Model Antrian Tunggal Pertemuan 20 Matakuliah : K0414 / Riset Operasi Bisnis dan Industri Tahun : 2008 / 2009 Model Antrian Tunggal Pertemuan 20

Learning Outcomes Mahasiswa akan dapat menghubungkan masalah antrian tunggal dalam berbagai metode atau teori yang telah dipelajari. Bina Nusantara University

Outline Materi: Model Antrian Tunggal Analisis Pola Pelayanan Model Antrian dengan Prioritas Model antrian tunggal M/M/1 Contoh penerapan. Bina Nusantara University

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

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 = P0 System utilization =  Probability of more than k units in system = Pn > k Bina Nusantara University

General Queuing Equations =   S W s = 1  q + Given one of Ws , Wq , Ls, or Lq you can use these equations to find all the others. L s =   q + L q =  W L s =  W Bina Nusantara University

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

M/M/1 Model Equations Average # of customers in system: L =   -  Average time in system: W s = 1  -  Average # of customers in queue: L q =  2  ( -  ) Average time in queue: W q =   ( -  ) =    System utilization Bina Nusantara University

M/M/1 Probability Equations Probability of 0 units in system, i.e., system idle: 1 -   P = 1 -  = Probability of more than k units in system: ( ) l  k+1 P = n>k This is also probability of k+1 or more units in system. Bina Nusantara University

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) Q1: What is the average time between departures? 5 minutes? 6 minutes? Q2: What is the average wait in the system? 1 W = = 0.5 hour or 30 minutes s 12/hr-10/hr Bina Nusantara University

M/M/1 Example 1 Q3: What is the average wait in line?  = 10/hr and  = 12/hr Q3: What is the average wait in line? 10 W = O.41667 hours = 25 minutes = q 12 (12-10) Also note: so W s = 1  q + - 2 12 O.41667 hours Bina Nusantara University

M/M/1 Example 1  = 10/hr and  = 12/hr Q4: What is the average number of customers in line and in the system? 102 4.1667 customers L = = q 12 (12-10) 10 5 customers L = = s 12-10 Also note: L q =  W = 10  0.41667 = 4.1667 q L s =  W = 10  0.5 = 5 s Bina Nusantara University

M/M/1 Example 1  = 10/hr and  = 12/hr Q5: What is the fraction of time the system is empty (server is idle)? 1 -   10 P = 1 -  = = - = 16.67% of the time 1 12 Q6: What is the fraction of time there are more than 5 customers in the system? ( ) 6 10 P = = 33.5% of the time n>5 12 Bina Nusantara University

More than 5 in the system... n>5 Note that “more than 5 in the system” is the same as: “more than 4 in line” “5 or more in line” “6 or more in the system”. All are P n>5 Bina Nusantara University

M/M/1 Example 1  = 10/hr and  = 12/hr Q7: 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 more in line. n>5 0.335 x 480 min./day = 160.8 min. = ~2 hr 40 min. Q8: What fraction of time are there 3 or fewer customers in line? 5 = 1 - 1 - P n>4 ( ) 10 12 = 1 - 0.402 = 0.598 or 59.8% Bina Nusantara University

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 Q1: What is the number of “customers” in the system? 5/day = = 3.57 broken copiers L S 6.4/day-5/day Bina Nusantara University

M/M/1 Example 2 Q2: How long is the average wait in line?  = 5/day (or  = 0.625/hour)  = 6.4/day (or  = 0.8/hour) Q2: How long is the average wait in line? 5 W = = 0.558 days (or 4.46 hours) q 6.4(6.4 - 5) 0.625 W = = 4.46 hours q 0.8(0.8 - 0.625) Bina Nusantara University

M/M/1 Example 2  = 5/day (or  = 0.625/hour)  = 6.4/day (or  = 0.8/hour) Q3: 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 ( ) 3 5 P = = 0.477 (47.7% of the time) n>2 6.4 0.477x 480 min./day = 229 min. = 3 hr 49 min. Bina Nusantara University

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