ISyE 3232
Stochastic Manufacturing and Service Systems
Fall 2011
H. Ayhan
Solutions to Homework 12
1. The throughput in each case is equal to the e±ective arrival rate
λ
ef
=
λ
(1
−
P(Customer arrives when all three lines are busy))
To calculate the probability of a customer arriving when all three lines are busy, I will use
a Continuous Time Markov Chain (CTMC)
X
K
(
t
) to represent the number of people in the
system at time
t
when there are
K
servers. Because the system has only three phone lines, the
state space is
S
=
{
0
,
1
,
2
,
3
}
.G
i
v
e
n
K
, the stationary distribution can be calculated using
the cut method. The average number of people in the queue can also be calculated from the
stationary distribution of the DTMC. With the throughput and average number of people in
the queue, Little’s Law allows us to calculate the average waiting time.
(a) The rate diagram for the case with one server (
K
= 1) is
±
²³
´
±±±
±²³
±²³
±²³
Making a cut between 0 and 1, 1 and 2, and 2 and 3 gives the equations:
η
0
=
1
2
η
1
η
1
=
1
2
η
2
η
2
=
1
2
η
3
Setting
η
0
= 1 gives the invariant measure
η
=
{
1
,
2
,
4
,
8
}
,w
ith
°
3
i
=0
η
i
= 15. Using the
relation between stationary distributions and invariant measures,
π
n
=
η
n
±
°
3
i
=0
η
i
²
results in the stationary distribution
π
=
{
1
/
15
,
2
/
15
,
4
/
15
,
8
/
15
}
. The probability of a
caller arriving to see all three lines busy equals
π
3
,so
λ
ef
=
λ
(1
−
π
3
)=1
³
1
−
8
15
´
=
7
15
per minute
Because there is one server, the number in the queue is one less than the state of the
Markov chain. The average number of people in the queue is
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 Fall '07
 Billings
 Probability theory, Markov chain, stationary distribution, Πn

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