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Matakuliah : K 0414 / Riset Operasi Bisnis dan Industri Tahun : 2008 / 2009 Model Antrian Tunggal Pertemuan 20 Bina Nusantara University 2
Learning Outcomes • Mahasiswa akan dapat menghubungkan masalah antrian tunggal dalam berbagai metode atau teori yang telah dipelajari. Bina Nusantara University 3
Outline Materi: • • • Model Antrian Tunggal Analisis Pola Pelayanan Model Antrian dengan Prioritas Model antrian tunggal M/M/1 Contoh penerapan. Bina Nusantara University 4
Types of Queuing Models • Simple (M/M/1). – Example: Information booth at mall. • Multi-channel (M/M/S). – Example: Airline ticket counter. • Constant Service (M/D/1). – Example: Automated car wash. • Limited Population. – Example: Department with only 7 drills that may break down and require service. Bina Nusantara University 5
Performance Measures • Average queue time = Wq • Average queue length = Lq • Average time in system = Ws • Average number in system = Ls • Probability of idle service facility = P 0 • System utilization = • Probability of more than k units in system = Pn > k Bina Nusantara University 6
General Queuing Equations = S 1 W = Wq + s Ls = Lq + Lq = W q Given one of Ws , Wq , Ls, or Lq you can use these equations to find all the others. Ls = W s Bina Nusantara University 7
M/M/1 Model • Type: Single server, single phase system. • Input source: Infinite; no balks, no reneging. • Queue: Unlimited; single line; FIFO (FCFS). • Arrival distribution: Poisson. • Service distribution: Negative exponential. Bina Nusantara University 8
M/M/1 Model Equations Average # of customers in system: L Average time in system: W Average # of customers in queue: L s s = - 1 = - 2 q = ( - ) Average time in queue: = W q ( - ) Bina Nusantara University System utilization = 9
M/M/1 Probability Equations Probability of 0 units in system, i. e. , system idle: P 0 = 1 - = 1 Probability of more than k units in system: P = n>k () k+1 This is also probability of k+1 or more units in syste Bina Nusantara University 10
M/M/1 Example 1 Average arrival rate is 10 per hour. Average service time is 5 minutes. = 10/hr and = 12/hr (1/ = 5 minutes = 1/12 hour) Q 1: What is the average time between departures? 5 minutes? 6 minutes? 1 = Q 2: W = 12/hr-10/hr 0. 5 hour or the system? What is the average wait in 30 s minutes Bina Nusantara University 11
M/M/1 Example 1 = 10/hr and = 12/hr Q 3: What is the average wait in line? W = q 10 = O. 41667 hours = 25 minutes 12 (12 -10) 1 Also note: W = Wq + s 1 1 1 so W = Ws = 2 - 12 = O. 41667 q hours Bina Nusantara University 12
M/M/1 Example 1 = 10/hr and = 12/hr Q 4: What is the average number of customers in line and in the system? 102 = 4. 1667 Lq = 12 (12 -10) customers 10 = =5 Ls 12 -10 customers Also note: L q = W = 10 0. 41667 = q L s = W =4. 1667 5 = 5 s 10 0. Bina Nusantara University 13
M/M/1 Example 1 = 10/hr and = 12/hr Q 5: What is the fraction of time the system is empty (server is idle)? 10 = 16. 67% of the =1 P 0 = 1 - = 1 12 time Q 6: What is the fraction of time there are more than 5 customers in the system? P = n>5 Bina Nusantara University 10 6 = 33. 5% of the 12 time ( ) 14
More than 5 in the system. . . Note that “more than 5 in the system” is the same as: ¨ “more than 4 in line” ¨ “ 5 or more in line” ¨ “ 6 orn>5 in the system”. more All are P Bina Nusantara University 15
M/M/1 Example 1 = 10/hr and = 12/hr Q 7: How much time per day (8 hours) are there 6 or more customers in line? P = 0. 335 so 33. 5% of time there are 6 or n>5 more in line. 0. 335 x 480 min. /day = 160. 8 min. = ~2 hr 40 min. Q 8: What fraction of time are there 3 or fewer customers in line? 10 5 = 1 - 0. 402 = 0. 598 or 59. 8% 1 -P =1 n>4 12 Bina Nusantara University ( ) 16
M/M/1 Example 2 Five copy machines break down at UM St. Louis per eight hour day on average. The average service time for repair is one hour and 15 minutes. = 5/day ( = 0. 625/hour) 1/ = 1. 25 hours = 0. 15625 days = 1 every 1. 25 hours = 6. 4/day Q 1: What is the number of “customers” in the 5/day = 3. 57 broken copiers LS = system? 6. 4/day-5/day Bina Nusantara University 17
M/M/1 Example 2 = 5/day (or = 0. 625/hour) = 6. 4/day (or = 0. 8/hour) Q 2: How long is the average wait in line? W = q 5 = 0. 558 days (or 4. 46 6. 4(6. 4 - 5) hours) W = q 0. 625 = 4. 46 hours 0. 8(0. 8 - 0. 625) Bina Nusantara University 18
M/M/1 Example 2 = 5/day (or = 0. 625/hour) = 6. 4/day (or = 0. 8/hour) Q 3: How much time per day are there 2 or more broken copiers waiting for the repair person? 2 or more “in line” = more than 2 in the system 5 3 P = 6. 4 = 0. 477 (47. 7% of the time) n>2 ( ) 0. 477 x 480 min. /day = 229 min. = 3 hr 49 min. Bina Nusantara University 19
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