Question

At a fashion retailer, there are multiple cashiers providing checkout service to customers simultaneously. On average, customers arrive at the checkout area every 6 minutes. The standard deviation of the inter-arrival time is 6 minutes. The average checkout time for each customer is 12 minutes, with its standard deviation equal to 12 minutes. Suppose that customers form a single line. What is the minimum number of cashiers required, in order to guarantee an average customer waiting time below 5 minutes?

Answer 1

Answer:-

Given That:-

At a fashion retailer, there are multiple cashiers providing checkout service to customers simultaneously. On average, customers arrive at the checkout area every 6 minutes. The standard deviation of the inter-arrival time is 6 minutes. The average checkout time for each customer is 12 minutes, with its standard deviation equal to 12 minutes. Suppose that customers form a single line. What is the minimum number of cashiers required, in order to guarantee an average customer waiting time below 5 minutes

ta = 6 minutes
sa = 6 minutes

So, ca = sa / ta = 1

te = 12 minutes
se = 12 minutes

So, ce = se / te = 1

Utilization, u = te / (m.ta) = 2/m where m = number of servers which is unknown here,

The average waiting time, CTq will be computed using the following formula:

The minimum value of 'm' is greater than te / ta i.e. greater than 2. So, start with m=3 and check where CTq is less than 5 minutes.

For m=3, minutes > 5 minutes

For m=4, minutes < 5 minutes

So, at least 4 cashiers will be required to attain the goal of 5 minutes waiting time.