Question

Please complete the following.

2. A cafeteria serving line has a single coffee urn from which customers serve themselves. Arrivals at the urn follow a Poisson distribution at the rate of three per minute. In serving themselves, customers take on average about 15 seconds with exponentially distributed serving times. a. How much time, on average, does it take before a customer can drink coffee? Prepare for VUT calculation a (avg. inter p (avg. arrival time) service time) m (number of servers) u (avg. utilization) CV. (coef. of CV. (coef. of var.- arrival) var.- service) Show VUT calculation and final answer to the question: b. How many customers on average would you expect to be waiting in line for their coffee? c. Instead of the customers serving themselves from the coffee urn, suppose that the cafeteria installs an automatic vendor that dispenses a cup of coffee in a constant time of 15 seconds. How does this change your answer to part b?

Answer 1

a ( avg interarrival time)	avg service time(p)	Number of servers m	Utilisation u	Cva	CVs
0.33 min	0.25	1	0.75	0	0

avg service time = 15/60=0.25 min.

Utilisation u= a/p = 0.25/0.33  = 0.75

Time in line = avg service time x (u/1-u) ( CVa²+CVs²)/2]

Here, time in line = 15 sec x 0.75/(1-0.75) (0+0/2)

= 45 sec.

Number of persons in line = Time in line /a = 45 sec /( 0.33x60) =2.25

In a constant service systems

Time in line = [avg service time x (u/1-u) ( CVa²+CVs²)/2)]/2

=22.5 sec.

Number of persons in line = Time in line /a = 22.5/20 =1.125