SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

61
⎛
P = ⎜
⎝
0.92 0.08 0 0
0 0 0.97 0.03
0.92 0.08 0 0
0 0 0.97 0.03
⎞
⎟
⎠
(b) π 1 ≈ 0.85
(c)
2000(500π 1 + 625(π 2 + π 3 ) + 800π 4 )
≈ 2000(500(0.8499) + 625(0.0739 + 0.0739) + 800(0.0023))
≈ $1, 038, 298
22. The proof is by induction.
For i ∈Aand j ∈B,f (1)
ij = p ij .Thus,
F (1)
AB = P (0)
AAP AB
Now suppose that the result is true for all n ≤ k, forsomek>1.
f (k+1)
ij = ∑ l∈A
Pr{S k+1 = j, S k ∈A,...,S 2 ∈A,S 1 = l | S 0 = i}
= ∑ l∈A
Pr{S k+1 = j, S k ∈A,...,S 2 ∈A|S 1 = l, S 0 = i}
× Pr{S 1 = l | S 0 = i}
= ∑ l∈A
Pr{S k = j, S k−1 ∈A,...,S 1 ∈A|S 0 = l}p il
= ∑ l∈A
f (k)
lj p il = ∑ p il f (k)
lj
l∈A
Therefore, in matrix form and using the induction hypothesis
F (k+1)
AB = P AA F (k)
AB
= P AA P (k−1)
AA P AB
= P AAP (k)
AB
Therefore, the result is true for n = k + 1, and is thus true for any n.