STA3007
Applied probability (08-09)
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
X
k
=0
kπ
k
=
210
109
= 1
.
9266
.
(b) The equipment utilization is
5
X
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