Some results on. Transient Queues. Shan XU

Transcription

1 Some results on Transient Queues by Shan XU A Major Paper Submitted to the Faculty of Graduate Studies through the Department of Mathematics and Statistics in Partial Fulfillment of the Requirements for the Degree of Master of Science at the University of Windsor Windsor, Ontario, Canada 21 c Shan XU

2 Some Results on Transient Queues by Shan Xu APPROVED BY: A. Hussein Department of Mathematics and Statistics M. Hlynka Department of Mathematics and Statistics November, 21

3 Author s Declaration of Originality iii I hereby certify that I am the sole author of this major paper and that no part of this major paper has been published or submitted for publication. I certify that, to the best of my knowledge, my major paper does not infringe upon anyone s copyright nor violate any proprietary rights and that any ideas, techniques, quotations, or any other material from the work of other people included in my thesis, published or otherwise, are fully acknowledged in accordance with the standard referencing practices. Furthermore, to the extent that I have included copyrighted material that surpasses the bounds of fair dealing within the meaning of the Canada Copyright Act, I certify that I have obtained a written permission from the copyright owner(s) to include such materials in my major paper and have included copies of such copyright clearances to my appendix. I declare that this is a true copy of my major paper, including any final revisions, as approved by my committee and the Graduate Studies office, and that this major paper has not been submitted for a higher degree to any other University or Institution.

4 Abstract iv This paper studies two jobs which must be performed in tandem, when the order of service of the two jobs is unimportant to the customer. One of the jobs requires waiting in an M/M/1 queue while the other job requires a fixed amount of time with no queueing. The paper gives a graphical illustration as to when it is better (in the sense of minimizing total system time for an arriving customer) to choose the queue and when it is better to choose the fixed time job. The paper extends and expands on results given by van de Coevering (1995).

5 Contents Author s Declaration of Originality iii Abstract iv List of Figures vi Chapter 1. Literature Review 1 Chapter 2. van de Coevering s result and extensions 7 Chapter 3. Task versus Queue? 38 Bibliography 42 v

6 List of Figures 2.1 p i2 (t) for i =... 4, λ = 3, µ = 4, < t < p i2 (t) for i =... 4, λ = 1, µ = 4, < t < p 22 (t) for i = 2, j = 2, λ = 3, µ = p i2 (t) for i =, 1, 3, 4, λ = 3, µ = 4, < t < p i2 (t) for i =, 1, 3, 4, λ = 1, µ = 4, < t < p 2 (t) for i =, j = 2, λ = 3, µ = 4, < t < p 2j (t) for j =... 4, λ = 3, µ = 4, < t < p 2j (t) for j =... 4, λ = 1, µ = 4, < t < p 2j (t) for i = 2, j =, 1, 3, 4,λ = 3, µ = 4, < t < p 2j (t) for j =, 1, 3, 4, λ = 1, µ = 4, < t < EL i (t) for λ = 3,µ = 4, i =, 1, 2, 3, 4, < t < EL i (t) for λ = 1, µ = 4, < t < EL 1 (t) for λ = 3,µ = 4, < t < EL 1 (t) for λ = 1, µ = 4, < t < EL 2 i(t) for λ = 3,µ = 4, < t < V ar i [L(t)] for λ = 3,µ = 4, < t < skew i [L(t)] for λ = 3,µ = 4,.1 < t < 1 29 vi

7 LIST OF FIGURES vii 3.1 Which is better? EL 1 (t) for λ = 3,µ = 4, < t <.5 41

8 CHAPTER 1 Literature Review To obtain limiting probability results for M/M/1 queues is relatively easy. To obtain transient results is much harder. Assume that the arrival rate is λ and the service rate is µ. Let ρ = λ µ be the traffic intensity. The earliest known formulas for transient queueing probabilities are given by Clarke (1953) and by Ledermann and Reuter (1954). Gross and Harris (1985) and Kleinrock (1975) present the Ledermann and Reuter result in terms of modified Bessel functions. Their expression is: p ij (t) = e (λ+µ)t [ρ (j i)/2 I j i (2t λµ) + ρ (j i 1)/2 I j+i+1 (2t λµ) + (1 ρ)ρ j ρ k/2 I k (2t λµ) k=j+i+2 where I k (x) = m= (x/2) k+2m (k + m)!m! is the modified Bessel function of the first kind of order k. (From Wikipedia, Daniel Bernoulli defined the Bessel functions, and they were then generalized by Friedrich Bessel. [Modified] Bessel functions of the first kind, denoted as J α (x), are solutions of the Bessel s differential equation that are finite at the origin (x = ) for non-negative integer α, and diverge as x approaches zero for 1

9 1. LITERATURE REVIEW 2 negative non-integer α. The solution type (e.g. integer or non-integer)and normalization of J α (x) are defined by its properties.... It is possible to define the function by its Taylor series expansion around x = : J α (x) = m= ( 1) m (m!)γ(m + α + 1) (x/2)(2m+α) where Γ(z) is the Gamma function. ) To obtain Clarke s expression, let q t be the queue length at time t. Let f t (q) be the probability that q t = q when q o is given. Then, for q q, Clarke obtains f t (q) = F t (q) F t (q + 1) = e (1+λ)t {λ 1 2 (q q) I q q (2λ 1 2 t) + λ 1 2 (q q 1) I q+q +1(2λ 1 2 t) + (1 λ)λ q λ 1 2 r I r (2λ 1 2 t)} r=q+q +2 For q < q f t (q) = F t (q) F t (q + 1) = e (1+λ)t {λ 1 2 (q q) I q q (2λ 1 2 t) + λ 1 2 (q q+1) I q+q +1(2λ 1 2 t) + (1 λ)λ q λ 1 2 r I r (2λ 1 2 t)} r=q+q +2 where I r (x) is the modified Bessel function of the first kind of order r. Champernowne (1956) gives a combinatorial argument to obtain the same expression as Clarke.

10 1. LITERATURE REVIEW 3 Sharma (199) presents some new formulas for transient queueing probabilities. p i (t n at time ) = (1 ρ)ρ i + e (λ+µ)t ρ i k= k= ( (λt) k k! i+k+n m= (k m) (µt)m 1 m! ( ) + e (λ+µ)t (λt) i+k n (µt) k 1 k!(i + k n)! 1. (i + k)!(k n)! ) We present Conolly and Langaris (1993) version of Sharma s result. Let ν = λ+µ. The probability of having n customers in an M/M/1 system at time t, given that there are customers at time, is given by p () n (t) = (1 ρ)ρ n + exp( νt)ρ n (λt) m m+n m= m! k= (m k) (µt)k 1 k!, where λ is the arrival rate, µ is the service rate, ρ = λ/µ, ν = λ + µ. Conolly and Langaris (1993) developed their own formula for which they claim computational improvements. It is given by p () n (λ + µ) (λ µ) (t) = (1 )ρ n + exp( νt) c (n) m 2µ tm. m where S(t) = ν 2µ m 1 ( 1/2 )( w2 2m 2 m ν )m 2 k= (νt) k, k! c () m is the coefficient of t m in S(t), c (1) m = m + 1 µ c() m 1 c () m, c (n) m = m + 1 µ (c(n 1) m+1 ρc (n 2) m ), for n = 2, 3,....

11 1. LITERATURE REVIEW 4 Karlin and McGregor (1958) and van Doorn (198) give a result for the M/M/1 queue and more general birth-and-death processes, which is called the spectral representation of the probability transition function P ij (t). It is given by p ij (t) = π j e xt q i (x)q j (x)dψ(x) where π = 1, π j = (λ λ j 1 /(µ 1 µ j ), where λ j is the birth rate in state j, µ j is the death rate in state j, q i (x) is a recursively defined system of polynomials in x satisfying orthogonality relations, ψ(x) is a non decreasing real valued function on [, ). Details can be found in van Doorn (198). Abate and Whitt (1988) give a spectral representation for the M/M/1 queue, namely: p in (t) = (1 ρ)ρ n + τ 1 1 τ 1 2 ρ n+1 q i (x)q n (x)φ(x)e xt dx. Here q i (x) are orthogonal polynomials for the M/M/1 system and the q i (x) form a recursively defined system of polynomials in x satisfying orthogonality relations q n (x) = ρ n/2 U n (α(x)) ρ (n+1)/2 U n 1 (α(x)), where U n (α) represents a polynomial in α of the form c n + c n1 α + + c nn α n for c = 1, c 1 =, c 11 = 2, c 2 = 1, c 21 =, c 22 = 4, c n+1,j = 2c n,j 1 c n 1,j for j n + 1 and n 3. U n (α) are the Chebyshev polynomials of the second kind, with U 1 (α) =, U (α) = 1, U 1 (α) = 2α, U 2 (α) = 4α 2 1, U 3 (α) = 8α 3 4α and U n+1 (α) = 2αU n (α) Un 1(α), n 1;

12 1. LITERATURE REVIEW 5 α(x) = 1 + ρ 2θ2 x 2. ρ Here where τ 1 = (1 ρ) 2 2 ; τ 2 = (1 + ρ) 2, 2 φ(x) = 1 ρ (1 τ1 x)(τ 2 x 1), τ2 1 x τ1 1. 2πρ x Leguesdron, Pellaumail, Rubino, and Sericola (1993) give a transient analysis of the M/M/1 queueing system. They present a new method based on the uniformization technique and on generating functions. For i j, the transient probabilities are: ( (j i)/2 λ p i,j (t) = e µ) (λ+µ)t [I j i (2t λµ) I i+j+2 (2t ( µ ) i+1 λµ)] + P,i+j+1 (t) λ for I k (x) as above, and the transient probabilities of the M/M/1 queue, given that the queue is empty at time t =, are: p,j (t) = pj q j n=j e (λ+µ)t(λ + µ)n t n n! (n j)/2 k= n + 1 2k n + 1 ( n + 1 k ) p k q n k Parthasarathy (1987) considers the M/M/1 queue system with Poisson arrivals and exponential service times by deriving the time-dependent solution in a direct way. It is given by: p n (t) = e (λ+µ)t µ n q k (t) k=1 ( ) n k λ + µ ( ) n λ p (t) µ

13 1. LITERATURE REVIEW 6 for p (t) = q 1 (y)e (λ+µ)y dy + δ a, α = 2 (λµ) and β = (λµ) where q n (t) = µβ(n α)(1 δ a )[I n a (αt) I n+a (αt)] + λβ n α 1 [I n+a+1 (αt) I n a 1 (αt)], and δ is the Kronecker delta.

14 CHAPTER 2 van de Coevering s result and extensions In this chapter, we consider results given in van de Covering (1995). We add additional explanations, correct one typo, add graphs, and extent his results. For an M/M/1 queue, van de Coevering presents the result p ij (t) = 2 π ρ(j i)/2 π e µtγ(y) γ(y) a i(y)a j (y)dy + (1 ρ)ρ j ρ < 1 ρ 1, (1) for γ(y) = 1 + ρ 2 ρcos(y) and a k (y) = sin(ky) ρsin((k + 1)y). He credits Morse (1955) and Takacs (1962, page 23) for this trigonometric version. It is difficult to find illustrations of this result. The closest (but not the same) seems to be in Morse (1958, p.66). For illustration, we fix j = 2. We present graphs of p i2 (t) for i =... 4, λ = 3 or 1, µ = 4, < t < 1 in Figures 2.1, 2.2. We stop at i = 4 arbitrarily. We note that one of the p i,2 (t) curves has p i,2 () = 1 while the others have p i,2 () =. Clearly p 2,2 (t) = 1. See Figure 2.3 We eliminate that curve and consider the other curves p i,2 (t), i =, 1, 2, 3, 4, and < t < 3 (for λ = 3) and < t < 1 (for λ = 1). This gives Figure 2.4,

15 2. VAN DE COEVERING S RESULT AND EXTENSIONS 8 Figure 2.1. p i2 (t) for i =... 4, λ = 3, µ = 4, < t < 1 Figure 2.2. p i2 (t) for i =... 4, λ = 1, µ = 4, < t < 1

16 2. VAN DE COEVERING S RESULT AND EXTENSIONS 9 Figure 2.3. p 22 (t) for i = 2, j = 2, λ = 3, µ = 4 Figure 2.4. p i2 (t) for i =, 1, 3, 4, λ = 3, µ = 4, < t < 3

17 2. VAN DE COEVERING S RESULT AND EXTENSIONS 1 Figure 2.5. p i2 (t) for i =, 1, 3, 4, λ = 1, µ = 4, < t < 1 Can we determine which curve corresponds to i =, 1, 3, 4 from Figures 2.4 and 2.5? The answer is Yes but some additional derivation is required. We could simply draw the graphs individually, and compare them. For example, Figure 2.6 shows p,2 (t) in the λ = 3, µ = 4 case.

18 2. VAN DE COEVERING S RESULT AND EXTENSIONS 11 Figure 2.6. p 2 (t) for i =, j = 2, λ = 3, µ = 4, < t < 1 However, we can get even more information without drawing the individual graphs. Property 2.1. (new result) For an M/M/1 queueing system with λ < µ, for small t, and fixed j >, we have p j+1,j ( t) > p j 1,j ( t) > p j+2,j ( t) > p j 2,j ( t). Proof. p j 1,j ( t) = λ t + o( t). p j+1,j ( t) = µ t + o( t). p j 2,j ( t) = o( t). Similarly, p j+2,j ( t) = o( t).

19 2. VAN DE COEVERING S RESULT AND EXTENSIONS 12 Thus, p j+1,j ( t) µ t + o( t) = p j 1,j ( t) λ t + o( t) = µ + o( t) t λ + o( t) t µ + λ + > 1, so, p j+1,j ( t) > p j 1,j ( t) for small t. Similarly, p j+2,j ( t) > 1 for small t. p j 2,j ( t) Also, p j 2,j ( t) p j+2,j ( t) = λ t + o( t) µ 2 ( t) 2 + o(( t) 2 ) = λ + o( t) t ( t) 2 µ + o(( t)2 ) ( t) 2 as t +. So, p j 2,j ( t) > p j+2,j ( t) for small t. Thus the graphs, in Figures 2.4 and 2.5, in order from largest to smallest, for small t, correspond to i = 3, 1, 4,. Note in addition that all of the graphs in Figure 2.1 must have the same limit π 2 as t. We compute π 2 = (1 ρ)ρ 2 = (1 3/4)(3/4) 2 = 9/64.

20 2. VAN DE COEVERING S RESULT AND EXTENSIONS 13 Figure 2.7. p 2j (t) for j =... 4, λ = 3, µ = 4, < t < 1 Next we consider the graphs p 2j (t) for λ = 3 or 1, µ = 4, and < t < 1 in Figures 2.7, 2.8. We know that the limits of the different curves will be π j for j =, 1, 2,.... We also know that π j = (1 ρ)ρ j for j =, 1, 2,.... Further π > π 1 > π 2 >.... Thus, if we look at the right hand side of Figure 2.7, we see that the curves, from largest to smallest, correspond to j =, 1, 2, 3,....

21 2. VAN DE COEVERING S RESULT AND EXTENSIONS 14 Figure 2.8. p 2j (t) for j =... 4, λ = 1, µ = 4, < t < 1 Figure 2.9. p 2j (t) for i = 2, j =, 1, 3, 4,λ = 3, µ = 4, < t < 1

22 2. VAN DE COEVERING S RESULT AND EXTENSIONS 15 Figure 2.1. p 2j (t) for j =, 1, 3, 4, λ = 1, µ = 4, < t < 1 Theorem 2.2. (van de Coevering) For an M/M/1 queueing system, let EL i (t) be the expected number of customers at time t if there are i customers at time. Then EL i (t) = 2 π ρ(j i)/2 π e µtγ(y) γ(y) 2 a i(y)sin(y)dy + ρ/(1 ρ) ρ < 1 i + (λ µ)t + ρ 1 /(ρ 1) ρ > 1. (2) Rather than the above version, we use the following (also given in van de Coevering) for computational purposes. Theorem 2.3. (van de Coevering) For an M/M/1 queueing system, for ρ < 1, EL i (t) = (λ µ)t + µ p i (x)dx + i. (3)

23 2. VAN DE COEVERING S RESULT AND EXTENSIONS 16 Proof. EL i (t) = jp ij (t). j= So, using Kolmogorov s forward differential equations, d dt EL i(t) = = jp ij(t) j= j(µp ij+1 (t) + λp ij 1 (t) (λ + µ)p ij (t)) j= = µ (j + 1)p ij+1 (t) µ p ij+1 (t) + λ (j 1)p ij 1 (t) + λ p ij 1 (t) j= (λ + µ) j= jp ij (t) j= j= = µ jp ij (t) µ(1 p i (t)) + λ jp ij (t) + λ(1) (λ + µ)el i (t) = µel i (t) µ(1 p i (t)) + λel i (t) + λ (λ + µ)el i (t) j= j= j= = λ µ + µp i (t). Integrate with respect to t. EL i (t) = (λ µ)t + µ p i (x)dx + K. Take t =. Then i = EL i () = (λ µ) + µ = EL i (t) = (λ µ)t + µ p i (x)dx + K = K p i (x)dx + i.

24 2. VAN DE COEVERING S RESULT AND EXTENSIONS 17 Figure EL i (t) for λ = 3,µ = 4, i =, 1, 2, 3, 4, < t < 1 Our results appear in Figure A graph of this type appears in Abate and Whitt (1988). They indicate that the only graphs of this type will be either (a) strictly increasing, (b) strictly decreasing, or (c) decreasing to a minimum and then increasing. Graphs of this type also appear in Yu et al. (26). Note that all of the curves EL(i, t) for i = 1, 2, 3, 4 have negative slope for t =. This is not so clear for i = 1 so we magnify that curve in Figure We prove the general result as follows.

25 2. VAN DE COEVERING S RESULT AND EXTENSIONS 18 Figure EL i (t) for λ = 1, µ = 4, < t < 1 Figure EL 1 (t) for λ = 3,µ = 4, < t <.5

26 2. VAN DE COEVERING S RESULT AND EXTENSIONS 19 Figure EL 1 (t) for λ = 1, µ = 4, < t < 1 Property 2.4. (new result) If λ < µ, then del i (t) dt = λ µ <, (4) t= for i = 1, 2... Proof. EL i () = i. EL i ( t) = all j jp ij ( t) = (i + 1)(λ t + o( t)) + i(1 λ t µ t + o( t)) + (i 1)(µ t + o( t)) + o( t)

27 2. VAN DE COEVERING S RESULT AND EXTENSIONS 2 d(el i (t)) dt = i + (λ µ) t + o( t) = lim t= t = λ µ <. EL i ( t) EL i () t Next we consider curves for EL 2 i (t). Theorem 2.5. (van de Coevering) For an M/M/1 queueing system, EL 2 i(t) = 2 π ρ(j i)/2 π e µtγ(y) γ(y) 3 a i(y)sin(y)dy 2ρ(1 ρ) 2 EL i (t) ρ < 1 + 2(ρ 1)it + (ρ 1) 2 t 2 + 2ρ i t + 2ρt EL i (t) + i + i 2 (5) [ρ (2i + 1)(ρ 1)]ρ i (ρ 1) 2 ρ 1. For computational purposes, we use the following. Theorem 2.6. (corrected version of van de Coevering s result) For an M/M/1 queueing system, with ρ < 1, EL 2 i(t) = 2(λ µ) EL i (t)dx + (λ + µ)t µ p i (x)dx + i 2. (6) Proof. EL 2 i(t) = j 2 p ij (t) j=

28 2. VAN DE COEVERING S RESULT AND EXTENSIONS 21 so, d dt EL2 i (t) = j 2 p ij (t) j= = j 2 (µp ij+1 (t) + λp ij 1 (t) (λ + µ)p ij (t)). j= Now if we simplify the summands j= j2 µp ij+1 (t) and j= j2 λp ij 1 (t), then j 2 µp ij+1 (t) = µ (j + 1) 2 p ij+1 (t) 2λ p ij+1 (t) + µ p ij+1 (t)) j= j= j= j= j= = µel 2 i (t) 2µ (j + 1)p ij+1 (t) + µ p ij+1 (t) = µel 2 i(t) 2µEL i (t) + µ(1 p i (t)) j= and j 2 λp ij 1 (t) = λ (j 1) 2 p ij 1 (t) + 2λ jp ij 1 (t) λ p ij 1 (t)) j= j= j= j= = λel 2 i (t) + 2λEL i (t) + λ p ij 1 (t) = λel 2 i (t) + 2λEL i(t) + λ. j= Thus, d dt EL2 i(t) = µel 2 i(t) 2µEL i (t) + µ µp i (t) + λel i (t) + λ (λ + µ)el 2 i(t) = 2(λ µ)el i (t) + (λ + µ) µp i (t).

29 2. VAN DE COEVERING S RESULT AND EXTENSIONS 22 Figure EL 2 i (t) for λ = 3,µ = 4, < t < 1 Integrate with respect to t to get EL 2 i(t) = 2(λ µ) EL i (t)dx + (λ + µ)t µ p i (x)dx + K. Take t = to show K = i 2 = i 2 = EL 2 i() = 2(λ µ) EL i (t)dx + (λ + µ) µ = EL 2 i (t) = 2(λ µ) EL i (t)dx + (λ + µ)t µ p i (x)dx + K = K p i (x)dx + i 2. A graph of E[L i (t) 2 ] appears in Figure 2.15 above.

30 2. VAN DE COEVERING S RESULT AND EXTENSIONS 23 Figure V ar i [L(t)] for λ = 3,µ = 4, < t < 1 From our results for E[L i (t) 2 ] and EL i (t), we can compute var i [L(t)] = E[L(i, t) 2 ] (EL i (t)) 2. The result appears in Figure 2.16 above. As expected, the variance is small for small t and increases as t gets larger. However, since the probabilities p ij (t) π j as t, we expect that the variance should approach a constant limit as t. This limit should be var(l) = E(L 2 ) (E(L)) 2 where E(L) = j= jπ j and E(L 2 ) = j= j2 π j. The result is well known to be var(l) = ρ (1 ρ) 2.

31 2. VAN DE COEVERING S RESULT AND EXTENSIONS 24 Next we consider results for EL 3 i(t)]. Theorem 2.7. (new result) For an M/M/1 queueing system, with ρ < 1, EL 3 i (t) = 3(λ µ) EL 2 i (x)dx + 3(λ + µ) EL i (t)dx + (λ µ)t + µ p i (x)dx + i 3. (7) Proof. EL 3 i (t) = j 3 p ij (t) j= so, d dt EL3 i(t) = j 3 p ij(t). j= Now we simplify the summands j= j3 µp ij+1 (t) and j= j3 λp ij 1 (t), then j 3 µp ij+1 (t) = µ (j + 1) 3 p ij+1 (t) µ 3(j + 1) 2 p ij+1 (t) j= + µ j= 3(j + 1)p ij+1 (t) j= j= p ij+1 (t) j= = µel 3 i (t) 3µEL2 i (t) + 3µEL i(t) µ(1 p i (t)) and j 3 λp ij 1 (t) = λ (j 1) 3 p ij 1 (t) + 3λ (j 1) 2 p ij 1 (t) j= j= j=

32 2. VAN DE COEVERING S RESULT AND EXTENSIONS λ (j 1)p ij 1 (t) + λ p ij 1 (t) j= j= = λel 3 i (t) + 3λEL2 i (t) + 3λEL i(t) + λ. Thus, d dt EL3 i (t) = j 3 (µp ij+1 (t) + λp ij 1 (t) (λ + µ)p ij (t)) j= = µel 3 i(t) 3µEL 2 i(t) + 3µEL i (t) µ(1 p i (t)) + λel 3 i(t) + 3λEL 2 i (t) + 3λEL i (t) + λ (λ + µ)el 3 i(t) = 3(λ µ)el 2 i(t) + 3(λ + µ)el i (t) + (λ µ) + µp i (t). Integrate with respect to t to get EL 3 i(t) = 3(λ µ) EL 2 i(x)dx+3(λ+µ) EL i (t)dx+(λ µ)t+µ p i (x)dx+k. Take t = to show K = i 3 = i 3 = EL 3 i(t) = 3(λ µ) + (λ µ)t + µ p i (x)dx + K = K = EL 3 i(t) = 3(λ µ) + µ p i (x)dx + i 3. EL 2 i(x)dx + 3(λ + µ) EL 2 i(x)dx + 3(λ + µ) EL i (t)dx EL i (t)dx + (λ µ)t

33 2. VAN DE COEVERING S RESULT AND EXTENSIONS 26 We can now obtain a graph for the skewness of L(i, t). (See Wikipedia.) Recall that for a random variable X, skewness(x) = E[(X E[X])3 ] (E[(X E[X]) 2 ]) 3/2. Thus skewness i (L(t)) = E i[(l(t) E i [L(t)]) 3 ] (E i [(L(t) E i [L(t)]) 2 ]) 3/2. First we find lim t + skewness i (L(t)) for i > and i =. Property 2.8. (new result) For an M/M/1 queueing system, with λ < µ, (a) lim t + skewness i(l(t)) = for i >, (b) lim t + skewness i (L(t)) = + for i =. Proof. (a) E(X( t)) = (i 1)p ii 1 ( t) + ip ii ( t) + (i + 1)p ii+1 ( t) + o( t) = (i 1)µ t + i(1 λ µ t) + (i + 1)(λ t) + o( t) = i (µ λ) t + o( t). E[(X( t) E(X( ))) 2 ] = i+1 j=i 1 (j (1 (µ λ) t)) 2 p ij ( t) + o( t)

34 2. VAN DE COEVERING S RESULT AND EXTENSIONS 27 = (1 (µ λ) t) 2 µ t + ((µ λ) t) 2 (1 λ t µ t) + (1 + (µ λ) t) 2 λ t + o( t) E[(X( t) E(X( t))) 3 ] = = (µ + λ) t + o( t). i+1 j=i 1 (j (1 (µ λ) t)) 3 p ij ( t) + o( t) = (1 (µ λ) t) 3 µ t + ((µ λ) t) 2 (i λ t µ t) + (1 + (µ λ) t) 3 λt + o( t) = ( µ + λ) t + o( t). So, E[X( t) E(X( t)) 3 ] (E[(X( t) E(X( )) 2 ]) 3/2 = = ( µ + λ) t + o( t) (µ + λ) 3/2 t 3/2 + o( t) µ + λ + o( t) t (µ + λ) 3/2 t 1/2 + o( t) t as t +. (b) If i =, E(X( t)) = λ t + o( t). E[(X( t) E(X( ))) 2 ] = i (j λ t) 2 p j ( t) + o( t) j= = (λ t) 2 (1 λ t) + (1 λ t) 2 λ t + o( t) = λ t + o( t).

35 2. VAN DE COEVERING S RESULT AND EXTENSIONS 28 E[(X( t) E(X( t))) 3 ] = 1 (j λ t) 3 p ij ( t) + o( t) j= = (λ t) 3 λ t + (1 λ t) 3 λ t + o( t) So, skewness i (L(t)) = = λ t + o( t). λ t + o( t) (λ t + o( t)) 3/2 + as t +.

36 2. VAN DE COEVERING S RESULT AND EXTENSIONS 29 Figure skew i [L(t)] for λ = 3,µ = 4,.1 < t < 1 Hence, we illustrate our graph on the interval (.1,1) rather than on the interval (,1). Skewness measures the lack of symmetry of the pdf. A pdf which is positively skewed has a long tail on the right. If the number of customers at time zero is i =, then the function must be positively skewed (and in fact is + according to our property). If t is large, it is possible that the number of customers is large, but the number of customers has a minimum of zero. In this case we expect the skewness to be positive, and this happens. If t is small and the initial number of customers is i >, then the algebra gives us a negative skewness for small values of t and this appears on the graph.

37 2. VAN DE COEVERING S RESULT AND EXTENSIONS 3 Van de Coevering obtains additional expressions for EL i (t), E i (L(t) 2 ), involving the functions A i (t) and B i (t) which are defined below. We add a function C i (t) so we can deal with E i (L(t) 3 ) as well. Define A i (t) = 2 π ρ(j i)/2 π e µtγ(y) sin(iy)sin(y)dy, (8) γ(y) 2 B i (t) = 2 π ρ(j i)/2 π C i (t) = 2 π ρ(j i)/2 π e µtγ(y) sin(iy)sin(y)dy, (9) γ(y) 3 e µtγ(y) sin(iy)sin(y)dy. (1) γ(y) 4 Property 2.9. For an M/M/1 queue, with λ < µ, EL i (t) = A i (t) ρa i+1 (t) + ρ/(1 ρ). (11) Proof. Recall from (1) that p ij (t) = 2 π ρ(j i)/2 π e µtγ(y) γ(y) a i(y)a j (y)dy + (1 ρ)ρ j ρ < 1 ρ 1, for γ(y) = 1 + ρ 2 ρcos(y) and a k (y) = sin(ky) ρsin((k + 1)y). Substituting the expression for p i (t) from (1) into (3) gives: EL i (t) = (λ µ)t + µ = (λ µ)t + µ p io (x)dx + i [ 2 π ρ i/2 π e µxγ(y) (sin(iy) γ(y)

38 2. VAN DE COEVERING S RESULT AND EXTENSIONS 31 ρ 1/2 sin((i + 1)y))( ρ 1/2 siny)dy + 1 ρ] dx + i = (λ µ)t + i + µ(1 ρ)t + µ 2 π ρ i/2 π e µxγ(y) γ(y) (sin(iy) ρ 1/2 sin((i + 1)y))( ρ 1/2 siny)dydx = (λ µ)t + i + µ(1 ρ)t µ 2 π ρ(1 i)/2 π ρ 1/2 sin((i + 1)y)siny)dxdy = (λ µ)t + i + µ(1 ρ)t 2 π ρ(1 i)/2 π [sin(iy)siny ρ 1/2 sin((i + 1)y)siny] dy = (λ µ)t + i + µ(1 ρ)t 2 π ρ(1 i)/2 π 2 π π ρ (2 i)/2e µtγ(y) 1 γ 2 (y) sin(iy)siny dy + 2 π ρ(2 i)/2 e µxγ(y) (sin(iy) sin y γ(y) [ e µtγ(y) γ 2 (y) + 1 γ 2 (y) ] e µtγ(y) γ 2 (y) γ 2 (y) sin((i + 1)y)siny dy 2 π ρ(1 i)/2 π sin(iy)siny dy π 1 sin((i + 1)y)siny dy γ 2 (y) = A i (t) ρa i+1 (t) A i () + ρa i+1 () + i. (12) As A i (t) as t we have EL i (t) ρ/(1 ρ) for all i, so EL i (t) = A i (t) ρa i+1 (t) A i () + ρa i+1 () + i ρ/(1 ρ) = A i + ρa i+1 () + i ρa i+1 () A i () = ρ/(1 ρ) + i. (13)

39 Substitute (12) into (11) to get 2. VAN DE COEVERING S RESULT AND EXTENSIONS 32 EL i (t) = A i (t) ρa i+1 (t) + ρ/(1 ρ) i (λ µ)t + i + (λ µ)t = A i (t) ρa i+1 (t) + ρ/(1 ρ). Lemma 2.1. For an M/M/1 queue with ρ < 1, lim t, EL 2 i(t) = (λµ + λ2 ) (µ λ) 2 = ρ2 + ρ (1 ρ) 2. Proof. As t, EL 2 i (t) E(L2 ). E(L 2 ) = i 2 π i = i= i 2 (1 ρ)ρ i = (1 ρ)ρ i 2 ρ i 1 i= = ρ(1 ρ)[ i(i 1)ρ i 2 ρ + = i= (i 2 i)ρ i 2 ρ + i= i= iρ i 1 2ρ = ρ(1 ρ)[ (1 ρ) (1 ρ) 2] = ρ2 + ρ (1 ρ) 2 iρ i 1 ] i= i=

40 2. VAN DE COEVERING S RESULT AND EXTENSIONS 33 Property For an M/M/1 queue, with λ < µ, EL 2 i (t) = 2(ρ 1)[ B i(t) + ρb i+1 (t)] EL i (t) + 2ρ (1 ρ) 2. (14) Proof. EL i (x)dx = A i (x) ρa i+1 (x) A i () + ρ/(1 ρ)dx = ρt/(1 ρ) + = ρt/(1 ρ) + ρt/(1 ρ) + A i (x) ρa i+1 (x)dx 2 π π ρ(1 i)/2 π 2 π ρi/2 e µtγ(y) γ 2 (y) e µtγ(y) γ 2 (y) sin(iy)siny dydx sin((i + 1)y)sin y dydx = ρt/(1 ρ) 1 µ (B i(t) B i ()) + ρ µ B i+1(t) 1 µ B i() = ρt/(1 ρ) + 1 µ [ Bi(t) + ρb i+1(t) + B i () ρb i+1 ()]. (15) By (3),(1) and (14) get: EL 2 i(t) = 2(λ µ) EL i (x)dx + (λ + µ)t µ p i (x)dx + i 2 = 2(λ µ) 1 µ [ B i(t) + ρb i+1 (t) + B i () ρb i+1 ()] + 2(λ µ)ρt/(1 ρ) + (λ + ρ)t + (λ µ)t + i EL i (t) + i 2 = 2(ρ 1) 1 µ [ B i(t) + ρb i+1 (t) + B i () ρb i+1 ()] ρ + 2(ρ 1) 1 ρ t + 2λt EL i(t) + i + i 2 ρ 1 ρ = λ/µ 1 (λ µ) = λ/µ (µ λ)/µ = λ µ λ

41 2. VAN DE COEVERING S RESULT AND EXTENSIONS 34 = 2(ρ 1) 1 µ [ B λ i(t) + ρb i+1 (t) + B i () ρb i+1 ()] 2(ρ 1) λ µ t + 2λt EL i (t) + i + i 2 = 2(ρ 1)[ B i (t) + ρb i+1 (t) + B i () ρb i+1 ()] EL i (t) + i + i 2. (16) Letting t and using that EL 2 i(t) (λµ + λ2 ) (µ λ) 2, (λµ + λ 2 ) (µ λ) 2 = 2(ρ 1)[ + + B i () ρb i+1 ()] ρ 1 ρ + i + i2 2(ρ 1)[B i () ρb i+1 ()] = (λµ + λ2 ) (µ λ) 2 + ρ 1 ρ i i2 = 2λµ (µ λ) 2 i i2 Thus, 2(ρ 1)[B i () ρb i+1 (t)()] = 2λµ (µ λ) 2 i i2. (17) Substituting (16) into (15) gets: EL 2 i(t) = 2(ρ 1)[ B i (t) + ρb i+1 (t)] + 2µλ (µ λ) 2 i i2 EL i (t) + i + i 2 = 2(ρ 1)[ B i (t) + ρb i+1 (t)] EL i (t) + 2ρ (1 ρ) 2. Lemma (new result) For an M/M/1 queue, with λ < µ, lim t EL 3 i(t) = E(L 3 ) = ρ3 + 4ρ 2 + ρ (1 ρ) 3.

42 2. VAN DE COEVERING S RESULT AND EXTENSIONS 35 Proof. E(L 3 ) = = i 3 π i i= i 3 (1 ρ)ρ i i= = (1 ρ)[ρ 3 i(i 1)(i 2)ρ i 3 + 3ρ 3 i 2 ρ i 3 2ρ 3 iρ i 3 ] i= = (1 ρ)[ρ 3 i(i 1)(i 2)ρ i i= i= i= i= i= i 2 ρ i 2ρ iρ i ] = (1 ρ)ρ 3 i(i 1)(i 2)ρ i 3 + 3(1 ρ) i 2 ρ i 2ρ(1 ρ) iρ i = (1 ρ)ρ 3 6 (1 ρ) 4 + (1 ρ)rho2 + ρ (1 ρ) 3 2ρ(1 ρ) 1 (1 ρ) 2 = ρ3 + 4ρ 2 + ρ (1 ρ) 3 i= i= i= Property (new result) For an M/M/1 queue, with λ < µ, EL 3 i (t) = 6µ(ρ 1)2 [ C i (t) + ρc i+1 (t)] + 5ρ3 ρ 2 + 2ρ (1 ρ) 3 3 ρ 1 EL2 i (t) 4 ρ ρ 1 EL i(t). (18) Proof. Together with (3), (5) and (13) get, EL 3 i(t) = 3(λ µ) + µ EL 2 i(x)dx + 3(λ + µ) p i (x)dx + i 3 EL i (x)dx + (λ µ)t

43 = 3(λ µ) 2. VAN DE COEVERING S RESULT AND EXTENSIONS (µ + λ) = 6(λ µ)(ρ 1) 2(ρ 1)[ B i (x) + ρb i+1 (x)] EL i (x) + 2ρ(1 ρ) 2 dx EL i (x)dx + (λ µ)t + EL i (t) (λ µ)t i + i 3 + 6(λ µ)ρ(1 ρ) 2 t + 3(µ + λ) (λ µ)t i + i 3 [ B i (x) + ρb i+1 (x)] dx 3(λ µ) = 6µ(ρ 1) 2 [ B i (x) + ρb i+1 (x)] dx 6µ + 6µ(ρ 1)ρ(1 ρ) 2 t + EL i (t) i + i 3 EL i (x)dx EL i (x)dx + (λ µ)t + EL i (t) EL(i, x)dx = 6µ(ρ 1) 2 [ B i (x) + ρb i+1 (x)] dx 6µ/µ[ B i (t) + ρb i+1 (t) + B i () ρb i+1 ()] 6µρt/(1 ρ) + 6µρt/(1 ρ) + EL i (t) i + i 3 = 6µ(ρ 1) 2 [ B i (x) + ρb i+1 (x)] dx 3(EL2 i (t) + EL i (t) 2ρ(1 ρ) 2 ) ρ 1 + EL i (t) i + i 3 = 6µ(ρ 1) 2 [ C i (t) + C i () + ρc i+1 (t) ρc i+1 ()] 3 ρ 1 EL2 i(t) 4 ρ ρ 1 EL i(t) i + i 3, when t, the equation becomes ρ 3 + 4ρ 2 + ρ = 6µ(ρ 1) 2 [C (1 ρ) 3 i () ρc i+1 ()] 3 λµ + λ 2 ρ 1 (µ λ) 2 4 ρ ρ ρ 1 1 ρ i + i3

44 2. VAN DE COEVERING S RESULT AND EXTENSIONS 37 6µ(ρ 1) 2 [C i () ρc i+1 ()] = ρ3 + 4ρ 2 + ρ 3ρ + 3ρ2 (1 ρ) 3 (1 ρ) 2 + 4ρ ρ2 (1 ρ) + i 2 i3 = ρ3 + 4ρ 2 + ρ 4ρ + ρ2 + (1 ρ) 3 (1 ρ) + i 2 i3 = 5ρ3 ρ 2 + 2ρ (1 ρ) 3 + i i 3. Thus, EL 3 i(t) = 6µ(ρ 1) 2 [ C i (t) + ρc i+1 (t)] + 5ρ3 ρ 2 + 2ρ (1 ρ) 3 + i i 3 3 ρ 1 EL2 i(t) 4 ρ ρ 1 EL i(t) i + i 3 = 6µ(ρ 1) 2 [ C i (t) + ρc i+1 (t)] + 5ρ3 ρ 2 + 2ρ (1 ρ) 3 3 ρ 1 EL2 i(t) 4 ρ ρ 1 EL i(t).

45 CHAPTER 3 Task versus Queue? MODEL: We have an M/M/1 queueing system. A customer arrives and sees i people in the system. This customer must receive service from the server and has to complete an additional task of fixed length D. The customer has two choices perform the task first or join the queue first. D S D S Figure 3.1. Which is better? 38

46 3. TASK VERSUS QUEUE? 39 (a) Let T QD be the total system time if the customer join Q first, then does the task (b) Let T DQ be the total system time if the customer join D first, then does the queue Assume i is the number of customers initially observed. Let p ij (t) = Prob(j customers at time t there are i customers at time ) Let E(T QD ) be the expected system time if the queue is done first; E(T DQ ) be the expected system time if the task is done first. This question was addressed by Hlynka and Molinaro (21). However, no graphical analysis. A graphical analysis provides an easy way of describing the results. Theorem 3.1. Given the setting above, (a) E(T QD ) = i + 1 µ + D (b) E(T DQ ) = D + EL i(d) + 1 µ Proof. Let X i, i = 1, 2, 3,... be independent and identically distributed service time random variables. (a) If the arriving customer joins the queue first, then the expected total time to complete the queue part is E( i+1 j=1 X j) because of the memoryless property and because the arriving customer must also be served. Then the customer must complete the task with time length D. Thus i+1 E(T QD ) = E( j=1 i+1 X i + D) = E(X i ) + D = j=1 i+1 j=1 1 µ + D = i + 1 µ + D. (b) If the arriving customer performs the task first, then upon completion of the task, the number of customers in the queueing system will be L i (D). Therefore, the time

47 to complete the queue part is L i (D)+1 j=1 X j. Hence, 3. TASK VERSUS QUEUE? 4 EL DQ = E(D + L i (D)+1 j=1 X j ) = D + E(L i (D) + 1)E(X j ) = D + E(L i(d) + 1). µ Theorem 3.2. (new result) For the setting above, E(T DQ ) < E(T QD ) iff EL i (D) < i. Proof. This follows immediately from the previous theorem. PROCEDURE: Suppose an arriving customer observes i customers in the M λ /M µ /1 queueing system upon arrival. (a) Draw the curve y = E(L i (t)). (b) Draw the line segment from (, i) until it intersects the curve (if it does) at (D, i). (c) If the task has time length D which is less than D, then perform the task first (in order to minimize the expected total system time for the arriving customer. Proof. From the previous theorem, E(T DQ ) < E(T QD ) iff EL i (D) < i. From Abate and Whitt (1987), the curve must have one of the three forms as illustrated by Diagram Hence if there is an intersection point (D, i), then EL i (D) < i iff D < D. We illustrate this in Figure 3.2.

48 3. TASK VERSUS QUEUE? 41 Figure 3.2. EL 1 (t) for λ = 3,µ = 4, < t <.5 All the diagrams are plotted by maple.

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