waiting line management problem solution, writer jacobs (1-15)
DESCRIPTION
This problem solution has been prepared by Abu Zafor, Abdus Salam and Imran Hossain of Islamic University, Kushtia of Management Department, Session: 2010-2011.TRANSCRIPT
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.
04/13/2023
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
Problem-15.(continued)
Requirement-c.Average time of customer is in shop in the system, Ws =
=
= 0.25 per hour