Answer to Question #134804 in Statistics and Probability for Moff T

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 (

-A customer arriving at the cashier area sees all cashiers serving other

customers and no one else is waiting. Let Tw be the time the customer waits before

getting served. Express Tw in terms of T1, T2, T3, and T4. How long the waiting time will

be on average?

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:

$\rho_0 = (1+\dfrac{\rho}{1!}+\dfrac{\rho^2}{2!}+\dfrac{\rho^3}{3!}+\dfrac{\rho^4}{4!}+\dfrac{\rho^5}{4!*0!})$

$\rho_0 = \dfrac{1}{(1+\dfrac{4}{1!}+\dfrac{4^2}{2!}+\dfrac{4^3}{3!}+\dfrac{4^4}{4!}+\dfrac{4^5}{4!*0!})}=0.013$

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

$p_4=\dfrac{\rho^4}{4^1*4!}*\rho_0=0.18$

the expected service time for the customer is

$TW=\dfrac{\dfrac{3*\rho^4}{(3-\rho)^2*4!}*\rho_0+\rho}{\lambda}=0.058 = 3.5$ minutes