In a drug store customers arrive at the counter with one ser

In a drug store customers arrive at the counter (with one server per counter) in a Poisson process at the rate of 48 per hour. The service time can be assumed to be exponential with an average of 1 minute and the service is provided at a counter attended by a server. The number of counters is varied depending on the number of customers waiting or being served as follows:

0–4 customers 1 counter

5–9 customers 2 counters

10–14 customers 3 counters

15 or more 4 counters

Assume that this policy is used to increase or decrease the number of servers. Determine the following:

a. Probability of system idleness.

b. How often would the store need more than one counter?

c. What is the average number of customers either waiting for service or being served?

d. What is the average waiting time in the queue?

Solution

The problem will be solved as per M/M/1 queuing theory of Kendall’s notation

Given.

Arrival rate = a =48 / hour

Service rate ( @ 1 minute per customer ) = S = 60 / hour

Therefore system utilization ratio = a/s = 48/60 = 0.8

Hence, probability of system idleness = 1 – a/s = 1 – 0.80 = 0.20

PROBABILITY OF SYSTEM IDLENESS = 0.20

Probability that there will be Zero customers waiting = Po = ( 1 – a/s) = 1 – 48/60 = 1 – 0.80 = 0.20

Probability that there will be 1 customer waiting = P1 = ( a/s) x Po = 0.80 x 0.20 = 0.16

Probability that there will be 2 customers waiting = P2 = (a/s)^2 x Po = 0.80 x 0.80 x 0.20 = 0.128

Probability that there will be 3 customers waiting = P3 = ( a/s)^3 x PO = 0.80 x 0.80 x 0.80 x 0.20 = 0.1024

Probability that there will be 4 customers waiting = P4 = ( a/s)^4 x Po = 0.80 x 0.80 x 0.80 x 0.80 x 0.20 = 0.0819

Therefore, Probability that there will be maximum 4 customers waiting = 0.20 +0.16 + 0.128 +0.1024 +0.0819 = 0.6723

Therefore , probability that there will be more than 4 customers waiting for which more than one counter will be needed = 1 – 0.6723 = 0.3277

Thus, 0.3277 x 100 = 32.77% of time store would need more than one counter

STORE WOULD NEED MORE THAN ONE COUNTER 32.77% OF TIME

Average number of customers either waiting for service or being served ( i.e average number of customers in the system )

= a^2/s x ( s – a ) + a/s

= ( a^2 +as – a^2) / s x ( s -a )

= as/s x ( s -a )

= a / ( s – a )

= 48/ ( 60 – 48)

= 48/12

= 4

AVERAGE NUMBER OF CUSTOMERS EITHER WAITING FOR SERVICE OR BEING SERVED = 4

Average waiting time in the queue

= a/ s x ( s -a ) hours

= 48 / 60 x ( 60 – 48) hours

= 48 / ( 60 x 12) hours

= 1/15 hours

= 4 minutes

AVERAGE WAITING TIME IN THE QUEUE = 4 MINUTES