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IEOR 4106: Introduction to Operations Research: Stochastic Models
Spring 2005, Professor Whitt, Second Midterm Exam
Chapters 56 in Ross, Thursday, March 31, 11:00am1:00pm
Open Book: but only the Ross textbook plus one
8
×
11
page of notes
Justify your answers; show your work.
1. The IEOR Department Ricoh Printer (
30
points)
The Columbia IEOR Department has a versatile Ricoh printer that can rapidly print one
sided and twosided copies, but unfortunately it often goes down. Ricoh is alternately up
(working) and down (waiting for repair or under repair). The average up time (time until
breakdown) is 4 days, while the average down time (time until repair) is 3 days. Assume
continuous operation. Let
X
(
t
) = 1 if the Ricoh is working at time
t
, and let
X
(
t
) = 0
otherwise.
(a) What do we need to assume about the successive up and down times in order to make
the stochastic process
{
X
(
t
) :
t
≥
0
}
a continuoustime Markov chain (CTMC)?
——————————————————————————
We need the successive up times and down times to be mutually independent random
variables. In addition, these random variables should have exponential distributions. The
lackofmemory property of the exponential distribution is critical for getting the Markov
property for the stochastic process
{
X
(
t
) :
t
≥
0
}
. The exponential distribution has a single
parameter, which can be taken to be its mean. Since the means are already speciﬁed, nothing
more needs to be assumed, beyond assuming that the means agree with the speciﬁed averages.
——————————————————————————
Henceforth assume that these extra assumptions are in place, so that indeed the stochastic
process
{
X
(
t
) :
t
≥
0
}
is a CTMC.
(b) Construct the CTMC; i.e., specify the model.
——————————————————————————
A CTMC can be speciﬁed by its
rate matrix
,
Q
. That takes a simple form here because
there are only two states: 0 and 1. The rate matrix here, ordering the two states 0 and 1, is
Q
=
±

1
/
3
1
/
3
1
/
4

1
/
4
¶
where the rates are expressed per day. That is, the repair rate is 1
/
3 per day, while the failure
rate is 1
/
4 per day.
In this case we have a birthanddeath (BD) process, so that
λ
0
=
Q
0
,
1
= 1
/
3, while
μ
1
=
Q
1
,
0
= 1
/
4. That is an equivalent speciﬁcation; i.e., we say that it is BD and we specify
these single
λ
i
and
μ
i
values.
——————————————————————————
(c) Assuming that Ricoh has been working continuously for 7 days, what is the probability
that it will remain working at least 8 more days?
——————————————————————————
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View Full Document The holding time in each state is exponentially distributed. The exponential up time has
mean 4 days, and thus rate 1
/
4, as indicated above. Let
T
be the failure time. Then
P
(
T >
8 + 7

T
= 7) =
P
(
T >
8) =
e

(1
/
4)8
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This note was uploaded on 04/08/2008 for the course IEOR 4106 taught by Professor Whitt during the Spring '08 term at Columbia.
 Spring '08
 Whitt
 Operations Research

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