Answer to Question #267136 in Operations Research for Sam

Question

A bank has one drive-in counter. It is estimated that cars arrive according to Poisson distribution at

the rate of 2 every 5 minutes and that there is enough space to accommodate a line of 10 cars. Other

arriving cars can wait outside this space, if necessary. It takes 1.5 minutes on an average to serve a

customer, but the service time actually varies according to an exponential distribution. You are

required to find:

a) The probability of time, the facility remains idle.

b) The expected number of customers waiting but currently not being served at a particular point

of time.

c) The expected time a customer spends in the system

d) The probability that the waiting line will exceed the capacity of the space leading to the drive-

in counter

Answer 1

Mean arrival rate:

$\lambda=2/5$

Mean service rate:

$1/\mu=1.5$

$\mu=1/1.5$

a)

The probability of time, the facility remains idle:

$1-\lambda/\mu=1-2/7.5=0.7333=73.33\%$

b)

The expected number of customers waiting:

$\frac{\lambda^2}{\mu(\mu-\lambda)}=\frac{0.4^2}{0.67(0.67-0.4)}=0.9$

c)

expected time a customer spends in the system:

$\frac{1}{\mu-\lambda}=\frac{1}{0.67-0.4}=3.7$ min

d)

the probability that n customers waiting in a queue for service:

$p_n=(1-\lambda/\mu)(\lambda/\mu)^n$

then probability that the waiting line will exceed 10 cars:

$p(n>10)=1-\displaystyle{\sum^{10}_{i=0}}p_i$

$p_0=0.4,p_1=0.24,p_2=0.144,p_3=0.0864,p_4=0.0519,p_5=0.0311$

$p_6=0.0187,p_7=0.0112,p_8=0.0067,p_9=0.0040,p_{10}=0.0024$

$p(n>10)=1-0.9964=0.0036$