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HMWK 5
Problems from the Text
Ch. 6, Page 409: 12, 13, 22
1. Chapter 6, Problem 12.
λ
n
=
nλ
+
θ, n < N,
λ
n
=
nλ, n
≥
N,
μ
n
=
nμ, n
≥
0
.
3. (b) Fix
N
= 3. Let
P
i
be the limiting probability that the system is in state i. We are
looking for
∑
∞
i
=3
P
i
.
We have
λ
k
P
k
=
μ
k
+1
P
k
+1
.
This yields
P
1
=
θ
μ
P
0
P
2
=
λ
+
θ
2
μ
P
1
=
θ
(
λ
+
θ
)
2
μ
2
P
0
P
3
=
2
λ
+
θ
2
μ
P
2
=
θ
(
λ
+
θ
)(2
λ
+
θ
)
6
μ
3
P
0
For
k
≥
4, we get
P
k
=
(
k

1)
λ
kμ
P
k

1
,
which implies
P
k
=
(
k

1)(
k

2)
···
3
k
(
k

1)
···
4
(
λ
μ
)
k

3
P
3
=
3
k
(
λ
μ
)
k

3
P
3
;
Therefore
∑
∞
k
=3
P
k
= 3(
μ
λ
)
3
P
3
∑
∞
k
=3
1
k
(
λ
μ
)
k
.
But
∑
∞
k
=1
1
k
(
λ
μ
)
k
=
log
(
1
1

λ
μ
) if
λ
μ
<
1
.
So
∑
∞
k
=3
P
k
= 3(
μ
λ
)
3
P
3
[
log
(
μ
μ

λ
)

λ
μ

1
2
(
λ
μ
)
2
]
= 3(
μ
λ
)
3
[
log
(
μ
μ

λ
)

λ
μ

1
2
(
λ
μ
)
2
]
θ
(
λ
+
θ
)(2
λ
+
θ
)
6
μ
3
P
0
.
Now
∑
∞
i
=0
(
P
i
) = 1 implies
P
0
= [1 +
θ
μ
+
θ
(
λ
+
θ
)
2
μ
2
+
θ
(
λ
+
θ
)(2
λ
+
θ
)
2
λ
3
[
log
(
μ
μ

λ
)

λ
μ

1
2
(
λ
μ
)
2
]]

1
.
And ﬁnally,
∑
∞
k
=3
P
k
= [[
1
2
λ
3
][
log
(
μ
μ

λ
)

λ
μ

1
2
(
λ
μ
)
2
]
θ
(
λ
+
θ
)(2
λ
+
θ
)
/
[1+
θ
μ
+
θ
(
λ
+
θ
)
2
μ
2
+
θ
(
λ
+
θ
)(2
λ
+
θ
)
2
λ
3
[
log
(
μ
μ

λ
)

λ
μ

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 Fall '08
 Whitt
 Operations Research

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