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Page 1: Waiting Line Management Problem Solution, Writer Jacobs (1-15)

04/13/2023

A Presentation onOperations ManagementCourse Code: MGT-312

Topics: Waiting Line Management

Group: Campus

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Group Members Roll No.Authorized By,

MD. Abu Zafor

Md. Abdus Salam

Md. Imran Hossain

Session: 2010-2011

Department of Management

Islamic University, Kushtia-Bangladesh.

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Problem-1.Given that,

Arrival rate of student, λ = 4 per hour [60/15]

Service rate of clerk, µ =6 per hour [60/10]

Requirement-a.

Percentage of Time Judy Idle, P0 =1-

= 1-

= 0.33 or, 33%

continue

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Problem-1.(continued)Requirement-b.Average time a student spend waiting in line, Wq =

=

=

= 0.33 per hour

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Problem-1.(continued)Requirement-c.Average waiting line in the system, Ws=

=

= 0.5 per hour

Requirement-d.Probability, P =

=

= 0.67 or, 67%

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Problem-3Given that,

Arrival rate of customer, λ = 10 per hour [60/6]

Service rate of customer, µ =15 per hour [60/4]

Requirement-a.Each service desk idle, Po = 1-

=1-

=1-0.67

= 0.33 or, 33%

continue

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Problem-3.(continued)Requirement-b.Probability that both service clerks are busy, =

=

= 0.67 or, 67%

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Problem-3.(continued)Requirement-c.If two service desk are performed, then service rate will be,

µ =

= 7.5

The probability that both service clerk are idle, Po = 1-

= -0.33,

or -33%

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Problem-3.(continued)Requirement-d.Average number customer are waiting in line, Lq =

=

=

=

=1.33 per hour

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Problem-3.(continued)

Requirement-e.Average time a customer waiting in the system, Ws =

=

= 0.2 per hour

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Problem-4.Given that,

Arrival rate of customer, λ = 10 per hour [60/6]

Service rate of customer, µ =15 per hour [60/4]

Requirement-a.The probability of waiting in line, =

=

= 0.67 or, 67%

continue

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Problem-4.(continued)

Requirement-b.Average customer are waiting in line, Lq =

=

=

=

=1.33 per hour

continue

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Problem-4.(continued)

Requirement-c.Average time a customer waiting in the system, Ws =

=

= 0.2 per hour

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Problem-5.Given that,

Arrival rate of car, λ = 72 per hour [ (60/50)*60 ]

Service rate of Burrito king, µ =80 per hour [ (60/45)*60 ]

Requirement-a.Average time in the system, Ws =

=

= = 0.13 per hour

continue

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Problem-5.(continued)Requirement-b-.Average number of car in the line, Lq =

=

= = 8.1 car

continue

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Problem-5.(continued)

Requirement-c.Average number of cars in the system, Ls =

=

=

= 9 cars

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Problem-6.Given that,

Arrival rate of customer, λ = 100 per hour

Service rate of customer, µ =120 per hour [ (60/30)*60 ]

Requirement-a.Average number of customer time ion the system, Ws =

=

=

= 0.05 customer.continue

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Problem-6.(continued)Requirement-b.

The effect of customer time would be in the system of having a second ticket taker doing nothing but validation and card punching, thereby cutting the average service time to 20 seconds.

Then service rate of customer will be µ = 180 per hour [ (60/20)*60]

continue

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Problem-6.(continued)

Average number of customer in the system, Ls =

=

=

= 1.25 customer

continue

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Problem-6.(continued)Requirement-c.If the second window will open, then service time will be, = 360 per hour [ 120*3 ]

Average waiting time in the system, Ws =

=

=

= 0.0038 per hour

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Problem-7.Given that,

Arrival rate of person, λ = 10 per hour

Service rate of person, µ =12 per hour [60/5]

Requirement-a.Average number of person in the line, Lq =

=

=

= 4.17 personcontinue

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Problem-7.(continued)Requirement-b.Average number of person in the system, Ls =

=

= 5 person

continue

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Problem-7.(continued)Requirement-c.Average time spend in the line, Wq =

=

=

=

= 0.42 per hour

continue

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Problem-7.(continued)

Requirement-d.Average time spend in the system, Ws =

=

= = 0.5 per hour

continue

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Problem-7.(continued)Requirement-e.

If arrival rate is, = 12 per hour,

then number of person waiting in line, Lq =

=

=

=

= 144 person

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Problem-8.Given that,

Arrival rate of customer, λ = 180 per hour [ 60*3 ]

Service rate of customer, µ =240 per hour [ (60/15)*60 ]

Requirement-a.Average number of customer expect to see in the system, Ls =

=

=

= 3 customer

continue

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Requirement-b.Average time expect it to take to get a cup of coffee, Wq =

=

=

= 0.013 per hour

continue

Problem-8.(continued)

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Problem-8.(continued)

Requirement-c.Percentage of time is the urn being used, P =

=

= 0.75 or, 75%

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Problem-10.Given that,

Arrival rate of patient, λ = 20 per hour [ 60/3 ]

Service rate of patient, µ =30 per hour [ 60/2 ]

Requirement-a.Perform by one Nurse,

Average number of patient in the system, Ls =

=

=

= 2 per hour

continue

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Problem-10.(continued)

Requirement-b.Average time in the system, Ws

=

=

= 0.10 per hour

continue

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Problem-10.(continued)

Requirement-c.Probability of the system being busy, =

=

= 0.67 or, 67%

continue

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Problem-10.(continued)Requirement-d.

Perform by three Nurse,

Service rate will be, = 90 per hour [ 30*3 ]

Average time of patient spend in the system, Ws =

=

= 0.014 per hour

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Problem-11.Given that,

Arrival rate of customer, λ = 5 per hour [ 60/12 ]

Service rate of customer, µ =6 per hour [ 60/10 ]

Requirement-a.

Average time of customer spend in the system, Ws =

=

= 1 per hour

continue

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Problem-11.(continued)Requirement-b.Average number of room in waiting area, Lq =

=

=

= 4.17 room

continue

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Problem-11.(continued)Requirement-c.Probability of the system in being busy, =

=

= 0.83 or, 83%

Requirement-d.Probability that the system is idle, Po = 1-

= 1-

= 1- 0.83

= 0.17 or, 17%

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Problem-13.Given that,

Arrival rate of customer, λ = 2 per hour

Service rate of customer, µ = 3 per hour [ 60/20 ]

Requirement-a.Average number of customer waiting in the line, Lq =

=

=

= 1.33 customer

continue

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Problem-13.(continued)

Requirement-b.Average time a customer waiting in line, Wq =

=

=

= 0.67 per hour

continue

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Problem-13.(continued)

Requirement-c.Average time a customer in the system, Ws =

=

= 1 per hour

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Problem-14.Given that,

Probability idle, Po = 45% or, 0.45

Arrival rate of customer, λ = 1.5 per hour [ 60/40 ]

Service rate of customer, µ = ??

We know that,

Po = 1-

=> 0.45 = 1-

=> 0.45 =

=> 0.45 = 1- 1.5

=>1.5 = 1 - 0.45

=> 1.5 = 0.55

=> = 2.73continue

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Problem-14.(continued)Probability system being busy, = 85% or, 0.85

Service rate of customer, = 2.73

Arrival rate of customer, λ = ??

We know that,

=

=> 0.85 =

=> = 0.85*2.73

=> = 2.31 per hour

The arrival rate is need 2.31 per hour.

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Problem-15.Given that,

Arrival rate of customer, λ = 2 per hour

Service rate of customer, µ = 6 per hour [ (60/20) *2 ]

Requirement-a.

Average number customer waiting in line, Lq =

=

=

= 0.17 customer

continue

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Problem-15.(continued)Requirement-b.Average time of customer waiting in line, Wq =

=

= 0.08 per hour

continue

Page 43: Waiting Line Management Problem Solution, Writer Jacobs (1-15)

Problem-15.(continued)

Requirement-c.Average time of customer is in shop in the system, Ws =

=

= 0.25 per hour