mAssignments: MANA-53xyD chapter 7 Homework -/ C | ezto.mheducation.com/hm.tpx e

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mAssignments: MANA-53xyD chapter 7 Homework -/ C | ezto.mheducation.com/hm.tpx e operations Managemen× x 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 12 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 at the coffee urn? (Do not round intermediate calculations. Round your answer to 1 decimal place.) Probability

Explanation / Answer

Given are :

Arrival rate = a = 4 / minute

Service rate ( @ 12 seconds per customer ) = s = 60 / 12    per minute = 5 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

= 4x4/5x(5-4) + 4/5

= 16/5 + 4/5

= 20/5

= 4

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

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

= a/ s x ( s- a) +   1/s

= 4/5 x ( 5-4) + 1/ 5

= 4/ 5 + 1/ 5

= 1 minute

Percentage of time Urn is being used

= a/s x 100

= ( 4/5 ) x 100

= 80 %

Probability that there are ZERO people at the coffee urn

= Po

= 1 – a/s

= 1 – 4/5

= 1 – 0.80

= 0.20

Probability of 1 customer in the system

= P1

= ( a/s) x Po

= 0.80 x 0.20

= 0.16

Probability of 2 customers in the system

= ( 4/5)^2 x 0.20

= 0.128

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

= 0.20 + 0.16 + 0.128

= 0.488

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 – 0.488

= 0.512

AVERAGE NUMBER OF CUSTOMERS = 4

EXPECTED TIME = 1 MINUTE

PERCENTAGE OF TIME = 80 %

PROBABILITY THAT THREE OR MORE PEOPLE ARE AT THE COFFEE URN = 0.512

AVERAGE NUMBER OF CUSTOMERS = 4

EXPECTED TIME = 1 MINUTE

PERCENTAGE OF TIME = 80 %

PROBABILITY THAT THREE OR MORE PEOPLE ARE AT THE COFFEE URN = 0.512

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