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Stochastic Manufacturing and Service Systems
Spring 2011
H. Ayhan and J. Dai
Solutions to Homework 11
1. The arrival rate is
λ
(
t
) =
10
9
≤
t
≤
11
10(
t

10)
11
< t
≤
12
15 + 5(13

t
)
12
< t
≤
13
15
13
< t
≤
17
(a)
λ
(12
.
5) = 17
.
5.
(b)
E
[
N
(8)] =
R
17
9
λ
(
s
)
ds
= 112
.
5
.
(c) The probability of k arrivals between 11:30 and 11:45 is just
(d)
(
R
11
.
75
11
.
5
λ
(
s
)
ds
)
k
e

R
11
.
75
11
.
5
λ
(
s
)
ds
k
!
=
(
R
11
.
75
11
.
5
10(
s

10)
ds
)
k
e

R
11
.
75
11
.
5
20(
s

10)
ds
k
!
=
(4
.
0625)
k
e

4
.
0625
/k
!
.
(e) To do this, note that (here minutes need to be changed into hours, e.g. 10
mins
=
1
6
hours)
P
(
T >
1
6
) =
P
(
N
(12
1
6
)

N
(12) = 0)
and
P
(
T >
1
3
) =
P
(
N
(12
1
3
)

N
(12) = 0)
.
Then
P
(
T >
1
6
) =
e

R
12
1
6
12
15+5(13

t
)
dt
=
e

235
/
72
.
Here
T
is not an exponential random variable, since it’s not memoryless (try comparing
P
(
T >
15

T >
5) with
P
(
T >
10)).
2.
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 Spring '07
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