EE 465 : Homework 8 Solutions
1. Let the state be the idle server. The corresponding CTMC is shown in Figure 1.
1
2
3
ü
2
ü
1
ü
3
ü
2
ü
1
ü
3
Figure 1:
The balance equations are
(
μ
2
+
μ
3
)
P
1
=
μ
1
P
2
+
μ
1
P
3
(
μ
1
+
μ
3
)
P
1
=
μ
2
P
2
+
μ
2
P
3
P
1
+
P
2
+
P
3
=
1
→
P
1
=
μ
1
μ
2
+
μ
3
(
P
2
+
P
3
)
→
P
1
(1 +
μ
2
+
μ
3
μ
1
) = 1
→
P
1
=
μ
1
μ
1
+
μ
2
+
μ
3
and
P
2
=
μ
2
μ
1
+
μ
2
+
μ
3
and
P
3
=
μ
3
μ
1
+
μ
2
+
μ
3
2. This model is mathematically equivalent to the M/M/1 queue with finite capacity
k
. The pro
duced items constitute the arrivals to the queue and the arriving customers constitute the service,
i.e. if we take the state of the system to be the number of items presently available, then we
just have the model of section 8.3.2.
(a) The proportion of customers that go away emptyhanded is equal to
P
0
, the proportion of
time there are no items on the shelves. From section 8.3.2, we have:
P
0
=
1

λ/μ
1

(
λ/μ
)
k
+1
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(b)
W
=
L
λ
a
=
L
λ
(1

P
k
)
Again, by using the results obtained in section 8.3.2, we have
L
=
λ
£
1 +
k
(
λ/μ
)
k
+1

(
k
+ 1)(
λ/μ
)
k
/
(
μ

λ
)(1

(
λ/μ
)
k
+1
)
P
k
=
(
λ/μ
)
k
(1

λ/μ
)
1

(
λ/μ
)
k
+1
Substituting these values in the equation for
W
, one gets the answer.
(c) The average number of items in stock is
L
, given in part (b)
(d) The state space is
J
such that

n
≤
J
≤
k
. The state is
J, J
≥
0, when there are
J
items
in stock and no waiting customers; and it is
J, J
≤
0, when there are no items in stock and

J
waiting customers. The state transition diagram is shown in Figure 2.
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 Fall '04
 Chow
 Queueing theory, AirTrain Newark, 0 1 J, 1 m

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