SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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8. See Exercise 7 for an analysis of the (3,7) policy.
For the (2,7) policy
⎛
G =
⎜
⎝
−1/2 0 0 0 0 1/2 0 0
1 −3/2 0 0 0 0 1/2 0
0 1 −3/2 0 0 0 0 1/2
0 0 1 −3/2 0 0 0 0
0 0 0 1 −1 0 0 0
0 0 0 0 1 −1 0 0
0 0 0 0 0 1 −1 0
0 0 0 0 0 0 1 −1
⎞
⎟
⎠
⎛
π =
⎜
⎝
0.2394366197
0.05633802817
0.08450704225
0.1267605634
0.1901408451
0.1901408451
0.07042253521
0.04225352113
⎞
⎟
⎠
λπ 0 ≈ 0.239
∑ 7j=0
jπ j ≈ 3.03
Therefore, (2,7) carries a slightly lower inventory level but with more than double the
lost sales rate.
9. To solve this problem we must evaluate p 02 (t) fort = 17520 hours (2 years). Using
the uniformization approach with n ∗ = 1000, and carrying 10 digits of accuracy in all
calculations,
˜p 02 (t) =0.56
with truncation error ≤ 0.72 × 10 −8 .
The reason that this is dramatically larger than the simulation estimate of 0.02 obtained
in Chapter 4 is the difference in repair-time distribution. In Chapter 4 the distribution
is uniform on [4, 48] hours. Here the distribution is exponential on [0, ∞) hours.The
exponential model is pessimistic because it allows repair times longer than 48 hours.
10. Let M = {0, 1, 2} represent the number of failed machines, so that {Y t ; t ≥ 0} is the
number of failed machines at time t hours. When {Y t =0}, the failure rate is 2(0.01),
since each machine fails at rate 0.01/hour and there are two machines. When {Y t =1}