Homework 7: Solutions
Chapter 4
8.38 In this case the equations are
r
1
=
λ
+ (1

p
)
r
2
r
2
=
r
1
which solves to give that each
r
i
=
λ/p
. For stability we must have that
λ/p <
min
{
μ
1
, μ
2
}
.
8.42 In this case the equations are
r
1
= 10 + 2
r
2
/
5
r
2
= 5 +
r
1
/
4
r
3
= 3
r
1
/
4 + 3
r
2
/
5
Inserting the second equation in the first gives
r
1
= 12 +
r
1
/
10 so
r
1
=
40
/
3. Using the second and third equations now we have
r
2
= 5 + 10
/
3 = 25
/
3
r
3
= 10 + 5
The problem told us that
μ
1
= 25,
μ
2
= 30, and
μ
3
= 20. Thus
r
i
< μ
i
for all
i
and there is a stationary distribution consisting of independent
shifted geometrics with failure probabilities
r
i
/μ
i
.
8.44 This is a special case of Example 7.1 in which the service rates are
μ
1
=
1
/
4,
μ
2
= 1
/
8, and the routing matrix is
p
(
i, j
) =
0
1
1
/
2
1
/
2
The stationary distribution has
π
1
= 1
/
3 and
π
2
= 2
/
3. These queues are
single servers so
ψ
1
(
n
) = 1
/
4
n
ψ
2
(
n
) = 1
/
8
n
Plugging into the formula in (7.3) we have
π
(
n,
4

n
) =
c
(4
/
3)
n
(16
/
3)
4

n
=
c
0
(1
/
4)
n
where
c
0
is chosen to make the sum equal to 1.
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 Spring '10
 DURRETT
 Equations, Probability, Probability theory, SEPTA Regional Rail, EUI, long run fraction

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