Solution manual for system modeling and analysis by kobayashi

Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ “Vo 200 pag ✐ ✐ ✐ C H A P T E R Basic Queueing Models 2.2 LITTLE’S FORMULA AND ITS GENERALIZATION 2.2-1 Little’s formula (a) (b) (c) (d) False : Little’s formula holds for arbitrary arrival processes False : For the same reason as part (a) True: Little’s formula holds for arbitrary work-conserving disciplines True: For the same reason as part (a) 2.2-2 Little’s formula for multiple type jobs Little’s law can be be generalized to Qr = λr E[Wr ], (2.1) where Qr is the mean number of type r jobs in queue and E[Wr ] is the mean waiting time for type r jobs in queue, for r = 1, , R The average queue size for a type r job is given by (2.1) for the FCFS queue discipline or any other work-conserving discipline 2.2-3 Distributions seen by arrivals and departures In the interval [0, T ], for each arrival which causes Q(t) to increase from n to n + (n = 0, 1, · · · ), there must be a corresponding departure that causes Q(t) to decrease from n + to n (since Q(0) = Q(T ) = 0) This implies that the average queue size seen by an arrival is the same as that seen by a departure in the interval [0, T ] BIRTH-AND-DEATH PROCESSES 2.3-1 Superposition of Poisson processes (a) We have P [Y ≥ y] = P [X1 ≥ y, X2 ≥ y, , Xm ≥ y] = P [X1 ≥ y] · P [X2 ≥ y] · · · P [Xm ≥ y] = e−λ1 y · e−λ2 y · · · e−λm y = e− where λ m i=1 m i=1 λi y = e−λy , λi Therefore, FY (y) = − e−λy , y ≥ 0, so Y is exponentially distributed with parameter λ ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ ✐ Chapter “Vo 200 pag Basic Queueing Models (b) Let Xj denote the inter-arrival time of the jth arrival stream Since the arrival streams are independent, so to are the random variables Xj , j = 1, , m Furthermore, since the jth arrival stream is Poisson, Xj is an exponentially distributed random variable with parameter λj The inter-arrival time of the aggregate stream is given by Y = min{X1 , , Xm } From the result of part (a), we can conclude that Y is exponentially dism tributed with parameter λ = i=1 λi Therefore, the aggregate stream is a Poisson process with rate λ 2.3-2 Consistency check of the Poisson process (a) Let Ih denote a small interval of length h From (2.3-2) we have P [N (h) = 0] = e−λh (λh)2 (λh)3 (λh)4 = − λh + − + + ··· 2! 3! 4! = − λh + o(h), P [1 arrival in Ih ] = P [N (h) = 1] = λhe−λh = λh(1 − λh + o(h)) = λh + o(h), P [no arrival in Ih ] = ∞ P [≥ arrivals in Ih ] = ∞ P [N (h) = j] = j=2 j=2 (λh)j −λh ·e = o(h) j! (b) Let X1 denote the time of the first arrival after the time origin (say t = 0) and X2 denote the inter-arrival time between the first arrival and the second arrival The RVs X1 and X2 are both exponentially distributed with parameter λ and have a common cdf: FX (x) = − e−λx , x ≥ The event of no arrival in the interval Ih is equivalent to the event {X1 > h} Therefore, P [no arrival in Ih ] = P [X1 > h] = e−λh = − λh + o(h) The event of two or more arrivals in the interval Ih is equivalent to the event {X1 + X2 ≤ h} Let Y = X1 + X2 Then, P [≥ arrivals in Ih ] = P [Y ≤ h] = FY (h) (2.2) There are several ways of determining the cdf FY (y) Since X1 and X2 are independent, the pdf of Y is given by fY (y) = fX1 (y) fX2 (y) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag One can further show that in general, the cdf of Y is given by FY (y) = FX1 (y) fX2 (y) = fX1 (y) FX2 (y) (2.3) Therefore, y FY (y) = FX (y) fX (y) = y λe−λy = (1 − e−λx ) · λe−λ(y−x) dx (eλx − 1)dx = − e−λy − λye−λy Returning to (2.2), we obtain P [2 or more arrivals in Ih ] = FY (h) = − e−λh − λhe−λh = − (1 − λh + o(h)) − (λh + o(h)) = o(h) Finally, P [one arrival in Ih ] = − P [no arrival in Ih ] − P [≥ arrivals in Ih ] = − (1 − λh + o(h)) − o(h) = λh + o(h) To prove (2.3), note that y FY (y) = y fX1 (x) y = 0 x fX1 (x − t)fX2 (t) dt dx y = fX2 (x) dx = t y y fX1 (x − t)fX2 (t) dx dt = y−t FX1 (y − t)fX2 (t) dt = FX1 (y) fX1 (α) dα fX2 (t) dt fX2 (y) 2.3-3 Decomposition of a Poisson process (a) We are given that {Xj } is a sequence of i.i.d random variables, exponentially distributed with parameter λ Then for fixed n, Sn = X1 + · · · + Xn and has an Erlang-n distribution The cdf of Sn is given by FSn (x) = P [Sn ≤ x] = − P [< n arrivals in an interval of length x] n−1 = n−1 P [A(x) = j] = − e−λx 1− j=0 j=0 (λx)j j! Therefore, n−1 n−1 P [A(x) = j] = − e−λx P [Sn > x] = j=0 j=0 (λx)j j! Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ ✐ Chapter “Vo 200 pag Basic Queueing Models Hence, ∞ P [SN > x] = P [SN > s|N = n]P [N = n] n=1 ∞ n−1 e−λx = n=1 ∞ j=0 n (λx)j · (1 − r)n−1 r j! = re−λx n=0 j=0 ∞ ∞ = re−λx j=0 n=j ∞ ∞ = re−λx (λx)j · (1 − r)n j! (λx)j · (1 − r)m+j j! j=0 m=0 ∞ = re (λx)j · (1 − r)n j! −λx j=0 ∞ [(λ(1 − r)x]j · (1 − r)m j! m=0 = re−λx · eλ(1−r)x r = e−λrx which shows that SN has an exponential distribution with parameter λr (b) In decomposing the Poisson stream into m substreams, each arrival is assigned independently to the kth substream with probability rk , where m k=1 rk = Consider an arrival that is assigned to the kth substream The number of subsequent arrivals of the original Poisson stream until the next arrival that is assigned to the kth substream is a random variable Nk with distribution P [Nk = n] = (1 − rk )n−1 rk , n = 0, 1, · · · Therefore, the inter-arrival time between arrivals assigned to the kth substream is a random variable SNk = X1 + · · · XNk , where Xi are inter-arrival times of the original Poisson process Hence, the Xi are i.i.d and exponentially distributed with parameter λ By the result from part (a), SNk is exponentially distributed with parameter rk λ Therefore, the kth substream is Poisson with rate rk λ 2.3-4 Alternate decomposition of a Poisson stream Let Xi represent the interarrival time between the ith and the (i + 1)-st arrival For substream 1, the time between the first and the second arrival is given by Y = X1 + X2 + · · · Xm Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag The event {Y ≤ y} is equivalent to the event that there are fewer than m arrivals of the original Poisson stream in an interval of length y, i.e., FY (y) P [Y ≤ y] = − P [< m arrivals in an interval of length y] m−1 = 1− m−1 P [A(y) = j] = − j=0 j=0 (λy)j −λy e , j! which is an Erlang-m distribution with mean m/λ 2.3-5 Derivation of the Poisson distribution (a) Equation (2.3-9) for n = can be written as: d ln(P0 (t)) = −λ, dt which is a simple, separable first-order differential equation Integrating both sides and solving for P0 (t) yields P0 (t) = Ke−λt , where the constant K is determined by the initial condition P0 (0) = Hence, K = and P0 (t) = e−λt (2.4) Substituting (6.5) into (2.3-9) for n = 1, we obtain P1 (t) + λP1 (t) = λe−λt (2.5) Equation (2.5) is a first-order differential equation that can be reduced to a separable form by multiplying both sides by an integrating factor More generally, let us re-write (2.5) as follows: P1 (t) + R(t)P1 (t) = Q(t), (2.6) where in this case, R(t) = λ and Q(t) = λe−λt The integrating factor can be obtained by supposing that the left-hand side of (2.6) to be the derivative of a product φ(t)P1 (t), given by φ(t)P1 (t) + φ (t)P1 (t) (2.7) Multiplying the left-hand side of (2.6) by φ(t), we have φ(t)P1 (t) + φ(t)R(t)P1 (t) = φ(t)Q(t) (2.8) Equating the left-hand side of (2.8) with (2.7), we see that they can be made equal by choosing φ(t) such that φ (t) = φ(t)R(t) (2.9) This is a simple separable equation that has the solution φ(t) = e R(t)dt , (2.10) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ ✐ Chapter “Vo 200 pag Basic Queueing Models which is the integrating factor we seek After multiplying (2.5) by the integrating factor φ(t), we obtain: d e dt R(t)dt R(t)dt P1 (t) = Q(t)e (2.11) The left-hand side is an exact derivative that can be integrated directly In particular, we have d λt [e P1 (t)] = dt λ (2.12) Hence, we obtain (using the fact that P1 (0) = 1): P1 (t) = λte−λt (2.13) To proceed by induction, we postulate the result Pn (t) = (λt)n −λt e n! (2.14) and show that the result holds for Pn+1 (t) From (4.61), we have: Pn+1 (t) + λPn+1 (t) = λPn (t) = λ (λt)n −λt e n! Multiplying both sides by the integrating factor φ(t) = eλt , we have: (λt)n d λt [e Pn+1 (t)] = λ dt n! Integrating both sides and using the fact that Pn+1 (0) = 0, we obtain the required result: Pn+1 (t) = (λt)(n+1) −λt e (n + 1)! Thus, by induction, we have shown the validity of (2.14) for all n ≥ (b) Taking the Laplace transform of the system of differential equations (4.61) and (4.62), we have: sPn∗ (s) − Pn (0) = sP0∗ (s) − P0 (0) = ∗ −λPn∗ (s) + λPn−1 (s), n ≥ ∗ −λP0 (s) From (2.16) and the fact that P0 (0) = 1, we obtain P0∗ (s) = that Pn (0) = in (2.15) we have, in particular for n = 1, s+λ (2.15) (2.16) Noting (s + λ)P1∗ (s) = λP0∗ (s) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Therefore, P1∗ (s) = show that λ (s+λ)2 Basic Queueing Models “Vo 200 pag Using induction, it is straightforward to Pn∗ (s) = λn (s + λ)n+1 (2.17) Inverting (2.17), we obtain the desired result: (λt)n −λt e , n! where we have used the following Laplace transform properties: Pn (t) = L−1 {f ∗ (s + a)} = L−1 = n+1 s f (t)e−at tn n! (2.18) (2.19) 2.3-6 Uniformity of Poisson arrivals (a) Suppose there are n arrivals in the interval (0, T ] The joint probability that there are i arrivals in a subinterval (0, t], one arrival in (t, t + h], and n − i − arrivals in (t + h, T ] is the product of three factors obtained from the Poisson distribution: (λt)i −λt e i! = λn he−λT (n − 1)! λhe−λh [λ(T − t − h)]n−i−1 −λ(T −t−h) e (n − i − 1)! n−1 i t (T − t − h)n−i−1 i Summing the above expression over the possible values of i, we find that the joint probability that there are n arrivals in (t, T ] with one arrival in (t, t + h] is λn he−λT (n − 1)! n n−1 i [λ(T − h)]n−1 t (T − t − h)n−i−1 = λhe−λT , i (n − 1)! i=0 (2.20) where we used the binomial formula k i=0 k i k−i xy = (x + y)k i Since h is an infinitesimal interval, we rewrite (2.20) as P [n arrivals in (0, T ], arrival in (t, t + h]] = (λT )n−1 λhe−λT + o(h), (n − 1)! obtaining the conditional probability P [1 arrival in (t, t + h]|n arrivals in (0, T ]] = = (λT )n−1 −λT + (n−1)! λhe n (λT ) −λT n! e o(h) nh + o(h) T Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ ✐ Chapter “Vo 200 pag Basic Queueing Models (b) Since the n arrivals are independent, any one of them will fall into the interval (t, t + h] with equal chance Thus, the conditional probability that a call arrives in (t, t + h], given that it is one of n arrivals in (0, T ] is Th This final expression is independent of n, hence the conditional probability is unconditional Thus, we have proved (2.3-34) 2.3-7 Pure birth process When λ(n) = λ and µ(n) = for all n ≥ 0, the differential-difference equations of the B-D process become: pn (t) = p0 (t) = −λpn (t) + λpn−1 (t), n = 1, 2, · · · , −λp0 (t) (2.21) (2.22) Using the same procedure as in Exercise 2.3-6, these equations can be solved to obtain the Poisson distribution: pn (t) = (λt)n −λt e , n = 0, 1, · · · n! 2.3-8 Time-dependent solution When µ(n) = for all n ≥ 0, but statedependent birth rates λ(n) are permitted, the differential-difference equations of the B-D process become: pn (t) = p0 (t) = −λ(n)pn (t) + λ(n − 1)pn−1 (t), n = 1, 2, · · · , −λp0 (t) (2.23) (2.24) If we multiply both sides of (2.23) by the integrating factor eλ(n)t , we obtain: d λ(n)t [e pn (t)] = λ(n − 1)pn−1 (t)eλ(n)t dt (2.25) After integrating both sides from to t and re-arranging, we obtain: t pn (t) = e−λ(n)t λ(n − 1) pn−1 (x)eλ(n)x dx + K , (2.26) where K is a constant determined by the initial condition pn (0) = K 2.3-9 Pure death process When λ(n) = and µ(n) = µ for all n ≥ 0, the differential-difference equations of the B-D process become: pN0 (t) pn (t) p0 (t) = −µpN0 (t), = −µpn (t) + µpn+1 (t), n = 1, · · · , N0 − = µp1 (t) (2.27) (2.28) (2.29) Solving (2.27), we obtain pN0 (t) = e−µt (2.30) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag Similar to the approach in Problem 4.7, we can obtain from (2.28) and (2.29), the following result: t pn (t) = µe−µt eµx pn+1 (x)dx, n = 1, · · · , N0 − (2.31) Applying (2.31) successively for n = 1, 2, · · · , N0 − 1, we obtain: pn (t) = (µt)N0 −n −µt e , n = 1, · · · N0 (N0 − n)! For each t we have: N0 p0 (t) + pn (t) = n=1 Thus, we find that N0 p0 (t) = − (µt)N0 −n −µt e =1− (N0 − n)! n=1 N0 −1 n=0 (µt)n −µt e n! 2.3-10 The time-dependent PGF Multiply both sides of (2.21) and (2.22) by z n and sum from n = to ∞ to obtain: ∞ ∞ pn (t)z n = −λ n=0 ∞ n pn−1 (t)z n , pn (t)z + λ n=0 (2.32) n=1 which is equivalent to ∂ G(z, t) ∂t = −λG(z, t) + λzG(z, t) (2.33) = −λ(1 − z)G(z, t) (2.34) Equation (2.34) is a simple, separable first order differential equation whose solution is: ∞ G(z, t) = e−λ(1−z)t = e−λt n=0 Hence, pn (t) = e−λt (λt)n n z n! (2.35) (λt)n n! 2.3-11 Time-dependent solution for a certain BD process When λn = λ and µn = nµ for all n, equations (2.3-46) become pn (t) = −(λ + nµ)pn (t) + λpn−1 (t) + (n + 1)µpn+1 (t), p0 (t) = −λp0 (t) + µp1 (t) n = 1, 2, 3, · · · , (2.36) (2.37) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 10 ✐ Chapter “Vo 200 pag Basic Queueing Models Multiply both sides of (2.36) by z n , sum from n = to ∞ and then add (2.37) to obtain ∞ ∞ ∞ pn (t)z n = −λ n=0 pn (t)z n − µz n=0 ∞ npn (t)z n−1 n=0 ∞ pn (t)z n + µ + λz n=0 npn (t)z n−1 (2.38) n=0 Noting that ∞ ∞ ∂ ∂ G(z, t) = G(z, t) = pn (t)z n and npn (t)z n−1 , ∂t ∂z n=0 n=0 we can rewrite (2.38) as ∂ ∂ G(z, t) = (z − 1) λG(z, t) + µ G(z, t) ∂t ∂z (2.39) One can easily verify by direct substitution that G(z, t) = exp λ (1 − e−µt )(z − 1) µ (2.40) is the unique solution to (2.39) Since pn (t) is the coefficient of z n in the power series expansion of G(z, t), we have pn (t) = ∂n G(z, t) |z=0 n! ∂z n (2.41) From (2.40), we obtain that ∂ λ G(z, t) = (1 − e−µt )G(z, t), ∂z µ which implies that ∂n λ G(z, t) = (1 − e−µt ) ∂z n µ n G(z, t) (2.42) From (2.40), (2.41), and (2.42), we obtain pn (t) = λ µ (1 − e−µt ) n! n G(z, t)|z=0 = λ µ (1 − e−µt ) n! n exp −λ (1 − e−µt ) µ 2.4 BIRTH-AND-DEATH QUEUEING MODELS 2.4-1 Splitting a Poisson stream Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag 11 (a) As shown in Section 2.3, the k-th substream is a Poisson process with rate pk λ, k = 1, 2, , K, and the K Poisson streams are statistically independent (b) The interarrival time T of each substream is K-stage Erlangian distributed with mean K/λ K−1 FT (t) = − e−λt j=0 (λt)j , t ≥ j! 2.4-2 Erlangian distribution (a) The LT of an exponential random variable with mean µ is given by f ∗ (s) = µ s+µ fY∗ (s) = nλ s + nλ Therefore, the LT of Yi is Since X = Y1 + · · · + Yn , nλ s + nλ ∗ (s) = [fY∗ (s)]n = fX n (2.43) (b) The pdf of X can be obtained by inverting the LT given in (2.43) Using properties of the LT we have 1 = (nλ)n e−λnx L−1 n n (s + nλ) s n−1 n x (nλx) = (nλ)n e−λnx · = e−λnx (n − 1)! x(n − 1)! fX (x) = (nλ)n L−1 (c) The mean of X is E[X] = nE[Yi ] = n 1 = nλ λ The variance of X is Var[X] = nVar[Yi ] = n 1 = (nλ)2 nλ2 2.4-3 Erlangian distribution (continued) The service completions of customers through n may be considered as arrivals of a Poisson process, since the service times are exponentially distributed and i.i.d The event that the Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 12 ✐ Chapter “Vo 200 pag Basic Queueing Models total time to serve n customers, W , exceeds some value x is equivalent to the event that there are n − or fewer arrivals in the interval [0, x], i.e., n−1 P [W > x|n] = P [n − or fewer arrivals in [0, x)] = j=0 (µx)j −µx e j! Hence, n−1 FW (x|n) = P [W ≤ x|n] = − j=0 (µx)j −µx e j! 2.4-4 Balance equation of M/M/1 (a) The detailed balance equation for the M/M/1 queue equates the probability flow rate from state n − to state n with that from state n to state n − The flow rates must be equal if the queue reaches a stable equilibrium Therefore, the balance equations are given by µpn = λpn−1 , n = 1, 2, · · · , (2.44) or equivalently, pn = ρpn−1 , n = 1, 2, · · · , (2.45) where ρ = λ/µ (b) Multiply both sides of (2.45) by z n and sum from n = to ∞ to obtain ∞ ∞ pn z n = ρ n=1 pn−1 z n (2.46) n=1 Simplifying the above equation, we have G(z) − p0 = ρzG(z) (2.47) Solving for G(z), we have G(z) = p0 − ρz Using the fact that G(z) = 1, we find that p0 = − ρ Hence, G(z) = 1−ρ − ρz (c) Expanding G(z) as a power series, we find that ∞ G(z) = (1 − ρ)ρn z n n=0 Hence, we see that pn = (1 − ρ)ρn for n = 0, 1, · · · Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter Basic Queueing Models “Vo 200 pag 13 2.4-5 PASTA and related properties in the M/M/1 queue ✐ (a) Due to the uniformity property of the Poisson process, the proportion of arriving calls in the interval (0, T ) that find n in the system can be expressed as the following ratio: an = expected number of arrivals in (0, T ) that find n in the system expected number of arriving calls in (0, T ) The expected number of calls during (0, T ) that find exactly n calls in the system is λpn T Thus, an λpn T ∞ i=0 λpi T = = λpn T = pn λT (b) From Problem (2.2-1), we know that an = dn holds for any workconserving queueing discipline Hence, in this case pn = dn , i.e., the probability distribution of the number of system seen by departing customers is {pn } (c) If the arrival process is state-dependent, i.e., the arrival rate λ(n) depends on the state of the system, then an = λ(n)pn T = ∞ i=0 λ(i)pi T λ(n)pn , ∞ i=0 λ(i)pi which does not equal pn in general 2.4-6 Derivation of the waiting time distribution From (2.4-25) we have ∞ = − (1 − ρ)e ∞ n−1 ρn − (1 − ρ)e−µx FW (x) = − ρ + (1 − ρ) n=1 ∞ −µx = − (1 − ρ)e−µx = − ρe n=1 j=0 ∞ ρn j=0 n=j+1 ∞ j+1 j=0 −µ(1−ρ)x ρn (µx) j! (µx)j j! j ∞ ρ (µx)j (ρµx)j = − ρe−µx − ρ j! j! j=0 2.4-7 Laplace transform method The waiting time experienced by a call that arrives with n ≥ calls ahead of it in the system is given by: W = R1 + S2 + · · · + Sn , (2.48) where R1 , S2 , · · · , Sn are i.i.d according to an exponential distribution of parameter µ If there are calls ahead of the arriving call, its waiting time will be That is, the conditional pdf of W given N = is given by fW (t|0) = δ(t), Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 14 ✐ Chapter “Vo 200 pag Basic Queueing Models where δ(t) is the Dirac delta function Therefore, the Laplace transform of the conditional pdf of W given N = n is given by: ∗ fW (s|n) µ s+µ = n 1, , n≥1 n=0 (2.49) The Laplace transform of the pdf of W is obtained by unconditioning as follows: ∞ ∗ fW (s) ∞ ∗ an fW (s|n) = = n=0 ∞ n=0 µ s+µ (1 − ρ)ρn = n=1 = ∗ pn fW (s|n) ∞ n + (1 − ρ) = (1 − ρ) n=1 µρ s+µ µρ(1 − ρ) + (1 − ρ) s + µ(1 − ρ) n +1 (2.50) Taking the inverse transform of (2.50), we have that the pdf of W is given by: fW (t) = µρ(1 − ρ)e−µ(1−ρ)t + (1 − ρ)δ(t), t ≥ (2.51) The cdf of W is then given by: t FW (t) = fW (x) = − ρe−µ(1−ρ)t , t ≥ 0 The time in system, T is related to the waiting time W by T = W + S, where S is the service time, which is exponentially distributed with parameter µ and independent of W Hence, the Laplace transform of the pdf of T is given by: fT∗ (s) = = ∗ fW (s)fS∗ (s) µ µ(1 − ρ) (1 − ρ) = , µρ − s+µ s+µ s + µ(1 − ρ) which implies that T is exponentially distributed with parameter µ(1 − ρ) 2.4-8 For an M/M/1 system, πn lim P [N (t) = n] = ρn (1 − ρ), n = 0, 1, 2, · · · t→∞ Hence, ∞ lim P [N (t) > n] = t→∞ ∞ ρj (1 − ρ) = ρn+1 πn = j=n+1 j=n+1 2.4-9 Cumulative Poisson distribution Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag 15 (a) The left-hand side (LHS) is given by k k ak−j e−a1 (k − j)! P (k − j; a1 )Q(j; a2 ) = LHS = j=0 k j=0 l = l=0 i=0 k ai2 −(a1 +a2 ) al−i e = (l − i)! i! l=0 j i=0 ai2 −a2 e i! (a1 + a2 )l −(a1 +a2 ) e l! (2.52) = Q(k; a1 + a2 ) = RHS, where the variable substitution l = i + k − j is made in (2.52) (b) We can write d −y y k P (k; y) = e dy k! Using integration by parts, we can write ∞ P (k; y)dy = e−y a ∞ yk ∞ − | k! y=a ∞ = P (k; a) − a e−y y k−1 dy (k − 1)! P (k − 1; y)dy a By applying this recursive relation repeatedly, we find ∞ P (k; y)dy = P (k; a) + P (k − 1; a) + · · · + P (0; a) = Q(k; a) a We can write ∞ ∞ P (k; y)dy = a a e−y yk dy = k! k! ∞ e−y y k dy = a Γ(k + 1; a) k! (c) The identity to be proven is equivalent to (k + a + 1)Q(k; a) = aQ(k − 1; a) + (k + 1)Q(k + 1; a) (2.53) The LHS of (2.53) can be written as LHS = aQ(k; a) + (k + 1)Q(k; a) (2.54) We can also write the following relations: Q(k; a) = Q(k − 1; a) + P (k; a) Q(k; a) = Q(k + 1; a) − P (k + 1; a) (2.55) (2.56) Substituting (2.55) and (2.56) into (2.54), we obtain LHS = a[Q(k − 1; a) + P (k; a)] + (k + 1)[Q(k + 1; a) − P (k + 1; a)] = [aP (k; a) − (k + 1)P (k + 1; a)] + [aQ(k − 1; a) + (k + 1)Q(k + 1; a)] (2.57) It is easy to verify that the first term in square brackets in (2.57) is zero Hence, (2.53) is established Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 16 ✐ Chapter “Vo 200 pag Basic Queueing Models (d) The result in part (c) can be rewritten in the form (2.53) Substituting k for k + in (2.53), we have (k + a)Q(k − 1; a) = aQ(k − 2; a) + kQ(k; a) (2.58) Rearranging terms, we can write kQ(k; a) − aQ(k − 1; a) = kQ(k − 1; a) − aQ(k − 2; a) (2.59) Using the relation kQ(k − 1; a) = Q(k − 1; a) + (k − 1)Q(k − 1; a) (2.60) in (2.59), we obtain kQ(k; a) − aQ(k − 1; a) = Q(k − 1; a) + [(k − 1)Q(k − 1; a) − aQ(k − 2; a)] (2.61) (2.62) Applying the above recursive relation repeatedly, we obtain kQ(k; a) − aQ(k − 1; a) = Q(k − 1; a) + Q(k − 2; a) + [(k − 2)Q(k − 2; a) − aQ(k − 3; a)] = Q(k − 1; a) + Q(k − 2; a) + Q(k − 3; a) + · · · + Q(1; a) + [Q(1; a) − aQ(0; a)] k−1 = Q(j; a) j=0 2.4-10 Queue with finite waiting room M/M/1(B) (a) Let {πn } denote equilibrium-state distribution of the M/M/1(B) queue The BD balance equations (2.3-52) reduce to (cf (2.4-5)) πn = ρπn−1 = ρn π0 , n = 0, 1, , B, where ρ = λ/µ The normalization requirement implies that B B ρn π0 = 1, πn = n=0 n=0 from which we obtain π0 = B n=0 ρn The summation in the denominator is a geometric series that can be expressed as B − ρB+1 ρn = 1−ρ n=0 Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag 17 Hence, 1−ρ − ρB+1 π0 = and πn = ρn (1 − ρ) , − ρB+1 n = 0, 1, , B (b) The conditional waiting time distribution for the M/M/1(B) is the same as for the M/M/1 as given by (2.4-18) and (2.4-24): n−1 (µx)j , x ≥ 0, n = 1, , B, j! FW (x|n) = − e−µx j=0 1, x ≥ 0, 0, x < FW (x|0) = Hence, the waiting time distribution is given by B FW (x) = πn FW (x|n) n=0    n−1 B j (µx) 1−ρ   1+ ρn 1 − e−µx = − ρB+1 j! n=1 j=0   B B n−1 j 1−ρ  (µx)  = 1+ ρn − e−µx ρn − ρB+1 j! n=1 n=1 j=0   B n−1 j (µx) 1 − ρB+1 − (1 − ρ)e−µx  = ρn − ρB+1 j! n=1 j=0 2.4-11 Time-dependent solution of M/M/1 (a) Equations (2.4-31) and (2.4-32) follow by direct substitution of λn = λ and µn = µ into (2.3-46) (b) Taking the Laplace transform of (2.4-31) yields (2.4-33) directly (c) Multiplying both sides of (2.4-33) by z n and summing from n = to ∞ yields ∞ ∞ Pn∗ (s)z n s = −(λ + µ) n=1 ∞ Pn∗ (s)z n n=1 ∗ Pn−1 (s)z n +λ n=1 ∞ ∗ Pn+1 (s)z n +µ n=1 Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 18 ✐ Chapter “Vo 200 pag Basic Queueing Models Using the definition of G∗ (z, s), we obtain s(G∗ (z, s) − P0∗ (s)) = −(λ + µ)(G∗ (z, s) − P0∗ (s)) + µz −1 (G∗ (z, s) − zP1∗ (s) − P0∗ (s)) (2.63) Taking the Laplace transform of (2.4-32) yields P0∗ (s) − P0 (0) = −λP0∗ (s) + µP1∗ (0) Rearranging terms and using the fact that P0 (0) = 1, we obtain P1∗ (s) = µ−1 [(λ + µ)P0∗ (s) − 1] (2.64) Substituting (2.64) into (2.63) and rearranging terms, we obtain [(λ + µ + s)z − µ − λz2]G∗ (z, s) = z − µ(1 − z)P0∗ (s), from which (2.4-36) follows (d) (2.4-37) and (2.4-38) follow directly from applying the quadratic formula to the equation λz2 − (λ + µ + s)z + µ = (e) The lefthand side of (2.4-34) is given by LHS = A(s)z1−n (s) + B(s)z2−n (s) The righthand side of (2.4-34) is given by RHS = −(n−1) −(n−1) (s)) (s) + B(s)z2 [λ(A(s)z1 λ+µ+s −(n+1) + µ(A(s)z1 −(n−1) = A(s)z1 −(n+1) + B(s)z2 µz1−2 (s) λ+ λ+µ+s )] −(n−1) + B(s)z2 λ + µz2−2 (s) λ+µ+s Note that z1 (s) and z2 (s) are roots of the quadratic equation λz2 − (λ + µ + s)z + µ = (2.65) Using (2.65), one can show that λ + µz1−2 (s) = z1 (s) and λ+µ+s λ + µz2−2 (s) = z2 (s), λ+µ+s and it then follows that LHS = RHS (f) Using (2.4-41) and (2.4-42), we can compute the following limits: z1−1 (s) =0 s→∞ s −1 z (s) = lim s→∞ s lim (2.66) (2.67) From (2.66), we conclude that z1−1 (s) remains bounded as s → ∞ From (2.67), we conclude that z2−1 (s) tends to infinity as s → ∞ Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag 19 (g) When n = 1, (2.4-40) becomes P1∗ (s) = P0∗ (s)z1−1 (s) (2.68) Equating (2.64) and (2.68), and solving for P0∗ (s) yields (2.4-43) (h) We can rewrite (2.4-41) as follows: z1−n (s) = λ µ n/2 s+λ+µ √ λµ s+λ+µ √ λµ − n 2−1 (2.69) We make use of the following property of the inverse Laplace transform : L−1 {G∗ (a(s + b))} = g a t a e−bt , (2.70) where G∗ (s) and g(t) form a Laplace transform pair Applying (2.70) in (2.69), we obtain L−1 {z1−n (s)} = λ µ n/2 e−(λ+µ)t (2 λµ) n √ In (2 2t λµ λµt) , (2.71) from which (2.4-46) follows (i) Applying the inverse Laplace transform to (2.4-43), we obtain s L−1 {P0∗ (s)} = L−1 z1−1 (s) s − L−1 (2.72) t = u(t) − f1 (u)du, (2.73) from which (2.4-47) follows (j) Substitution of (2.4-49) into (2.4-46) yields (2.4-50) after some algebra (k) 2.4-12 Mean queue length and mean waiting time (a) ∞ ¯ Q = ∞ n=m ∞ ρn−m pm = pm (n − m)pn = n=m ∞ = ρpm k=0 k d dρ = ρpm dρ dρ ∞ kρk = pm ρ k=0 ∞ ρk = ρpm k=0 kρk−1 k=0 d dρ − ρ ρ pm ρC(m, a) = · = , = ρpm (1 − ρ)2 1−ρ 1−ρ 1−ρ since C(m, a) = pm /(1 − ρ) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 20 ✐ Chapter “Vo 200 pag Basic Queueing Models (b) Applying Little’s formula: ¯ ¯ = Q = (ρ/λ)C(m, a) = C(m, a) W λ 1−ρ mµ(1 − ρ) 2.4-13 Erlang’s delay formula and Erlang’s loss formula From (2.4-64), (2.461), and (2.4-63), we have am m m! m−a m−1 an am n=0 n! + m! 1−ρ C(m, a) = m am n=0 m! Dividing the numerator and denominator by the definition of B(m, a), we obtain C(m, a) = m B(m, a) m−a = m − B(m, a) + B(m, a) m−a (2.74) and using ρ = mB(m, a) m − a(1 − B(m, a)) a m and (2.75) 2.4-14 Derivation of waiting time distribution In this scenario, all m exponential servers are busy Let Xj , ≤ j ≤ m be the interval from an arbitrarily chosen instant until the completion of a job in service at the jth server The Xj ’s are independent and identically distributed with complementary distribution function c FX (t) = P {Xj ≥ t}e−µt j The distribution of Ti , ≤ i ≤ n + is equivalent to that of the random variable T defined by T min{X1 , X2 , , Xm } The complementary distribution function of T satisfies FTc (t) = P {T ≥ t} = P {Xj ≥ t : ∀j, ≤ j ≤ m} m m = e−µt = e−mµt P {Xj ≥ t} = j=1 j=1 Therefore, Ti , ≤ i ≤ n + are exponentially distributed with parameter mµ Further, it should be clear that the Ti ’s are independent 2.4-15 Derivation of waiting time distribution (continued) In (2.4-70), the double summation over (n, j) can be rewritten as a double summation over (k, j), where k = n − j, as follows: ∞ n ρn n=0 j=0 (mµx)j = j! = ∞ ∞ ρk+j k=0 j=0 ∞ k ρmµx ρ e k=0 (mµx)j j! = ρmµx e 1−ρ (2.76) Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Basic Queueing Models “Vo 200 pag 21 Using (2.76)in (2.4-70) yields c c c FW (x) = FW (0)e−mµx emρµx = FW (0)e−mµ(1−ρ)x ∗ 2.4-16 System time distribution Let fW (s|m + n) denote the Laplace transform of the pdf of the waiting time experienced by a call when it arrive to the system with m + n active calls The waiting time completes after n + job completions occur The complementary distribution function of the time for a call completion was computed in (2.4-14) to be FTc (t) = e−mµt Therefore, the Laplace transform of T is given be fS∗ (t) = mµ , s + mµ and the Laplace transform of the waiting time given m + n calls in the system is given by n+1 mµ ∗ (s|m + n) = fW , n ≥ s + mµ Therefore, ∞ ∗ (s|n + m)am+n fW ∗ (s) = P {W = 0} + fW n=0 ∞ = P {W = 0} + n=0 mµ s + mµ (2.77) n+1 ρn π m , (2.78) where we have used an+m = πn+m = ρn πm Rearranging terms in (2.78), we obtain ∞ ∗ fW (s) ∗ fW (s|n + m)am+n = P {W = 0} + n=0 ∞ n+1 mµ ρn πm s + mµ n=0 πm mµ = P {W = 0} + s + mµ(1 − ρ) mµ(1 − ρ) c c = [1 − FW (0)] + FW (0) s + mµ(1 − ρ) = P {W = 0} + (2.79) (2.80) (2.81) (2.82) where we made use of (2.4-64) Taking the inverse Laplace transform, we obtain c c c FW (x) = − FW (0) + FW (0)(1 − e−mµ(1−ρ)x ) = FW (0)e−mµ(1−ρ)x Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ 22 ✐ Chapter “Vo 200 pag Basic Queueing Models 2.4-17 Discouraged arrivals We are given a birth-and-death process with birth and death rates given as: λ , n = 0, 1, 2, · · · n+1 µ, n = 1, 2, 3, · · · λ(n) = µ(n) = For this system, the detailed balance equation is given by λ(n − 1)pn−1 = µ(n)pn , n = 1, 2, 3, · · · , which implies pn = λ(n − 1) λ/n λ λ∗ (n − 1) pn−1 = pn−1 = pn−1 = pn−1 , µ(n) µ nµ µ∗ (n) where λ∗ (n) = λ and µ∗ (n) = nµ are the birth and death rates corresponding to the M/M/∞ queue Therefore, the distribution {pn } takes the same form as that of the M/M/∞ queue 2.4-18 Pressured server and discouraged arrivals From (2.3-56), n n πn = π0 Using ∞ n=0 λ (λ/µ)n λi−1 = π0 · c = π0 , n ≥ b µi i i µ (n!)c i=1 i=1 (2.83) πn = 1, we can solve for π0 to obtain ∞ π0 = j=0 ρj (j!)b+c When b + c = 1, we have π0 = e−ρ and πn = ρn −ρ e , n! which is the same result as the M/M/∞ queue 2.4-19 Queueing system with finite storage (a) The birth rates are given by λn = λ, ≤ n ≤ K − The death rates are given by µn = µ, ≤ n ≤ K (b) From (2.3-56), we have n πn = π0 λ = π ρn , µ i=1 ≤ n ≤ K Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full file at https://TestbankDirect.eu/ ✐ ✐ Chapter ✐ Using K n=0 Basic Queueing Models “Vo 200 pag 23 πn = 1, we solve for π0 to obtain −1 K ρk π0 = = n=0 Hence, πn = ρn − ρK+1 1−ρ −1 = 1−ρ − ρK+1 1−ρ , ≤ n ≤ K − ρK+1 2.4-20 Queueing system with finite storage (continued) (a) The balance equation is λπn = µπn+1 , ≤ n ≤ K − (2.84) (b) The PGF of {πn } is given by K πn z n G(z) = (2.85) n=0 Multiplying both sides of (2.84) by z n and summing from to K − yields K−1 K−1 πn z n = µ λ n=0 πn+1 z n (2.86) n=0 Applying (2.85) into (2.86), we obtain λ[G(z) − πK z K ] = µz −1 [G(z) − π0 ] Solving for G(z) and using πK = ρK π0 , we obtain G(z) = π0 ρK+1 z K+1 − ρz − (c) Using G(1) = 1, we can solve for π0 to obtain π0 = Hence, πn = ρn and G(z) = ρ−1 ρK+1 − ρ−1 , ≤ n ≤ K, −1 ρK+1 ρ−1 ρK+1 − (ρz)K+1 − ρz − 2.4-21 Two-stage cyclic queueing system Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ ... distributed and i.i.d The event that the Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi. .. pn = (1 − ρ)ρn for n = 0, 1, · · · Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full... given N = is given by fW (t|0) = δ(t), Copyright c 2009 by H Kobayashi and B.L Mark ✐ Full file ✐at https://TestbankDirect.eu/ ✐ Solution Manual for System Modeling and Analysis by Kobayashi Full