ISyE 3232
Stochastic Manufacturing and Service Systems
Spring 2011
J. Dai and H. Ayhan
Solutions to Homework 7
1.
Since
α
is a distribution, the sum of entries should be 1.
α
= (1
/
5
,
2
/
5
,
2
/
5)
.
The transition matrix
P
has the property that elements in each row of
P
sum up to one.
Therefore we have
P
=
1
/
5
4
/
5
0
2
/
5
1
/
2
1
/
10
0
1
/
10
9
/
10
(a)
Pr(
X
1
= 0

X
0
= 1) =
P
1
,
0
= 2
/
5
.
(b)
The row vector
αP
= (0
.
2
,
0
.
4
,
0
.
4) describes distribution of
X
1
.
For
i
= 0
,
1
,
2
Pr(
X
1
=
i
)
=
Pr(
X
1
=
i

X
0
= 0) Pr(
X
0
= 0) + Pr(
X
1
=
i

X
0
= 1) Pr(
X
0
= 1)
+ Pr(
X
1
=
i

X
0
= 2) Pr(
X
0
= 2)
=
α
0
P
0
,i
+
α
1
P
1
,i
+
α
2
P
2
,i
=
i
th element of row vector
αP.
Note that
αP
=
α,
so
α
is actually a stationary distribution.
(c)
The row vector
αP
15
= (0
.
2
,
0
.
4
,
0
.
4) describes distribution of
X
15
.
For
i
= 0
,
1
,
2
Pr(
X
15
=
i
)
=
Pr(
X
15
=
i

X
0
= 0) Pr(
X
0
= 0) + Pr(
X
15
=
i

X
0
= 1) Pr(
X
0
= 1)
+ Pr(
X
15
=
i

X
0
= 2) Pr(
X
0
= 2)
=
α
0
P
15
0
,i
+
α
1
P
15
1
,i
+
α
2
P
15
2
,i
=
i
th element of row vector
αP
15
.
2.
In the solutions below we use the following formulas: for two events
A
and
B
P
(
A

B
) =
P
(
A
T
B
)
P
(
B
)
and for two random variables
Y
and
Z
and a constant
k
E
[
Y

Z
=
k
] =
∞
X
n
=0
n
P
{
Y
=
n

Z
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 Spring '07
 Billings
 Markov chain, Stochastic Manufacturing, row vector αP

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