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12
/ department of mathematics and computer scienceJJ J N I II 1/21JJ J N I II 1/21
Performance Analysis of Assembly Systems
Marcel van Vuuren
Joint work with Ivo Adan
June 12, 2006
12
/ department of mathematics and computer scienceJJ J N I II 2/21JJ J N I II 2/21
Presentation Outline
• Assembly system
• Literature
• The analysis
– Decomposition
– Subsystem analysis
– Iterative algorithm
• Numerical results and conclusions
12
/ department of mathematics and computer scienceJJ J N I II 3/21JJ J N I II 3/21
An assembly station
Model:
• General arrival processes
• Finite buffers
• Blocking after service
• General service process
• All components have to be available before service starts
• A component can wait in the server for other components
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/ department of mathematics and computer scienceJJ J N I II 4/21JJ J N I II 4/21
Model description
S: Service time of assemblyserver
Ai: Arrival process at buffer i
bi: Buffer size of buffer i
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/ department of mathematics and computer scienceJJ J N I II 5/21JJ J N I II 5/21
Literature
• Fork-join queue in an open network(Hemachandra and Eedupuganti)
• Fork-join queue in a closed network(Rao and Suri, and Krishnamurti et al.)
• An exact analysis of system with two parts(Gold)
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/ department of mathematics and computer scienceJJ J N I II 6/21JJ J N I II 6/21
Analysis approach
• Decomposition
• Subsystem Analysis
• Iterative Algorithm
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/ department of mathematics and computer scienceJJ J N I II 7/21JJ J N I II 7/21
Decomposition in subsystems
WAi: Wait to assembly atbuffer i
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/ department of mathematics and computer scienceJJ J N I II 8/21JJ J N I II 8/21
Wait to assembly at buffer i
The wait to assembly time consists of the waiting time time for theother components.So,WAi = maxj 6=i RAj
Note:
• The RAj’s can be obtained from the subsystems
• Both the WA’s and RA’s have mass in zero
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/ department of mathematics and computer scienceJJ J N I II 9/21JJ J N I II 9/21
Wait to assembly at buffer i
max1≤j≤k
RAj = max{RAk, max1≤j≤k−1
RAj}
P ( max1≤j≤k−1
RAj = 0) =∏
1≤j≤k−1
(1− pe,j)
pne,i =∏j 6=i
(1− pe,j)
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/ department of mathematics and computer scienceJJ J N I II 10/21JJ J N I II 10/21
Maximum of two random variables (1)
E1 and E2 Erlang distributed with parameters ki and µi (i = 1, 2)
First a number op phases with parameter µ1 + µ2 then either µ1 or µ2
qk,j: k wins and the other finished j phases
q1,j =
(k1 − 1 + j
k1 − 1
) (µ2
µ1 + µ2
)j (µ1
µ1 + µ2
)k1
, 0 ≤ j ≤ k2 − 1
q2,i =
(k2 − 1 + i
k2 − 1
) (µ1
µ1 + µ2
)i (µ2
µ1 + µ2
)k2
, 0 ≤ i ≤ k1 − 1
12
/ department of mathematics and computer scienceJJ J N I II 11/21JJ J N I II 11/21
Maximum of two random variables (2)
EM1,j =k1 + j
µ1 + µ2+
k2 − j
µ2
EM 21,j =
(k1 + j)(k1 + j + 1)
(µ1 + µ2)2+
(k1 + j)(k2 − j)
(µ1 + µ2)µ2+
(k2 − j)(k2 − j + 1)
µ22
E(max{E1, E2}) =
k2−1∑j=0
q1,jEM1,j +
k1−1∑i=0
q2,iEM2,i
E(max{E1, E2}2) =
k2−1∑j=0
q1,jEM 21,j +
k1−1∑i=0
q2,iEM 22,i
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/ department of mathematics and computer scienceJJ J N I II 12/21JJ J N I II 12/21
Subsystem analysis
• Fit aEk−1,k orC2 distribution on the first twomoments ofAi, S andWAi
• Construct MAP ’s of the arrival and departure processes
• Construct a QBD of the subsystem
• Analyse the QBD by using matrix analytic methods
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/ department of mathematics and computer scienceJJ J N I II 13/21JJ J N I II 13/21
Constructing the QBD (1)
Q =
B00 B01
B10 A1 A0
A2. . . . . .. . . . . . A0
A2 A1 C10
C01 C00
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/ department of mathematics and computer scienceJJ J N I II 14/21JJ J N I II 14/21
Constructing the QBD (2)
Construct a MAP of the arrival process:AR0 and AR1
Construct a MAP of the departure process (WA and S):DE0 and DE1
Construct a MAP of the departure process in level 0 (WA):D̃E0
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/ department of mathematics and computer scienceJJ J N I II 15/21JJ J N I II 15/21
Constructing the QBD (3)
A0 = AR1 ⊗ Inwac+nsa
A1 = AR0 ⊗ Inwac+nsa+ Ina ⊗DE0
A2 = Ina ⊗DE1
B01 = AR1 ⊗ Inwac+nsa
B00 = AR0 ⊗ Inwac+nsa+ Ina ⊗ D̃E0
B10 = Ina ⊗DE1
C01 = AR1 ⊗ Inwac+nsa
C00 = Ina ⊗DE0
C10 = Ina ⊗DE1
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/ department of mathematics and computer scienceJJ J N I II 16/21JJ J N I II 16/21
Analyzing the QBD
πi = x1Ri−1 + xbR̂
b−i, i = 1, . . . , b (1)
0 = A0 + RA1 + R2A2
0 = A2 + R̂A1 + R̂2A0
0 = π0B00 + π1B10
0 = π0B01 + π1A1 + π2A2
0 = πb−1A0 + πbA1 + πb+1C01
0 = πbC10 + πb+1C00
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/ department of mathematics and computer scienceJJ J N I II 17/21JJ J N I II 17/21
Characteristics of RA
pe: The probability that the queue is empty on departure
pe =π1B10e
T,
α: The distribution of the phase of the inter-arrival time A just after adeparture to level 0
α =π1B10
π1B10e,
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/ department of mathematics and computer scienceJJ J N I II 18/21JJ J N I II 18/21
Iterative algorithm
1. Choose initial characteristics for the wait to assembly times at eachbuffer
2. For each subsystem (from subsystem 1 to n):
• Determine the wait to assembly time at buffer i
• Determine the queue-length probabilities of the subsystem
• Determine a new RAi
Repeat step 2 until the characteristics of the WA’s of the subsystems haveconverged.
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/ department of mathematics and computer scienceJJ J N I II 19/21JJ J N I II 19/21
Numerical results
The following parameters are varied:
• number of parts: 2, 4, 8
• buffersize: 0, 2, 4, 8
• SCV of the arrivals: 0.2, 0.5, 1, 2
• SCV of the service process: 0.5, 1
• Occupation rate: 0.75, 1
• Imbalance in the arrival rates
• Imbalance in the SCV’s of the arrivals
A total of 768 test cases.
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/ department of mathematics and computer scienceJJ J N I II 20/21JJ J N I II 20/21
Numerical results (2)
Perf. char. Avg. 0-5 % 5-10 % > 10 %Throughput 1.5 % 97.4 % 2.6 % 0.0 %
Mean sojourn time 2.8 % 84.9 % 13.4 % 1.7 %
Most sensitive for:
• different buffer sizes
• different number of parts
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/ department of mathematics and computer scienceJJ J N I II 21/21JJ J N I II 21/21
Conclusions
Conclusions:
• Very good results
• Fast computation
• A good method for analyzing assembly systems
Future research:
• Incorporate the method in a network setting