2905 Queueing Theory and Simulation PART III: HIGHER DIMENSIONAL AND NON-MARKOVIAN QUEUES

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1 295 Queueing Theory and Simulation PART III: HIGHER DIMENSIONAL AND NON-MARKOVIAN QUEUES 16 Queueing Systems with Two Types of Customers In this section, we discuss queueing systems with two types of customers. The queueing system has no waiting space. There are two possible cases: infinite-server case and finite-server case Infinite-Server Queue Consider the infinite-server queue with two types of customers. The arrival process of customers of type i (i = 1, 2) is Poisson with rate λ i and their service times are independent, identically distributed, exponential random variables with mean µ 1 i (i = 1, 2). We define the 2-dimensional states {E j1,j 2 }, where j i is the number of customers of type i in the system, with corresponding equilibrium distribution {p(j 1, j 2 )}, then clearly the Markov property still holds. Here p(j 1, j 2 ) is the steady state probability that there are j 1 type 1 customers and j 2 type 2 customers in the system. 113

2 By equating expected rate out to expected rate in for each state, the equilibrium state equations are (λ 1 + λ 2 + j 1 µ 1 + j 2 µ 2 )p(j 1, j 2 ) = λ 1 p(j 1 1, j 2 ) + (j 1 + 1)µ 1 p(j 1 + 1, j 2 ) +λ 2 p(j 1, j 2 1) + (j 2 + 1)µ 2 p(j 1, j 2 + 1) (1) and [p( 1, j 2 ) = p(j 1, 1) = ; j 1 =, 1,... ; j 2 =, 1,....] E j1 1,j 2 λ 1 j 1 µ 1 λ 2 E j1,j 2 +1 (j 2 + 1)µ 2 E j1,j 2 E j1 +1,j 2 (j 1 + 1)µ 1 λ 2 j 2 µ 2 E j1,j 2 1 λ 1 Figure 16.1 The Markov Chain of the System at State E j1,j 2. In addition, the probabilities must satisfy the normalization equation: p(j 1, j 2 ) = 1. j 1 = j 2 = 114

3 In this case, we already know the answer. Since the number of servers is infinite, the two types of customers do not affect one another. Thus the marginal distribution of the number of customers of each type is that which would be obtained by solving the corresponding one-dimensional problem, namely the Poisson distribution: p 1 (j) = p 2 (j) = p(j, k) = (λ 1/µ 1 ) j e (λ 1/µ 1 ), j! p(k, j) = (λ 2/µ 2 ) j e (λ 2/µ 2 ). j! k= k= (2) Since the number of customers present of each type is independent of the number present of the other type, therefore p(j 1, j 2 ) = p 1 (j 1 )p 2 (j 2 ) = (λ 1/µ 1 ) j 1 (λ 2 /µ 2 ) j 2 e [(λ 1/µ 1 )+(λ 2 /µ 2 )]. j 1! j 2! (3) 115

4 Solution of Product Form The fact that the solution p(j 1, j 2 ) can be decomposed into a product of two factors has enabled us to solve the problem with ease. In fact one may try p(j 1, j 2 ) = (λ 1/µ 1 ) j 1 j 1! (λ 2 /µ 2 ) j 2 C (4) j 2! where the constant C is determined from the normalization equation. In this case (λ 1 /µ 1 ) j 1 (λ 2 /µ 2 ) j 2 p(j 1, j 2 ) = C = Ce λ 2/µ 2 (λ 1 /µ 1 ) j 1 = Ce [(λ 1/µ 1 )+(λ 2 /µ 2 )]. j 1! j 2! j 1! j 1 = j 2 = j 1 = j 2 = Hence we have the constant C = e [(λ 1/µ 1 )+(λ 2 /µ 2 )]. In practice, a good strategy for finding solutions to equations of the form (1) is to assume a product solution of the form (4); and see if such a solution satisfies Eq. (1). If it works, then the solution has been obtained. If it doesn t, then try a different approach. In this case it works! j 1 = 116

5 16.2 Multiple-server Queue with Blocked Customers Cleared The situation is similar to that of the previous example except now the system has finitely many servers. The system is again described by Eq. (1), which is valid now only for j 1 + j 2 < s. When j 1 + j 2 = s, then the states E j1 +1,j 2 and E j1,j 2 +1 cannot occur and the equation becomes (j 1 µ 1 + j 2 µ 2 )p(j 1, j 2 ) = λ 1 p(j 1 1, j 2 ) + λ 2 p(j 2, j 2 1). (5) Observe that Eq. (5) can be obtained from Eq. (1) by deleting the first two terms on the left and the last two terms on the right. The product-form solution (4) satisfies Eq. (1) and also the equation with only the deleted terms (λ 1 + λ 2 )p(j 1, j 2 ) = (j 1 + 1)µ 1 p(j 1 + 1, j 2 ) + (j 2 + 1)µ 2 p(j 1, j 2 + 1), therefore it also satisfies Eq. (5). Thus the product solution in Eq. (4) is a solution of this problem. In particular, if we don t distinguish the two types of customers, then the probability p(j) that there are j customers (including type 1 and type 2) in service is given by p(j) = j p(j 1, j 2 ) = p(j 1, j j 1 ). j 1 +j 2 =j j 1 = 117

6 With the help of binomial theorem, we have p(j) = C j j 1 = The normalization equation implies that We conclude that ( λ 1 µ 1 ) j 1 ( λ 2 µ 2 ) j j 1 j 1! (j j 1 )! = C j! p(j) = C = aj /j! s a k /k! k= j j 1 = { s k= j! j 1!(j j 1 )! (λ 1 ) j 1 ( λ 2 ) j j 1 = C 1 ( λ1 + λ ) 2 j. µ 1 µ 2 j! µ 1 µ 2 s p(k) = 1 k= ( 1 λ1 + λ ) } k 1 2. k! µ 1 µ 2 (j =, 1,..., s) where a = λ 1 µ 1 + λ 2 µ 2. We note that for the case of infinite-server queue (we let s ) we have p(j) = aj e a j! (j =, 1,..., ). 118

7 17 Queues in Tandem Consider two sets of servers arranged in tandem, so that the output from the first set of servers is the input of the second set of servers. Assume that the arrival process at the first stage of this tandem queueing system is Poisson with rate λ, the service times in the first stage are exponentially distributed with mean µ 1 1, and the queueing discipline is blocked customers delayed. The customers completing service in the first stage will enter the second stage (and wait if all servers in second stage are busy), where the service times are assumed to be exponentially distributed with mean µ 1 2. The number of servers in stage i is s i(i = 1, 2)... Second Set of Servers First Set of Servers Figure 17.1 Two Queues in Tandem. 119

8 Let p(j 1, j 2 ) be the probability that there are j 1 customers in stage 1 and j 2 customers in stage 2. Let { jµi (j =, 1,..., s µ i (j) = i ) s i µ i (j = s i + 1, s i + 2,...) be the departure rates. Then equating rate in to rate out for each state, we obtain (λ + µ 1 (j 1 ) + µ 2 (j 2 )) p(j 1, j 2 ) = λp(j 1 1, j 2 ) + µ 1 (j 1 + 1)p(j 1 + 1, j 2 1) +µ 2 (j 2 + 1)p(j 1, j 2 + 1) [p( 1, j 2 ) = p(j 1, 1) = ; j 1 =, 1,... ; j 2 =, 1,...] Since the first stage of this system is precisely an Erlang delay system, so that the marginal distribution of the number of customers in stage one is given by the Erlang delay probabilities. Let us try (hopefully it would work) a product solution of the form p(j 1, j 2 ) = p 1 (j 1 )p 2 (j 2 ) (7) (6) with the factor p 1 (j 1 ) given by the Erlang delay probabilities: (λ/µ 1 ) j 1 C 1 (j 1 =, 1,..., s 1 1), j p 1 (j 1 ) = 1! (λ/µ 1 ) j 1 C 1 s 1!s j 1 s 1 (j 1 = s 1, s 1 + 1,...). 1 (8) 12

9 We shall substitute the assumed product solutions (7) and (8) into the equilibrium state equations (6), with the hope that the system of equations will be reduced to a one-dimensional set of equations that can be solved for the remaining factor p 2 (j 2 ). Indeed, (6) is reduced to [λ + µ 2 (j 2 )]p 2 (j 2 ) = λp 2 (j 2 1) + µ 2 (j 2 + 1)p 2 (j 2 + 1) (9) [p 2 ( 1) = ; j 2 =, 1,...]. Eq. (9) is the equilibrium state equations that define the Erlang delay probabilities. We conclude that p 2 (j 2 ) is given by p 2 (j 2 ) = Using p i (j) = 1 (i = 1, 2), we get j= C 2 (λ/µ 2 ) j 2 j 2! C 2 (λ/µ 2 ) j 2 s 2!s j 2 s 2 2 (j 2 =, 1, 2,..., s 2 1), (j 2 = s 2, s 2 + 1,..., ). (1) C i = ( s i 1 k= ρ k i k! + ρ si ) 1 i where ρ i = λ (i = 1, 2). s i!(1 ρ i /s i ) µ i This equation implies that a proper joint distribution exists only when λ/µ 1 < s 1 and λ/µ 2 < s

10 It is a remarkable result that the number of customers in the second stage also follows the Erlang delay distribution, that is, the distribution of customers in the second stage is the same as if the first stage were not present and the customers arrived according to a Poisson process directly at the second stage. It also suggests that the output from the first stage is a Poisson process. This is in fact true in general and we state without proof the Burke s theorem as follows. Proposition 1 (Burke s Theorem) The statistical equilibrium output of an M/M/s queue with arrival rate λ and mean service times µ 1 is a Poisson process with the same rate λ. Remark: In 1956 Burke showed that the departure process of an M/M/s queue was Poisson. [P. Burke, (1956) The Output of a Queueing System, Oper. Res. (4) ] In the following, we are going to prove the Burke s Theorem. We consider an M/M/s queue in steady state. Let λ be the arrival rate of customers and µ be the service rate and q i (i =, 1, 2,...) be the steady state probability that there are j customers in the system. We recall that q i can be obtained by using the Erlang s delay formula. 122

11 Let τ be the inter-departure interval and let N(t) be the state of the system at time t after a departure (so that there have been no further departure by time t) and we define We note that S n (t) = P (N(t) = n and τ > t). S (t + h) = P (no arrival of customer in (t, t + h)) S (t) = (1 λh + o(h))s (t); S n (t + h) = P (an arrival of customer in (t, t + h)) S n 1 (t) +P (no arrival of customer in (t, t + h)) S n (t) = (λh + o(h))(1 nµh + o(h))s n 1 (t) + (1 λh + o(h))(1 nµh + o(h))s n (t), n s; S n (t + h) = P (an arrival of customer in (t, t + h)) S n 1 (t) +P (no arrival of customer in (t, t + h)) S n (t) = (λh + o(h))(1 sµh + o(h))s n 1 (t) + (1 λh + o(h))(1 sµh + o(h))s n (t), n > s. By re-arranging the terms, we have the following equations: S (t + h) S (t) = λs (t) + o(h) h h ; S n (t + h) S n (t) h S n (t + h) S n (t) h = λs n 1 (t) (λ + nµ)s n (t) + o(h) h, n s; = λs n 1 (t) (λ + sµ)s n (t) + o(h) h, n > s. 123

12 By letting h one can obtain the following differential equations. We note that ds (t) = λs (t); dt ds n (t) = λs n 1 (t) (λ + nµ)s n (t) n s; dt ds n (t) = λs n 1 (t) (λ + sµ)s n (t) n > s. dt S n () = q n for n =, 1,... This is because we assume that the system is in steady state. (11) Moreover, we have q n = λ min{n, s}µ q n 1. By solving the differential equations inductively one may obtain S n (t) = q n e λt. 124

13 Alternative one may compute We then have for n s For n > s we have We then have S n(t) = q n λe λt. λs n 1 (t) (λ + nµ)s n (t) = λq n 1 e λt (λ + nµ)q n e λt = (λq n 1 (λ + nµ)q n )e λt = (λq n 1 nµq n )e λt λq n e λt = + S n(t). λs n 1 (t) (λ + sµ)s n (t) = λq n 1 e λt (λ + sµ)q n e λt P (τ > t) = = (λq n 1 (λ + sµ)q n )e λt = (λq n 1 sµq n )e λt λq n e λt = + S n(t). S n (t) = e λt n= n= q n = e λt. This implies that the p.d.f. of τ is given by λe λt, i.e. the inter-departure time is exponentially distributed with rate λ. Finally this implies that the departure process of an M/M/s queue in steady state is a Poisson process with the same rate of the input process. 125

14 18 Queues in Parallel.. λ 1 λ 2 λ 1 λ 2 Case 1: No Inter-action Figure 18.1 Two Queues in Parallel. Case 2: With Overflow Case (i) We assume there is no inter-action between the two queues. There are s i (i = 1, 2) servers in queue i(i = 1, 2) and there is infinite many waiting spaces in each queue and we assume blocked customers are delayed. The arrival rate of queue i is λ i and the service completion rate of a server in queue i is µ i. 126

15 Hence we have the steady state probability that queue i has j customers given by ρ j i j! p i (j) = C i (j =, 1,, s i ) k= ρ j i s i!s j s i i where ρ i = λ i /µ i. Moreover if ρ i < s i then s i 1 C i = ρ k i k! + ρk i s i!s k s i i k=s i C i (j = s i + 1,...) 1 = ( si 1 k= ) 1 ρ k i k! + ρ si i. (s i 1)!(s i ρ i ) If ρ i s i, the infinite geometric sum diverges then C i = and hence p i (j) = for all finite j. If ρ i < s i then the steady state probability that there are i customers in Queue 1 and j customers in Queue 2 is given by p(i, j) = p 1 (i)p 2 (j) i, j =, 1, 2,.... Case (ii) There are s i (i = 1, 2) servers in queue i(i = 1, 2) and there are finite many waiting spaces in each queue and we assume blocked customers are cleared. The arrival rate of queue i is λ i and the service completion rate of a server in queue i is µ i. We assume there is inter-action between the two queues as follows. Whenever Queue 2(1) is full, an arrived customer of type 2(1) will overflow to Queue 1(2) provided that Queue 1(2) is not yet full. Let us consider a simple example as follows. We assume that Queue 1 and 2 are M/M/1/1 queue. The following figure gives the Markov chain of the queueing system. 127

16 λ = λ 1 + λ 2 λ = λ 1 + λ 2 E,2 E 1,2 E µ 2,2 1 µ 1 µ 2 λ 2 λ 1 µ 2 λ 2 λ 1 µ 2 λ = λ 1 + λ 2 E,1 E µ 1 1,1 µ 1 E 2,1 µ 2 λ 2 λ µ 2 λ 2 µ 2 λ = λ1 1 λ 1 + λ 2 E, µ 1 E 1, µ 1 E 2, Figure 17.2 The Markov Chain of the Two Queue Overflow System. The generator matrix for this queueing problem is given by A 8 = (, ) (1, ) (2, ) (, 1) (1, 1) (2, 1) (, 2) (1, 2) (2, 2) * µ 1 µ 2 λ 1 * µ 1 µ 2 λ 1 * µ 2 λ 2 * µ 1 µ 2 λ 2 λ 1 * µ 1 µ 2 λ λ1 µ 2 λ 2 * µ 1 λ 2 λ * µ1 λ λ *. (12) Here is such that the column sum is zero. Unfortunately there is no analytic solution for the steady state probabilities of this system. Direct method or numerical method are common methods for solving the steady state probabilities. 128

17 19 The M/G/1 Queue: Imbedded Markov Chain Technique In the preceding sections we will concentrate mainly on queues with Poisson input and exponential service times. These assumptions imply that the future evolution of the system from some time t depends only on the state of the system at time t, and is independent of the history of the system prior to time t (i.e. a Markov process). In fact, the systems that we have considered are birth-and-death processes, which are special cases of Markov processes. In this section, we shall consider models which do not constitute Markov processes. One of the most powerful methods for the analysis of certain queueing models is the Imbedded Markov Chain technique (IMC). We shall concentrate on queueing models, to which the IMC. analysis is applicable. We shall assume, unless specified otherwise, that all queues are BCD and customers are served in order of arrival. 129

18 19.1 Imbedded Markov Chain Consider a general queueing system. Let N(t) be the number of customers in the system at time t. Then, by the law of total probability, P {N(t + x) = j} = P {N(t + x) = j N(t) = i}p {N(t) = i} (t, x ; j =, 1,...). i= (13) To obtain the state probabilities, all that is required is to evaluate the conditional probabilities P {N(t + x) = j N(t) = i} called the transition probabilities, and then solve Eq. (13). For a birth-and-death process, the transition probabilities can be easily obtained by choosing h small and letting h ; λ i h + o(h) j = i + 1 µ P {N(t + h) = j N(t) = i} = i h + o(h) j = i 1 (14) 1 (λ i + µ i )h + o(h) j = i o(h) otherwise. With the transition probabilities Eq. (14), the state probabilities can then be derived: p j (t + h) = λ j 1 hp j 1 (t) + µ j+1 hp j+1 (t) + [1 (λ j + µ j )h]p j (t) + o(h). 13

19 But for a M/G/1 queue (and many other queueing models) the transition probabilities depend not only on the state of the system at time t but also on the past history of the system (such as how long a server has been serving a particular customer); that is, the system does not have the Markov property at arbitrary time t. However, there are time epochs at which the Markov property holds. Such a point is called a renewal point (In a Markov process, every point t is a renewal point). If we focus on a discrete set of renewal points then the states of the system can be considered as a Markov chain, this is called a Imbedded Markov Chain (IMC). With regard to the M/G/1 queue, the departure epochs are renewal points. For whenever a customer leaves the system, either the system becomes empty or a previously waiting customer starts service, in either case the transition probabilities (and therefore the future evolution of the system) depend only on the number of customers in the system at the departure epoch. We proceed to study the M/G/1 queue by using the IMC defined on the set of departure epochs. Let Nk be the number of customers in the system at the instant the kth customer completes service and departs. Then P {Nk+1 = j} = P {Nk+1 = j Nk = i}p {Nk = i} (j =, 1,... ; k = 1, 2,...). (15) i= 131

20 Observe that P {Nk+1 = j N k = } = P {j customers arrive during the service time of the (k + 1)th customer} (λt) j = e λt f(t) dt (j ) } j! }{{}{{} p.d.f. of service time Poisson arrival where f(t) is the service time density function. Similarly for i > and j i 1 and of course, P {N k+1 = j N k = i} = P {(j i + 1) customers arrive during the service time of the (k + 1)th customer} (λt) j i+1 = (j i + 1)! e λt f(t)dt P {N k+1 = j N k = i} = if i >, j < i 1. Let us denote the right-hand-side of Eq. (16) by p j. Then the transition probabilities are P {N k+1 = j N k = i} = p j if i = p j i+1 if i >, and j i 1 if i >, and j < i 1. Note that the transition probabilities are independent of the value of the index k. (16) (17) (18) 132

21 If we introduce the notations a (k) j = P {Nk = j}, a(k) = (a (k), a(k) 1, a(k) 2,...)T p ij = P {Nk+1 = i N k = j}, and P = (p ij ) T (P is called the transition matrix) where A T denotes the transpose of the matrix A. Then where P is a transition probability matrix given by p p p 1 p 1 p P = p 2 p 2 p 1 p. p 3 p 3 p 2 p 1 p..... a (k+1) = P a (k) (19) It can be shown by using the theory of Markov chains that a unique probability distribution π j = lim k a (k) j (j =, 1, 2,...) (2) independent of the initial conditions, exists if and only if ρ < 1. (If ρ 1 then π j = for all finite j). 133

22 Furthermore, {πj } is a stationary distribution, i.e. if then π = (π, π 1,..., ) T π = P π. (21) The proofs of (2) and (21) are the only place where the formal theory of Markov chains is used. We shall accept without proof that the existence of the limiting distribution (2). Eq. (21) now becomes The normalization equation is π j = p j π + j+1 i=1 p j i+1 π i (j =, 1,...). (22) πj = 1. (23) j= We want to solve πi given that p i is known. Now let us define the probability generating function g(z) = πj z j. (24) j= 134

23 Substituting Eq. (22) into Eq. (24) we have g(z) = j= ( p j π + j+1 i=1 ) p j i+1 πi z j = π } {{ } πj p j z j + j= j= j+1 i=1 p j i+1 π i z j. (25) If we also define the probability generating function of p i as h(z) = p j z j. (26) j= We obtain after changing the order of summation, g(z) = h(z)π + p j (i 1) z j πi = h(z)π + = h(z)π + i=1 i=1 j=i 1 p k z k+(i 1) πi k= h(z)πi z i 1 i=1 = h(z)π + 1 z h(z) (g(z) π ). 135

24 Re-arranging the terms we get g(z) = To determine π we have to make use of the result that g(1) = πj = 1. From Eq. (27), we have 1 = g(1) = lim z 1 h(z)(z 1) z h(z) π h(z)(z 1) z h(z) π. (27) j= h (z)(z 1) + h(z) = lim z 1 1 h (z) 1 = 1 h (1) π (h(1) = 1). What is h (1)? We shall investigate this. (/ form) π (L Hospital Rule) 136

25 Hence We remark that π = 1 h (1) = 1 jp j = 1 = 1 = 1 = 1 = 1 j=1 ( (λt) j ) j e λt f(t)dt j! ( (λt) j ) (j 1)! e λt f(t)dt (λt) j e λt f(t)dt (j 1)! j=1 j=1 j=1 ( λte λt ) e λt f(t)dt λtf(t)dt = 1 λ µ (µ 1 is the expected service time) = 1 ρ. π = 1 ρ holds for any service probability distribution and < ρ <

26 2 Laplace-Stieltjes Transform Let us now introduce the method of Laplace-Stieltjes transform and show that the generating function h(z) (and hence g(z)) can be expressed in terms of the Laplace-Stieltjes transform of the service time distribution function. Definition: A Stieltjes integral of a function g(x) with respect to a function h(x) from a to b is b n g(x)dh(x) = g(ξ k )[h(x k ) h(x k 1 )] where and Note: a lim k=1 x k 1 ξ k x k (k = 1, 2,..., n), = max(x 1 x, x 2 x 1,..., x n x n 1 ). (1) If h(x) is continuously differentiable on [a, b], then b a g(x)dh(x) = b where the RHS is the ordinary Riemann integral. a g(x)h (x)dx, 138

27 (2) If then h(x) = b a { 1 for x c for x < c g(x)dh(x) = g(c). (28) 2.1 Definition and basic properties of Laplace-Stieltjes transform Definition: The ordinary Laplace transform of L(f(t)) of a distribution function, f(t), for a continuous nonnegative random variable is given by L[f(t)] = e st f(t)dt (s ). The Laplace-Stieltjes transform L [F (t)] of a distribution function, F (t), for a continuous nonnegative random variable is given by L [F (t)] = ϕ(s) = e st df (t) (s ). For a discrete probability distribution {f n }, the Laplace-Stieltjes transform L [{f n }] = ϕ(s) = e ns f n. n= 139

28 Some Basic properties: (1) If g(z) is the generating function of {f n }, then L [{f n }] = f n (e s ) n = g(e s ). n= (2) If F (t) is differentiable, F (t) = f(t), then ϕ(s) is the ordinary Laplace transform L[f(t)] of f(t): (3) L [F (t)] = ( 1) n dn ds nϕ(s) s= e st df (t) = e st f(t)dt = L[f(t)]. = ( 1) n d n s= df (t) ds ne st = ( 1) n ( 1) n t n df (t) = t n df (t). 14

29 (4) For the Laplace transformation, it is well known that if then ϕ 1 (s)ϕ 2 (s) = (Let t = x + y) = ϕ 1 (s) = L(f 1 (t)) and ϕ 2 (s) = L(f 2 (t)) e sy f 1 (y)dy e sx f 2 (x)dx = e s(x+y) f 1 (y)f 2 (x)dxdy t ( t ) e st f 2 (x)f 1 (t x)dxdt = L f 2 (x)f 1 (t x)dx. If F 1 (t) and F 2 (t) have Laplace-Stieltjes transforms ϕ 1 (s) and ϕ 2 (s) respectively then the Laplace-Stieltjes transforms of ϕ(s) of the convolution: is the product ϕ(s) = F (t) = t e st ( t F 1 (t x)df 2 (x) = t ) f 2 (x)f 1 (t x)dx dt = F 1 (t x)f 2 (x)dx e st df (t) = ϕ 1 (s)ϕ 2 (s). Given X 1 and X 2 are independent nonnegative random variables with distribution functions F 1 (t) and F 2 (t) and Laplace-Stieltjes transform ϕ 1 (s) and ϕ 2 (s). Then the distribution function of the sum (X 1 + X 2 ) has Laplace-Stieltjes transform equal to the product of the component transforms. 141

30 2.2 Applications We consider and p j = (λt) j e λt df (t) (j =, 1,...) j! h(z) = p j z j. j= Then h(z) = = = = ( (λt) j j= ) e λt df (t) z j j! (λtz) j e λt df (t) j! j= e λzt e λt df (t) e λ(1 z)t df (t). The last integral is the Laplace-Stieltjes transform with argument λ(1 z) of the distribution function F (t): h(z) = ϕ(λ λz). (29) 142

31 Let us now come back to the generating function g(z) in Eq. (27). By Eq. (29) g(z) = Since π = (1 ρ), we have the final result (z 1)ϕ(λ λz) z ϕ(λ λz) π. g(z) = (z 1)ϕ(λ λz) (1 ρ) for (ρ < 1, z 1). (3) z ϕ(λ λz) Remark: From Eq. (3), we can, in principle, find the mean number of customers in the system E(N ) = g (1). and π1 = g (),..., πk = g(k) (). k! 143

32 21 M/G/1 Queue We now turn to the waiting time (in the queue) distribution W (t) = P {W t} for the M/G/1 queue with service in order of arrival. Let W (t) have Laplace-Stieltjes transform ω(s), ω(s) = e st dw (t), (31) and denote by η(s) the Laplace-Stieltjes transform of the distribution function of the sojourn time S(t) (the sum of the waiting time and the service time). Since the waiting time and service time of a customer are independent random variables, we have η(s) = ω(s)ϕ(s). (32) Since service is in arrival order, the customers left behind by an arbitrary customer must all have arrived during his sojourn time. Therefore the probability generating function g(z) of the number of customers left behind by an arbitrary customer is given by ( (λt) j ) g(z) = e λt ds(t) z j = η(λ λz). (33) j! j= 144

33 Using Eqs. (3), (32) and (33) we have g(z) = η(λ λz) = ω(λ λz)ϕ(λ λz) and g(z) = Putting s = λ λz we have (z 1)ϕ(λ λz) (1 ρ). z ϕ(λ λz) ω(s) = s λ (1 ρ) ( λ s λ ) ϕ(s) = s(1 ρ) s λ[1 ϕ(s)]. (34) Using the fact that E(W ) = ω () [Refer to basic property (3) of Section 2.1], it is a straightforward (an exercise) but tedious task (requiring two applications of L Hôpital s rule) to establish that ) ρτ E(W ) = (1 + σ2. (35) 2(1 ρ) τ 2 Here τ and σ 2 are respectively the mean and variance of the service time. By Little s formula ρ 2 ) E(Q) = (1 + σ2 (36) 2(1 ρ) τ 2 where Q is the mean number of customers waiting in the queue. Remark: When the service times are exponential, then τ 2 = σ 2 and (35) becomes E(W ) = ρτ 1 ρ, in agreement with formula for Erlang delay systems. 145

34 21.1 Special Case: M/D/1 Queue Assuming all service times are constant and equal to 1/µ. function is given by Then the service time distribution H(t) = { t < 1 µ 1 t 1 µ. We have [using (26)] Hence (3) becomes Thus we have h(z) = ϕ(λ λz) = g(z) = e λ(1 z)t dh(t) = e (1 z)ρ, (ρ = λ/µ). (1 z)(1 ρ). (37) 1 ze (1 z)ρ { π = g() = (1 ρ) π 1 = g () = (1 ρ)(e ρ 1) To find π n for n = 2, 3,... it is more convenient to employ the following method. Considering (37), we expand 1 (1 ze ρ(1 z) ) as a geometric series giving 146 (38)

35 g(z) = (1 z)(1 ρ) = (1 z)(1 ρ) n= (ze ρ(1 z) ) k for ze ρ(1 z) < 1 k= e kρ z k e kρz k= = (1 z)(1 ρ) e kρ ( kρz) m z k m! k= m= (39) = (1 z)(1 ρ) e kρ n k (kρ)n k ( 1) (n k)! zn (Let n = m + k) { k= n=k } n = (1 z)(1 ρ) e kρ n k (kρ)n k ( 1) (n k)! zn { n= k= } n = (1 ρ) e kρ n k (kρ)n k n 1 ( 1) (n k)! zn e kρ ( 1) n k 1 (kρ)n k 1 (n k 1)! zn. Thus for n 2 π n = (1 ρ) k= ( n k=1 n=1 n 1 e kρ n k (kρ)n k ( 1) (n k)! + k=1 k= (kρ)n k 1 e kρ ( 1) n k (n k 1)! ). (4) 147

36 A Summary on Higher Dimensional and Non-Markovian Queueing Systems Product form solutions of higher dimensional queueing systems. Queues in tandem and queue in parallel. The statement of Burke s theorem on the M/M/s queue. The Imbedded Markov chain technique. The M/G/1 Queue: h(z)(z 1)(1 ρ) g(z) = z h(z) where g(z) is the generating function of the steady state probabilities of the number of customers in the system and h(z) is the generating function of the transition probabilities. The definition and properties of the Laplace-Stieltjes transform. The expected queueing time in the M/G/1 queue: E(W ) = ρτ 2(1 ρ) ) (1 + σ2. τ 2 148