Question

Problem 1

There are four checkout line in the deli, each of which is served by an employee who collects payment. Lately, we noticed that a checkout line was reserved for customers who used “dinning dollars”. (Customers with cash payments can only join the remaining three lines).

The original setup without the dining – dollars line. During the busy lunch period from 11:30 am to 2 pm, customers arrive at the deli at an average rate of 20 customers per minute. Suppose the customers are equally split among the four checkout lines. (In probability lingo, each customer randomly chooses a line to join, with the probability of choosing a particular line equal to the probability of choosing any other line). For simplicity, assume that each customer, once joins a line, does not switch to another line. An employee spends an average of 10 seconds per customer to determine the payment amount and to collect the payment with proper changes, if necessary. Assume that customer arrivals follow a Poisson process, and the service times are exponentially distributed.

1. What is the average time a customer spends in the checkout process (from joining a checkout line to departing )

2. What is the average number of customers in the checkout area (including all four checkout lines, waiting or being served)?

Now suppose one of the four checkout lines is dedicated to customers with dinning dollars. Because collecting dinning dollars in simpler than collecting cash (e.g, no changes are necessary), the average service time for dinning dollars customer is only 6 seconds. Suppose 25% of customers use dinning dollars. The remaining 75% of customers pay by cash, and thus must join one of the three cash lines. Again, these customers randomly pick a cash line to join (with equal probabilities). The average service time for a cash customer is 11 seconds. Assume that the arrival processes of the two customer classes are both Poisson, and their service times are exponentially distributed.

3. What is the average time a dinning-dollars customer spends in the checkout process (from joining the line to departing)? What about a cash customer?

4. What suggestions do you have to improve service without increasing staffing levels?

Answer 1

no of servers	4
mean arrival rate, λ	20	customers/min
total service time	10	seconds
	0.17	min
mean service rate	6	customers/min/server
mean system service rate, μ	24	customers/min
a) average time in the system, W=1/(μ-λ)
avg time in the checkout process=	0.25	minutes
b) average number in the system, L = λW
the average number in the checkout process =	5	customers
c)
	Cash customer	dining dollars
no of servers	3	1
mean arrival rate, λ	15	5	customers/min
total service time (in min)	0.18	0.10	min
mean service rate	5.45	10.00	customers/min/server
mean system service rate, μ	16.36	10.00	customers/min
average time in the checkout process (W)	0.73	0.20	min

formula used: