Question

Problem 10-10

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 4.0 per minute. In serving themselves, customers take about 8 seconds, exponentially distributed.

a. How many customers would you expect to see on the average at the coffee urn? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Average no of customers ____

b. How long would you expect it to take to get a cup of coffee? (Round your answer to 2 decimal places.)

Expected time ____ minute(s)

c. What percentage of time is the urn being used? (Do not round intermediate calculations. Round your answer to 1 decimal place.)

Percentage of time ____ %

d. What is the probability that three or more people are in the cafeteria? (Round your intermediate calculations to 3 decimal places and final 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 8 seconds, how many customers would you expect to see at the coffee urn (waiting and/or pouring coffee)? (Do not round intermediate calculations. 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 8 seconds, how long would you expect it to take (in minutes) to get a cup of coffee, including waiting time? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Expected time ____ minute(s)

Answer 1

arrival rate = 4 per minute

Service rate = 60/8 = 7.5 per minute

a)

Number of customers = arrival rate/(Service rate -arrival rate) = 4/(7.5-4) = 1.142857143 = 1.14 (Rounded to 2 decimal places)

b)

Expected time = 1/(service rate-arrival rate) = 1/(7.5-4) = 1/3.5 = 0.285714286 = 0.29 (Rounded to 2 decimal places) minute(s)

c)

Percentage of time =arrival rate/service rate = 4/7.5 = 53.33 %

d)

probability that three or more people are in the cafeteria = 1-(P0+P1+P2) where P0 = probability of no customers in the system, P1 = probability of 1 customer in the system

P0 = (1-arrival rate/service rate) = (1-(4/7.5))
P1 = (arrival rate/service rate) *(1-arrival rate/service rate) = (4/7.5)*0.467
P2 = (arrival rate/service rate)^2 *(1-arrival rate/service rate) = (4/7.5)^2*0.467

probability that three or more people are in the cafeteria = 1-((1-(4/7.5)) +(4/7.5)*0.467 +(4/7.5)^2*0.467 ) = 0.151431111 = 15.1% (Rounded to 1 decimal places)