Question

Problem 7-19

A cafeteria serving line has a 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 about 15 seconds, exponentially distributed.

a. How many customers would you expect to see on the average at the coffee urn?

Average no of customers            

b. How long would you expect it to take to get a cup of coffee?

Expected time             minute(s)

c.

What percentage of time is the urn being used?

Percentage of time             %

d. What is the probability that three or more people are at the coffee urn? (Do not round intermediate calculations. Round your answer to 1 decimal place.)

Probability             %

e. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 15 seconds, how many customers would you expect to see at the coffee urn (waiting and/or pouring coffee)? (Round your answer to 2 decimal places.)

Average no of customers            

f. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 15 seconds, how long would you expect it to take (in minutes) to get a cup of coffee, including waiting time? (Round your answer to 2 decimal places.)

Expected time             minute(s)

Answer 1

PLEASE FIND BELOW ANSWERS TO FIRST 4 QUESTIONS :

1. Arrival rate =a = 3 / minute

Service rate ( @ 15 seconds per customer ) = S = 4 / minute

Average number of customers expected to be seen at coffee urn = a^2 / S x ( S – a ) = 3 x 3 / 4 x ( 4 -3 ) = 9/4 =2.25

1. Waiting time to get a cup of coffee = a/ S x ( S -a ) minutes = 3 / 4 x ( 4- 3 ) = ¾ minute =45 seconds

1. Percentage of time urn being busy = a/s x 100 = ¾ x 100 = 75 %

1. Probability that there is ZERO customer at coffee urn = PO = ( 1 – a/S ) = 1 – ¾ = 0.25

Probability that there is 1 customer at the coffee urn = ( a/s ) x Po = ( ¾) x 0.25 = 0.75 x 0.25 =0.1875

Probability that there are 2 customers at the coffee urn = ( a/s)^2 x Po = ( ¾)^2 x 0.25 = 0.75 x 0.75 x 0.25 = 0.1406

Thus, probability that there will be maximum 2 customers at the coffee urn

= Probability that there is zero customer at the coffee urn + probability that there is 1 customer at the coffee urn + Probability that there is 2 customers at the coffee urn

= 0.25 + 0.1875 + 0.1406

= 0.5781

Probability that there are 3 or more people at the coffee urn

= 1 – Probability that there will be maximum 2 customers waiting in the coffee urn

= 1 – 0.5781

= 0.4219 ( 0.4 rounded to 1 decimal place )

PROBABILITY THAT THERE WILL BE 1 OR MORE PEOPLE AT THE COFFEE URN