Basic Algorithms and Software for the Layout Problem Chapter 5

Basic Algorithms and Software for the Layout Problem

Chapter 5

Algorithms

• Optimal• Heuristic

Algorithms

• Construction– MST– Graph Theoretic

Method

• Improvement– 2-opt

• Greedy• Steepest Descent

– 3-opt• Greedy• Steepest Descent

• Hybrid– Modified Penalty

Algorithm

MST Algorithm

• Step 1: Given the flow matrix [fij], clearance matrix [dij] and machine lengths li, compute an adjacency weight matrix where:f’ij = (fij)(dij+0.5(li+lj)).

• Step 2: Find the largest element in [f’ij] and the corresponding i, j. Denote this pair of i, j as i*, j*. Connect machines i*, j*. Set f’i*j* =f’i *i* =-infinity

MST Algorithm

• Step 3: Find the largest element f’i*k,f’j*l in row i*, j* of matrix If f’i*k*>f’j*l* connect k to i*, remove row i*, column i* from matrix and set i* = k. Otherwise, connect l to j*, remove row j*, column j* from matrix and set j* = l. Set f’i*j* =f’i *i* =-infinity

• Step 4: Repeat step 3 until all machines are connected. The sequence of machines obtained determines the arrangement of machines.

Example 1

M a c h i n e Lengths (in feet)

M 1 2 3 4 5 6

a 1 - 12 3 6 0 20 20

c 2 12 - 5 5 5 0 10

h 3 3 5 - 10 4 2 16

i 4 6 5 10 - 2 12 20

n 5 0 5 4 2 - 6 10

e 6 20 0 2 12 6 - 10

Example 1 Solution

M 1 2 3 4 5 6

a 1 - 204 60 132 0 340

c 2 204 - 75 85 60 0

h 3 60 75 - 200 60 30

i 4 132 85 200 - 34 204

n 5 0 60 60 34 - 72

e 6 340 0 30 204 72 -

Example 1 Solution

5 2 1 6 4 3

Graph Theoretic Method

• Terminology– Graph– Complete graph– Planar Graph– Maximal Planar Graph

9

2

5

8

1

4

7

3

6 6

1 2 3

54

7 8

Graph Theoretic Method

• Layout…. • And its dual…

9

2

5

8

1

4

7

3

6 6

1 2 3

54

7 8

Graph Theoretic Method*

Step 1: Identify the department-pair in the flow matrix with the maximum flow. Place the corresponding nodes in a new PAG and connect them.

Step 2: From the rows corresponding to the connected nodes in the flow matrix, select the node which is not yet in the PAG and has the largest flows with the connected nodes.

Step 3: Update PAG by connecting the selected node to those in Step 2. This forms a triangular face in the PAG.

Step 4: For each column of the flow matrix corresponding to a node not present in the PAG, examine the sum of flow entries in the rows corresponding to the nodes of the triangular face selected in step 3. Select the column for which this sum is the largest. Update PAG by placing the corresponding node within the selected face and connect it to nodes of the face. This forms three new triangular faces.

Step 5: Arbitrarily select one of the faces formed and go to Step 4. Repeat Step 5 until all the nodes have been included in the PAG.

* Based on the result that the maximum number of arcs in a planar graph with n nodes 3n-6

Graph Theoretic Method

Do Example 2

1 2 3 4 5 6 7 8 9 10 1

1

12

1 - 1 0 8 0 2 3 0 0 0 0 0

2 1 - 0 1 1 1 0 0 0 0 0 0

M 3 0 0 - 0 2 0 0 0 0 0 0 0

a 4 8 1 0 - 0 4 14 11 0 0 0 0

c 5 0 1 2 0 - 1 0 0 0 0 0 0

h 6 2 1 0 4 1 - 3 0 0 3 0 0

i 7 3 0 0 14 0 3 - 5 5 9 8 2

n 8 0 0 0 11 0 0 5 - 8 0 0 0

e 9 0 0 0 0 0 0 5 8 - 0 0 0

10 0 0 0 0 0 3 9 0 0 - 6 0

11 0 0 0 0 0 0 8 0 0 6 - 4

12 0 0 0 0 0 0 2 0 0 0 4 -

Graph Theoretic Method

Example 2

4 714

14

8

4 7

511

14

511

8

4 7

9 50

8

Graph Theoretic Method

Example 2

32

10

9

11

74

1

6 12

5

13

8

32

10

9

11

74

1

6 12

5

13

8

Graph Theoretic Method

Example 2

10

2

5

3

1

9

1211

7

68

4

2-opt algorithm

• Step 1: Let S be the initial solution provided by the user and z its OFV. Set i=1; j=i+1=2.

• Step 2: Consider the exchange between the positions of departments i and j in the solution S. If the exchange results in a solution S’ that has an OFV z’< z, set z*=z’ and S*=S’. If j < mn, set j=j+1; otherwise, set i=i+1, j=I+1. If i < mn, repeat step 2; otherwise, go to step 3.

• Step 3: If S not =S*, set S=S*, z=z*, i=1, j=i+1=2 and go to step 2. Otherwise, return S* as the best solution to the user. Stop.

• Do Example 3 using SINROW or MULROW

2-opt algorithm

. . . . . mn

. . .

. . . . . .

. . . . . .

m+1 m+2 . . . .

1 2 . . . m

Example 3

O 1 2 3 4 1 2 3 4

f 1 - 17 12 11 S 1 - 1 1 2

[fij]= f 2 17 - 12 4 [dij]= i 2 1 - 2 1

i 3 12 12 - 4 t 3 1 2 - 1

c 4 11 4 4 - e 4 2 1 1 -

e

3-opt algorithm

Step 1: Let S be the initial solution and z its OFV; Set S*=S, z*=z, i=1; j=i+1; k=j+1.

Step 2: Consider changing the position of department i to that of j, j to that of k, and k to that of i, simultaneously. If the resulting solution SN has OFV z’ < z, set z*=z’ and S*=S’.

Step 3: If k < mn, set k = k +1, and repeat step 2. Otherwise, set j=j+1 and check if j < mn-1.

If j < mn-1, set k=j +1, and repeat step 2. Otherwise, set i = i +1, j=i+1, k = j +1, and check if i < mn-2.

If i < mn -2, repeat step 2. Otherwise, go to step 4.Step 4: If S not = S*, set S=S*, z=z*, i=1, j=i+1, k=j+1 and go to

step 2. Otherwise, return S* as the best solution to the user. Stop.

Layout Software

- CRAFT

- BLOCPLAN

- PFAST

- FactoryFLOW

- Layout-iQ

- VIP-PLANOPT

- Flowpath Calculator

CRAFT in Excel

• Do Example 4

BLOCPLAN

• Adjacency score

• Rel-dist score• R-score = 1- (rel-dis score- lower bound)/(upperbound-lower

bound)

Rij numeric value assigned to the relationship code between departments i and j,

• n total number of departments, and

• dij rectilinear distance between the centers of departments i and j

1

1 1

1

1 1

n n

ij iji j i

n n

iji j i

R D

R

1

1 1

n n

ij iji j i

d R

1 if departments and are on the same floor and adjacent

0 otherwiseij

i jD

BLOCPLAN• Do Example 5

PFAST

Layout-iQ

VIP-PLANOPT

Flowpath Calculator

BLOCPLAN• Re-layout

– CRAFT-M

• Multi-Floor Layout– BLOCPLAN