ON THE SERIES EXPANSION FOR THE STATIONARY PROBABILITIES OF AN M/D/1 QUEUE

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1 Journal of the Operations Research Society of Japan 2005, Vol. 48, No. 2, The Operations Research Society of Japan ON THE SERIES EXPANSION FOR THE STATIONARY PROBABILITIES OF AN M/D/1 QUEUE Kenji Nakagawa Nagaoka University of Technology (Received August 27, 2003; Revised July 20, 2004) Abstract In this paper, we give the series expansion for the stationary probabilities π n of the queue length of an M/D/1 queue based on analytic properties of the probability generating function π(z). We determine the poles and their associated residues of π(z), then give the partial fraction expansion of π(z). The series expansion of π n are given by the poles and residues. We also give an upper bound and a lower bound for π n. Keywords: Queue, M/D/1, stationary distribution, series expansion, complex function theory 1. Introduction We study the stationary probabilities π n of the queue length of an M/D/1 queue. We determine the poles and their associated residues of the probability generating function π(z) ofπ n. Then we have the series expansion of π n represented by the poles and residues. Moreover, we give an upper and lower bounds for π n. 2. The Stationary Distribution of an M/D/1 Queue The state transition probability matrix P of an M/D/1 queue with arrival rate and service rate 1 is given by P = a 0 a 1 a 2 a 3... a 0 a 1 a 2 a a 0 a 1 a a 0 a ,a n = n n! e,n=0, 1,... (2.1) For the stability of the queue, <1 is assumed. Let π =(π 0,π 1,...) denote the stationary distribution of P, and define the probability generating function π(z) of π by π(z) π n z n. (2.2) n=0 By the Pollaczek-Khinchin formula [2], we have π(z) = (1 )(z 1) exp((z 1)). (2.3) z exp((z 1)) 111

2 112 K. Nakagawa It is well known [1] that the explicit form of π n is given by the Taylor expansion of π(z), i.e., π 0 =1 π 1 =(1 )(e 1) ( n 1 π n =(1 ) e n + k=1 [ (k) e k ( 1) n k n k ]) (n k)! + (k)n k 1,n 2. (2.4) (n k 1)! But, it is not good to use this formula for calculation because it includes alternating additions of positive and negative numbers of very large absolute value. We will have a series expansion of π n by investigating analytic properties of π(z), especially the poles and their associated residues. Our series expansion is interesting from both theoretical and numerical point of view. Theoretically, all the poles and their associated residues of π(z) are first determined in this paper. Numerically, our formula gives very stable computation of π n because each term of the series decreases very quickly. We first determine all the poles of π(z) in z <. Denotebyψ(z) z exp((z 1)) the denominator of π(z). We show in Figure 1 the zeros of ψ(z) with =0.5obtained by numerical computation. In Figure 1, the x-coordinate is the real part of z and the y-coordinate is the imaginary part y Figure 1: The zeros of ψ(z) =z exp((z 1)), =0.5, z= x + iy We can see that there are infinitely many zeros and among them the real zeros are z =1 and z = Since ψ(z) is a function of real coefficients, if a complex number is a zero of ψ(z), then so is its complex conjugate. Further, the difference of the imaginary part of adjacent zeros seems nearly constant. x

3 On the Series Expansion for the Stationary Probabilities of M/D/1 Queue On the Position of the Zeros of ψ(z) In order to study the position of the zeros of ψ(z) =z exp((z 1)), we define sets S k,k Zby S k = {z = x + iy <x<, 2πk π y< 2πk + π }, k Z, (3.1) where Z is the set of integers. S k is a horizontal strip. We see S k S k = φ, k k and k Z S k = C, the whole finite complex plane. We have the following theorem. Theorem 3.1 (i) ψ(z) has two zeros z =1and z = ζ 0 > 1 in S 0, both of which are simple zeros. (ii) For k 0, ψ(z) has a unique zero ζ k in S k, which is a simple zero. ζ k and ζ k are complex conjugate. Proof: (i) From the graphs of Y = x and Y =exp((x 1)), we see that ψ(z) hastworeal zeros x =1andx = ζ 0 > 1. We show that these are the only zeros in S 0. We will apply the argument principle of complex function theory. For sufficiently large R>0, consider four points z 1 = R + i π, z 2 = R + i π, z 3 = R i π, z 4 = R i π on the boundary of S 0. (z j belongs to the jth orthant, j =1,...,4.) Let C 1 denote the line segment whose starting point is z 1 and ending point z 2. Similarly, let C j denote the line segment from z j to z j+1,j=1,...,4, with regarding z 5 = z 1.LetC denote the closed curve which is made by connecting C 1,...,C 4. The images of z j,c j,andc by the mapping ψ(z) are denoted by z j,c j,andc, respectively. We have z 1 = R + e (R 1) + i π, z 2 = R + e ( R 1) + i π, z 3 = R + e ( R 1) i π, z 4 = R + e (R 1) i π. (z j belongs to the jth orthant, j =1,...,4.) We see that C 1 is the line segment with starting point z 1 and ending point z 2,andC 3 is the line segment with starting point z 3 and ending point z 4. For sufficiently large R, ψ(z) is nearly the identity mapping on C 2, hence C 2 is contained in a sufficiently small neighborhood of the line segment connecting z 2 and z 3. Thus, C 1C 2C 3 is a curve which starts from z 1 in the first orthant, passes through z 2 in the second orthant and z 3 in the third orthant, and finally ends at the z 4 in the fourth orthant. In other words, C 1C 2C 3 nearly goes round about the origin (see Figure 2). For large R, the difference of the arguments between z 1 and z 4 is small, hence, the change of the argument along the curve C 1C 2C 3 is nearly 2π. In other words, for any ɛ>0and sufficiently large R, wehave d arg ψ(z) 2π <ɛ. (3.2) C 1 C 2 C 3

4 114 K. Nakagawa Figure 2: The image C of C by the mapping ψ(z) ( =0.5, k=0,r=8) Next, we consider the change of the argument along C 4. Represent a point z t on C 4 by z t =(1 t)z 4 + tz 1 (2t 1)π = R + i, 0 t 1. Denoting by z t the image of z t by the mapping ψ(z), we have z t =exp((r 1) + i2πt)(1 + δ), (3.3) (2t 1)π where δ =(R + i )/ exp((r 1) + i2πt). If R is large, then δ is small and hence the change of arg(1 + δ) alongc 4 is small. From (3.3), we have arg z t =2πt + arg(1 + δ) and thus 1 d arg z t = 2πdt + d arg(1 + δ). (3.4) C 4 0 Therefore, for any ɛ>0 and sufficiently large R, wehave d arg ψ(z) 2π <ɛ. (3.5) C 4 From (3.2), (3.5), we have 1 d arg ψ(z) 2 < ɛ 2π C π. (3.6) (1/2π) d arg ψ(z) is the number of rotation of the closed curve C C = ψ(c) around the origin z =0,soitisaninteger.From(3.6),wehave 1 d arg ψ(z) =2. (3.7) 2π C C 4

5 On the Series Expansion for the Stationary Probabilities of M/D/1 Queue 115 By the argument principle, we see that ψ(z) has exactly two zeros in the region surrounded by the closed curve C, whicharez =1andz = ζ 0. Letting R tend to infinity, we know that z =1,ζ 0 are the only zeros in S 0 and they are both simple zeros. (ii) Let k>0. Consider four points z 1 = R + i 2πk + π,z 2 = R + i 2πk + π, z 3 = R + i 2πk π,z 4 = R + i 2πk π on the boundary of S k, and define C j as the line segment from z j to z j+1, j =1,...,4, with regarding z 5 = z 1.LetCdenote the closed curve connecting C 1,...,C 4. The image of z j, C j,andc by the mapping ψ(z) are denoted by z j, C j,andc, respectively. Similarly to the proof of (i), for sufficiently large R, we see that the curve C 1C 2C 3 is included in the upper half plane Iz >0, where Iz denotes the imaginary part of z. The difference of the argument between the starting point z 1 of C 1C 2C 3 and the ending point z 4 is very small (see Figure 3). Figure 3: The image C of C by the mapping ψ(z) ( =0.5, k=1,r=8) Therefore, for any ɛ>0 and sufficiently large R, wehave d arg ψ(z) <ɛ. (3.8) C 1 C 2 C 3 Similarly to the proof of (i), the change of the argument of ψ(z) alongc 4 satisfies d arg ψ(z) 2π <ɛ (3.9) C 4 for large R. From (3.8),(3.9), we have 1 d arg ψ(z) =1, (3.10) 2π C

6 116 K. Nakagawa hence by the argument principle, we see that ψ(z) has a unique zero ζ k in the region surrounded by the closed curve C. Letting R, we know that z = ζ k is the unique zero in S k,k>0, and it is a simple zero. We can prove similarly for k<0. We will also use the following lemma for determining the position of the zeros of ψ(z). Rz denotes the real part of z. Lemma 3.1 z =1is the unique zero of ψ(z) in the region Rz <ζ 0. Proof: We apply Rouché s theorem. Write z = x + iy. For sufficiently small ɛ>0and large R>0, we consider an open set D = { z <R, x<ζ 0 ɛ} and its boundary C = D. For an arbitrary z C, wehave z ζ 0 ɛ > exp((ζ 0 ɛ 1)) exp((x 1)) = exp((z 1)). By Rouché s theorem, z has the same number of zeros as ψ(z) =z exp((z 1)) does in D. Sincez has one simple zero z =0inD, z = 1 is also the unique zero of ψ(z) ind and is simple. By letting ɛ 0, R,weseethatz = 1 is the unique zero of ψ(z) inthe region Rz <ζ Coordinates of the zeros of ψ(z) We will approximate the x-coordinate and y-coordinate of the kth zero ζ k of ψ(z). Write ζ k = x k + iy k,k Z. By comparing the real and imaginary parts of both sides of the equation we have ζ k =exp((ζ k 1)), (3.11) x k =exp((x k 1)) cos y k, (3.12) y k =exp((x k 1)) sin y k. (3.13) Lemma 3.2 The sequences {x k } k=0 and {y k} k=0 are monotonically increasing, i.e., x k < x k+1, y k <y k+1, k =0, 1,..., thus we have ζ k < ζ k+1, k =0, 1,... Proof: We see from Theorem 3.1 that {y k } k=0 is monotonically increasing. We next show that {x k } k=0 is monotonically increasing, too. From (3.12), (3.13), we have x 2 k + y 2 k =exp(2(x k 1)), k=0, 1,... (3.14) Define f(x) exp(2(x 1)) x 2. We can write (3.14) as f(x k )=y 2 k. We have f (x) = 2{ exp(2(x 1)) x}. By the graphs of Y = exp(2(x 1)) and Y = x, wesee that f (x) = 0 has exactly two solutions α 1, α 2, with 0 < α 1 < 1 < α 2. α 1 gives a local maximum of f(x) andα 2 gives a local minimum. f(x) is monotonically increasing in x>α 2 and f(x) goes to infinity as x does. The maximum of f(x) in0 x α 2 is f(α 1 )=exp(2(α 1 1)) α 2 1 < 1. Therefore, for any β>1, the equation f(x) =β has a

7 On the Series Expansion for the Stationary Probabilities of M/D/1 Queue 117 unique solution in x>0. From Theorem 3.1 (ii), we have y k y 1 > π hence, x = x k is the unique solution of > 1, k =1, 2,..., f(x) =y 2 k,k=1, 2,... and, by the increasing property of f(x), y k <y k+1 leads to x k <x k+1, k =1, 2,... The inequality x 0 <x 1 also holds because f(x 0 )=0=y0 2 and x 0 >α 2. We will show an approximation of ζ k.fork>0, we have y k > 0 from Theorem 3.1, and x k > 0 from Lemma 3.1, hence from (3.12), (3.13), cos y k > 0, sin y k > 0. Thus, 2πk <y k < 2πk + π/2. (3.15) From Theorem 3.1 or (3.15), when k goes to infinity, y k goes to infinity, thus from (3.13) x k does, too. Therefore, from (3.12), we have cos y k 0, and thus from (3.15), From (3.13), (3.16), we have and then ζ k can be represented approximately as ζ k 1 ln 2πk + π/2 2πk y k π 2. (3.16) y k exp((x k 1)), (3.17) +1+i 2πk + π/2,k. (3.18) For k<0, we have an approximation for ζ k by (3.18) and ζ k = ζ k. Among the zeros of ψ(z), z = 1 is not a pole of π(z) because z = 1 is a simple zero of ψ(z) and the numerator of π(z) has a factor z 1. Therefore, the poles of π(z) are{ζ k } k Z. In summary, Theorem 3.2 The poles of π(z) are {ζ k } k Z, all of which are simple poles. We have ζ k < ζ k+1 for k =0, 1,... The pole ζ k of π(z) can be approximated as follows. ζ ±k 1 ln 2πk + π/2 +1± i 2πk + π/2, k. (3.19) We show, in the case of =0.5, a comparison between the exact value of ζ k and the approximation (3.19) in Figure Partial Fraction Expansion of π(z) The principal part of π(z) atz = ζ k is α k (z ζ k ) 1 with α k the residue of π(z) atz = ζ k. If π(z) can be represented by a partial fraction of a form like π(z) = k α k(z ζ k ) 1, the coefficients π n of π(z) has a series expansion. But, unfortunately, the partial fraction k α k(z ζ k ) 1 does not converge, hence we need some idea to have a convergent series. For this purpose, we apply the following theorem.

8 118 K. Nakagawa poles of π (z) approximation in Theorem y Figure 4: Comparison between the exact value of ζ k and the approximation (3.19) x Theorem A (see [3]) Let f(z) be holomorphic at z = 0 and meromorphic in z < with poles {ζ k } k=1 all of which are of order 1. There is a sequence of rectifiable simple closed curves {C m } m=1 and every C m includes the origin and poles {ζ k } nm k=1 in its interior. n m+1 n m are assumed to be bounded. Let l m denote the length of C m and ρ m the distance between the origin and C m. The residue of f(z) atz = ζ k is denoted by α k. For an integer q 1, assume α k = o(ζ q+1 k ), k,andρ m,l m = O(ρ m ),f(z) =o(ρ q m), m,z C m. Then, for any z ζ k,wehave f(z) = = q 1 j=0 q 1 j=0 f (j) (0) z j z q j! f (j) (0) z j + j! k=1 k=1 α k ζ q k (ζ k z) ( q 1 1 α k + z ζ k j=0 z j ζ j+1 k ) (4.1) (4.2) The series (4.1),(4.2) converges absolutely and uniformly in wide sense in the region C {ζ k } k. The residue α k of our function π(z) atz = ζ k is obtained by So, applying the Theorem A, we have α k = lim z ζk (z ζ k )π(z) = (1 )ζ k(ζ k 1) ζ k 1

9 On the Series Expansion for the Stationary Probabilities of M/D/1 Queue 119 Theorem 4.1 π(z) has the partial fraction expansion π(z) =π 0 + π 1 z z 2 k= α k ζ 2 k (ζ k z), (4.3) α k = (1 )ζ k(ζ k 1), k Z. (4.4) ζ k 1 The series (4.3) converges absolutely and uniformly in wide sense in the region C {ζ k } k. Proof: Let q = 2 in Theorem A. For m =1, 2,...,letC m be the square with vertices z 1 = 2πm z 3 = 2πm + i2πm,z 2 = 2πm + i2πm, i2πm,z 4 = 2πm i2πm. From Theorem 3.1, the poles {ζ k } m 1 k= (m 1) are included in the interior of C m,thusn m = 2m 1. Denoting by l m the length of C m and ρ m the distance between the origin and C m, we have l m =16πm/, ρ m =2πm/, so,ρ m,l m = O(ρ m ),m.wehave α k = (1 )ζ k(ζ k 1) = o(ζ 3 ζ k 1 k), k. We next show π(z) =o(ρ 2 m), z C m, m. We first consider the case that z is on the line segment z 1 z 2. z = z t is represented as We can write z t = t + i 2πm, 2πm t 2πm. (1 )(z 1) π(z) = z exp( (z 1)) 1, then, defining g(z) z exp( (z 1)) 1, we show g(z t ) 2 γ, z t C m, where γ is a positive constant which does not depend on m. In fact, g(z t ) 2 =(t exp( (t 1)) 1) 2 +(2πm 1 exp( (t 1))) 2 (t exp( (t 1)) 1) 2 +(2π 1 exp( (t 1))) 2 γ This implies that π(z) =o(ρ 2 m)ifz is on the line segment z 1 z 2.Inthecasethatz is on z 3 z 4, we can show that π(z) =o(ρ 2 m) in a similar way. Moreover, we can easily show π(z) =o(ρ 2 m) when z is on z 2 z 3 and z 4 z 1. In consequence, π(z) satisfies all the assumptions in Theorem A with q = 2. Then π(z) has the following expression; π(z) =π 0 + π 1 z z 2 α k ζ 2 k= k (ζ k z) = π 0 + π 1 z +(1 )z 2 ζ k 1 ζ k (ζ k 1) k= 1 ζ k z.

10 120 K. Nakagawa 5. Series Expansion of π n From the partial fraction expansion (4.3), we have π(z) =π 0 + π 1 z +(1 )z 2 k= ζ k 1 ζ 2 k (ζ k 1) ( z ) n (5.1) ζ k for z with sufficiently small absolute value. The summation in n converges uniformly in k, hence we can change the order of the summations to obtain the following series expansion of π n,n 2. π n =(1 ) k= ζ k 1 1 ζ k 1 { ζ0 1 1 =(1 ) ζ 0 1 ζ n 0 ζ n k +2 k=1 ( ζk 1 Re ζ k 1 n=0 1 ζ n k (5.2) )},n 2. (5.3) 5.1. Upper and lower bounds for π n For the principal part p 0 (z) =α 0 /(z ζ 0 )ofπ(z) atz = ζ 0, define f(z) π(z) p 0 (z). (5.4) z = ζ 0 is a removable singularity of f(z). The poles of f(z) with the smallest absolute value are z = ζ 1,ζ 1.Letf(z) = n=0 d nz n be the power series expansion of f(z) at the origin, then from (5.4) and we have p 0 (z) = α 0 ζ 0 ( z ) n, ζ 0 n=0 π n = α 0 ζ0 n+1 + d n, n =0, 1,... (5.5) For arbitrary r with ζ 0 <r< ζ 1, defining M 1 (r) =max z =r f(z), we obtain by Cauchy s estimate that d n M 1(r),n=0, 1,... r n Thus from (5.5) we have the following estimates for π n ; α 0 ζ0 n+1 M 1(r) π r n n α 0 0 ζ n+1 + M 1(r) r n,n=0, 1,... In a similar way, we have Theorem 5.1 Define f K (z) π(z) K k= K α k z ζ k, K =0, 1,... (5.6)

11 On the Series Expansion for the Stationary Probabilities of M/D/1 Queue 121 and for r with ζ K <r< ζ K+1.Thenπ n is evaluated as follows; K k= K α k ζ n+1 M K(r) k r n M K (r) =max f K (z) (5.7) z =r π n K k= K α k ζ n+1 + M K(r), n =0, 1,... (5.8) k r n 6. Numerical Result We show in Figure 5 the comparison of the exact value of π n and the proposed upper bound obtained by Theorem 4 with K = 1, i.e., the upper bound is π n α 0 ζ n M 1(r),n=0, 1,... (6.1) r Three cases =0.1, 0.5 and0.9 are shown in Figure 5. The parameters are written in Table 1. In Figure 5, the black circle indicates the exact value and the white one the upper bound (6.1). The calculation of π n is due to the recursive formula given by π = πp. The computational complexity of this recursive formula is O(n 2 ), whereas that of our upper bound is O(1). The computational complexity of the direct expression (2.4) of π n is O(n), but it is numerically unstable because it includes alternating additions of positive and negative numbers of very large absolute value. Table 1: The parameters of upper bound (6.1) ζ 0 α 0 r M 1 (r) Conclusion We studied the stationary probabilities π n of the queue length of an M/D/1 queue. We determined the series expansion of π n with the poles and their associated residues of π(z). Upper and lower bounds for π n are also provided. An M/D/1 is a simple queueing model, but the proofs of the theorems are complicated. We would like to improve and simplify our proof to attack more complex Markov chains. Acknowledgement The author thanks the reviewers for their valuable comments and suggestions to improve the results of this paper.

12 122 K. Nakagawa 10 0 = π n = = Figure 5: The comparison of π n and the upper bound References [1] D. Gross and C. Harris: Fundamentals of Queueing Theory (Wiley, 1985). [2] J.F. Hayes and T.V.J. Ganesh Babu: Modeling and Analysis of Telecommunications Networks (Wiley, New Jersey, 2004). [3] Y. Komatsu: Theory of Functions (Asakura, 1960 (in Japanese)). n Kenji Nakagawa Nagaoka University of Technology Nagaoka, Niigata , Japan Tel&Fax nakagawa@nagaokaut.ac.jp