Motivating Example (1/2)
GROUP TECHNOLOGY
PART II
BRANCHING ALGORITHMS
1 2 3 4 5 6
Printed circuit
board
Andrew Kusiak,
Intelligent Systems Laboratory
2139 Seamans Center
The University of Iowa
Iowa City, Iowa 52242 - 1527
1
2
21
36
125
Component 125
Component 36
Component 21
Component 2
290
Component 1
Tel: 319 - 335 5934
Fax: 319 - 335 5669
[email protected]
http://www.icaen.uiowa.edu/~ankusiak
The University of Iowa
1000
1
1
1
1
1
B2
Intelligent Systems Laboratory
878
The University of Iowa
Intelligent Systems Laboratory
Motivating Example (2/2)
M
se ac
N tup hin
o.
e
2
1 2 3 4 5 6
1
2
21
36
125
Classroom Exercise 1
ne
hi
ac p
M tu
se o. 1
N
1 2 3 4 5 6
1000
1
nt 2
ne
po . 1 21
m o
Co t N 36
se
125
1
1
1
1
1
Decompose the manufacturing system
1
1
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1
1
t
en
on
mp . 2
o
o
C N
set
290
878
P1
1
M1
M2
M3
M4
M5
P2
P3
1
1
P4
P5
1
P6
1
1
1
1
1
1
1
o n ts
mm nen
Co mpo
co
The University of Iowa
Intelligent Systems Laboratory
The University of Iowa
Exercise 1 Solution
M1
M2
M3
M4
M5
M1
M2
M3
M4
M5
The University of Iowa
P1
1
P2
P3
1
1
P4
P5
1
1
Exercise 1 Solution
P6
M1
M2
M3
M4
M5
1
1
1
1
1
P1
1
P2
1
P3
1
1
P4
P5
1
1
P6
1
Reorganized
matrix
1
1
1
1
Intelligent Systems Laboratory
M1
M3
M4
M2
M5
P1
1
P2
P3
1
1
P4
P5
1
1
P6
1
1
1
1
1
P1
1
P5
1
1
1
1
P3
1
1
P2
P4
P6
1
1
1
1
1
1
Intelligent Systems Laboratory
The University of Iowa
Intelligent Systems Laboratory
Page 1
1
Exercise 2 Solution
Classroom Exercise 2
Consider this data
Exercise 1
M1
M2
M3
M4
M5
P1
1
P2
P3
1
P4
1
P5
1
1
1
1
1
1
1
1
P6
M1
M2
M3
M4
M5
P1
1
P2
P3
1
1
P4
P5
1
1
P6
P2
P3
1
P4
1
P5
1
P6
1
1
1
1
1
1
1
1
Delete
1
1
1
1
1
1
1
M1
M2
M3
M4
M5
One operation added
The University of Iowa
P1
1
M1
M2
M3
M4
M5
Intelligent Systems Laboratory
P1
1
P2
P3
1
P4
1
P5
1
P6
1
1
1
1
1
1
1
1
The University of Iowa
Intelligent Systems Laboratory
Algorithm 1
Step 0. (Initialization): Begin with the incidence matrix [aij]
at level 0.
Solve the GT problem represented with [aij] with the
CI algorithm.
Step 1. (Branching): Using the breadth-first search strategy,
select an active node (not fathomed) and solve the
corresponding GT problem with the CI algorithm.
Step 2. (Fathoming): Exclude a new node from further
consideration if:
Test 1: cluster size is not satisfactory
Test 2: cluster structure is not satisfactory
Step 3. (Backtracking): Return to an active node.
Step 4. (Stopping rule): Stop when there are no active nodes
remained; the current incumbent solution is optimal;
otherwise go to Step 1.
BRANCHING ALGORITHM
• Explicit Enumeration
• Implicit Enumeration
The University of Iowa
Intelligent Systems Laboratory
The University of Iowa
Intelligent Systems Laboratory
Part number
Example:
Bottleneck
Parts
Branching
Branching is performed in one of the following two ways:
(1) the CI algorithm in case when the incidence matrix
partitions into mutually separable submatrices.
(2) removing one column at a time from the
corresponding incidence matrix in case when the
matrix does not partition into mutually separable
submatrices.
1
2
3
4
5
6
7
8
Machine
number
2
6
1
3
4
5
7
8
The fathoming is based on the following tests:
Test 1: cluster size is not satisfactory
Test 2: cluster structure is not satisfactory
Intelligent Systems Laboratory
The University of Iowa
2
1
3
4
1
5
6
7
1
1
1
1
1
1
1
1
1
1
1
Step 0: CI algorithm
Fathoming
The University of Iowa
1
2
6
1
1
1
1
1
1
3
1
1
4
5
1
1
1
1
1
1
1
7
1
1
1
Intelligent Systems Laboratory
Page 2
2
2
6
1
3
4
5
7
8
2
6
1
1
1
1
3
1
4
5
1
7
Partial Solution
1
1
1
1
1
1
6
1
1
3
1
1
4
5
7
1
1
1
1
1
1
1
1
1
1
Machine cells
MC-1 = {2, 6}, MC'-2 = {1, 3, 4, 5, 7, 8}
1
1
1
2
1
1
2
6
1
3
4
5
7
8
1
Part families
PF-1 = {2, 6}, PF'-2 = {1, 3, 4, 5, 7}
CI continued
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Intelligent Systems Laboratory
The branching algorithm
enumeration tree
1 2 3
1 1
1
2
1
3
4
5
6
1
7 1
8
1
4 5 6 7
1
1
1
1
1
1
1
1
0
2 6
2 1 1
6 1
0
3 5
1 1 1
3
1
8 1 1
MC - 2
1
-7
-3
1 4 5 7
1 1
1
3
1
4
1
1
5
1
7 1 1
1
8
1
Level 1
1
-1
3 4 5 7
1 1
1
3
1
4
1
1
5
1
7
1
1
8 1
1
MC - 1
4 5 7
1
1
1
1
1
1
-4
1 3 5 7
1 1 1 1
3
1
4
1
1
7 1
1
8
1 1
MC - 3
-5
1 3 4 7
1 1 1
4
1 1
5
1
7 1
1 1
8
1
1 3
1 1 1
3
4
5
7 1
8
1
4 5
1
1
1
1
1
1
Level 2
4 7
4 1 1
5 1
7 1 1
2
6
1
3
8
4
5
7
PF - 2
PF - 1
2 6
1 1
1
Level 3
Intelligent Systems Laboratory
1
1
1
1
PF - 3
4
7
1
1
1
1
1
The University of Iowa
1
1
1
Intelligent Systems Laboratory
1
5
1
Machine
number
1
• Assuming that machine 1 is a bottleneck
• Assign one copy of machine 1 to each part
performed on machine 1
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5
1
Algorithm 2: Bottleneck Machines
Part number
Branching Scheme
1 2 3 4
1 1
1
Machine 2
1
1
number 3
1 1
1
1 1
4
3
Tree machine cells:
MC-1 = {2, 6}, MC-2 = {1, 3, 8}, MC-3 = {4, 5, 7},
Three part families:
PF-1 = {2, 6}, PF-2 = {3, 5}, PF-3 = {4, 7},
and the bottleneck part 1
0
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Intelligent Systems Laboratory
The final solution
Level 0
1
0
1 3
1 1 1
3
4
5
7 1
8
1
The University of Iowa
1(1)
1(2)
1(3)
2
3
4
Part number
2 3 4
5
1
1
1
1
1
1
1
1
1
1
1
• Applying the CI algorithm, this proves that
machine 1 is not a bottleneck machine.
The same is true for machines 2 and 3.
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Page 3
3
Considering machine 4 as the bottleneck machine
Considering machines 2 and 3 as
bottleneck machines
1
1
2
3
4(1)
4(2)
4(3)
.
.
.
2
1
3
1
1
1
1
4
5
1
1
1
1
1
1
In the first iteration
1
1
2
3
4(1)
4(2)
4(3)
No decomposition takes place
The University of Iowa
2
1
3
1
1
1
1
5
1
h2
1
1
h4(1)
1
1
v1
Intelligent Systems Laboratory
4
1
v4
The University of Iowa
h4(3)
Intelligent Systems Laboratory
• For the first cell it is enough to consider
only one machine (the first copy)
?
Question:
2
4(1)
4(2)
1
3
1
4
1
1
1
1
The University of Iowa
2
3
5
Can one determine bottleneck parts
and bottleneck machines at the same time?
1
1
1
1
1
1
1
Intelligent Systems Laboratory
The University of Iowa
Intelligent Systems Laboratory
Mathematical Programming
Approach
m
dij =
∑
d(aki , akj)
k =1
1
if aki ≠ akj
where d(aki , akj ) = {
0 otherwise
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Intelligent Systems Laboratory
The University of Iowa
Intelligent Systems Laboratory
Page 4
4
The p-Median Model
Example
m
dij = ∑ d (aki , akj )
k =1
P1 =
Two parts
1
1
0
1
0
1
Notation
m = number of machines
n = number of parts
p = number of part families
1 if part i belongs to part
xij = {
family j
0 otherwise
dij = Hamming distance measure between
parts i and j
0
1
1
0
1
1
P5 =
Hamming distance
d15 = 1 + 0 + 1 + 1+ 1 + 0 = 4
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min
Intelligent Systems Laboratory
The University of Iowa
Part number
xij
n
n
∑
∑ dij xij
1
1
i =1 j =1
Given
p=2
Part Number
subject to:
Part Number
Part Family Number
n
∑ xij = 1 for all i = 1, ..., n
∑ xjj = p
xij <= xjj for all i = 1, ..., n ,
j = 1, ..., n
xij = 0,1 for all i = 1, ..., n ,
j = 1, ..., n
The University of Iowa
Intelligent Systems Laboratory
The University of Iowa
MIN 4X12+4X14+3X15+4X21+4X23+1X25+4X32+4X34+
3X35+4X41+4X43+1X45+3X51+1X52+3X53+1X54
n
for all i = 1, ..., n
j =1
X11+X12+X13+X14+X15=1
X21+X22+X23+X24+X25=1
X31+X32+X33+X34+X35=1
X41+X42+X43+X44+X45=1
X51+X52+X53+X54+X55=1[dij] =
X11+X22+X33+X44+X55=2
The University of Iowa
Part number
1
2
3
4
5
3
4
1
4
5
1
1
1
1
1
1
Part number
1
2
3
4
5
0
4
0
4
3
4
0
4
0
1
0
4
0
4
3
4
0
4
0
1
3
1
3
1
0
Part
number
Intelligent Systems Laboratory
Constraint xij <= xjj
Input file
∑ xij = 1
1
1
2
[dij] = 3
4
5
j =1
2
1
Machine 2
number
3
m
dij = ∑ d (aki , akj )
k =1
j =1
n
SUBJECT TO
Intelligent Systems Laboratory
1
2
3
4
5
0
4
0
4
4
4
0
4
0
1
0
4
0
4
3
4
0
4
0
1
3
1
3
1
0
Part
number
Intelligent Systems Laboratory
X21-X11<=0
X31-X11<=0
X41-X11<=0
X51-X11<=0
X12-X22<=0
X32-X22<=0
X42-X22<=0
X52-X22<=0
X13-X33<=0
X23-X33<=0
X43-X33<=0
X53-X33<=0
The University of Iowa
X14-X44<=0
X24-X44<=0
X34-X44<=0
X54-X44<=0
X15-X55<=0
X25-X55<=0
X35-X55<=0
X45-X55<=0
END
INTEGER 25
Intelligent Systems Laboratory
Page 5
5
Solution x11 = 1, x31 = 1
?
x24 = 1, x44 = 1, x54 = 1
2
Based on the definition of xij , two part families are
formed:
PF-1 = {1, 3}
MC-1 = {2, 4}
Part number
1
2
3
1
1
PF -1
4
5
1
1
Can the problem with bottleneck parts
(machines) be modeled by the p-median
formulation ?
PF-2 = {2, 4, 5}
MC-2 = {1, 3}
1
3
2
1
1
4
1
1
PF-2
2
4
5
1
Yes
MC-1
Machine 2
number
3
1
4
1
1
1
1
1
1
1
3
1
1
MC-2
1
The University of Iowa
Intelligent Systems Laboratory
1
2
1
1
1
2
1
1
3
4
The University of Iowa
Intelligent Systems Laboratory
Generalized p-median Model
5
1
n = the total number of process plans
MC-1 {
3
1
1
4
1
1
Fk = set of process plans for part number k , k = 1, .., l ,
1
l
MC-2 {
where | ∪Fk | = n
k=1
Solution:
p = the minimum required number of part (process)
families
dij = distance measure between process plans i and j
cj = cost of process plan j
x13=1, x23 = 1
x35 =1, x45 =1, x55=1
2
The University of Iowa
Model
n
Intelligent Systems Laboratory
n
The University of Iowa
Intelligent Systems Laboratory
n
Min ∑ ∑ dij xij +
i =1 j =1
∑cj xjj
Part number
j =1
1
subject to:
n
n
∑
∑ xij = 1
for all k = 1 ,..., l
1
[aij] = 2
3
4
i ∈ Fk j =1
n
∑ xjj >= p
2
3
4
Process plan number
1 2 3
4 5
1
1 1
1
1
1 1
5
6 7
8 9
10 11
1
1 1
1
1
1 1
1
1
1
1
1
1
1
Machine
number
j =1
xij <= xjj
for all i =1, ..., n ,
j = 1, ..., n
Costs
xij = 0, 1
The University of Iowa
for all i =1, ..., n ,
cj are not considered here
j = 1, ..., n
Intelligent Systems Laboratory
The University of Iowa
Intelligent Systems Laboratory
Page 6
6
m
dij = ∑ d (aki , akj)
k =1
Hamming distances
n
n
dij xij + ∑ cj xjj
i =1 j =1
j =1
MIN
999X12+999X13+2X14+2X15+2X16+2X17+2X18+
2X19+3X1a11+999X21+999X23+2X24+4X25+2X26+
2X28+2X210+3X211+999X31+999X32+2X34+2X35+
Not
2X36+2X37+2X38+2X39+4X310+1X311+2X41+2X42+
considered
2X43+999X45+4X46+2X47+4X48+2X49+2X410+
3X411+2X51+4X52+2X53+999X54+2X56+4X57+
2X58+2X510+1X511+2X61+2X62+2X63+4X64+
2X65+999X67+2X69+2X610+1X611+2X71+2X73+
2X74+4X75+999X76+2X78+4X79+2X710+3X711+
2X81+2X82+2X83+4X84+2X85+2X87+999X89+
2X810+1X811+2X91+2X93+2X94+2X96+4X97+
999X98+2X910+1X911+2X102+4X103+2X104+
2X105+2X106+2X107+2X108+2X109+999X1011+
3X11a1+3X112+1X113+3X114+1X115+1X116+3X117+
1X118+1X119+999X1110
Input file
Process plan number
1
1
2
3
4
5
[dij] = 6
7
8
9
10
11
2
3
4
5
6
7
8
9
0 ∞
∞ 0
∞
∞
2
4
2
2
2
2
2
2
2
0
3
0
2
2
2
2
2
2
4
1
2
2
2
0
2
2
2
4
2
2
0
2
2
4
0
∞
2
2
2
4
2
0
2
0
2
0
2
2
0
2
4
∞ ∞
2
4
2
0
2
0
2
3
∞
4
2
4
2
2
3
∞
0
2
4
2
0
2
1
∞
0
2
2
1
The University of Iowa
0
2
4
2
3
10
∞
0
0
2
4
2
2
2
2
2
2
2
1
2
1
∞
∞
11
3
3
1
3
1
1
3
1
1
0
∞
P
P
n
u
m
b
e
r
0
Intelligent Systems Laboratory
Min
The University of Iowa
n
∑ ∑
Intelligent Systems Laboratory
Process plan number
1
1
2
3
4
5
[dij] = 6
7
8
9
10
1
11
[aij] = 2
3
4
2
3
0 ∞
∞ 0
∞
∞
2
2
2
2
2
2
0
3
0
2
2
2
12
2
22 3
4 5
6 7
4
2
2 2
2
1
1
1 3
1
3
1 1
∞ ∞
2
4
2
0
2
10
2
3
1 1
1
1 1
4
5
2
2
2
0
2
4
2
6
7
2
2
2
4
2
8
2
0
2
2
4
2 2
2
0
2
2
4
2
∞
0
0
∞ Part
number 2
4
2 0
2
∞ 0
2
42 ∞ 3 0 2 4 4
∞
4Process
2 0plan2number
0
0
2
0 2
4 ∞
1
1 1
1
The University of Iowa
8 9
2
2
1 11
1
1
1
1
The University of Iowa
X21-X11<=0
X31-X11<=0
X41-X11<=0
X51-X11<=0
X61-X11<=0
X71-X11<=0
X81-X11<=0
X91-X11<=0
X101-X11<=0
X11a1-X11<=0
X52-X22<=0
X62-X22<=0
X72-X22<=0
X82-X22<=0
9
1
10
0
2
4
2
2
2
25
2
2
n
11
3
3
1
3
1
1
3
1
1
10 11
0 ∞
∞ 10
1
1
n
∑
i ∈ Fk j =1
P
P
X11+X12+X13+X14+X15+X16+X17+X18+X19+X110+X1a11+
X21+X22+X23+X24+X25+X26+X27+X28+X29+X210+X211+
X31+X32+X33+X34+X35+X36+X37+X38+X39+X310+X311=1
n
u
m
b
e
r
X41+X42+X43+X44+X45+X46+X47+X48+X49+X410+X411+
X51+X52+X53+X54+X55+X56+X57+X58+X59+X510+X511=1
X61+X62+X63+X64+X65+X66+X67+X68+X69+X610+X611+
X71+X72+X73+X74+X75+X76+X77+X78+X79+X710+X711=1
X81+X82+X83+X84+X85+X86+X87+X88+X89+X810+X811+
X91+X92+X93+X94+X95+X96+X97+X98+X99+X910+X911=1
X101+X102+X103+X104+X105+X106+X107+X108+X109+
X1010+X1011+X11a1+X112+X113+X114+X115+X116+X117+
n
X118+X119+X1110+X1111=1
Machine
number
X11+X22+X33+X44+X55+X66+X77+X88+X99+X1010+
X1111<=2
Intelligent Systems Laboratory
X12-X22<=0
X32-X22<=0
X42-X22<=0
X92-X22<=0
X102-X22<=0
X112-X22<=0
X13-X33<=0
X23-X33<=0
X43-X33<=0
X53-X33<=0
X63-X33<=0
X73-X33<=0
X83-X33<=0
X93-X33<=0
∑
SUBJECT TO
The University of Iowa
X15-X55<=0
X25-X55<=0
X35-X55<=0
X45-X55<=0
X65-X55<=0
X75-X55<=0
X85-X55<=0
X95-X55<=0
X105-X55<=0
X115-X55<=0
xij <= xjj
X103-X33<=0
X113-X33<=0
X14-X44<=0
X24-X44<=0
X34-X44<=0
X54-X44<=0
X64-X44<=0
X74-X44<=0
X84-X44<=0
X94-X44<=0
X104-X44<=0
X114-X44<=0
Intelligent Systems Laboratory
The University of Iowa
xij = 1
∑ xjj >= p
j =1
Intelligent Systems Laboratory
X16-X66<=0
X26-X66<=0
X36-X66<=0
X46-X66<=0
X56-X66<=0
X76-X66<=0
X86-X66<=0
X96-X66<=0
X106-X66<=0
X116-X66<=0
X17-X77<=0
X27-X77<=0
X37-X77<=0
X47-X77<=0
X57-X77<=0
X67-X77<=0
X87-X77<=0
X97-X77<=0
X107-X77<=0
X117-X77<=0
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The University of Iowa
X1a10-X1010<=0
X211-X1111<=0
X311-X1111<=0
X411-X1111<=0
X511-X1111<=0
X611-X1111<=0
X711-X1111<=0
X811-X1111<=0
X911-X1111<=0
X1011-X1111<=0
END
INTEGER 121
X110-X1010<=0
X210-X1010<=0
X310-X1010<=0
X410-X1010<=0
X510-X1010<=0
X610-X1010<=0
X710-X1010<=0
X810-X1010<=0
X910-X1010<=0
X1110-X1010<=0
X19-X99<=0
X29-X99<=0
X39-X99<=0
X49-X99<=0
X59-X99<=0
X69-X99<=0
X79-X99<=0
X89-X99<=0
X109-X99<=0
X119-X99<=0
X18-X88<=0
X28-X88<=0
X38-X88<=0
X48-X88<=0
X58-X88<=0
X68-X88<=0
X78-X88<=0
X98-X88<=0
X108-X88<=0
X118-X88<=0
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Intelligent Systems Laboratory
Process
family
PF - 1
2
7
Part number
1
1
[aij] = 2
3
4
2
3
4
Process plan number
1 2 3
4 5
1
1 1
1
1
1 1
5
6 7
8 9
10 11
1
1 1
1
1
1 1
1
1
1
1
1
1
1
Machine
number
1
1
1
1
1
1
1
1
1
This solution can be interpreted as follows:
PF-1 = {2, 7}
(relabeled PF-7)
PF-2 = {5, 9, 11} (relabeled PF-9)
For these two process families, two machine
cells are obtained:
MC-1 = {2, 4}
MC-2 = {1, 3}
Solution: x27 = 1, x77 = 1
x59 = 1, x99 = 1, x11,9 = 1
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2
4
1
3
Machine cell
MC - 1
Machine cell
MC - 2
Process
family
PF - 2
5
9
11
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Intelligent Systems Laboratory
Tool-Part Grouping
Example
?
1
Question:
Tool
number
Name a few non-traditional applications
of group technology?
1
2
3
4
5
6
7
8
Part number
2 3 4 5
1
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Tool
number
2
5
8
3
1
7
4
6
Part number
4 5 1 3
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Solution: TF-1 = {2, 5, 8, 3} and TF-2 = {1, 7, 4, 6},
and two corresponding part families,
PF-1 = {4, 8, 3} and PF-2 = {5, 1}.
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Page 8
8
Assignment of Parts to the Existing
Machine Cells
OTHER EXAMPLES
Non-Traditional GT Applications
MACHINE
Object
Person
Requirement
Subprogram
Production rule
PART
Attributes
Function
Function
Function
Element
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MC-1
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Assignment of Parts to the Existing
Machine Cells
n
maximum number of newly assigned parts per cell
number of parts
number of machines
number of cells
set of cells that include machine l
set of machines that have to be visited by part i
(included in the process plan of part i)
⎧1 if part i is assigned to machine cell j
x ij = ⎨
⎩0 otherwise
∑ xij ≥ 1
j ∈Kl
n
∑ x ij ≤ N
all l in Mi; i = 1, ..., n
(Each part assigned to at least
one machine cell)
i =1
j = 1, ..., k
(Max No. of parts per cell)
xij = 0, 1
i = 1, ..., n; j = 1,..., k
1
3
1
2
1
3
1
Machine
number
1
1
1
Existing
2
machine
3
cell
4
1
1
1
1
5
1
4
6
4
Intelligent Systems Laboratory
The Set of Existing Four Cells
Part
5
The University of Iowa
Minimize traffic among cells
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5 parts are to be assigned to the existing
4 cells
2
otherwise
i=1 j=1
Intelligent Systems Laboratory
1
if part i is assigned to machine cell j
k
Min ∑ ∑ x ij
N
n
m
k
Kl
Mi
Machine
Intelligent Systems Laboratory
⎧1
x ij = ⎨
⎩0
Notation
Processing
requirements
for new parts
Assign parts to
cells MC-1 - MC-3
to minimize traffic
among cells
MC-3
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MC-2
1
1
1
1
2
1
2
3
4 6
2 5
1
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Part 1
Part 2
M1 = {2, 3, 6} Set of machines to be visited by Part 1
Part
K2 = {1, 3, 4} Set of cells that include
machine 2
1 2 3 4 5
K3 = {2}
1
1
1
∑ xij ≥ 1
2
1 1
K6 = {3}
j ∈Kl
Machine
3
1
4
1
6
of cells that include
K1 = {1, 2, 4} Set
1
machine 1
2
K2 = {1, 3, 4}
3
Machine
4
K5 = {4}
∑ xij ≥ 1
5
1
1
5
Constraints for Part 1
1
M2 = {1, 2, 5} Set of machines to be visited by Part
2
Part
1
1
1
j ∈Kl
1
Machine
number
x11 + x13 + x14 >=1
>=1
x12
x13 >=1
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1
1
2
Machine 2
cell
3
1
3
2
4
6
4
1
2
5
x21 + x22 + x24
x21 + x23 + x24
x24 >= 1
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>= 1
>= 1
>= 1
>= 1
>= 1
>= 1
>= 1
>= 1
>= 1
>= 1
>= 1 for Part 2
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Min x11 + x12 + ... + x14 + x21 + ... + x24 +
x31 + ... +x34 + 41 + ... + x44 + x51 + ... + x54
>= 1
Model x11 + x13 + x14
x12
>= 1 for part 1
x13
>= 1
x21 + x22 + x24
x21 +
x23 + x24
x24
x32
x34
x41 + x42 +
x44
x43
x52
x53
6
1
2
3
1
1
4
5
1
1
1
1
1
1
1
1
1
1
1
Machine
number
1
1
2
Machine 2
cell
3
1
3
2
4
6
4
1
2
5
Intelligent Systems Laboratory
n
∑ xij ≤ N
i =1
Part
Limit on number
of parts/cell
Cell
x11 + x21 + x31 + x41 + x51 <=
x12 + x22 + x32 + x42 + x52 <=
x13 + x23 + x33 + x43 + x53 <=
x14 + x24 + x34 + x44 + x54 <=
for part 2
for part 3
3
3
3
3
for part 4
xij = 0, 1 i = 1, ..., 5 and j = 1,..., 4
for part 5
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Solution
⎧1
x ij = ⎨
⎩0
if part i is assigned to machine cell j
otherwise
x12 = x13 = x24 = x32 = x34 = x43
= x44 = x52= x53 = 1
Machine cell 2: Parts 1, 3, 5
Machine cell 3: Parts 1, 4, 5
Machine cell 4: Parts 2, 3, 4
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