starts working under one of the n! possible permutations of the servers. That is, under this rule a possible permutation is chosen at random and the model assigns arrivals to the idle eligible servers according to the chosen ordering. Therefore, to nd the rate of lost customers for the General model under this policy, we rst derive the rate of the lost customers for each specic ordering and then we take the average of the rate of lost customers for all the possible ordering permutations as the rate of lost customers for the random priority ordering policy. For n = 2, we rst nd the limiting distribution for the ordering permutation (1; 2). By using this ordering an arrival is assigned to the idle eligible server that appears earlier on the ordering list, in this case server 1 is rst and server 2 is second. If both of the servers are ineligible or busy the arrival will be lost. Let (1; 2), (2), (1), and (0) be the states of the resulting Markov chain under this chosen permutation when the set of idle servers isf1; 2g,f2g,f1g, andfg, respectively. We have State Rate at which the process leaves = Rate at which it enters 1; 2 (p 1 +p 2 (1p 1 ))P (1; 2) = 2 P (1) + 1 P (2) 1 ( 2 +p 1 )P (1) = 1 P (0) +p 1 P (1; 2) 2 ( 1 +p 2 )P (2) = 2 P (0) +p 2 (1p 1 )P (1; 2) 0 ( 1 + 2 )P (0) =p 1 P (1) +p 2 P (2) Solving these balance equations yields in the following stationary probabilities P (1; 2) = 1 2 2 p 1 p 2 (p 1 +p 2 ) + 1 + 2 (p 1 +p 2 (1p 1 )) + 1 (1p 1 ) + 2 P (0); P (2) = 2 p 2 (p 1 +p 2 (1p 1 )) + ( 1 + 2 ) (p 1 +p 2 (1p 1 )) + 1 (1p 1 ) + 2 P (0); P (1) = 1 p 1 (p 1 +p 2 (1p 1 )) + ( 1 + 2 )(1p 1 ) (p 1 +p 2 (1p 1 )) + 1 (1p 1 ) + 2 P (0): where P (0) is such that P (1; 2) +P (2) +P (1) +P (0) = 1. Similarly, we can nd the limiting distribution for the ordering (2; 1). Now, let R ij , be the rate of lost customers when the chosen permutation is (i;j) , i;j = 1; 2 and i6=j. General Case, Idle Time Ordering Policies 43 R 12 = h (1p 1 )(1p 2 )P (12) + (1p 2 )P (2) + (1p 1 )P (1) +P (0) i ; R 21 = h (1p 1 )(1p 2 )P (21) + (1p 2 )P (2) + (1p 1 )P (1) +P (0) i : and nally, we use R 12 and R 21 to nd R RPO = 1 2 [R 12 +R 21 ] Now that we have the limiting distribution and the rate of lost customers of our queue- ing loss model under all these four rules, in the rest of this chapter, we would compare the performance of these assignment policies under dierent assumptions on service rates and eligibility vectors. 5.3 i = for all i As it was mentioned in the previous chapter, assigning arrivals to the idle eligible server with lowest p i decreases the rate of the lost customers. Intuitively, it is more likely that the servers with fewer requests have been idle longer. Thus, it is reasonable to expect that the longest idle rule naturally gives more priority to servers with lower p i and the shortest idle rule gives more priority to servers with higher p i . As a result, we expect the longest idle rule to have the lowest and the shortest idle rule to have the highest rate of lost customers among idle time ordering policies. For n = 2, we are able to analytically compare the longest idle rule with the random rule and the shortest idle rule and show that the longest idle idle has the lowest rate of lost customers. Forn = 3, we also provide some numerical examples to support our conjecture. For n = 2, by analyzing our model under each policy as a continuous time Markov chain we derive the limiting distribution and the rate at which customers are lost. Let R LIR , R SIR , and R RR be the rate of lost customers for the longest idle, the shortest idle, and the random rule, respectively. Also, let = , 1 = 1 2 [p 1 +p 1 (1p 2 )], and 2 = 1 2 [p 2 +p 2 (1p 1 )]. With some intense calculations for the longest idle rule we have R LIR = 2 (p 1 +p 2 )(1p 1 )(1p 2 ) +(p 1 +p 2 2p 1 p 2 )( 1 + 2 ) +p 1 p 2 ( 1 + 2 ) 2 (p 1 +p 2 ) +(p 1 +p 2 )( 1 + 2 ) +p 1 p 2 ( 1 + 2 ) ; (5.1) General Case, Idle Time Ordering Policies 44 and for the shortest idle rule we have R SIR = 2 (C 1 p 2 +C 2 p 1 )(1p 1 )(1p 2 ) 2 (C 1 p 2 +C 2 p 1 ) +(C 1 p 2 +C 2 p 1 )( 1 + 2 ) +Kp 1 p 2 ( 1 + 2 ) +((1p 1 )C 1 p 2 + (1p 2 )C 2 p 1 )( 1 + 2 ) +Kp 1 p 2 ( 1 + 2 ) ; (5.2) where C 1 = ( 1 + 2 ) + 2(1p 1 ), C 2 = ( 1 + 2 ) + 2(1p 2 ), and K = ( 1 + 2 ) + (2p 1 p 2 ). Finally, for the random rule we have R RR = 2 [(p 1 +p 2 + 2)(1p 1 )(1p 2 ) + (p 2 p 1 ) 2 + (2p 1 p 2 )( 1 + 2 )] 2 [p 1 +p 2 + 2( 1 + 2 +)] +[(p 1 +p 2 )( 1 + 2 ) + 1 p 2 + 2 p 1 ] +p 1 p 2 ( 1 + 2 ) +[(p 1 +p 2 2p 1 p 2 )( 1 + 2 ) + 1 p 2 + 2 p 1 ] +p 1 p 2 ( 1 + 2 ) : (5.3) First we show that the longest idle rule has lower rate of lost customers than the random rule and later we show that the random rule has lower rate of lost customers than the shortest idle rule. Proposition 1: For a model with two servers and Poisson arrivals, when 1 = 2 and p 1 6= p 2 , the longest idle rule results in a lower rate of lost customers than the random rule. Proof: Using (5.1) and (5.3), to prove this proposition we need to show that [ 2 (p 1 +p 2 )(1p 1 )(1p 2 ) +(p 1 +p 2 2p 1 p 2 )( 1 + 2 ) +p 1 p 2 ( 1 + 2 )] [ 2 [p 1 +p 2 + 2( 1 + 2 +)] +[(p 1 +p 2 )( 1 + 2 ) + 1 p 2 + 2 p 1 ] +p 1 p 2 ( 1 + 2 )] [ 2 [(p 1 +p 2 + 2)(1p 1 )(1p 2 ) + (p 2 p 1 ) 2 + (2p 1 p 2 )( 1 + 2 )] +[(p 1 +p 2 2p 1 p 2 )( 1 + 2 ) + 1 p 2 + 2 p 1 ] +p 1 p 2 ( 1 + 2 )] [ 2 (p 1 +p 2 ) +(p 1 +p 2 )( 1 + 2 ) +p 1 p 2 ( 1 + 2 )] General Case, Idle Time Ordering Policies 45 By canceling equal terms from both sides of the inequality we can rewrite the preceding as follows 2 4 (p 1 +p 2 )(1p 1 )(1p 2 )( 1 + 2 ) + 3 (p 1 +p 2 )(1p 1 )(1p 2 )( 1 p 2 + 2 p 1 ) +2 3 ( 1 + 2 )(p 1 +p 2 2p 1 p 2 )( 1 + 2 +) 2 2 ( 1 + 2 )p 1 p 2 ( 1 p 2 + 2 p 1 ) +2 2 ( 1 + 2 )p 1 p 2 ( 1 + 2 +) +( 1 + 2 )p 1 p 2 ( 1 p 2 + 2 p 1 ) 2 4 (p 1 +p 2 )(1p 1 )(1p 2 )( 1 + 2 ) + 2 3 (1p 1 )(1p 2 )( 1 + 2 )p 1 p 2 + 4 (p 1 +p 2 )(2p 1 p 2 )( 1 + 2 ) + 4 (p 1 +p 2 )(p 2 p 1 ) 2 + 3 (p 1 +p 2 )( 1 p 2 + 2 p 1 ) + 3 (p 1 +p 2 )(2p 1 p 2 )( 1 + 2 ) 2 + 3 (p 1 +p 2 )(p 2 p 1 ) 2 ( 1 + 2 ) + 2 p 1 p 2 (2p 1 p 2 )( 1 + 2 ) 2 + 2 p 1 p 2 (p 2 p 1 ) 2 ( 1 + 2 ) +p 1 p 2 ( 1 + 2 )( 1 p 2 + 2 p 1 ) this can be further simplied in the following form 3 (p 1 +p 2 )(1p 1 )(1p 2 )( 1 p 2 + 2 p 1 ) 4 3 ( 1 + 2 ) 2 p 1 p 2 4 4 ( 1 + 2 )p 1 p 2 2 2 ( 1 + 2 )p 1 p 2 ( 1 p 2 + 2 p 1 ) + 2 2 ( 1 + 2 )p 1 p 2 ( 1 + 2 +) 2 3 (1p 1 )(1p 2 )( 1 + 2 )p 1 p 2 + 4 (p 1 +p 2 )((p 2 p 1 ) 2 (p 1 +p 2 )( 1 + 2 )) (5.4) + 3 (p 1 +p 2 )( 1 p 2 + 2 p 1 ) + 3 (p 1 +p 2 )( 1 + 2 )((p 2 p 1 ) 2 (p 1 +p 2 )( 1 + 2 )) +2 2 p 1 p 2 ( 1 + 2 ) 2 + 2 p 1 p 2 ( 1 + 2 )((p 2 p 1 ) 2 (p 1 +p 2 )( 1 + 2 )) since (p 1 p 2 ) 2 (p 1 +p 2 )( 1 + 2 ) = (p 1 p 2 ) 2 (p 1 +p 2 )(p 1 +p 2 p 1 p 2 ) =2p 1 p 2 (2 1 2 (p 1 +p 2 )) =2p 1 p 2 ( 1 p 2 + 2 p 1 ) and (1p 1 )(1p 2 ) = 1 ( 1 + 2 ) we can simplify (5.4) and write General Case, Idle Time Ordering Policies 46 3 (p 1 +p 2 )( 1 + 2 )( 1 p 2 + 2 p 1 ) 2 3 ( 1 + 2 ) 2 p 1 p 2 4 4 ( 1 + 2 )p 1 p 2 2 4 (p 1 +p 2 )( 1 p 2 + 2 p 1 ) 2 3 (p 1 +p 2 )( 1 + 2 )( 1 p 2 + 2 p 1 ) moving all terms to the left hand side of the inequality we have 2 4 [(p 1 +p 2 )( 1 p 2 + 2 p 1 ) 2( 1 + 2 )p 1 p 2 ] + 3 [(p 1 +p 2 )( 1 + 2 )( 1 p 2 + 2 p 1 ) 2( 1 + 2 ) 2 p 1 p 2 ] 0 nally we have 4 p 1 p 2 (p 1 p 2 ) 2 1 2 3 p 1 p 2 ( 1 + 2 )(p 1 p 2 ) 2 0 where it is clear that the left hand side is always less or equal to zero therefore, our proof is completed. Proposition 2: For a model with two servers and Poisson arrivals, when 1 = 2 and p 1 6= p 2 , the random rule results in a lower rate of lost customers than the shortest idle rule. Proof: We follow the same approach that has been used in Proposition 1. Using (5.2) and (5.3), to prove this proposition we need to show that [ 2 [(p 1 +p 2 + 2)(1p 1 )(1p 2 ) + (p 2 p 1 ) 2 + (2p 1 p 2 )( 1 + 2 )] +[(p 1 +p 2 2p 1 p 2 )( 1 + 2 ) + 1 p 2 + 2 p 1 ] +p 1 p 2 ( 1 + 2 )] [ 2 (C 1 p 2 +C 2 p 1 ) +(C 1 p 2 +C 2 p 1 )( 1 + 2 ) +Kp 1 p 2 ( 1 + 2 )] [ 2 (C 1 p 2 +C 2 p 1 )(1p 1 )(1p 2 ) +((1p 1 )C 1 p 2 + (1p 2 )C 2 p 1 )( 1 + 2 ) +Kp 1 p 2 ( 1 + 2 )] [ 2 [p 1 +p 2 + 2( 1 + 2 +)] +[(p 1 +p 2 )( 1 + 2 ) + 1 p 2 + 2 p 1 ] +p 1 p 2 ( 1 + 2 )] General Case, Idle Time Ordering Policies 47 By eliminating identical terms on both sides of the inequality we can rewrite it as 2 3 (C 1 p 2 +C 2 p 1 )p 1 p 2 ( 1 + 2 ) 2 2 (C 1 p 2 +C 2 p 1 )p 1 p 2 ( 1 + 2 ) 2 4 (C 1 p 2 +C 2 p 1 )(p 1 +p 2 )( 1 + 2 ) 3 (C 1 p 2 +C 2 p 1 )(p 1 +p 2 )( 1 + 2 ) 2 + 4 (C 1 p 2 +C 2 p 1 )(p 2 p 1 ) 2 + 3 (C 1 p 2 +C 2 p 1 )(p 2 p 1 ) 2 ( 1 + 2 ) 3 (C 1 p 2 +C 2 p 1 )( 1 p 2 + 2 p 1 ) + 2 (p 1 +p 2 + 2)(1p 1 )(1p 2 )Kp 1 p 2 ( 1 + 2 ) +( 1 + 2 ) 2 (2p 1 p 2 +(p 1 +p 2 ))Kp 1 p 2 + 2 (p 2 p 1 ) 2 ( 1 + 2 )Kp 1 p 2 +( 1 + 2 )( 1 p 2 + 2 p 1 )Kp 1 p 2 +( 1 + 2 ) 2 (C 1 p 2 +C 2 p 1 )Kp 1 p 2 3 (1p 1 )(1p 2 )(C 1 p 2 +C 2 p 1 )( 1 p 2 + 2 p 1 ) 2 (C 1 p 2 +C 2 p 1 )( 1 + 2 )p 1 p 2 3 (C 1 +C 2 )( 1 + 2 )(p 1 +p 2 + 2)p 1 p 2 2 3 (C 1 +C 2 )( 1 + 2 ) 2 p 1 p 2 2 (C 1 +C 2 )( 1 + 2 ) 2 (p 1 +p 2 )p 1 p 2 2 (C 1 +C 2 )( 1 + 2 )( 1 p 2 + 2 p 1 )p 1 p 2 +( 1 + 2 ) 2 (C 1 p 2 +C 2 p 1 )p 1 p 2 (C 1 +C 2 )( 1 + 2 ) 2 (p 1 p 2 ) 2 + 2 ( 1 + 2 )(p 1 +p 2 + 2)Kp 1 p 2 +( 1 + 2 )( 1 p 2 + 2 p 1 )Kp 1 p 2 Since (1p 1 )(1p 2 ) = 1 ( 1 + 2 ) and 2K = C 1 +C 2 . We can further simplify our inequality and then by moving all terms to the left side of the inequality we can write 4 [(C 1 p 2 +C 2 p 1 )[(p 2 p 1 ) 2 (p 1 +p 2 )( 1 + 2 )] + 2(C 1 +C 2 )p 1 p 2 ( 1 + 2 )] + 3 [(C 1 p 2 +C 2 p 1 )[(p 2 p 1 ) 2 ( 1 + 2 ) (p 1 +p 2 )( 1 + 2 ) 2 2( 1 + 2 )p 1 p 2 + ( 1 + 2 )( 1 p 2 + 2 p 1 )] + (C 1 +C 2 )( 1 + 2 )(p 1 +p 2 )p 1 p 2 +2(1 ( 1 + 2 ))( 1 + 2 )Kp 1 p 2 + 2(C 1 +C 2 )( 1 + 2 ) 2 p 1 p 2 2( 1 + 2 )Kp 1 p 2 ] 2 [(C 1 p 2 +C 2 p 1 )( 1 + 2 ) 2 p 1 p 2 + ( 1 + 2 )(p 2 p 1 ) 2 Kp 1 p 2 +(C 1 +C 2 )( 1 + 2 )( 1 p 2 + 2 p 1 )p 1 p 2 ] 0 which is 4 [( 1 + 2 )(p 2 p 1 ) 2 p 1 p 2 2(p 2 p 1 ) 2 p 1 p 2 ] + 3 [ 1 2 ( 1 + 2 ) 2 (p 2 p 1 ) 2 p 1 p 2 3( 1 + 2 )(p 2 p 1 ) 2 p 1 p 2 ] 2 [( 1 + 2 ) 2 (p 2 p 1 ) 2 p 1 p 2 ] 0 which is the same as 2 5 (p 2 p 1 ) 2 p 1 p 2 2 4 ( 1 + 2 )(p 2 p 1 ) 2 p 1 p 2 General Case, Idle Time Ordering Policies 48 1 2 3 ( 1 + 2 ) 2 (p 2 p 1 ) 2 p 1 p 2 0 The last inequality is indeed true hence, our proof is completed. It is immediate from the results of Proposition 1 and Proposition 2 that the model of this section has lower rate of lost customers using the longest idle rule as its assignment policy than the shortest idle rule. Thus, we showed that, for n = 2, R LIR R RR R SIR : In Table 5.1 we see examples for n = 3, where = 1 and 1 = 2 = 3 = 1. In this case, the longest idle rule has the lowest and the shortest idle rule has always the highest rate of lost customers. As we mentioned previously, analytically comparing the rate of Tab. 5.1: Comparison between LIR, RR, SIR, and RPO (p 1 ;p 2 ;p 3 ) LIR RR SIR RPO Ordering (0:43; 0:23; 0:78) 0:26810 0:27594 0:28755 0:27577 LIRRPORRSIR (0:4; 0:4; 0:84) 0:21842 0:22563 0:23777 0:22546 LIRRPORRSIR (0:99; 0:38; 0:46) 0:18040 0:19177 0:21100 0:19147 LIRRPORRSIR (0:15; 0:89; 0:75) 0:20050 0:20899 0:21938 0:20840 LIRRPORRSIR (0:26; 0:9; 0:4) 0:23916 0:25070 0:26817 0:25050 LIRRPORRSIR (0:63; 0:43; 0:31) 0:28297 0:28597 0:29072 0:28590 LIRRPORRSIR (0:16; 0:91; 0:48) 0:24690 0:25844 0:27468 0:25820 LIRRPORRSIR (0:11; 0:78; 0:76) 0:23113 0:23785 0:24513 0:23741 LIRRPORRSIR (0:46; 0:21; 0:88) 0:24326 0:25390 0:26951 0:25367 LIRRPORRSIR (0:51; 0:83; 0:63) 0:15767 0:16033 0:16567 0:16018 LIRRPORRSIR For each vector (p 1 ;p 2 ;p 3 ) we have calculated the rate of lost customers for the longest idle rule (LIR), the random rule (RR), the shortest idle rule (SIR), and the random priority ordering policy (RPO). lost customers for the random priority ordering policy with the other three assignment policies is computationally dicult. Hence, we have also considered the random priority ordering policy in our numerical examples to examine how this policy performs. Recall that for nding the rate of lost customers for the random priority ordering policy we take the average of the rate of lost customers for all the possible permutations ordering policies. As it can be seen in Table 5.1, the random priority rule performs better than the random rule. To see why this result seems reasonable, assume that our model has 3 servers which all are idle. Upon arrival of a new customer we nd that server 1 and 2 are eligible and server 3 is ineligible. Intuitively, you want the ordered list of idle servers to give priority to the servers with lowerp i , and in our example it is more likely that server 3 has a lower p i comparing to the other servers. By using the random rule one of the servers 1 or 2 General Case, Idle Time Ordering Policies 49 is chosen at random. If the random priority ordering policy is used the ordering is one of the 3! possible permutations of the servers. Again both of these two eligible servers have equal chance of serving the arrival. However, if the next customer arrives before the departure of the customer already in service and both idle servers are eligible to serve the new customer, the random priority ordering rule assigns the new arrival to server 3 with probability 2 3 while the random rule chooses one of the servers at random. Therefore, it seems reasonable to expect that the random priority ordering policy results in a lower rate of lost customers than the random rule. 5.4 p i =p for all i In this case, our model under the longest idle rule, the random rule, and the shortest idle rule is a special case of the exchangeable model introduced in chapter 2, where it was shown that all these three rules result in the same rate of lost customers. Thus, the only remaining policy which we have to investigate is the random priority ordering policy. However, computing the rate of lost customers under this policy, even for the case with n = 2, is cumbersome. Furthermore, even if we nd the rate of lost customers for this policy, comparing it analytically with the same for the rest of the assignment policies is a formidable task. Therefore, in order to compare these dierent assignment policies, we have generated some numerical results for n = 2 and 3. Our numerical results indicate that using random priority ordering policy (RPO) results in lower rate of lost customers comparing to idle time ordering policies (ITOP ). Table 5.2 shows the rate of lost cus- Tab. 5.2: Comparison between ITOP and RPO ( p, 1 ; 2 ) ITOP RPO (p, 1 ; 2 ) ITOP RPO (0:01; 1:77; 0:94) 0:8341625 0:8341616 (0:51; 0:67; 1:35) 0:542655 0:542454 (0:96; 0:35; 1:93) 0:412642 0:409730 (0:64; 1:71; 1:75) 0:3633097 0:3633092 (0:83; 0:93; 0:06) 0:675896 0:675614 (0:41; 0:78; 0:49) 0:662630 0:662594 (0:49; 0:04; 1:86) 0:662680 0:662366 (0:95; 0:52; 1:8) 0:391278 0:389155 (0:86; 0:29; 0:18) 0:796545 0:796536 (0:19; 0:71; 0:05) 0:855296 0:855264 (0:53; 1:5; 1:8) 0:431477 0:431456 (0:32; 0:59; 1:23) 0:656657 0:656588 (0:4; 1:01; 0:91) 0:594066 0:594063 (0:38; 0:94; 0:81) 0:619306 0:619302 (0:69; 1:33; 1:69) 0:371031 0:370963 (0:55; 1:34; 0:5) 0:555041 0:554660 (0:37; 1:03; 0:06) 0:759324 0:759189 (1:15; 1:3; 0:28) 0:643291 0:643290 (0:05; 1:95; 1:14) 0:908523 0:908522 (0:32; 1:48; 1:34) 0:596362 0:596361 For each pair of (p; 1 ; 2 ) we have calculated the rate of lost customers for the idle time ordering policies and the random priority ordering policy. tomers for some numerical examples for the case wheren = 2. These results are driven by General Case, Idle Time Ordering Policies 50 Matlab by directly solving the balance equations for each numerical example. For all the examples in this table = 2,p is generated according to a uniform [0,1] random variable, and i ; i = 1; 2, according to a uniform [0,2]. As it is shown in Table 5.2, in all of the numerical examples, the random priority ordering policy performs better than the idle time ordering policies in terms of the rate of lost customers. In Table 5.3, we assume that n = 3 and = 3. We calculate the rate of lost customers for each vector (p; 1 ; 2 ; 3 ). Where p is generated according to a uniform [0,1] random variable and i ;i = 1; 2; 3, according to a uniform [0,2]. For the numerical examples in Table 5.3, again we see that the random ordering priority policy performs better than the idle time ordering policies in the sense of the rate of lost customers. Tab. 5.3: Comparison between ITOP and RPO ( p, 1 ; 2 ; 3 ) ITOP RPO (p, 1 ; 2 ; 3 ) ITOP RPO (0:44; 0:75; 0:64; 1:64) 0:5052 0:5048 (0:42; 1:66; 0:47; 1:39) 0:4883 0:4878 (0:84; 1:79; 0:3; 0:31) 0:5005 0:4985 (0:83; 0:82; 1:58; 0:54) 0:3957 0:3947 (0:19; 1:06; 0:17; 0:3) 0:7752 0:7751 (0:64; 1:97; 1:12; 0:76) 0:3603 0:3595 (0:35; 1:33; 0:27; 1:36) 0:5735 0:5732 (0:92; 1:64; 0:82; 0:73) 0:3475 0:3465 (0:61; 1:5; 1:71; 0:35) 0:4097 0:4086 (0:03; 0:33; 1:32; 0:75) 0:9234 0:9233 (0:12; 1:16; 0:52; 0:13) 0:81284 0:81281 (0:64; 1:02; 1:7; 0:15) 0:4783 0:4774 (0:82; 1:22; 0:41; 1:98) 0:3429 0:3410 (0:4; 1:55; 1:29; 0:99) 0:4633 0:4632 (0:34; 1:83; 1:42; 1:76) 0:46022 0:46019 (0:54; 1:95; 1:11; 0:78) 0:3999 0:3993 (0:84; 1:76; 1:78; 1:59) 0:20557 0:20554 (0:23; 1:57; 1:59; 0:57) 0:6156 0:6155 (0:21; 1:37; 1:98; 1:44) 0:60070 0:60069 (0:45; 1:9; 1:5; 1:66) 0:37545 0:37542 For each pair of (p; 1 ; 2 ; 3 ) we have For calculated the rate of lost customers for the idle time ordering policies and the random priority ordering policy. We are able to show that this is true for a special case where p 1 = p 2 = 1. Before studying this model, let us dene the following relation between certain pair of subsets. Denition : LetS 1 andS 2 be two subsets of the setf1; 2g. SayS 1 dominatesS 2 if either S 1 =f1g and S 2 =f2g or S 1 S 2 . Proposition 3 : For a model with two servers, where 1 2 and p 1 = p 2 = 1, the rate of lost customer is lower under the random priority rule than the idle time ordering policies. Proof : We prove this by showing that for any xed amount of time t, t> 0, the num- ber of customers lost is stochastically smaller under the random priority ordering policy. Consider two dierent scenarios, where in scenario one the arrivals are assigned to servers General Case, Idle Time Ordering Policies 51 according to the list policy (1; 2) and in scenario two according to the random rule. Cou- ple the arrival processes in both scenarios so that arrivals occur at identical times. Any time both scenarios make the same assignment, couple the service time of the newly busy servers so that they are identical. If an arrival is assigned to server 1 in scenario one and is assigned to server 2 in scenario two, one of the following cases is true: Case 1 : Server 1 is busy in scenario two. For this case couple the service time of server 1 in both scenarios so that they are identical. Because of the memoryless property of the exponential random variables we can assume that the remaining service time after the arrival is also exponential with the same rate, so it is the same as the server has just started serving an arrival. Case 2 : Server 1 is idle in scenario two. For this case couple the service time of server 1 in scenario one, using lemma 1 in [16], with server 2 in scenario two so that they are either identical or the service time of server 2 exceeds that of server 1 by a random amount that is independent and exponential with rate 2 . Consider a time period that starts when server 1 is the only busy server in scenario one and server 2 is the only busy server in scenario two, and it continues while the set of busy servers in scenario one dominates the set of busy servers in scenario two. Call this time period the \advantage period". During this period there is a chance that an arrival nds both servers busy in the scenario under the random rule, while there is an idle server under policy (1; 2). Note that after such an arrival the advantage period ends. It is easy to see that only these time periods cause a dierence between the two scenarios. Let P (12 vs RR) be the probability that during an advantage period the random rule loses a customer P (12vsRR) = 1 1 + + 2 + 1 + 2 1 X i=1 1 + 1 + 2 + 2 i1 ! : Now dene another scenario, scenario three, which employs the list policy (2:1). Similarly, we can nd P (RRvs 21), the probability that during an advantage period the priority list policy (2; 1) loses a customer, P (RRvs 21) = 1 1 + + 2 + 1 + 2 1 X i=1 1 2 h 1 + 1 + 2 + 2 + 2 + 1 + 2 + 1 i i1 ! : Note that this time scenario two, which uses the random rule, has advantage over scenario General Case, Idle Time Ordering Policies 52 three. Since 1 + 1 + 2 + 2 2 + 1 + 2 + 1 ; one can easily show that P (12vsRR) P (RRvs 21). In order to complete our proof we only need to show that N a , the number of times that the random rule has advantage over (2; 1), is stochastically no greater than N d , the number of times it has disadvantage over (1; 2). Note that before an advantage period begins, both scenarios have no busy servers. Suppose that we are using our coupling argument for all three scenarios at the same time. That is, both scenarios one and three are coupled with scenario two which uses the random rule. Any time that both servers are idle in scenario three, which uses (2; 1), they are also idle in the scenarios one and two. Consequently, upon the next arrival, the random rule is equally likely to either have advantage over (2; 1) or disadvantage over (1; 2). Note that at only these moments the random rule has a chance to have advantage over (2; 1). This suces to show that N a is stochastically no greater than N d and estab- lishes the result. 5.5 General Model Analytical comparison of the behavior of the General model under all four dierent as- signment policies considered in this chapter is computationally intractable even forn = 2. In Table 5.4 and Table 5.5, we have numerical examples for the cases n = 2 and n = 3, respectively. In both tables = 1. Results in these tables indicate that there is no dominance by any of the rules in terms of the rate of lost customers. As shown in these examples for each of the longest, the shortest, and the random priority ordering rules there exists an example where that rule results in the lowest rate of lost customers among all of the four considered rules. As we mentioned previously, intuitively, you want to assign customers to idle eligible server with lower p i . Using longest idle rule it is more likely that servers with low p i ap- pear earlier on the ordered list of idle customers. Therefore, when the dierence between p i 's are large and service rates are not that far apart, one would expect that the longest idle rule outperforms the rest of the rules. It seems that our numerical results support this idea. General Case, Idle Time Ordering Policies 53 Tab. 5.4: Comparison between LIR, RR, SIR, and RPO i ( 1 ; 2 ;p 1 ;p 2 ) LIR RR SIR RPO Lowest Rate 1 (1; 2; 0:5; 0:25) 0:50290 0:50593 0:50950 0:50581 LIRRPORRSIR 2 (2; 1; 0:25; 0:5) 0:46526 0:46857 0:46925 0:46860 LIRRRRPOSIR 3 (2:5; 2; 0:5; 0:45) 0:35253 0:35264 0:35261 0:35263 LIRSIRRPORR 4 (3; 2; 0:5; 0:45) 0:34592 0:34600 0:34586 0:34598 SIRLIRRPORR 5 (3:4; 2; 0:5; 0:49) 0:32575 0:32575 0:32568 0:32569 SIRRPORRLIR 6 (2; 1; 0:5; 0:45) 0:39306 0:39327 0:39289 0:39313 SIRLIRRPORR 7 (0:05; 2; 0:5; 0:45) 0:59686 0:59681 0:59729 0:59574 RPORRLIRSIR 8 (0:05; 2; 0:5; 0:51) 0:55887 0:55890 0:55880 0:55758 RPOSIRLIRRR 9 (6; 2; 0:5; 0:49) 0:31247 0:31245 0:31233 0:31225 RPOSIRRRLIR 10 (0:85; 1; 0:5; 0:49) 0:434879 0:434880 0:434900 0:434864 RPOLIRRRSIR 11 (1:2; 1; 0:5; 0:49) 0:410074 0:410089 0:410066 0:410075 SIRLIRRPORR 12 (3; 3:05; 0:5; 0:01) 0:51086 0:513029 0:51488 0:513030 LIRRRRPOSIR 13 (0:1; 2; 0:5; 0:1) 0:82384 0:82610 0:82845 0:82600 LIRRPORRSIR 14 (0:1; 2; 0:5; 0:45) 0:56815 0:56809 0:56885 0:56660 RPORRLIRSIR 15 (0:1; 2; 0:5; 0:7) 0:42166 0:42556 0:42230 0:42280 LIRSIRRPORR For each vector ( 1 ; 2 ;p 1 ;p 2 ) we have calculated the rate of lost customers for the longest idle rule (LIR), the random rule (RR), the shortest idle rule (SIR), and the random priority ordering policy (RPO). Tab. 5.5: Comparison between LIR, RR, SIR, and RPO i ( 1 ; 2 ; 3 ;p 1 ;p 2 ;p 3 ) LIR RR SIR RPO Lowest Rate 1 (0:1; 0:5; 3; 0:1; 0:2; 0:3) 0:58384 0:58219 0:58005 0:58192 SIRRPORRLIR 2 (1:2; 2; 3; 0:05; 0:5; 0:95) 0:12357 0:13000 0:14180 0:13032 LIRRRRPOSIR 3 (1:5; 2; 3; 0:2; 0:5; 0:8) 0:14926 0:15336 0:16013 0:15344 LIRRRRPOSIR 4 (2; 2; 3; 0:45; 0:5; 0:55) 0:17538 0:17535 0:17528 0:17533 SIRRPORRLIR 5 (:05; 2; 9; 0:49; 0:5; 0:51) 0:27550 0:27528 0:27484 0:27163 RPOSIRRRLIR 6 (9; 2;:05; 0:49; 0:5; 0:51) 0:28542 0:28564 0:28608 0:28205 RPOLIRRRSIR 7 (3:2; 2; 3; 0:1; 0:3; 0:95) 0:15109 0:16203 0:17831 0:16216 LIRRRRPOSIR 8 (0:2; 0:5; 1; 0:1; 0:6;:95) 0:29383 0:29819 0:30381 0:29847 LIRRRRPOSIR 9 (1; 2; 3; 0:18; 0:2; 0:22) 0:54508 0:54496 0:54481 0:54494 SIRRPORRLIR 10 (3; 2; 3; 0:1; 0:24; 0:95) 0:16483 0:17627 0:1923 0:17636 LIRRRRPoSIR For each vector ( 1 ; 2 ; 3 ;p 1 ;p 2 ;p 3 ) we have calculated the rate of lost customers for the longest idle rule (LIR), the random rule (RR), the shortest idle rule (SIR), and the random priority ordering policy (RPO). 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Abstract (if available)

Abstract My PhD thesis studies a queuing loss model having Poisson arrivals and n heterogeneous skill based servers with arbitrary service distributions. Arrivals are discriminating in the sense that each arrival has a binary vector indicating which of the servers are eligible to serve it. Arrivals can only be assigned to servers that are both idle and eligible, with those not finding an idle eligible server being lost. It is of interest to study the behavior of these models under a class of no memory policies which we refer to as ""idle time ordering policies"". These policies are considered as fair policies in the sense that there is no special priority given to any of the servers. Letting the ""idle servers vector"" be i₁,…, ik if there are currently k idle servers, with i₁ having been idle the longest, i₂ the second longest, and so on, an idle time ordering policy is one whose assignment decisions are based solely on the number of idle servers and the positions in the idle servers vector of those servers that are eligible to be assigned. Examples of such policies would be to assign to a randomly chosen idle and eligible server, or to assign to idle eligible server that has been idle the longest, or that has been idle the shortest. Call centers provide a strong motivating example of a service organization that cares about the issue of server fairness. In particular, many call centers follow the longest idle policy. We are interested in measuring the performance of policies within this class with respect to their rate of lost customers. ❧ Our work resulted in two published papers, one of which assumes that the eligibility vectors are exchangeable and its objective is to find the idle time ordering policy which minimizes the total rate of lost customers. Utilizing the method of stages enables us to analyze this model as a continuous time Markov chain. Then using a conjecture concerning the reverse chain helps us find, up to a multiplicative constant, the limiting probabilities for this model. Surprisingly these limiting probabilities are the same no matter which idle time ordering policy is employed. It is shown that limiting probabilities of the set of idle servers depend on the service distributions only through their means. Moreover, it is shown, conditional on the set of idle servers, that the remaining service times (as well as provided service times) of the busy servers are independent and have their respective equilibrium service distributions. ❧ The second paper relaxes the exchangeable eligibility assumption and studies the behavior of the model under the longest idle assignment policy. The preceding model was introduced by Adan-Weiss. Using a supplementary variable approach to make their model Markovian, Adan and Weiss derived the limiting distribution for this model, and in doing so showed that the limiting distribution of the ordered list of idle servers depend on the service distributions only through their means. In this paper we obtain their results using the method of stages. Again using a conjecture concerning the reverse chain the limiting probabilities of the ordered list of idle servers are found up to a multiplicative constant. Given the set of busy servers, the remaining service times are independent with their respective equilibrium distributions (This result is only implicitly noted by Adan and Weiss). In practice, finding the limiting probabilities or any other quantity of interest becomes computationally intractable when n is large. Hence, for each of the papers using Markov chain Monte Carlo methods we provide an algorithm to simulate a Markov chain whose limiting probabilities enables us to obtain desired quantities for our model. ❧ Finally, we consider a related model which assumes that the elements of the eligibility vectors are independent and servers have known exponential service distributions. Our goal is to determine the best policy, or when it is not computationally feasible, to determine good heuristic policies for making the assignments. Preliminary results show that finding the best policy analytically is computationally intensive. Hence, we first show how some of these policies can be eliminated from our search for the best policy which significantly reduces the complexity of our problem and allows us to find the best policy when n = 3. Later, we develop two different heuristic approaches for finding an optimal or close to optimal assignment policy. In the first heuristic, for each pair of servers we consider a system consisting of those 2 servers and find the best ordering among them. Let i <₂ j show that among servers i and j, the best ordering is (i, j). If the pairwise orderings are transitive, i <₂ j and j <₂ k imply i <₂ k, we use the ordering for our original problem that satisfies these pairwise transitivity relations, and if transitivity does not hold among all of the pairs, we use the ordering that in some sense violates the transitivity relations less frequently. The second heuristic sorts the servers in the decreasing order of the ratio μᵢ/pᵢ. For n = 3 we are able to compare these heuristic with the best stationary policy. For n > 3, we use simulation to find an estimate of the rate of the lost customers for our heuristics. In this case, we compare our heuristics with the random rule and we also define some lower bounds to examine their performance.

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Asset Metadata

Creator Haji, Babak (author)

Core Title Queueing loss system with heterogeneous servers and discriminating arrivals

School Viterbi School of Engineering

Degree Doctor of Philosophy

Degree Program Industrial and Systems Engineering

Publication Date 05/04/2015

Defense Date 03/19/2015

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag equilibrium distribution,Gibbs sampler,heterogeneous servers,limiting probabilities,method of stages,no-memory policies,OAI-PMH Harvest,queuing loss system,reverse chain

Format application/pdf (imt)

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Ross, Sheldon M. (committee chair), Dessouky, Maged M. (committee member), Ward, Amy R. (committee member)

Creator Email babak.61782@gmail.com,bhaji@usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-c3-567125

Unique identifier UC11299638

Identifier etd-HajiBabak-3433.pdf (filename),usctheses-c3-567125 (legacy record id)

Legacy Identifier etd-HajiBabak-3433.pdf

Dmrecord 567125

Document Type Dissertation

Format application/pdf (imt)

Rights Haji, Babak

Type texts

Source University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Access Conditions The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the a...

Repository Name University of Southern California Digital Library

Repository Location USC Digital Library, University of Southern California, University Park Campus MC 2810, 3434 South Grand Avenue, 2nd Floor, Los Angeles, California 90089-2810, USA