SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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8 CHAPTER 3. BASICS (b) Pr{Y >1 | Y>2/3} = e−2(1) e −2(2/3) = e −2(1/3) ≈ 0.51 (c) Pr{Y
9 (a) Pr{X 2 =1| X 1 =0} = Pr{X 2 =1,X 1 =0} Pr{X 1 =0} = 0.05 0.80 =0.0625 Pr{X 2 =1| X 1 =1} = 0.10 0.20 =0.50 Clearly X 1 and X 2 are dependent since Pr{X 2 =1| X 1 =0} ≠Pr{X 2 =1| X 1 =1}. { 100, a =0 (b) Let g(a) = −20, a =1 E[g(X 2 ) | X 1 =1] = g(0) Pr{X 2 =0| X 1 =1} + g(1) Pr{X 2 =1| X 1 =1} = 100(1 − 0.50) − 20(0.50) = 40 10. ̂µ =2.75 ̂σ =0.064 ŝe = √ ̂σ 6 ≈ 0.026 ̂µ + ŝe = 2.776 ̂µ − ŝe = 2.724 Assuming a mean of 2.776 Pr{X >2.90} = 1− Pr{X ≤ 2.90} = { X − 2.776 1− Pr ≤ 0.064 = 1− Pr{Z ≤ 1.938} ≈ 1 − 0.974 = 0.026 } 2.90 − 2.776 0.064 where Z is a standard-normal random variable. Assuming a mean of 2.724 { Pr{X >2.90} = 1− Pr Z ≤ } 2.90 − 2.724 0.064 = 1− Pr{Z ≤ 2.75} ≈1 − 0.997 = 0.003
Page 1 and 2: SOLUTIONS MANUAL for Stochastic Mod
Page 3 and 4: ii CONTENTS
Page 5 and 6: Chapter 2 Sample Paths 1. The simul
Page 7 and 8: 3 7. Inputs: Number of hamburgers d
Page 9 and 10: Chapter 3 Basics 1. (a) Pr{X =4} =
Page 11: (b) 6. (a) F Y (a) = = ∫ a ⎧
Page 15 and 16: 11 16. Let U be a random variable h
Page 17 and 18: 13 Y = ⎧ ⎪⎨ ⎪⎩ 1, 0 ≤ U
Page 19 and 20: 15 (d) E[X] = ∑ all a ap X (a) =0
Page 21 and 22: 17 ( ) ∞∑ d 2 = γ a=1 dq 2 qa+
Page 23 and 24: 19 = ∫ ∞ −∞ ∫ ∞ a 2 f X
Page 25 and 26: 21 (b) Let T ≡ number of trials u
Page 27 and 28: (b) f is maximized at a = β giving
Page 29 and 30: Chapter 4 Simulation 1. An estimate
Page 31 and 32: 27 S n+1 ← S n − FD −1 endif
Page 33 and 34: 29 (b) Ȳ 1 = {0(5 − 0) + 1(6 −
Page 35 and 36: Chapter 5 Arrival-Counting Processe
Page 37 and 38: 33 Pr{Y 2,6 > 30} = 1− Pr{Y 2,6
Page 39 and 40: 8. (a) Restricted to periods of the
Page 41 and 42: 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
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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
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67 = 1− Pr{X 1 >n} Pr{X 2 >n} = 1
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69 = ∑ h∈A Pr{S 1 = j | S 0 = h
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71 Therefore, the long-run expected
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73 31. We show (6.5) for k = 2. The
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Chapter 7 Continuous-Time Processes
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77 dp i3 (t) dt dp i4 (t) dt dp i5
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79 endif C 2 ← T n+1 + FH −1 (r
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81 8. See Exercise 7 for an analysi
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y conditioning on the first state v
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85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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87 or λµ 00 = 1 +λµ 10 λµ 01
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89 e 2 () (order arrives from manuf
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91 − λ M ∫ ∞ λ 1 + λ M t =
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Chapter 8 Queueing Processes 1. Let
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95 λ i = µ i = { 20(1 − i/16),
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97 (c) l q = ∞∑ π 2 + (j − 2
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We have assumed a 24-hour day, but
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101 ∑ 20 j=0 d j ≈ 453.388 ⎧
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103 (c) For the M/M/s/c with s ≤
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will be at the front of the queue o
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107 (b) For the M/M/1 For the M/D/1
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109 = (1− ρ 1 )(1 − ρ 2 ) ∞
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111 i a 0i µ (i) 1 20 30 2 0 30 (
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113 or or ⎛ Λ[(I − R) ′ ]
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115 i s i ρ i 1 19 0.88 2 4 0.75 3
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It appears that there will be very
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Chapter 9 Topics in Simulation of S
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9. To obtain a rough-cut model, fir
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123 w q =0.49hours l q = 3.2 trucks
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125 When we fix the initial state,
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