# Model Antrian Ganda Pertemuan 21 Matakuliah: K0414 / Riset Operasi Bisnis dan Industri Tahun: 2008 / 2009.

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Model Antrian Ganda Pertemuan 21 Matakuliah: K0414 / Riset Operasi Bisnis dan Industri Tahun: 2008 / 2009

Bina Nusantara University 3 Learning Outcomes Mahasiswa akan dapat menghitung penyelesaian model antrian tunggal dan ganda dalam berbagai contoh aplikasi.

Bina Nusantara University 4 Outline Materi: Model Antrian Ganda M/M/C Jaringan Antrian Contoh Penerapan

Bina Nusantara University 5 Type: Multiple servers; single-phase. Input source: Infinite; no balks, no reneging. Queue: Unlimited; multiple lines; FIFO (FCFS). Arrival distribution: Poisson. Service distribution: Negative exponential. M/M/S Model

Bina Nusantara University 6 M/M/S Equations Probability of zero people or units in the system: Average number of people or units in the system: Average time a unit spends in the system: Note: M = number of servers in these equations

Bina Nusantara University 7 M/M/S Equations Average number of people or units waiting for service: Average time a person or unit spends in the queue:

Bina Nusantara University 8 M/M/2 Model Equations  (2  + )(2  - ) W s = 44 4  2 - 2 L q = W q W q = 2 L s = W s P 0 = 2  - 2  + Average time in system: Average time in queue: Average # of customers in queue: Average # of customers in system: Probability the system is empty:

Bina Nusantara University 9 M/M/2 Example Average arrival rate is 10 per hour. Average service time is 5 minutes for each of 2 servers. = 10/hr,  = 12/hr, and S=2 Q1: What is the average wait in the system? W s = 4  12 4(12) 2 -(10) 2 = 0.1008 hours = 6.05 minutes

Bina Nusantara University 10 M/M/2 Example = 10/hr,  = 12/hr, and S=2 Q2: What is the average wait in line? Also note: so W s = 1  W q + W q = 1  W s -= 0.1008 - 0.0833 =0.0175 hrs W q = (10) 2 12 (2  12 + 10)(2  12 - 10) = 0.0175 hrs = 1.05 minutes

Bina Nusantara University 11 M/M/2 Example = 10/hr,  = 12/hr, and S=2 Q3: What is the average number of customers in line and in the system? L q = W q = 10/hr  0.0175 hr = 0.175 customers L s = W s = 10/hr  0.1008 hr = 1.008 customers

Bina Nusantara University 12 M/M/2 Example = 10/hr and  = 12/hr Q4: What is the fraction of time the system is empty (server is idle)? = 41.2% of the time P 0 = 2  12 - 10 2  12 + 10

Bina Nusantara University 13 M/M/1, M/M/2 and M/M/3 1 server2 servers3 servers W q 25 min.1.05 min.0.1333 min. (8 sec.) 0.417 hr0.0175 hr0.00222 hr W S 30 min.6.05 min.5.1333 min. L q 4.167 cust.0.175 cust.0.0222 cust. L S 5 cust.1.01 cust.0.855 cust. P 0 16.7% 41.2% 43.2%

Bina Nusantara University 14 Service Cost per Day = 10/hr and  = 12/hr Suppose servers are paid \$7/hr and work 8 hours/day and the marginal cost to serve each customer is \$0.50. M/M/1 Service cost per day = \$7/ hr x 8 hr/day + \$0.5/ cust x 10 cust/hr x 8 hr/day = \$96/ day M/M/2 Service cost per day = 2 x \$7/hr x 8 hr/day + \$0.5 /cust x 10 cust/hr x 8 hr/day = \$152/ day

Bina Nusantara University 15 Customer Waiting Cost per Day = 10/hr and  = 12/hr Suppose customer waiting cost is \$10/hr. M/M/1 Waiting cost per day = \$10/hr x 0.417 hr/cust x 10 cust/hr x 8 hr/day = \$333.33/ day M/M/1 total cost = 96 + 333.33 = \$429.33/day M/M/2 Waiting cost per day = \$10/hr x 0.0175 hr/cust x 10 cust/hr x 8 hr/day =\$14/ day M/M/2 total cost = 152 + 14 = \$166/day

Bina Nusantara University 16 Unknown Waiting Cost Suppose customer waiting cost is not known = C. M/M/1 Waiting cost per day = Cx 0.417 hr/cust x 10 cust/hr x 8 hr/day = 33.33C \$/ day M/M/1 total cost = 96 + 33.33 C M/M/2 Waiting cost per day = Cx 0.0175 hr/cust x 10 cust/hr x 8 hr/day =1.4C \$/ day M/M/2 total cost = 152 + 1.4 C M/M/2 is preferred when 152 + 1.4C < 96 + 33.33C or C > \$1.754/hr

Bina Nusantara University 17 M/M/2 and M/M/3 Q: How large must customer waiting cost be for M/M/3 to be preferred over M/M/2? M/M/2 total cost = 152 + 1.4 C M/M/3 Waiting cost per day = Cx 0.00222 hr/cust x 10 cust/hr x 8 hr/day = 0.1776C \$/ day M/M/3 total cost = 208 + 0.1776 C M/M/3 is preferred over M/M/2 when 208 + 0.1776C < 152 + 1.4C C > \$45.81/hr

Bina Nusantara University 18  = Mean number of arrivals per time period. –Example: 3 units/hour.  = Mean number of arrivals served per time period. –Example: 4 units/hour. 1/  = 15 minutes/unit. Remember: &  Are Rates If average service time is 15 minutes, then μ is 4 customers/hour

Bina Nusantara University 19 M/D/S –Constant service time; Every service time is the same. –Random (Poisson) arrivals. Limited population. –Probability of arrival depends on number in service. Limited queue length. –Limited space for waiting. Many others... Other Queuing Models

Bina Nusantara University 20

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