SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

48 CHAPTER 5. ARRIVAL-COUNTING PROCESSES
(
η 7c 2 )
δ = −300 + 2700
c(c 2 − 2700)
per day
To minimize the cost we take the derivative w.r.t. c and set it equal to 0, then solve
for c. Of the 3 solutions, the only one in the range [0, 30] is c = 15.9 for which
η
≈ $34.46 per day
δ
Compare this to c =30forwhich
η
= $50 per day
δ
Therefore, it is worthwhile to replace early.
30. Suppose we wait for n patrons before beginning a tour. Then δ = n minutes is the
expected time between tours. The expected cost for a tour of size n is
(a) The long-run cost/minute is
η = 10+(0+(0.50) + (2)(0.50) + (3)(0.50)
+ ···+(n − 1)(0.50))
(n(n − 1))
= 10+(0.50)
2
η
δ = 10 (0.50)(n − 1)
+
n 2
A plot as a function of n reveals that η/δ decreases until n = 6, then increases.
At n =6,η/δ =$2.92 per minute.
(b) δ =1(6)=6minutes
31. To prove the result for decomposition of m processes, do the decomposition in pairs,
as follows:
(i) Decompose the original process into 2 subprocesses with probabilities γ 1 and 1−γ 1 .
Clearly both subprocesses are Poisson, the first with rate λ 1 = γ 1 λ, and the latter with
rate λ ′ =(1− γ 1 )λ.
(ii) Decompose the λ ′ process into two subprocesses with probabilities γ 2 /(1 − γ 1 )and
1 − γ 2 /(1 − γ 1 ). Clearly both subprocesses are Poisson, the first with rate
( )
λ 2 = λ ′ γ2
= γ 2 λ
1 − γ 1