Team #2
Mingang Fu
Lin Ben
Kuowei Chen
Introduction
Traditional Manufacturing Processes
 Product Layout
 Functional Layout
L
G
M
(a) Product layout
Lathe
department
Milling
department
Drilling
department
Grinding
department
D
(b)Functional layout
Introduction
 Group Technology
(c) Group layout
Problem Description
 A part family may consist of groups of parts requiring
similar and sometimes identical operation processes,
materials, and tools.
 A manufacturing cell is formed by the machines
which are required to produce a part family.
 Goal: form a manufacturing system that consists of
cells
 to maximize the moves of parts processed within the
cells, at the same time, to minimize the parts flow
between cells
Problem Description
 Example:
parts
parts
1 2 3 4 5 6 7
1 7 3 4 6 2 5
1
2
1
1
1
1
2
1 1
5
1 1
3
1 1
1
3
1 1 1
machines 4
1 1
1
machines 4
1 1 1
6
1 1 1
5
1
1
6
7
1 1
1
1
1
1
1 1
7
1 1
Formulation 1
• Notations:
i and j are machine indexes (i, j = 1, 2,…, Nm).
k is a part index (k = 1, 2, 3,…, Np)
c is a cell index (c = 1, 2, 3,…,Nc)
pvk is the production volume of part k
cdk is available transfer units per trip for part k using a transfer device
U c is the upper limit of cell size
 pv 
ntijk is the number of trips made by part k between machines i and j: ntijk   k 
 cdk 
Where  w indicates the smallest integer value greater than or equal to w.
• Variables: xic 

1, if machine i is assigned to cell c
0, otherwise
Formulation 1
N
1 Nc N m N m p
 Objective function: max z  2  ntijk xic x jc
c 1 i 1 j 1 k 1
 Constraints:
Nc
s.t.
x
c 1
ic
 1,
i
Nm
1   xic  U c ,
i 1
x ic  0,1 ,
i,c
c
Formulation 1 - Example
Nm = 7, Np = 7, Nc = 3
parts
1 2 3 4 5 6 7
1
2
1
1
1
1
3
1 1
1
machines 4
1 1
1
5
1
1
6
7
X11=1, X21=1
X32=1, X42=1, X52=1
X63=1, X73=1
1 1
1
1
1
Assume pvk  1 and cdk  1, k
Objective function value = 2
Formulation 1 - Example
parts
1 7 3 4 6 2 5
2
1 1
5
1 1
3
1 1 1
machines 4
1 1 1
6
1 1 1
X21=1, X51=1
X32=1, X42=1, X62=1
X13=1, X73=1
1
1 1
7
1 1
Objective function value = 10
Formulation 2
 m machines and n parts with k cells and there are a total of k(m+n) variables
and (m+n) constrains.
1, if machine i is assigned to cell l
xil  
0, otherwise
1, if part j is assigned to part family l
y jl  
0, otherwise
k
x
l 1
il
k
y
l 1
jl
 1, i  1...m
 1, j  1...n
k  the number of cells (families) specified
m  the number of machines
n  the number of parts
 With the size of problem increases, the model becomes too large to
handle. To overcome this problem, we can change the integer
programming model with following declaration:
xi  l , machine i is assigned to cell l
y j  l , part j is assigned to part family l
 Then we define “group efficiency” () as following and maximize it.
e0
e e 0
1
e



ev e  ev
1
1
e
e  the number of operations in the data matrix
1
ev  the number of viods in the diagonal blocks
e0  the number of exceptioan l elements
Formulation 2 - Example
parts
Cell 1,
Family 1
1 2 3 4 5 6 7
1
2
1
1
1
1
3
1 1
1
machines 4
1 1
1
5
1
1
6
7
Cell 2,
Family 2
1 1
1
1
1
Cell 3,
Family 3
X1=1, X2=1, X3=2, X4=2, X5=2, X6=3, X7=3;
Y1=1, Y2=1, Y3=2, Y4=2, Y5=2, Y6=3, Y7=3
X11=1, X21=1, X31=0, …X71=0
X12=0, X22=0, X32=1, X42=1, X52=1,
X62=0, X72=0
X13=0, …X53=0, X63=1, X73=1
Y11=1, Y21=1, Y31=0,
…Y71=0
Y12=0, Y22=0, Y32=1,
Y42=1, Y52=1, Y62=0, Y72=0
Y13=0, …Y53=0, Y63=1,
Y73=1
e=17, ev=10, eo=10,
Gamma=7/27
Formulation 2 - Example
parts
1 7 3 4 6 2 5
2
1 1
5
1 1
3
1 1 1
machines 4
1 1 1
6
1 1 1
X1=3, X2=1, X3=2, X4=2, X5=1, X6=2, X7=3;
Y1=1, Y2=3, Y3=2, Y4=2, Y5=3, Y6=2, Y7=1
e=17, ev=0, eo=0,
Gamma=17/17=1, best
1
1 1
7
1 1
Thank you !