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

Learning Outcomes
• Mahasiswa akan dapat menghitung penyelesaian model antrian tunggal dan ganda dalam berbagai contoh aplikasi.
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Outline Materi:
• Model Antrian Ganda M/M/C
• Jaringan Antrian
• Contoh Penerapan
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M/M/S Model
• 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.
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M/M/S Equations
Probability of zero people or units in the system:
P
0





M
 n

1

0
1 n!


 λ
μ 

 n





1
1
M!


 λ
μ 


M
Mμ
Mμ
 λ
Average number of people or units in the system:
L s


M

 

!

 

M


M
 
P 

 
Average time a unit spends in the system:
W s


M





 


M
M
  
P 


 
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Note:
M
= number of servers in these equations
6

M/M/S Equations
Average number of people or units waiting for service: L q

L s



Average time a person or unit spends in the queue:
W q

W s



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M/M/2 Model Equations
Average time in system:
Average time in queue:
Average # of customers in queue:
Average # of customers in system:
L q
=  W q
L s
=
 W s
Probability the system is empty:
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W s

 2  2
W q
=
 2
 (2  +  )(2   )
P
0
 
 + 
8

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
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M/M/2 Example
 = 10/hr,  = 12/hr, and S=2
Q2: What is the average wait in line?
W q
=
12 (2
(10) 2
 12 + 10)(2  12 - 10)
= 0.0175 hrs = 1.05 minutes
Also note: so W q
= W s
W s
= W q
+
1

-
1

= 0.1008 - 0.0833 =0.0175 hrs
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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
=
L s
=
 W
 W q s
= 10/hr  0.0175 hr
= 10/hr  0.1008 hr
= 0.175 customers
= 1.008 customers
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M/M/2 Example
 = 10/hr and  = 12/hr
Q4: What is the fraction of time the system is empty
(server is idle)?
P
0
 12 - 10
 12 + 10
= 41.2% of the time
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M/M/1, M/M/2 and M/M/3
W q
W
S
L q
L
S
P
0
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1 server
25 min.
0.417 hr
30 min.
2 servers
1.05 min.
0.0175 hr
6.05 min.
3 servers
0.1333 min. (8 sec.)
0.00222 hr
5.1333 min.
4.167 cust.
0.175 cust.
0.0222 cust.
5 cust.
1.01 cust.
0.855 cust.
16.7% 41.2% 43.2%
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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
= $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
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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
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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 + 33.33C
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 + 1.4C
M/M/2 is preferred when 152 + 1.4C
< 96 + 33.33C
or
C > $1.754/hr
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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 + 1.4C
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 + 0.1776C
M/M/3 is preferred over M/M/2 when
208 + 0.1776C
< 152 + 1.4C
C > $45.81/hr
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Remember:

&

Are Rates
 
= Mean number of arrivals per time period.
– Example: 3 units/hour.
If average service time is 15 minutes, then μ is 4 customers/hour
 
= Mean number of arrivals served per time period.
– Example: 4 units/hour.
• 1/ 
= 15 minutes/unit.
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Other Queuing Models
• 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...
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