SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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90 CHAPTER 7. CONTINUOUS-TIME PROCESSES 19. an exponential distribution. This result shows that we can represent the holding time in a state (which is exponentially distributed) as the sum of a geometrically distributed number of exponentially distributed random variables with a common rate. This is precisely what uniformization does. p ij (t) = e −g∗ t ∞∑ n=0 q (n) ij = ˜p ij (t)+e −g∗ t ≥ ˜p ij (t) because g ∗ > 0, t>0andq (n) ij ≥ 0. Notice that (g ∗ t) n n! ∞∑ n=n ∗ +1 q (n) ij (g ∗ t) n n! e −g∗ t ∞∑ n=n ∗ +1 q (n) ij (g ∗ t) n n! = 1− e −g∗ t ≤ 1 − e −g∗ t n∗ ∑ n=0 n∗ ∑ n=0 q (n) ij (g ∗ t) n n! (g ∗ t) n n! because q (n) ij ≤ 1. 20. Pr{Z 1 ≤ min j≠1 Z j,H >t} = Pr{Z 1 ≤ Z 2 ,...,Z 1 ≤ Z k ,Z 1 >t,...,Z k >t} = Pr{t
91 − λ M ∫ ∞ λ 1 + λ M t = e −λ1t e −λ M t − λ M e −(λ 1+λ M )t λ 1 + λ M ( = 1 − λ ) M e −(λ 1+λ M )t λ 1 + λ M = ( ) λ1 e −λ Ht λ H (λ 1 + λ M )e −(λ 1+λ M )a da 21. (a) ⎛ G = ⎜ ⎝ Notice that for row 2 − 1 12 0.95 12 0.1 12 0.4 12 0 − 73 1500 ( 73 0.63 )( 1500 0.73 ( 73 0.10 )( 1500 0.73 0 ( 1 .36 )( 50 .60 − 1 50 ( 1 0.24 )( 50 0.60 1 3 0 0 − 1 3 ⎞ ⎟ ⎠ 1/g 22 = 15 = 15 1 − p 22 0.73 = 1500 73 so g 22 =73/1500 to account for transitions from state 2 to itself. Similarly for row 3. The steady state probabilities are the same: ⎛ π = ⎜ ⎝ 0.08463233821 0.2873171709 0.6068924063 0.02115808455 (b) No answer provided. (c) The only change in the generator matrix is that the last row becomes (0, 0, 0, 0). 22. Let G 1 ,G 2 ,... be the times between visits to state 1. Because the process is semi- Markov, these are independent, time-stationary random variables. Let R n be the time spent in state 1 on the nth visit. Then R 1 ,R 2 ,... are also independent with E[R n ]=τ 1 . Therefore we have a renewal-reward process. From Exercise 25, Chapter 4, the expected number of states visited between visits to 1 is 1/ξ 1 . The fraction of those visits that are to state j has expectation ξ j /ξ 1 . Therefore, E[G n ] = τ 1 + j≠1(ξ ∑ j /ξ 1 )τ j ⎞ ⎟ ⎠ ∑ m = 1/ξ 1 ξ j τ j j=1
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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 >
Page 43 and 44: 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: 89 e 2 () (order arrives from manuf
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 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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