158
Introduction to Probability Models
(b) By conditioning upon whether the state was 0 or 1 B when he entered we get
that the desired probability is given by
1 12
4
+
=
2 26
6
7
9
(d) Again, condition on the state when he enters to obtain
(c) P1 A + P1 B + 2P2 =
1 1 21
7
1 1 1
+
+
+
=
2 4 2
2 4 62
12
This could also have been obtained from (a) and (c) by the formula W =
That is, W =
7
"92 #
2 3
=
7
.
12
L
.
λa
16. Let the states be (0, 0), (1, 0), (0, 1), and (1, 1), where state (i, j) means that there
are i customers with server 1 and j with server 2. The balance equations are as
follows:
λP00 = μ1 P10 + μ2 P01
(λ + μ1 )P10 = λP00 + μ2 P11
(λ + μ2 )P01 = μ1 P11
(μ1 + μ2 )P11 = λP01 + λP10
P00 + P01 + P10 + P11 = 1
Substituting the values λ = 5, μ1 = 4, μ2 = 2 and solving yields the solution
P00 = 128/513,
P11 = 175/513
P10 = 110/513,
P01 = 100/513,
(a) W = L/λa = [1(P01 + P10 ) + 2P11 ]/[λ(1 − P11 )] = 56/119
Another way is to condition on the state as seen by the arrival. Letting T denote
the time spent, this gives
W = E[T |00]128/338 + E[T |01]100/338
+ E[T |10]110/338
= (1/4)(228/338) + (1/2)(110/338)
= 56/119
(b) P01 + P11 = 275/513
17. The state space can be taken to consist of states (0, 0), (0, 1), (1, 0), (1, 1), where the
ith component of the state refers to the number of customers at server i, i = 1, 2.