ECEN 646 Homework 7 Solutions
1
C
¸inlar Exercise (8.6), Chapter 4
Each traffic flow process is Poisson described by the rate shown in the
figure.
2
C
¸inlar Exercise (8.8), Chapter 4
The rate of arrival of trucks at the restaurant is
λ
t
= 0
.
1(0
.
3
×
80+0
.
2
×
60) = 3
.
6. The rate of arrival of cars at the restaurant is
λ
c
= 0
.
1(0
.
7
×
80+
0
.
8
×
60) = 10
.
4. Therefore the rates of arrivals of cars with 1,2,3,4, and 5
passengers are 3.12, 3.12, 2.08, 1.04, and 1.04, respectively.
(a)
Then
Z
= 1
N
t
+ 1
N
1
+ 2
N
2
+ 3
N
3
+ 4
N
4
+ 5
N
5
, where
N
t
is the number
of arrivals of trucks in one hour, and
N
i
for
i
= 1
,
2
,
3
,
4
,
5 is the number of
arrivals of cars with
i
passengers. So,
E [
Z
] =
λ
t
+
λ
1
+2
λ
2
+3
λ
3
+4
λ
4
+5
λ
5
= 3
.
6+3
.
12+2
×
3
.
12+3
×
2
.
08+4
×
1
.
04+5
×
1
.
04 = 28
.
56
(b)
E
bracketleftbig
α
Z
bracketrightbig
= E
bracketleftbig
α
N
t
+
N
1
+2
N
2
+3
N
3
+4
N
4
+5
N
5
bracketrightbig
= E
bracketleftbig
α
N
t
bracketrightbig
E
bracketleftbig
α
N
1
bracketrightbig
E
bracketleftbig
α
2
N
2
bracketrightbig
E
bracketleftbig
α
3
N
3
bracketrightbig
E
bracketleftbig
α
4
N
4
bracketrightbig
E
bracketleftbig
α
4
N
4
bracketrightbig
Each term involves the MGF of a Poisson process at time 1 hour.
For
example,
E
bracketleftbig
α
3
N
3
bracketrightbig
=
∞
summationdisplay
k
=0
e

λ
3
λ
k
3
k
!
α
3
k
=
e
λ
3
α
3

λ
3
1
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Thus,
E
bracketleftbig
α
Z
bracketrightbig
=
e
3
.
6
α
+3
.
12
α
+3
.
12
α
2
+2
.
08
α
3
+1
.
04
α
4
+1
.
04
α
5

14
3
C
¸inlar Exercise (8.9), Chapter 4
The overall Poisson process
N
with rate
λ
is divided into a device failure
process with rate
pλ
, and a device doesn’t fail process with rate
qλ
. We’re
given that
T
=
T
K
, and
K
is the number of the shock that causes the failure.
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 Fall '11
 SerapSavari
 Exponential distribution, Poisson process

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