Stella can handle about ten customers per hour at her oneper

Stella can handle about ten customers per hour at her one-person comic book store. The customer arrival rate averages about six customers per hour. Stella is interested in knowing the operating characteristics of her single-channel, single phase queuing system.

Solution

The probability that no customer will be in waiting line

P=(1-/µ)

Where = mean arrival rate, µ = mean service rate, and n = customers in the waiting line system

Now as per the question

P=(1-6/10) =0.4

Average number of customers in the queuing system

L= / µ-

=6/4 =1.5

The probability that exactly n customers will be in the queuing system will be

P=( /µ)n(1- /µ)

=0.4(6/10)n

Average number of customers in the waiting line will be

L= 2 / µ( µ - )

=36/10(4)

=36/40

=0.9

Average time a customer spends in the queuing system will be

W= 1/( µ - )

=1/4

=0.25

Average time a customer spends waiting in line will be

W= / µ( µ - )

=6/10*4

=6/40

=0.15

The probability of the server being busy and a customer needs to wait, also called utilization factor will be

P= / µ

=6/10

=0.6

Probability of idle server a possibility of customer being served will be

=1-/µ

=(1-6/10) =0.4