SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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92 CHAPTER 7. CONTINUOUS-TIME PROCESSES Thus, the long-run reward rate (fraction of time spent in state 1) is π 1 = E[R n] E[G n ] = ξ 1 τ 1 ∑ mj=1 ξ j τ j A similar argument applies for all other states. 23. We need to show that satisfies π ′ G = 0. Select the jth equation π i = ξ i /g ii ∑ mj=1 = ξ i/g ii ξ j /g jj d ∑ π k g kj − π j g jj =0 k≠j (1) Substituting (1) ( ) ( ) ∑ ξk /g kk ξj /g jj g kj − g jj k≠j d d = 1 ( ) ∑ gkj ξ k − 1 d g k≠j kk d ξ j ⎛ ⎞ = 1 ⎝ ∑ p kj ξ k − ξ j ⎠ =0 ? d k≠j For the equality to hold, we must have ∑ p kj ξ k = ξ j (2) k≠j But for a Markov process p jj = 0, so (2) is equivalent to m∑ p kj ξ k = ξ j k=1 which is guaranteed because ξ is the steady-state distribution of the embedded Markov chain. 24. No answer provided.
Chapter 8 Queueing Processes 1. Let µ be the production rate. Sample average production time = 31.9/10 = 3.19 minutes. Therefore, ̂µ = 1 ≈ 0.31 parts/minute. With 2 machines the rate is 3.19 2̂µ =0.62 parts/minute. 2. F X (a) = { 0, a < 0 1 − e −0.31a , a ≥ 0 93
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SOLUTIONS MANUAL for Stochastic Mod
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ii CONTENTS
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Chapter 2 Sample Paths 1. The simul
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3 7. Inputs: Number of hamburgers d
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Chapter 3 Basics 1. (a) Pr{X =4} =
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(b) 6. (a) F Y (a) = = ∫ a ⎧
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9 (a) Pr{X 2 =1| X 1 =0} = Pr{X 2 =
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11 16. Let U be a random variable h
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13 Y = ⎧ ⎪⎨ ⎪⎩ 1, 0 ≤ U
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15 (d) E[X] = ∑ all a ap X (a) =0
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17 ( ) ∞∑ d 2 = γ a=1 dq 2 qa+
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19 = ∫ ∞ −∞ ∫ ∞ a 2 f X
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21 (b) Let T ≡ number of trials u
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(b) f is maximized at a = β giving
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Chapter 4 Simulation 1. An estimate
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27 S n+1 ← S n − FD −1 endif
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29 (b) Ȳ 1 = {0(5 − 0) + 1(6 −
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Chapter 5 Arrival-Counting Processe
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33 Pr{Y 2,6 > 30} = 1− Pr{Y 2,6
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8. (a) Restricted to periods of the
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37 Pr{Y (A) 1.5 > 1000,Y (B) 1.5 >
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39 Pr{Y (B) 2 − Y (B) 1 > 5,Y (B)
Page 45 and 46: 41 (b) 20. (a) Pr{Y 52 − Y 48 =20
Page 47 and 48: 43 ⎧ ⎪⎨ λ(t) = ⎪⎩ 144, 0
Page 49 and 50: while {b
Page 51 and 52: 27. We suppose that the time to bur
Page 53 and 54: and the latter with rate ( λ ′
Page 55 and 56: Chapter 6 Discrete-Time Processes 1
Page 57 and 58: 53 n =5 ⎛ ⎜ ⎝ 0.00001 0.68645
Page 59 and 60: 55 with ŝe = √ ̂p(1 − ̂p) 45
Page 61 and 62: 57 14. Case 6.3 S n represents the
Page 63 and 64: 59 Pr{≤ 1 defective} = a + b + c
Page 65 and 66: 61 ⎛ P = ⎜ ⎝ 0.92 0.08 0 0 0
Page 67 and 68: 63 25. (a) We first argue that µ 1
Page 69 and 70: 65 The expected number of spins is
Page 71 and 72: 67 = 1− Pr{X 1 >n} Pr{X 2 >n} = 1
Page 73 and 74: 69 = ∑ h∈A Pr{S 1 = j | S 0 = h
Page 75 and 76: 71 Therefore, the long-run expected
Page 77 and 78: 73 31. We show (6.5) for k = 2. The
Page 79 and 80: Chapter 7 Continuous-Time Processes
Page 81 and 82: 77 dp i3 (t) dt dp i4 (t) dt dp i5
Page 83 and 84: 79 endif C 2 ← T n+1 + FH −1 (r
Page 85 and 86: 81 8. See Exercise 7 for an analysi
Page 87 and 88: y conditioning on the first state v
Page 89 and 90: 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Page 91 and 92: 87 or λµ 00 = 1 +λµ 10 λµ 01
Page 93 and 94: 89 e 2 () (order arrives from manuf
Page 95: 91 − λ M ∫ ∞ λ 1 + λ M t =
Page 99 and 100: 95 λ i = µ i = { 20(1 − i/16),
Page 101 and 102: 97 (c) l q = ∞∑ π 2 + (j − 2
Page 103 and 104: We have assumed a 24-hour day, but
Page 105 and 106: 101 ∑ 20 j=0 d j ≈ 453.388 ⎧
Page 107 and 108: 103 (c) For the M/M/s/c with s ≤
Page 109 and 110: will be at the front of the queue o
Page 111 and 112: 107 (b) For the M/M/1 For the M/D/1
Page 113 and 114: 109 = (1− ρ 1 )(1 − ρ 2 ) ∞
Page 115 and 116: 111 i a 0i µ (i) 1 20 30 2 0 30 (
Page 117 and 118: 113 or or ⎛ Λ[(I − R) ′ ]
Page 119 and 120: 115 i s i ρ i 1 19 0.88 2 4 0.75 3
Page 121 and 122: It appears that there will be very
Page 123 and 124: Chapter 9 Topics in Simulation of S
Page 125 and 126: 9. To obtain a rough-cut model, fir
Page 127 and 128: 123 w q =0.49hours l q = 3.2 trucks
Page 129 and 130: 125 When we fix the initial state,
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