A Simple Solution for the M/D/c Waiting Time Distribution

Transcription

1 A Simple Solution for the M/D/c Waiting Time Distribution G.J.Franx, Universiteit van Amsterdam November 6, 998 Abstract A surprisingly simple and explicit expression for the waiting time distribution of the M/D/c queueing system is derived by a full probabilistic analysis, requiring neither generating functions nor Laplace transforms. Unlike the solutions known so far, this expression presents no numerical complications, not even for high traffic intensities. Finally, the result is proved explicitly to satisfy Erlang s integral equation for the M/D/c queue, which has been somewhat problematic for the expressions known so far. M/D/c waiting time distribution, numerically convenient explicit formula, full probabilistic analysis, Erlang s M/D/c integral equation. Introduction The M/D/c system is one of the classical queueing models that have already been studied since the beginning of this century. In this model we assume the customers to arrive according to a Poisson process with rate λ. There are c identical servers, serving each customer on a first come first serve basis during a constant time D. We assume the traffic intensity ρ λd/c <. Erlang [909] derived an explicit expression for the waiting time distribution F (y) of the M/D/ queue, by solving for c the following integral equation: F (y) 0 F (x + y D) λc x c (c )! e λx dx, y 0. () This equation is obtained by comparing the waiting time of an arbitrary customer A to the waiting time of the c th customer after A, who will arrive between x and x + dx time units after A with probability λc x c (c )! e λx dx. If this customer has to wait, he will be served exactly D time units after A. Since the M/D/c queue is equivalent to the E c /D/ queue with respect to the waiting time distribution, Erlang s equation can be interpreted as a special case of Lindley s integral

2 equation for the GI/G/ queueing system. Although his equation proved quite successful for the case c, Erlang [920] realized that for c > it would hardly lead to an explicit mathematical solution. By an ingenious probabilistic argument and the use of generating functions Crommelin [932] derived a general expression for the waiting time distribution of the M/D/c queue for all c N, which for c corresponds to Erlang s result. If P n denotes the stationary probability of the system containing no more than n customers, Crommelin s result reads: c P {W x} n0 P n m { λ(x md)} (k+)c n e λ(x md), md x < (m + )D (2) {(k + )c n}! Prabhu [962] proved that for c Erlang s integral equation yields a solution of type (2), where P n is replaced by some alternatively defined constant α n. However, it is unresolved how to interpret α n as the cumulative state probability. Apart from this, Crommelin s result is not really practical for numerical purposes, due to alternating terms which are in general much larger than their sum. As a way to get around the problem of round off errors, a recursion scheme based on Crommelin s argument is described in Tijms [994]. However, for increasing c and ρ this recursion scheme (too) will ultimately be hampered by round off errors, for which case an asymptotic expansion is recommended. This paper presents an alternative probabilistic approach, leading to a simple formula for the waiting time distribution, which is numerically stable for all ρ <. This formula (0 or ) is the main result of this paper, and is derived without the use of generating functions or Laplace transforms. Before doing so, some preliminary results about the number of customers in the system are introduced, and an essential lemma about the queue length at service initiations is proved. Obviously, the derived expression (0) must satisfy Erlang s integral equation (. In order to complete the circle, an explicit proof of this is given in section 5. 2 The number of customers present in the system Let p i (t) denote the probability of the system holding i customers at time t. Observe that all customers in service at time t will have left the system at time t + D. Consequently, customers present at time t + D either arrived during the time interval (t, t + D] or were already waiting for service at time t. Hence, by conditioning on the number of customers present at time t we find: p i (t + D) c p j (t) (λd)i e λd + i+c jc+ p j (t) (λd)i+c j (i + c j)! e λd, t R, i N 0. (3) 2

3 The stationary distribution p i lim t p i (t) is found by letting t in (3): p i c (λd) i p j e λd + i+c jc+ p j (λd) i+c j (i + c j)! e λd, for all i N 0. (4) Analogously, by conditioning on the queue length, we can derive a similar set of equations for q i, the stationary probability of the queue containing i customers: q 0 c c j q j m0 (λd) m i+c e λd, and q i m! q j (λd) i+c j (i + c j)! e λd for i > 0. (5) We can also derive (5) from (4) by using q 0 c p j, and q i p i+c for i > 0. Together with the normalization equation and the capacity utilization equation, (4) constitutes an infinite system of linear equations that can be solved in several ways. In Tijms [994] a Fast Fourier Transform method and a geometric tail approach to solving (4) are described, both of which are quite efficient. 3 The queue length at service initiations For our analysis we will define the following random variables: A n : the arrival time of the n th customer after t 0 S n W n : the starting time of the service of the n th customer : S n A n : the waiting time of the n th customer L + q (t) : the queue length at time t + (immediately after time t) q + i (t) : P {L+ q (t)i} : probability of finding a queue of length i at time t + ω i q i : lim q + i (S n n) lim P {L + q (S n )i}. n : lim P {L q (t)i} : stationary probability of finding a queue of length i t Since the probabilistic argument will concentrate on the service initiations S n, we are interested in the stationary probabilities ω i of having i waiting customers in the queue immediately after some arbitrary service start. Therefore we will prove the following lemma: Lemma. ω i lim n P {L + q (S n )i} lim t P {L q (t)i} q i, for all i N 0. Proof. With probability q + i (S n) there are i customers left in the queue immediately after the service start S n of the n th customer. During the service interval (S n, S n + D] there will 3

4 be j new arrivals with probability (λd)j e λd. We distinguish two cases. If i + j c, the server will start serving the (n + c) th customer immediately after finishing the service of the n th customer. So for this case we find L + q (S n+c ) L + q (S n + D) i + j c. On the other hand, if i + j < c, the (n + c) th customer has not yet arrived at time S n + D. So immediately after finishing the service of the n th customer, the server in question will be idle. Consequently, on arrival the (n + c) th customer will find at least one server idle, implying L + q (S n+c ) 0. Combining both cases, we conclude that L + q (S n+c ) max(0, L + q (S n )+j c), irrespective of the residual service times at t S n. Therefore, the probability q + i (S n+c) of having i waiting customers in the queue immediately after the service start of the (n+c) th customer can be determined by conditioning on L + q (S n ) : q + 0(S n+c ) c j c q + j (S (λd) m n) e λd, and q + i m! (S i+c n+c) q + j (S n) (λd)i+c j (i+c j)! e λd for i>0 (6) m0 The stationary distribution ω i lim j q + i (S j) is found by letting n in (6): ω 0 c c j ω j m0 (λd) m i+c e λd, and ω i m! ω j (λd) i+c j (i + c j)! e λd for i > 0. (7) This set of equations is identical to the set of equations (5) for the stationary probabilities q i. Since the corresponding embedded Markov chain is ergodic for ρ <, there is a unique solution to (7). Therefore ω i equals q i, for all i 0. Remark Observe that (6) is derived by conditioning on merely L + q (S n ). This is possible because of the crucial property that the residual service times of the busy servers at t S n have no influence on L + q (S n+c ). This is not a trivial property. For instance, it does not hold for L, the total number of customers in the system. Therefore lemma holds only for the queue length L q, not for L. 4 The Waiting Time Distribution Without loss of generality, we assume that all customers are assigned to the servers in cyclic order. This will not violate the FCFS discipline because of the deterministic service times. Our first objective will be to determine W n, the waiting time of the n th customer to arrive after t 0. This customer will be called the marked customer and will be served by the so called marked server. 4

5 For (k )D x < kd the analysis of P {W n < x} will concentrate on the time instant S n kc, when the marked server will start to serve the (n kc) th customer. Observe that the marked customer will be the k th customer to be served by the marked server from S n kc onwards. Let u be some positive time lapse D. If the marked customer arrives no earlier than t S n kc + u, the marked server has been serving the (n kc) th customer for at least u time units on the arrival instant A n. Therefore the marked customer will find the marked server with an amount of unfinished work kd u, implying W n kd u. (Here it is essential that u D, guaranteeing that the marked server is busy during all of the time interval (S n kc, S n kc + u]. ) On the other hand, if the marked customer arrives before t S n kc + u, he will find the marked server with an amount of unfinished work > kd u, implying W n > kd u. Combining both cases, we conclude: ( k N) ( u (0, D]) A n S n kc + u W n kd u The corresponding probabilities can be found by conditioning on L + q (S n kc ), the queue length immediately after S n kc. If at this instant the queue contains kc or more customers, the marked customer has already entered the queue and is waiting there in (kc) th position, implying A n < S n kc. On the other hand, if the queue contains i < kc customers at time S n kc, the marked customer will be the (kc i) th customer to arrive from S n kc onwards. With probability kc i Thus, conditioning on L + q (S n kc ) yields: (λu) j e λu this arrival takes place no earlier than t S n kc + u. P {W n kd u} P {A n S n kc + u} kc kc i q + i (S n kc) (λu) j e λu. (8) By letting n, using lemma, we find the stationary waiting time distribution P {W kd u} kc kc i q i (λu) j kc e λu e λu (λu) j kc j q i. (9) Defining the cumulative probability Q m : m 0 q i, we can summarize the result as: kc ( k N) ( u (0, D]) P {W kd u} e λu Q kc j (λu) j Substitution of x kd u gives us the waiting time distribution in terms of x :. (0) kc P {W x} e λ(kd x) Q kc j λ j (kd x) j, for (k )D x < kd. () As this expression contains only a finite number of positive terms, it does not present any numerical complications, regardless of the traffic intensity ρ. 5

6 5 Erlang s integral equation for the M/D/c queue Since it represents the waiting time distribution of the M/D/c queue, expression (0) must satisfy Erlang s integral equation. In order to complete the circle, this section presents an explicit proof, which will make use of the following two lemmas: Lemma 2. n+c (λd) i Q n Q n+c i e λd, for all n N 0. Proof. We define Q n (t) as the probability of the queue containing no more than n customers at time t. Note that Q n lim t Q n (t). Let (t, t + D] be an arbitrary time interval of length D, during which i new arrivals will take place with probability (λd)i e λd. By conditioning on i we find: n+c Q n (t + D) Q n+c i (t) (λd)i e λd, for all n N 0. (2) By letting t in (2) we find lemma 2. Lemma 3. k (j + i + )(k i)! (j + k + )!, for all j, k N 0. Proof. For j 0 lemma 3 holds for all k N 0, which can be verified as follows: k (i + )!(k i)! k+ ( ) l (k + l)! (k + )! k+ l (k + )! k+ ( ) l (k + )! k + ( ) l (k + l)! ( ) l k+ l (k + )!. For general j the validity of the lemma is proved by induction. Suppose the lemma holds for some j and all k N 0, then (j + k + )! k (k i)!(j + i + ) k i (k i)!(j + i + ) + k!(j+)!. (3) 6

7 Since the lemma has already been proved for j 0, we can rewrite the last term for any positive k as (j + )! k! (j + )! k { k n k (i + )!(k i)! k (i+)!(k i)!(j+i+2) + (i + )!(k i)!(j + ) } (j+i+2)(k i)!(j+) ( ) n n!(k n)!(j + n + ) + k (j+)! (j + i + 2)(k i)!. If this expression for k!(j+)! is substituted in the right hand side of (3) we find: k (j + )! (j + i + 2)(k i)! (j + k + )!. Thus we proved the lemma to be valid for j +, k, if it is valid for j, k N. Since there were no restrictions on the value of k in our starting supposition, this finishes our proof. See also exercise 5, p 29, Riordan [968]. In the formulation () of Erlang s integral equation it is required that F (t) 0 for all t < 0. We can get rid of this requirement by reformulating the equation as: F (y) max(0,d y) F (x + y D) λc x c (c )! e λx dx, for all y R +. (4) Because of the special structure of our result (), we will substitute: y md u, with m y D +, and 0 < u D. Abbreviating the integrand as H(x, y), we can rewrite the right hand side of (4): u max(0,u (m )D) H(x, md u)dx + u+(k+)d u+kd This format enables us to substitute expression () in the integrand: + u max(0,u (m )D) u+(k+)d (k+m)c u+kd (m )c λ j+c (u x) j x c Q (m )c j e λu dx (c )! H(x, md u)dx. (5) λ j+c {(k+)d+u x} j x c Q (k+m)c j e λ{(k+)d+u} dx. (6) (c )! 7

8 By carrying out the integration, and realizing that max(0, u (m )D) 0 except for m, in which case the first integral vanishes, we can write (6) as: + (m )c e λu e λ{(k+)d+u} (k+m)c Q (m )c j (λu) j+c λ j+c Q (k+m)c j c c (j + i + )(c i)! {(k+)d+u} c i D j+i+ (j + i + )(c i)!. (7) By expanding {(k + )D + u} c i and subsequently rearranging the summations, using c c i n0 A i,n c n0 A i,n, the second term of (7) can be rewritten as: with β n (k+m)c c e λu n0 (λd) j+c n Q (k+m)c j (λu) n β n, (8) n! e λd (k + ) c i n (j+i+)(c i n)!. (9) Using lemma 3, the last summation in (9) can be transformed as follows: (k + ) c i n (j + i + )(c i n)! i (j + i + ) k l (c n i l)! k l l (j + i + )(c n i l)! k l (j + c n l)!. Substitution of this result in (9) in combination with lemma 2, yields: β n (k+m)c Q (k+m)c j (k+m+)c n l (λd) j+c n l (j + c n l)! e λd (λd) i Q (k+m+)c n l i e λd ic n l { } c n l (λd) i Q (k+m)c n l Q (k+m+)c n l i e λd 8

9 e λ(k+)d Q (k+m)c n l jl Q (k+m+)c n j (λd) j l (j l)!. We can simplify the last term by observing that: jl Q (k+m+)c n j Q (k+m+)c n j (λd) j l (j l)! j (λd) j l (j l)! {λ(k + )D} j Q (k+m+)c n j, which enables us to rewrite β n as: β n 0 e λ(k+)d Q m. jo Q (k+m)c n l {λ(k + )D} j Q (k+m)c n l Q (k+m+)c n j Putting the parts together, using lemma 3, we can reconstruct (7) as follows: (m )c e λu { mc e λu ic mc e λu (λu) i Q mc i (λu) j+c Q (m )c j (j + c)! + c (λu) n e λu n! n0 } (λu) i c (λu) n Q mc i + Q m n! n0 This implies that the integral equation (4) reduces to. β n mc F (y) e λu Q mc i (λu) i, 9

10 which, for y md u, corresponds exactly to expression (0). Acknowledgements The author wishes to thank Onno Boxma, Nico van Dijk, Flora Spieksma, and Henk Tijms for their fruitful remarks and suggestions. References [] Erlang, A.K. (909, 920), The theory of probabilities and telephone conversations, and Telephone waiting times, first published in Nyt Tidsskrift for Matematik, B Vol. 20, p. 33, and Matematisk Tidsskrift, B Vol. 3, p 25. English translations of both articles in Brockmeyer, E. et. al. (948), The Life and Works of A.K.Erlang, The Copenhagen Telephone Company, Copenhagen. [2] Crommelin, C.D. (932), Delay probability formulas when the holding times are constant, P.O. Electr. Engr. J., 25, [3] Prabhu, N.U. (962), Elementary methods for some waiting time problems, Opns. Res. 0, [4] Prabhu, N.U. (965), Queues and Inventories, Wiley, New York. [5] Riordan, J. (968), Combinatorial Identities, Wiley, NewYork. [6] Tijms, H.C. (994), Stochastic Models, an Algorithmic Approach, Wiley, New York. 0