STA3007
Applied probability (0809)
Solution to Assignment 11
Chapter 6
Problem 4.1
If there is single repairman. As
N
=
M
= 5
, R
= 1
, λ
= 2
, μ
= 1, we have
λ
k
=
λ
= 2
, k
= 0
,
1
,
2
,
3
,
4
,
λ
5
= 0
,
μ
k
=
kμ
=
k, k
= 0
,
1
,
2
,
3
,
4
,
5
,
and
θ
0
= 1
, θ
1
= 2
, θ
2
= 2
, θ
3
=
4
3
, θ
4
=
2
3
, θ
5
=
4
15
.
Therefore, the limiting distribution is
π
i
=
15
109
θ
i
, i
= 0
,
1
,
2
,
3
,
4
,
5. i.e.
π
0
=
15
109
, π
1
=
30
109
, π
2
=
30
109
, π
3
=
20
109
, π
4
=
10
109
, π
5
=
4
109
.
(a) The average number of machines operating is
5
k
=0
kπ
k
=
210
109
= 1
.
9266
.
(b) The equipment utilization is
5
k
=0
kπ
k
/M
=
210
109
/
5 = 0
.
3853
.
(c) The average idle repair capacity is
π
5
=
4
109
= 0
.
0367
.
If there are two repairmen. As
N
=
M
= 5
, R
= 2
, λ
= 2
, μ
= 1, we have
λ
k
=
Rλ
= 4
, k
= 0
,
1
,
2
,
3
,
λ
4
=
λ
= 2
, λ
5
= 0
,
μ
k
=
kμ
=
k, k
= 0
,
1
,
2
,
3
,
4
,
5
,
and
θ
0
= 1
, θ
1
= 4
, θ
2
= 8
, θ
3
=
32
3
, θ
4
=
32
3
, θ
5
=
64
15
.
1
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Therefore, the limiting distribution is
π
i
=
15
579
θ
i
, i
= 0
,
1
,
2
,
3
,
4
,
5. i.e.
π
0
=
15
579
, π
1
=
60
579
, π
2
=
120
579
, π
3
=
160
579
, π
4
=
160
579
, π
5
=
64
579
.
(a) The average number of machines operating is
5
k
=0
kπ
k
=
1740
579
= 3
.
0052
.
(b) The equipment utilization is
5
k
=0
kπ
k
/M
=
1740
579
/
5 = 0
.
6010
.
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 Spring '11
 afdsa
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