Matakuliah
Tahun
: K0414 / Riset Operasi Bisnis dan Industri
: 2008 / 2009
Model Antrian Tunggal
Pertemuan 20
Learning Outcomes
• Mahasiswa akan dapat menghubungkan masalah antrian
tunggal dalam berbagai metode atau teori yang telah
dipelajari.
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Outline Materi:
•
•
•
•
•
Model Antrian Tunggal
Analisis Pola Pelayanan
Model Antrian dengan Prioritas
Model antrian tunggal M/M/1
Contoh penerapan.
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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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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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General Queuing Equations

=
S
1
Ws = Wq + 

Ls = Lq + 
Given one of Ws , Wq ,
Ls, or Lq you can use
these equations to
find all the others.
Lq =  W q
Ls =  W s
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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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M/M/1 Model Equations
Average # of customers in system: L
Average time in system: W
s
s
=

 -
1
=
 -
2
Average # of customers in queue: L
q

=
 ( -  )
Average time in queue: W q =
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System utilization

 ( -  )

=

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M/M/1 Probability Equations
Probability of 0 units in system, i.e., system idle:

P0 = 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 system.
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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
Ws =
= 0.5 hour or 30 minutes
12/hr-10/hr
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M/M/1 Example 1
 = 10/hr and  = 12/hr
Q3: What is the average wait in line?
Wq =
10
= O.41667 hours = 25 minutes
12 (12-10)
Also note:
1
Ws = Wq + 
1
1 1
= O.41667 hours
so Wq = Ws =

2 12
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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
Lq =
12 (12-10)
10
= 5 customers
Ls =
12-10
Also note:
L q =  W q = 10  0.41667 = 4.1667
L s =  W s = 10  0.5 = 5
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M/M/1 Example 1
 = 10/hr and  = 12/hr
Q5: What is the fraction of time the system is empty
(server is idle)?

10
= 1= 16.67% of the time
P0 = 1 -  = 1 
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Q6: What is the fraction of time there are more than 5
customers in the system?
=
P
n>5
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10 6 = 33.5% of the time
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( )
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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 or more in the system”.
n>5
All are P
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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?
= 0.335 so 33.5% of time there are 6 or more in line.
P
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?
1-P =1n>4
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10 5 = 1 - 0.402 = 0.598 or 59.8%
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( )
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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?
LS
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5/day
= 3.57 broken copiers
=
6.4/day-5/day
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M/M/1 Example 2
 = 5/day
(or  = 0.625/hour)
 = 6.4/day
(or  = 0.8/hour)
Q2: How long is the average wait in line?
Wq =
5
6.4(6.4 - 5)
Wq =
0.625
= 4.46 hours
0.8(0.8 - 0.625)
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= 0.558 days (or 4.46 hours)
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M/M/1 Example 2
 = 5/day
 = 6.4/day
(or  = 0.625/hour)
(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
P
n>2 =
5 3
= 0.477 (47.7% of the time)
6.4
( )
0.477x 480 min./day = 229 min. = 3 hr 49 min.
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