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

3. Learning Outcomes
Mahasiswa akan dapat menghubungkan masalah antrian tunggal dalam berbagai metode atau teori yang telah dipelajari.
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4. Outline Materi: Model Antrian Tunggal Analisis Pola Pelayanan
Model Antrian dengan Prioritas
Model antrian tunggal M/M/1
Contoh penerapan.
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5. 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.
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6. 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
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7. 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
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8. 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.
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9. 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
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10. 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.
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11. 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
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12. 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 hours = 25 minutes = q
12 (12-10)
Also note: so W s = 1  q + - 2 12
O hours
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13. 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
customers L = = q
12 (12-10) 10
5 customers L = = s
12-10
Also note: L q =
 W =
10  = q L s =
 W =
10  0.5 = 5 s
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14. 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
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15. 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
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16. 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 =
so 33.5% of time there are 6 or more in line.
n>5
0.335 x 480 min./day = 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 =
= or 59.8%
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17. 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 = 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
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18. 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 =
= days (or hours) q
6.4( )
0.625 W =
= 4.46 hours q
0.8( )
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19. 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 =
= (47.7% of the time)
n>2
6.4
0.477x 480 min./day = 229 min. = 3 hr 49 min.
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20. Terima Kasih
Semoga sukses
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