SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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124 CHAPTER 9. TOPICS IN SIMULATION OF STOCHASTIC PROCESSES ∑ ki=1 X i Z i = 644.99 ∑ kj=1 X j =80.3 ∑ kh=1 Z h =80.3 ̂σ XZ ≈ 0.02 16. k∑ (V i − ¯V ) 2 = i=1 = = = k∑ (X i − Z i − ( ¯X − ¯Z)) 2 i=1 k∑ ((X i − ¯X) − (Z i − ¯Z)) 2 i=1 k∑ { (Xi − ¯X) 2 − 2(X i − ¯X)(Z i − ¯Z) i=1 +(Z i − ¯Z) 2} k∑ (X i − ¯X) k∑ 2 + (Z i − ¯Z) 2 i=1 i=1 k∑ − 2 (X i − ¯X)(Z i − ¯Z) i=1 = (k − 1)(̂σ X 2 + ̂σ Z 2 − 2̂σ XZ ) 17. There are mostly disadvantages. • Unless we understand the bias quite well, we have no idea what overload approximately compensates the underload. Therefore, the bias may still be quite significant, and we will still have to study it. • When we attempt to study the bias, we must either study the underload and overload individually, which is twice as much work, or study them together, in which case the combined bias process may have very unusual behavior. An advantage is that we might be able to bound the bias by looking at convergence from above and below. 18. If ̂θ is an estimator of θ, then the bias of ̂θ is E[̂θ − θ] where the expectation is with respect to the distribution of ̂θ. Let S 0 be the initial state, as in the Markov chain. Then, if we sample the initial state from the steady-state distribution, we proved E[̂θ] = ∑ E[̂θ | S 0 = x]Pr{S 0 = x} = θ (1) x∈M
125 When we fix the initial state, the expectation of interest is E[̂θ | S 0 =1] a conditional expectation, only 1 term of (1). Remember that “expectation” is a mathematical averaging process over the possible outcomes. When we sample the initial state, all initial states are possible. When we fix the initial state, only 1 initial state is possible. Therefore, the probability distribution of the estimator changes, even though the results of the simulation (outcome) may be the same. ⎛ ⎞ 0.429 ⎜ ⎟ 19. π ≈ ⎝ 0.333 ⎠ 0.238 Therefore Pr{S =3} = π 3 ≈ 0.238 E[S] = π 1 +2π 2 +3π 3 ≈ 1.809 Var[S] = 3∑ (j − 1.809) 2 π j ≈ 0.630 j=1 A 1% relative error for Pr{S =3} implies that Therefore, | 0.238 − p (n) 13 |≤ 0.002. A 1% relative error for E[S] implies that | 0.238 − p (n) 13 | ≤ 0.01 0.238 | 1.809 − E[S n | S 0 =1]| 1.809 Therefore, | 1.809 − E[S n | S 0 =1]|≤ 0.018 where ≤ 0.01 E[S n | S 0 =1]= 3∑ j=1 jp (n) 1j A 1% relative error for Var[S] implies that | 0.630 − Var[S n | S 0 =1]| 0.063 ≤ 0.01
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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
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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
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 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: 123 w q =0.49hours l q = 3.2 trucks
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