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Cell formation considering sequence

of operations and workload

.

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Let Cijbe the number of parts that visit machine j immediately after visiting machine i

or visiting machine i immediately visiting machine j. The Cijvalues can be computed

from a given machine component incidence matrix.

Let Xij= 1 if machines i and j belong to the same machine group.

Here Xijare defined for values i = 1,...,M-1 and j = i+1,..,M.

1

1 1

1M M

ij ij

i j i

C X

Minimize

1

1 1

1 1,..,k M

ik kj

i j k

X X N k M

1

1

1 1,.., 2, 1,.., 1, 1,..,

ij ik jk

ij ik jk

ij ik jk

X X X

X X X

X X X i M j i M k j M

0,1ij

X

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The binary IP formulation has 15 variables and 63 constraints.

The optimal solution is given by X13

= X14

= X34

= X25

= X26

= X56

= 1

with objective function value = 11.

1 4 5 6 2 3 7 8

1 1 2 1 1

3 3 3 2

4 4 3 3

2 4 1 1 1

5 2 2 2 2 36 2 1 3 1 3 2

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Nair and Narendran (1998) introduced the following similarity coefficient between

machines i and j:

ij

ijij

AS

B

Here Aijrepresents the contribution of parts visiting both the machines i and j. Part k visiting

both i and j contributes 1 if it visits the machine as its first or last machine in the sequence and

contributes 2 if it is an intermediate machine. Similarly Bij

represents the contribution of parts

visiting both either machine i or machine j or both. Part k contributes 1 if it visits the machine as

its first or last machine in the sequence and contributes 2 if it is an intermediate machine. The

machine groups are formed based on the Sijvalues.

New Similarity Coefficient

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Machinesi-j

Jij Machinesi-j

Jij Machinesi-j

Jij

1-2 2/9 2-3 3/9 3-5 0

1-3 9/10 2-4 3/6 3-6 5/10

1-4 7/9 2-5 8/12 4-5 3/11

1-5 3/11 2-6 7/12 4-6 3/12

1-6 6/13 3-4 7/8 5-6 11/16

Considering machines 1 and 2, we observe that part 6 visits both. A total of 7

parts visit either machine 1 or machine 2 or both. A12= 2 because part 6 visits

machine 1 as first operation and visits machine 2 as fourth (last) operation. Each

contributes 1 to the A12value. The contributions of parts 1, 2, 4, 5, 6, 7, 8 are 1,

1, 2, 1, 2, 1, 1 respectively. Therefore B12= 9 and S12= 2/9.

The two machine groups are [1 3 4] and [2 5 6] respectively.

There are six inter cell moves.

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Cell load data

1 2 3 4 5 6 7 81 .3 .4 .2 .1

2 .3 .3 .2 .4

3 .5 .3 .3

4 .4 .1 .2

5 .3 .2 .2 .4 .3

6 .3 .2 .4 .1 .3 .5

1

1

M

ij ik

ikj M

ik

i

w X

L

X

Let us assume that there are K groups (k = 1,..K). Each machine should belong to one

machine group. Let Xik= 1 if machine i belongs to cell k.

The cell load in cell k from part j is given by

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The total cell load variation is given by

2

1 1 1

M K N

il ij jl i l jZ X w L

The mathematical programming formulation (Venugopal and Narendran, 1992)

is to Minimize Z subject to

1

1K

ikl

X

1

1M

ik

i

X

Xik= 0,1.

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[1 3 4] and [2 5 6]. Find the cell load variation

for these machine cells?

The average cell load in cell 1 through part 1 is 0.8/3. The sum of loads on part 1 in

machines 1, 2 and 3 is 0.8. The average load is .8/3 = -2666. The average cell load by

the parts on the two cells are given in Table

1 2 3 4 5 6 7 8

1 .266 .1 0 .233 .0666 .233 .0666 .133

2 .233 .166 .2 .033 .1 .0666 .233 .2666

The cell load variance is given by (.3-.266)2+(.5-.266)2+(.4-.233)2+(.3-.266)2+---- = 0.47222.

For the cells [1 3 4] and [2 5 6] the variance is 0.31444.

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Similarity coefficient

wjil= 1 if wji > 0 or wjl> 0.

1 2 3 4 5 61 -- .047 .42 .583 .125 .155

2 .047 -- .166 .166 .375 .355

3 .42 .166 -- .365 0 .2

4 .585 ,166 .365 -- .071 .1075 125 .375 0 .071 -- .313

6 .155 .355 .2 .107 .313 --

[1 3 4] and [2 5 6]

s12= .1/.3 7 = 0.00476.