SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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120 CHAPTER 9. TOPICS IN SIMULATION OF STOCHASTIC PROCESSES The utilization of the technicians is ρ (j) = λ(j) ≈ 0.93 for j =1, 2 µ (j) Treating each station as an M/M/1 queue the expected flow time is ( w q (1) + 1 + w(3) µ (1) q + 1 ) (1.25) µ (3) ≈ (85 + 6 + 47 + 3)(1.25) ≈ 141 minutes since w (1) q = w (2) q and the expected number of cycles through the system is 1/0.8 =1.25 cycles. • Proposed System The repair technicians now become a single station with λ = λ (1) + λ (2) =0.313 and µ =0.167. The utilization is ρ = λ ≈ 0.93, so it is unchanged. 2µ Treating the repair station as an M/M/2 queue, the expected flow time is ( w q + 1 µ + w(3) q + 1 ) (1.25) µ (3) ≈ (43 + 6 + 47 + 3)(1.25) ≈ 124 minutes which is a dramatic reduction. • Refinement We can refine the approximation by using the ideas in Section 8.10. Notice that ε (j) s j 1 0.0625 2 0.0625 3 0.0370 Since the arrival process is Poisson, ε a (j) = 1 throughout. Therefore, the preceding values of w q (j) can be modified by the factor (1 + ε s (j) )/2.
9. To obtain a rough-cut model, first note that the registration function is not really relevant to the issue of additional bed versus additional doctor, so we will ignore it. We also (a) ignore the interaction of beds and doctors, (b) treat treatment time as exponentially distributed, and (c) develop a composite patient and ignore patient type and priority. • Additional doctor. With an additional doctor there is a one-to-one correspondence between bed and doctor. λ = 1/20 patient/minute 1/µ = 72(0.15) + 25(0.85) ≈ 32 minutes Using an M/M/3 model l q = 0.3 patients waiting for a doctor and a bed w q = 6.3 minutes wait for a doctor and a bed 121 • Additional bed. In terms of time to wait to see a doctor, the beds are not a constraint. M/M/2 model l q = 2.8 patients waiting to see a doctor w q = 57 minutes to see a doctor The expected number of patients in the system is Using an l = λw = λ(w q +1/µ) ≈ 4.45 Thus, with 4 beds the expected number waiting for a bed is 4.45 − 4=0.45, which is larger than the 0.3 patients when adding another doctor. It appears that adding a doctor will be better in terms of time to see a doctor and time to reach a bed. 10. To obtain a rough-cut model we will replace all random variables by their expected values and treat it as deterministic. Suppose the company is open 8 hours per day. Then the period of interest is 90(8) = 720 hours. The number of lost sales during 1 day is (8 hours) (1 customer/hour) (6 bags/customer) = 48 bags. If the company orders s bags, they will run out in s bags = s/6 hours 6 bags/hour
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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
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 and 122: It appears that there will be very
Page 123: Chapter 9 Topics in Simulation of S
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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