Question

Customers arrive at a service facility with one server according to a Poisson process with a rate of 5 per hour. The service time are i.i.d. exponential r.v.´s, and on the average, the server can serve 7 customers per hour. Suppose that the system is in the stationary regime. (a) What is the probability that at a particular time moment, there will be no queue? (b) What is the probability that a particular time moment, there will be more than three customers waiting for service? (c) What is the mean proportion of time when the server is idle?

Answer 1

a)

There will no queue

when there will 0 or 1 customer

hence

p0 + p1

= (1 - 5/7)(1 + 5/7)

= 0.48979

b)

$p_n = \rho^n (1- \rho)$

P(X > 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X =3))

= 1 - 2/7( 1 + (5/7) + (5/7)^2 + (5/7)^3)

= 0.26030

c)

mean proportion of time when the server is idle

= p0

$p_0 =1- \frac{\lambda }{\mu }$

$p_0 = (1- \rho)$

=1- 5/7

=2/7