Production Cell Formation
D0394 Perencanaan Sistem
Manufaktur
Cell Formation
• Definisi : Penugasan komponen dan mesin
dalam suatu kelompok (group) dan cell.
• Tujuannya adalah :
– Aliran komponen antar Cell minimum.
– Pemanfaatan mesin dalan cell tinggi.
Production Flow Analysis
(PFA)
• Production Flow Analysis (Burbidge, 1989)
Dirancang untuk mendapatkan pembagian
komponen dan juga pembagian mesin secara
lengkap kedalam kelompoknya (famili) dengan
menganalisa route proses untuk individual
komponen.
• Rangka Pengkodean dan metoda
pengelompokan berdasarkan jarak
menekankan pada atribut komponen,
membantu membentuk kelompok komponen
yang serupa, akan tetapi tidak memberikan
informasi mengenai mesin yang digunakan
untuk proses.
Production Flow Analysis
(PFA)
• Gagasan dasarnya adalah membentuk
part-machine incidence matrix dan
mengidentifikasi komponen tersebut yang
memerlukan mesin serupa, kemudian
menjadikan clusters.
• Machine-Part incidence matrix
– Matrix elemen biner yang tiap baris
menyatakan mesin dan tiap kolom menyatakan
komponen. Nilai 1 untuk elemen mij
menunjukkan bahwa mesin i diperlukan
komponen j .
Part-Machine Incidence
Matrix Komponen 3 diproses
Komponen
1
2
Mesin (m)
3
4
pada mesin 1
1
1
0
1
0
2
0
1
0
0
3
1
1
0
1
4
0
0
1
0
5
0
1
0
1
• Note: The P-M matrix does not explicitly
consider the sequence of machine visits for
the individual parts (i.e., the individual part
routes)
Partitioning of the PartMachine
Incidence Matrix
Exceptional
Elements
1
2
3
4
Void
1
1
0
1
0
2
0
1
0
0
3
1
1
0
1
4
0
0
1
0
5
0
1
0
1
Voids and Exceptional
Elements
• Voids - indicate that a machine assigned to a cell
is not required by all parts assigned to that cell.
Leads to large, inefficient cells and can
potentially contribute to low utilizations.
• Exceptional elements - indicate that a part
assigned to a cell requires processing by a
machine not in the cell. This forces an inter-cell
handling step and requires coordination between
cells.
• The voids and exceptional elements created are
dependent on the number and size of the
diagonal blocks formed in the part-machine
incidence matrix.
Rearranged Partition
Exceptional
Elements
1
3
2
4
Void
1
1
1
4
0
1
3
1
5
2
1
1
1
1
0
1
Perfect Cluster
1
3
2
4
1
1
1
4
1
1
3
5
2
1
1
1
1
1
1
No voids or exceptional elements.
Bond Energy
• Consider a P-M matrix A. We define the
bond energy as a measure of
effectiveness
1 P Mwhere:
ME  A 
a a

2
p 1 m1
p,m
p.m1
 a p,m1  a p1,m  a p1,m
...
a p ,m 1
...
a p 1,m
a p ,m
a p 1, m
...
a p ,m 1
...

Contoh Perhitungan BE
1
2
3
4
1
1
0
1
0
2
0
1
0
0
3
1
1
0
1
4
0
0
1
0
ME(A) = ½ X 4 = 2
5
0
1
0
1
1
1
1
4
0
3
1
5
0
2
0
3
1
1
0
0
0
2
4
0
0
0
0
1
1
1
1
1
0
ME (A) = ½ X 14 = 7
Algoritma Bond Energy

1 P M
ME  A   ap,m a p.m1  a p,m1  a p1,m  a p1,m
2 p1 m1

• over all P!M! possible arrays obtained by
permuting the rows and columns of A.
M 1 P
P 1 M
m 1 p 1
p 1 m 1
ME  A   ap,m a p,m1   ap,m a p1,m
To maximize the bond energy, the problem
decomposes into two parts: one finding the optimal
column permutation, and the other finding the
optimal row permutation.
Algoritma Bond Energy
• Step 1. Select a part column arbitrarily and set i =1.
Try placing each of the remaining (P-i) part columns
in each of the (i+1) possible positions (to the left and
right of each of the i columns already placed) and
compute the contribution of each column to the ME:
i
M
ME Column   ap,m a p 1,m
p 1 m 1
Place the column that gives the largest ME in its best
position. In case of a tie, select arbitrarily. Increment i
by 1 and repeat until i=P. When all columns have
been placed, go to step 2.
Algoritma Bond Energy
• Step 2. Repeat the procedure for rows,
calculating the BE as
i
P
ME Row    ap,m a p ,m1
m 1 p 1
Binary Ordering Algorithm
• Block diagonal structure
– partition the matrix such that “boxes” on the main diagonal
contain 0’s and 1’s but the off-diagonal boxes contain only
0’s.
• Assuming that there are N parts and M machines.
Step 1. Order rows:
– For each column, k, assign weight 2N-k to column k. Evaluate
each row. Sort rows in nonincreasing order. If rows were
previously ordered, and no change just occurred, stop;
otherwise, goto step 2.
Step 2. Order columns:
– For each row, k, assign weight 2M-k to row k. Evaluate each
column. Sort columns in nonincreasing order. If no change,
stop; otherwise, goto step 1.
Contoh Binary Ordering
Matriks P-M
Mesin
A
Komponen
1 3 4 7
1 1
E
1
1
F
6
1
1
1
128 64
Value
200
1
17
102
1
1
1
8
1
1
D
2N-k
5
1
C
B
2
1
1
1
54
1
32
16
8
17
200
4
2
1
P-M Matrix
With row
reordered
Mesin
A
B
Contoh Binary Ordering
Komponen
1 3 4 7
1 1
2
1
1
1
C
1
1
D
1
1
1
5
6
8
2M-k
32
16
1
1
8
1
1
4
E
1
1
2
F
1
1
1
Value
48
56
17
7
48
12
12
3
P-M Matrix With
row reordered and
Contoh Binary Ordering
column reordered
Komponen
Mesin 3 1 2 4 5 6 7 8 Value
A
1 1 1
224
B
1
C
1
1
1
D
224
1
1
1
1
1
1
156
1
30
E
1
1
3
F
1
1
3
2
1
2N-k
128 64
32
16
8
4
Clustering
Mesin
A
3
1
1
1
2
1
B
1
1
1
C
1
D
4
5
6
7
8
Value
224
224
1
1
1
1
1
1
156
1
30
E
1
1
3
F
1
1
3
2
1
2N-k
128 64
32
16
8
4
Clustering
Mesin
A
3
1
1
1
2
1
B
1
1
1
C
1
D
4
5
6
7
8
Value
224
224
1
1
1
1
1
1
156
1
30
E
1
1
3
F
1
1
3
2
1
2N-k
128 64
32
16
8
4
Handling Exceptional
Elements
• Methods to handle exceptional
elements
– Add extra machines of the required types
so that the machines are included in both
groups.
– Redesign the parts so that they don’t
require the machines not in the group.
– Route the exceptional parts to both groups.
– Combine the groups into a larger group.
Askin and Standridge’s Single Pass
Heuristic
• Simple binary ordering ignores machine
utilization and machine group sizes.
• This heuristic starts with the binary ordered
matrix and sequentially assigns ordered parts
(along with the required machines to machine
groups).
• Replace the 1’s in the binary ordered P-M matrix
with a measure of the machine utilizations
associated with each part type:
Askin and Standridge’s Single Pass
Heuristic
uim
•
•
•
•
•
s im Di
tim Di
Qi

Rm
sim setup time for part i on machine m
Di demand for part i
tim processing time for part i on machine m
Rm available time on machine m
Qi batch size for part i
Ordered P-M Matrix with
Utilizations
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Askin and Standridge’s Single Pass
Heuristic
• Starting with part 1 and continuing to N
– Assign the next part to the first group that
has sufficient capacity on already allocated
machines
– If no group has capacity
• try adding machines to the most recently created
group
• if this would violate the upper limit on the
number of machines in a group, open a new
group and add the required machines.
Forming Cell
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Forming Cell
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Forming Cell
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Forming Cell
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Forming Cell
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Forming Cell
Komponen
1
2
3
A
.4
.6
.7
1.1
2
B
.4
.5
.2
1.7
2
1.5
2
1.0
1
D
.5
5
6
.4
.2
.4
.3
.1
.5
7
8
Min
Mesin
Mesin
C
4
Total
Utilization
.1
E
.5
.4
.9
1
F
.6
.3
.9
1
mu = 4
Evaluation of Groupings
Given a P-M matrix, what makes one partitioning
better than another?
1 2 3 4 5
1 1 1 1 1 1
2 1 1 1 1
3 1 1 0 1
4
1
0
1
5
6
1
A
6 7 8
1
0 1 1
1 1 1
1 1 1
1 2 3 4
1 1 1 1 1
5 6 7 8
1 1
2 1
3 1
4
0
1
1 1 1
1
1
1
5
6
B
0 1 1
1 1 1 1
1 1 1
Graph Definition
• Consider a graph G = (N, A),
where :
– The node n  N represent machine,
– The arc a  A represent part flow
between machine , and
– The arc cost cij is some measure of
part flow between the machine
represented by the node i and the
machine represented by node j
• Want to partition the graph into
a number subgraphs, where
each subgraph identifies a group
of machines and the parts assign
to those machines
A
cad
D
cab
B
E
C
F
Cut Costs
• A cut is defined as the set of
arcs connecting nodes in
different subgraphs.
– if G1={A,B,D} and G2={C,E,F},
the cut is {(B,E), (B,F)}, and
the cut cost is cBE+cBF
• If p is the maximum number
of nodes we want in a
partition, our problem is to
find a partitioning of G such
that each partition has a
maximum of p nodes and
the cut cost is minimized.
A
cad
D
B
E
C
F
Initial Graph Partitioning Problem
• Find a minimum cost partitioning of a graph of 2n
nodes into two subgraphs of n nodes each.
• Let S be a set of 2n nodes, and let C = [cij] be a
symmetric cost matrix where cii = 0.
• Objective: Partition S into subsets A and B such that
the external cost, T, is minimized.
T   Cab
AXB
Sum of all Cij where I  A
Initial Graph Partitioning Problem
• If A and B represent an
arbitrary partitioning of S, and
A* and B* represent an
optimal partitioning, then
there must be subsets x A
and y B such that
interchanging x and y
produces A* and B*.
• That is, A* = A - x + y, and
B* = B - y + x
• Given x and y, what is the
cost benefit in swapping
them?
A
A*
x
y
y
x
B
B*
Internal and External Costs for a
a
A
x
b
B
y
Ea   C ay
External cost asociated with a
I a   Cax
Internal cost asociated with a
Da  E a - Ia
Saving if a is moved from A to B
yB
xA
Internal and External Costs for b
a
b
B
A
x
y
Eb   C bx
External cost asociated b
I b   C by
Internal cost asociated with b
Db  E b - I b
Saving if b is moved from B to A
xa
yB
Internal and External Costs for b
b
a
B
A
Savings associated with swapping a and b = Da + Db 2c
ab to update
Need
'
D
x  D x  2C xa - 2C xb , x  A - a
the remaining Dx
'
and Dy term after
Dy  D y  2C yb - 2C ya , y  B - b
swapping a and b


