Question

Carrefour Hypermarket employs four cashiers in a certain store to serve its customers. Suppose that Cashier 1 has an average service time of 6 minutes, Cashier 2 has an average service time of 4 minutes, Cashier 3 has an average service time of 2 minutes, and Cashier 4 has an average service time of 3 minutes, respectively. Suppose also that the service times are exponentially distributed and independent from each

other.

Let T1 be the service time of Cashier 1, T2 be the service time of Cashier 2, T3 be the

service time of Cashier 3, and T4 be the service time of Cashier 4 (please use this notation

in your solutions).

Suppose on a given day Cashier 4 is on leave, and Cashier 1, Cashier 2, and Cashier 3 are serving their customers. A customer arriving at the cashier area finds all cashiers serving other customers and no one else is waiting. What is the expected service time for the customer?

A small company has a call center with two phone lines only.

Answer 1

Carrefour Hypermarket is four-channel queuing system.

The queue will not grow indefinitely if the number of cashiers is four.

Then utilization factor for the system($\rho$) is four:

$\rho = 4$

$\lambda$ - mean number of arrivals per time period(get an one hour as time period). Let's calculate it:

$\lambda_1=\dfrac{60}{T1}=\dfrac{60}{6}=10$ average number of customers served by Cashier 1.

$\lambda_2=\dfrac{60}{T2}=\dfrac{60}{4}=15$    average number of customers served by Cashier 2.

$\lambda_3=\dfrac{60}{T3}=\dfrac{60}{2}=30$ average number of customers served by Cashier 3.

$\lambda_4=\dfrac{60}{T4}=\dfrac{60}{3}=20$ average number of customers served by Cashier 4.

$\lambda = \lambda_1+\lambda_2+\lambda_3+\lambda_4$

$\lambda = 10+15+30+20=75$ average number of customers served by all cashiers

 $\mu$ mean number of people or items served per time period(get an one hour as time period)..

$\mu = \dfrac{\lambda}{\rho}= \dfrac{75}{4}= 18.25\approx19$

19 is the intensity of service for one customer.

Without Cashier 4 we get:

$\lambda = \lambda_1+\lambda_2+\lambda_3$

$\lambda_s = 10+15+30=55$ average number of customers served by Cashier 1, Cashier 2 and Cashier 3.

$\mu_s = \dfrac{\lambda_s}{\rho}= \dfrac{55}{4}= 13.75\approx14$

Let's find the probability that there are no customers at the checkout:

$\rho_0 = (1+\dfrac{\rho}{1!}+\dfrac{\rho^2}{2!}+\dfrac{\rho^3}{3!}+\dfrac{\rho^4}{3!*0!})$

$\rho_0 = \dfrac{1}{(1+\dfrac{4}{1!}+\dfrac{4^2}{2!}+\dfrac{4^3}{3!}+\dfrac{4^4}{3!*0!})}=0.029$

The probability that there is one customer in the queue is found by the formula:

$p_4=\dfrac{\rho^4}{3^1*3!}*\rho_0=0.41$

the expected service time for the customer is

$TS=\dfrac{\dfrac{3*\rho^4}{(3-\rho)^2*3!}*\rho_0+\rho}{\lambda}=0.14 = 8$ minutes