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Tabl e 4. Data for the test example 3 with fifteen machines [2]
M i
λ i 1
μ i 1
λ i 2
μ i 2
ω i
N i
Machine 1
0.0120
0.2200
0.0050
0.0400
1.3000
55
Machine 2
0.0100
0.0400
0.0800
0.1500
1.3000
40
Machine 3
0.1000
0.2000
0.0400
0.0900
1.3000
40
Machine 4
0.0100
0.0870
0.1160
0.2971
1.3000
55
Machine 5
0.1000
0.2500
0.0500
0.0800
1.3000
50
Machine 6
0.0141
0.2323
0.0059
0.0422
1.3000
50
Machine 7
0.0011
0.0348
0.0089
0.1306
1.3000
65
Machine 8
0.0143
0.1350
0.0057
0.0607
1.3000
55
Machine 9
0.0008
0.0349
0.0092
0.1192
1.3000
35
Machine 10
0.0133
0.1709
0.0067
0.0546
1.3000
40
Machine 11
0.0296
0.2323
0.0124
0.0422
1.3000
55
Machine 12
0.0074
0.0696
0.0596
0.2611
1.3000
40
Machine 13
0.0429
0.2024
0.0171
0.0910
1.3000
50
Machine 14
0.0037
0.0698
0.0423
0.2383
1.3000
55
Machine 15
0.0080
0.4273
0.0040
0.1366
1.3000
-
Tabl e 5. Numerical results for the test examples
Method
Example 1
Example 2
Example 3
Simulation
0.56508
0.52239
0.60723
Existing aggregation method
0.52748
0.48617
0.56691
Gap1(%)
-6.65%
-6.93%
-6,64%
Aggregation method proposed by
0.54350
0.49911
0.58498
Belmansour and Nourelfath [3]
Gap2(%)
-3.82%
-4.46%
-3.66%
Proposed method
0.56838
0.53259
0.62519
Gap3(%)
0.58%
1.95%
2.96%
6Con lu on
This paper addresses a new method to evaluate the system throughput of a
buffered production line with machines having multiple failure modes. This
method is based on the transformation of each machine to a single-failure ma-
chine. Then, the system throughput is evaluated based on the equivalent machine
method adapted to the context of transfer lines. The approach proposed in the
paper has been coded in an algorithm that has proven to be fast and accu-
rate. A comparative study based on both simulation experiments and existing
aggregation methods has shown the accuracy of the proposed approach.
The originality of this method is its reduced number of variables because
it considers only the empty and full states of each buffer and the processing
rates ratio related to each buffer. Therefore, to evaluate the throughput of a
K -machine ( K
1)-buffer production line we have to solve (4 K
3) equations.
 
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