SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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118 CHAPTER 8. QUEUEING PROCESSES
Chapter 9 Topics in Simulation of <strong>Stochastic</strong> Processes 1. Solve ̂σ/ √ k ≤ 20 which gives k = (110/20) 2 ≈ 31 replications. 2–7. No answer provided. 8. To obtain a rough-cut model we will (a) use a 0.5 probability that an item joins the queue of each inspector, rather than selecting the shortest queue, <strong>and</strong> (b) treat all processing times as exponentially distributed (later we will refine (b)). • Current System Λ = station j name a 0j µ (j) 1 repair 1 0.125 0.167 2 repair 2 0.125 0.167 3 inspect 0 0.333 ⎛ ⎜ ⎝ R = λ (1) ⎞ λ (2) ⎟ λ (3) ⎛ ⎜ ⎝ A = 0 0 1 0 0 1 0.1 0.1 0 ⎛ ⎜ ⎝ 0.125 0.125 0 ⎞ ⎟ ⎠ ⎞ ⎟ ⎠ ⎠ =[(I − R) ′ ] −1 A = 119 ⎛ ⎜ ⎝ 0.156 0.156 0.313 ⎞ ⎟ ⎠
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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)
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41 (b) 20. (a) Pr{Y 52 − Y 48 =20
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43 ⎧ ⎪⎨ λ(t) = ⎪⎩ 144, 0
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while {b
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27. We suppose that the time to bur
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and the latter with rate ( λ ′
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Chapter 6 Discrete-Time Processes 1
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53 n =5 ⎛ ⎜ ⎝ 0.00001 0.68645
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55 with ŝe = √ ̂p(1 − ̂p) 45
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57 14. Case 6.3 S n represents the
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59 Pr{≤ 1 defective} = a + b + c
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61 ⎛ P = ⎜ ⎝ 0.92 0.08 0 0 0
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63 25. (a) We first argue that µ 1
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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 and 96: 91 − λ M ∫ ∞ λ 1 + λ M t =
Page 97 and 98: Chapter 8 Queueing Processes 1. Let
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: It appears that there will be very
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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