This problem solution has been prepared by Abu Zafor, Abdus Salam and Imran Hossain of Islamic University, Kushtia of Management Department, Session: 2010-2011.

1. A Presentation on
Operations Management
Course Code: MGT-312
Topics: Waiting Line Management
Group: Campus4/7/2014

2. 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.
4/7/2014

3. 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-
λ
µ
= 1-
4
6
= 0.33 or, 33%
continue4/7/2014

4. Problem-1.(continued)
Requirement-b.
Average time a student spend waiting in line, Wq =
λ
µ(µ−λ)
=
4
6(6−4)
=
4
6∗2
= 0.33 per hour
4/7/2014

5. Problem-1.(continued)
Requirement-c.
Average waiting line in the system, Ws=
1
(µ−λ)
=
1
(6−4)
= 0.5 per hour
Requirement-d.
Probability, P =
λ
µ
=
4
6
= 0.67 or, 67%4/7/2014

6. Problem-3
Given 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-
λ
µ
=1-
10
15
=1-0.67
= 0.33 or, 33%
continue4/7/2014

7. Problem-3.(continued)
Requirement-b.
Probability that both service clerks are busy, 𝜌 =
λ
µ
=
10
15
= 0.67 or, 67%
4/7/2014

8. Problem-3.(continued)
Requirement-c.
If two service desk are performed, then service rate will be,
µ =
60
4∗2
= 7.5
The probability that both service clerk are idle, Po = 1-
10
7.5
= -0.33,
or -33%
4/7/2014

9. Problem-3.(continued)
Requirement-d.
Average number customer are waiting in line, Lq =
λ²
µ(µ−λ)
=
10²
15(15−10)
=
100
15∗5
=
100
75
=1.33 per hour
4/7/2014

10. Problem-3.(continued)
Requirement-e.
Average time a customer waiting in the system, Ws =
1
(µ−λ)
=
1
(15−10)
= 0.2 per hour
4/7/2014

11. 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, 𝜌 =
λ
µ
=
10
15
= 0.67 or, 67%
continue
4/7/2014

12. Problem-4.(continued)
Requirement-b.
Average customer are waiting in line, Lq =
λ²
µ(µ−λ)
=
10²
15(15−10)
=
100
15∗5
=
100
75
=1.33 per hour
continue4/7/2014

13. Problem-4.(continued)
Requirement-c.
Average time a customer waiting in the system, Ws =
1
(µ−λ)
=
1
(15−10)
= 0.2 per hour
4/7/2014

14. 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 =
1
(µ−λ)
=
1
(80−72)
=
1
8
= 0.13 per hour
continue4/7/2014

15. Problem-5.(continued)
Requirement-b-.
Average number of car in the line, Lq =
λ²
µ(µ−λ)
=
72²
80(80−72)
=
5184
80∗8
= 8.1 car
continue4/7/2014

16. Problem-5.(continued)
Requirement-c.
Average number of cars in the system, Ls =
λ
(µ−λ)
=
72
(80−72)
=
72
8
= 9 cars
4/7/2014

17. 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 =
1
(µ−λ)
=
1
(120−100)
=
1
20
= 0.05 customer.
continue
4/7/2014

18. 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
4/7/2014

19. Problem-6.(continued)
Average number of customer in the system, Ls =
λ
(µ−λ)
=
100
(180−100)
=
100
80
= 1.25 customer
continue4/7/2014

20. Problem-6.(continued)
Requirement-c.
If the second window will open, then service time will be, µ= 360 per
hour [ 120*3 ]
Average waiting time in the system, Ws =
1
(µ−λ)
=
1
(360−100)
=
1
260
= 0.0038 per hour4/7/2014

21. 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 =
λ²
µ(µ−λ)
=
10²
12(12−10)
=
100
12∗2
= 4.17 person
continue4/7/2014

22. Problem-7.(continued)
Requirement-b.
Average number of person in the system, Ls =
λ
(µ−λ)
=
10
(12−10)
= 5 person
continue4/7/2014

23. Problem-7.(continued)
Requirement-c.
Average time spend in the line, Wq =
λ
µ(µ−λ)
=
10
12(12−10)
=
10
12∗2
=
10
24
= 0.42 per hour
continue
4/7/2014

24. Problem-7.(continued)
Requirement-d.
Average time spend in the system, Ws =
1
(µ−λ)
=
1
(12−10)
=
1
2
= 0.5 per hour
continue4/7/2014

25. Problem-7.(continued)
Requirement-e.
If arrival rate is, λ= 12 per hour,
then number of person waiting in line, Lq =
λ²
µ(µ−λ)
=
12²
12(12−12)
=
144
12∗0
=
144
0
= 144 person
4/7/2014

26. 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 =
λ
(µ−λ)
=
180
(240−180)
=
180
60
= 3 customer
continue4/7/2014

27. Requirement-b.
Average time expect it to take to get a cup of coffee, Wq =
λ
µ(µ−λ)
=
180
240(240−180)
=
180
240∗60
= 0.013 per hour
continue
Problem-8.(continued)
4/7/2014

28. Problem-8.(continued)
Requirement-c.
Percentage of time is the urn being used, P =
λ
µ
=
180
240
= 0.75 or, 75%
4/7/2014

29. 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 =
λ
(µ−λ)
=
20
(30−20)
=
20
10
= 2 per hour
continue
4/7/2014

30. Problem-10.(continued)
Requirement-b.
Average time in the system, Ws =
1
(µ−λ)
=
1
(30−20)
= 0.10 per hour
continue
4/7/2014

31. Problem-10.(continued)
Requirement-c.
Probability of the system being busy, 𝜌 =
λ
µ
=
20
30
= 0.67 or, 67%
continue4/7/2014

32. Problem-10.(continued)
Requirement-d.
Perform by three Nurse,
Service rate will be, µ = 90 per hour [ 30*3 ]
Average time of patient spend in the system, Ws =
1
(µ−λ)
=
1
(90−20)
= 0.014 per hour
4/7/2014

33. 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
(µ−λ)
=
1
(6−5)
= 1 per hour
continue
4/7/2014

34. Problem-11.(continued)
Requirement-b.
Average number of room in waiting area, Lq =
λ²
µ(µ−λ)
=
5²
6(6−5)
=
25
6∗1
= 4.17 room
continue4/7/2014

35. Problem-11.(continued)
Requirement-c.
Probability of the system in being busy, 𝜌 =
λ
µ
=
5
6
= 0.83 or, 83%
Requirement-d.
Probability that the system is idle, Po = 1-
λ
µ
= 1-
5
6
= 1- 0.83
= 0.17 or, 17%
4/7/2014

36. 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 =
λ²
µ(µ−λ)
=
2²
3(3−2)
=
4
3
= 1.33 customer
continue4/7/2014

37. Problem-13.(continued)
Requirement-b.
Average time a customer waiting in line, Wq =
λ
µ(µ−λ)
=
2
3(3−2)
=
2
3
= 0.67 per hour
continue4/7/2014

38. Problem-13.(continued)
Requirement-c.
Average time a customer in the system, Ws =
1
(µ−λ)
=
1
(3−2)
= 1 per hour
4/7/2014

39. 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-
λ
µ
=> 0.45= 1-
1.5
µ
=> 0.45 =
1µ −1.5
µ
=> 0.45 µ = 1µ - 1.5
=>1.5 = 1µ - 0.45µ
=> 1.5 = 0.55µ
=> µ = 2.73 continue
4/7/2014

40. Problem-14.(continued)
Probability system being busy, 𝜌 = 85% or, 0.85
Service rate of customer, µ = 2.73
Arrival rate of customer, λ = ??
We know that,
𝜌 =
λ
µ
=> 0.85 =
λ
2.73
=> λ = 0.85*2.73
=> λ = 2.31 per hour
The arrival rate is need 2.31 per hour.
4/7/2014

41. 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 =
λ²
µ(µ−λ)
=
2²
6(6−2)
=
4
24
= 0.17 customer
continue4/7/2014

42. Problem-15.(continued)
Requirement-b.
Average time of customer waiting in line, Wq =
λ
µ(µ−λ)
=
2
6(6−2)
= 0.08 per hour
continue4/7/2014

43. Problem-15.(continued)
Requirement-c.
Average time of customer is in shop in the system, Ws =
1
(µ−λ)
=
1
(6−2)
= 0.25 per hour