SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

91
−
λ M
∫ ∞
λ 1 + λ M t
= e −λ1t e −λ M t − λ M
e −(λ 1+λ M )t
λ 1 + λ M
(
= 1 − λ )
M
e −(λ 1+λ M )t
λ 1 + λ M
=
( )
λ1
e −λ Ht
λ H
(λ 1 + λ M )e −(λ 1+λ M )a da
21. (a)
⎛
G =
⎜
⎝
Notice that for row 2
− 1 12
0.95
12
0.1
12
0.4
12
0 − 73
1500
( 73 0.63
)( 1500 0.73 ( 73 0.10
)( 1500 0.73
0 ( 1 .36
)( 50 .60 − 1 50
( 1 0.24
)( 50 0.60
1
3
0 0 − 1 3
⎞
⎟
⎠
1/g 22 = 15 = 15
1 − p 22 0.73 = 1500
73
so g 22 =73/1500 to account for transitions from state 2 to itself. Similarly for
row 3. The steady state probabilities are the same:
⎛
π = ⎜
⎝
0.08463233821
0.2873171709
0.6068924063
0.02115808455
(b) No answer provided.
(c) The only change in the generator matrix is that the last row becomes (0, 0, 0, 0).
22. Let G 1 ,G 2 ,... be the times between visits to state 1. Because the process is semi-
Markov, these are independent, time-stationary random variables.
Let R n be the time spent in state 1 on the nth visit. Then R 1 ,R 2 ,... are also independent
with E[R n ]=τ 1 . Therefore we have a renewal-reward process.
From Exercise 25, Chapter 4, the expected number of states visited between visits to 1
is 1/ξ 1 . The fraction of those visits that are to state j has expectation ξ j /ξ 1 . Therefore,
E[G n ] = τ 1 +
j≠1(ξ ∑ j /ξ 1 )τ j
⎞
⎟
⎠
∑ m
= 1/ξ 1 ξ j τ j
j=1