IEOR 6711: Stochastic Models I, Fall 2003, Professor Whitt
Solutions to Final Exam: Thursday, December 18.
Below are six questions with several parts. Do as much as you can. Show your work.
1.
TwoPump Gas Station
(12 points)
Potential customers arrive at a fullservice, twopump gas station according to a Poisson
process at a rate of 40 cars per hour. There are two service attendants to help customers, one
for each pump.
If the two pumps are busy, then arriving customers wait in a single queue,
to be served in the order of arrival by the first available pump. However, customers will not
enter the station to wait if there are already two customers waiting, in addition to the two in
service. Suppose that the amount of time required to service a car is exponentially distributed
with a mean of three minutes.
———————————————————————–
This problem is a minor variant of Problem 15 in Chapter 6, on page 393, of the easier
Ross book,
Introduction to Probability Models
. To answer the questions posed here, we need
to find the steadystate distribution of the CTMC. In general, the steadystate distribution
can be found by solving the equation
αQ
= 0, where
α
is the steadystate probability vector.
This CTMC is a birthanddeath process, so we can find the steadystate solution directly, by
solving the local balance equations
α
k
λ
k
=
α
k
+1
μ
k
+1
for all
k .
But, first, we need to read carefully to make sure we get the model right.
We need to
recognize that there are five states:
k
for 0
≤
k
≤
4. We thus can write down the steadystate
probabilities in terms of an unknown
x
, and then normalize to find
x
:
α
0
= 1
*
x,
α
1
=
λ
0
μ
1
*
x,
α
2
=
λ
0
μ
1
λ
1
μ
2
*
x ,
α
3
=
λ
0
μ
1
λ
1
μ
2
λ
2
μ
3
*
x ,
α
4
=
λ
0
μ
1
λ
1
μ
2
λ
2
μ
3
λ
3
μ
4
*
x .
A main issue is to be careful and get the rates right, using the same units. We could either
use cars per hour or cars per minute, but we should be consistent. We should realize that a
mean service time of 3 minutes means one service per 3 minutes or 20 services per hour.
So, here, inserting the birth rates and death rates, we get
α
0
= 1
*
x,
α
1
=
40
20
*
x,
α
2
=
40
20
40
40
*
x ,
α
3
=
40
20
40
40
40
40
*
x ,
α
4
=
40
20
40
40
40
40
40
40
*
x
or
α
0
= 1
*
x,
α
1
= 2
*
x,
α
2
= 2
*
x ,
α
3
= 2
*
x ,
α
4
= 2
*
x ,
so that
α
0
= 1
/
9
,
α
1
=
α
2
=
α
3
=
α
4
= 2
/
9
.
———————————————————————–
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(a) In the longrun, what rate are cars served?
———————————————————————–
Cars are served at rate
20
*
α
1
+ 40
*
(
α
2
+
α
3
+
α
4
) = 280
/
9 = 31
.
111
. . .
per hour
.
Alternatively, we could multiply the arrival rate times 1

α
4
getting 40
×
(7
/
9) = 280
/
9, as
above.
———————————————————————–
(b) In the longrun, what proportion of potential customers fail to be served?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 all
 Probability theory, Stochastic process, Markov chain, Markov Mouse

Click to edit the document details