Page 71 - A First Course In Stochastic Models

62 RENEWAL-REWARD PROCESSES

and so
∞
∞
k −λt(1−z)

P {A n = k}z = e b(t) dt.
0
k=0
Since the random variables Q n−1 −δ(Q n−1 ) and A n are independent of each other,
A n
E(z Q n ) = E(z Q n−1 −δ(Q n−1 ) )E(z ). (2.5.9)
We have

∞

E(z Q n−1 −δ(Q n−1 ) ) = P {Q n−1 = 0} + z j−1 P {Q n−1 = j}
j=1
1
= P {Q n−1 = 0} + [E(z Q n−1 ) − P {Q n−1 = 0}].
z
Substituting this in (2.5.9), we find

zE(z Q n ) = E(z Q n−1 ) − (1 − z)P {Q n−1 = 0} A(z).
Letting n → ∞, we next obtain the desired result (2.5.7). This completes the proof.
Before concluding this section, we give an amusing application of the Pol-
laczek–Khintchine formula.

Example 2.5.1 Ladies in waiting ∗

Everybody knows women spend on average more time in the loo than men. As
worldwide studies show, women typically take 89 seconds to use the loo—about
twice as long as the 39 seconds required by the average man. However, this does
not mean that the queue for the women’s loo is twice as long as the queue for
the men’s. The sequence for the women’s loo is usually far longer. To explain
this using the Pollaczek–Khintchine formula, let us make the following reasonable
assumptions:

1. Men and women arrive at the loo according to independent Poisson processes
with the same rates.
2. The expected amount of time people spend in the loo is twice as large for
women as for men.
3. The coefficient of variation of the time people spend in the loo is larger for
women than for men.
4. There is one loo for women only and one loo for men only.

∗ This application is based on the article ‘Ladies Waiting’ by Robert Matthews in New Scientist, Vol. 167,
Issue 2249, 29 July 2000.