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

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

  • 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, = 4 per hour [60/15] Service rate of clerk, =6 per hour [60/10] Requirement-a. Percentage of Time Judy Idle, P0 =1- = 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 = () = 4 6(64) = 4 62 = 0.33 per hour 4/7/2014 5. Problem-1.(continued) Requirement-c. Average waiting line in the system, Ws= 1 () = 1 (64) = 0.5 per hour Requirement-d. Probability, P = = 4 6 = 0.67 or, 67%4/7/2014 6. Problem-3 Given that, Arrival rate of customer, = 10 per hour [60/6] Service rate of customer, =15 per hour [60/4] Requirement-a. Each service desk idle, Po = 1- =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, = = 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, = 60 42 = 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 = () = 10 15(1510) = 100 155 = 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 () = 1 (1510) = 0.2 per hour 4/7/2014 11. Problem-4. Given that, Arrival rate of customer, = 10 per hour [60/6] Service rate of customer, =15 per hour [60/4] Requirement-a. The probability of waiting in line, = = 10 15 = 0.67 or, 67% continue 4/7/2014 12. Problem-4.(continued) Requirement-b. Average customer are waiting in line, Lq = () = 10 15(1510) = 100 155 = 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 () = 1 (1510) = 0.2 per hour 4/7/2014 14. Problem-5. Given that, Arrival rate of car, = 72 per hour [ (60/50)*60 ] Service rate of Burrito king, =80 per hour [ (60/45)*60 ] Requirement-a. Average time in the system, Ws = 1 () = 1 (8072) = 1 8 = 0.13 per hour continue4/7/2014 15. Problem-5.(continued) Requirement-b-. Average number of car in the line, Lq = () = 72 80(8072) = 5184 808 = 8.1 car continue4/7/2014 16. Problem-5.(continued) Requirement-c. Average number of cars in the system, Ls = () = 72 (8072) = 72 8 = 9 cars 4/7/2014 17. Problem-6. Given that, Arrival rate of customer, = 100 per hour Service rate of customer, =120 per hour [ (60/30)*60 ] Requirement-a. Average number of customer time ion the system, Ws = 1 () = 1 (120100) = 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 = 180 per hour [ (60/20)*60] continue 4/7/2014 19. Problem-6.(continued) Average number of customer in the system, Ls = () = 100 (180100) = 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, = 360 per hour [ 120*3 ] Average waiting time in the system, Ws = 1 () = 1 (360100) = 1 260 = 0.0038 per hour4/7/2014 21. Problem-7. Given that, Arrival rate of person, = 10 per hour Service rate of person, =12 per hour [60/5] Requirement-a. Average number of person in the line, Lq = () = 10 12(1210) = 100 122 = 4.17 person continue4/7/2014 22. Problem-7.(continued) Requirement-b. Average number of person in the system, Ls = () = 10 (1210) = 5 person continue4/7/2014 23. Problem-7.(continued) Requirement-c. Average time spend in the line, Wq = () = 10 12(1210) = 10 122 = 10 24 = 0.42 per hour continue 4/7/2014 24. Problem-7.(continued) Requirement-d. Average time spend in the system, Ws = 1 () = 1 (1210) = 1 2 = 0.5 per hour continue4/7/2014 25. Problem-7.(continued) Requirement-e. If arrival rate is, = 12 per hour, then number of person waiting in line, Lq = () = 12 12(1212) = 144 120 = 144 0 = 144 person 4/7/2014 26. Problem-8. Given that, Arrival rate of customer, = 180 per hour [ 60*3 ] Service rate of customer, =240 per hour [ (60/15)*60 ] Requirement-a. Average number of customer expect to see in the system, Ls = () = 180 (240180) = 180 60 = 3 customer continue4/7/2014 27. Requirement-b. Average time expect it to take to get a cup of coffee, Wq = () = 180 240(240180) = 180 24060 = 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 = = 180 240 = 0.75 or, 75% 4/7/2014 29. Problem-10. Given that, Arrival rate of patient, = 20 per hour [ 60/3 ] Service rate of patient, =30 per hour [ 60/2 ] Requirement-a. Perform by one Nurse, Average number of patient in the system, Ls = () = 20 (3020) = 20 10 = 2 per hour continue 4/7/2014 30. Problem-10.(continued) Requirement-b. Average time in the system, Ws = 1 () = 1 (3020) = 0.10 per hour continue 4/7/2014 31. Problem-10.(continued) Requirement-c. Probability of the system being busy, = = 20 30 = 0.67 or, 67% continue4/7/2014 32. 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 = 1 () = 1 (9020) = 0.014 per hour 4/7/2014 33. Problem-11. Given that, Arrival rate of customer, = 5 per hour [ 60/12 ] Service rate of customer, =6 per hour [ 60/10 ] Requirement-a. Average time of customer spend in the system, Ws = 1 () = 1 (65) = 1 per hour continue 4/7/2014 34. Problem-11.(continued) Requirement-b. Average number of room in waiting area, Lq = () = 5 6(65) = 25 61 = 4.17 room continue4/7/2014 35. Problem-11.(continued) Requirement-c. Probability of the system in being busy, = = 5 6 = 0.83 or, 83% Requirement-d. Probability that the system is idle, Po = 1- = 1- 5 6 = 1- 0.83 = 0.17 or, 17% 4/7/2014 36. Problem-13. Given that, Arrival rate of customer, = 2 per hour Service rate of customer, = 3 per hour [ 60/20 ] Requirement-a. Average number of customer waiting in the line, Lq = () = 2 3(32) = 4 3 = 1.33 customer continue4/7/2014 37. Problem-13.(continued) Requirement-b. Average time a customer waiting in line, Wq = () = 2 3(32) = 2 3 = 0.67 per hour continue4/7/2014 38. Problem-13.(continued) Requirement-c. Average time a customer in the system, Ws = 1 () = 1 (32) = 1 per hour 4/7/2014 39. Problem-14. Given that, Probability idle, Po = 45% or, 0.45 Arrival rate of customer, = 1.5 per hour [ 60/40 ] Service rate of customer, = ?? We know that, Po = 1- => 0.45= 1- 1.5 => 0.45 = 1 1.5 => 0.45 = 1 - 1.5 =>1.5 = 1 - 0.45 => 1.5 = 0.55 => = 2.73 continue 4/7/2014 40. Problem-14.(continued) Probability system being busy, = 85% or, 0.85 Service rate of customer, = 2.73 Arrival rate of customer, = ?? We know that, = => 0.85 = 2.73 => = 0.85*2.73 => = 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, = 2 per hour Service rate of customer, = 6 per hour [ (60/20) *2 ] Requirement-a. Average number customer waiting in line, Lq = () = 2 6(62) = 4 24 = 0.17 customer continue4/7/2014 42. Problem-15.(continued) Requirement-b. Average time of customer waiting in line, Wq = () = 2 6(62) = 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 () = 1 (62) = 0.25 per hour