Assignments: MANA-53xyD chapter 7 Homework 9Operations Managemer ×eproblem 10-21

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Assignments: MANA-53xyD chapter 7 Homework 9Operations Managemer ×eproblem 10-21f The Avex × Recycle Bi C ezto.mheducation.com/hm.tpx 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 2.5 per minute. In serving themselves, customers take about 22 seconds, exponentially distributed Acrobat Reader D 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.) Brother Utilities 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() 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 at the coffee urn? (Do not round intermediate calculations. Round your answer to 1 decimal place.) Probability ENG Norton 4:57 PM 11/20/2017

Explanation / Answer

Given are :

Arrival rate = a = 2.5 / minute

Service rate ( @ 22 seconds per customer ) = s = 60 / 22    per minute = 2.727 per minute

Average number of customers at coffee urn

= Average number of customers waiting to be served + Average number of customers taking service at coffee urn

= a^2 / s x ( s– a )     + a/ s

= 2.5 x 2.5 / 2.727 x ( 2.727 – 2.5)   + 2.5/2.727  

= 10.096 + 0.916

= 11.012

Time one would expect to get a cup of coffee ( in minutes )

= Average time for waiting in the queue + Average time for getting served

= 2.5 / 2.727x ( 2.727 – 2.5)   + 1/ 2.727

= 4.038 + 0.366

= 4.404 minutes

Percentage of time Urn is being used

= a/s x 100

= ( 2.5 x 22/60)    x 100

= 91.66%

Probability that there are ZERO people at the coffee urn

= Po

= 1 – a/s

= 1 – 2.5 x 22/60

= 1 – 0.916

= 0.084

Probability of 1 customer in the system

= P1

= ( a/s) x Po

= 0.916 x 0.084

= 0.077

Probability of 2 customers in the system

= ( 0.916)^2 x 0.084

= 0.070

Hence,

Probability that there are less than 3 people in the coffee urn

= Probability of zero customer at the coffee urn + Probability of 1 customer at the coffee urn + Probability of 2 customers at the coffee urn

= 0.084 + 0.077 + 0.070

= 0.231

Hence , Probability that three or more people are at the coffee urn

= 1 – Probability that there are less than 3 people at the coffee urn

= 1 – 0.231

= 0.769

AVERAGE NUMBER OF CUSTOMERS = 11.012

EXPECTED TIME = 4.404 MINUTES

PERCENTAGE TIME = 91.66 %

PROBABILITY THAT THREE OR MORE PEOPLE ARE AT THE COFFEE URN = 0.769

AVERAGE NUMBER OF CUSTOMERS = 11.012

EXPECTED TIME = 4.404 MINUTES

PERCENTAGE TIME = 91.66 %

PROBABILITY THAT THREE OR MORE PEOPLE ARE AT THE COFFEE URN = 0.769

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