1. Facility Layout 3 Non-computerized Block Layouts:
Graph Based Construction, SLP/DEO, and Pairwise Exchange

2. Graph-based construction method (p.317)
construction => new layout
Steps:
Build a graph that represents adjacency
Convert to block layout
Evaluate block layout (usually A-based)

3. Step 1: Construct the graph.
Select the two departments with the largest weights, break ties arbitrarily.
Select the third department based on the sum of the weights relative to the two departments already selected.
Select the fourth department based on maximizing the value to the first three departments in the graph.
Add departments maximizing value to a face until all departments have been added to the graph.

4. We begin with a relationship chart with weights, instead of letter ratings.1 2 3 4 5 - 9 8 10 12 13 7 20
Select the two departments with the largest weights.
(A direct link between two departments means that they will be adjacent)
Form graph with departments 3 & 4

5. Select the 3rd department to enter the graph.
REL 1 2 3 4 5 - 9 8 10 12 13 7 20
Select the next department with the largest sum of weights with 3 & 4.
Add department 2 to graph

6. Select the 4th department to enter the graph.
REL 1 2 3 4 5 - 9 8 10 12 13 7 20
Select the next department with the largest sum of weights with 2, 3 & 4.
Add department 1 to graph

7. Select the 5th department to enter the graph.
REL 1 2 3 4 5 - 9 8 10 12 13 7 20
Assign the next department with the largest sum of weights to a face.
Add department 5 to face 1-2-4

8. Construct Block Layout from graph
Total Weight =
( ) + ( ) = 81
To get total weight, sum all weights – they are all adjacent, so it is the adjacency-based score.

9. SLP Construction Approach
Determine Department Entry Order (DEO) from REL chart
Add blocks to layout based on DEO
Maximize Adjacency score

10. Transform REL Chart
REL 1 2 3 4 5 - I E U A O

11. Generate DEO

12. Create Block Layout from DEO and Relationships3
168
Adjacency Score =
Efficiency =
Score = ( ) + ( ) + (64 + 0) + 0 = 168
Effecicency = score / [( ) + ( ) + (64+0) + 0] = 172 4 2
168/173 = 97% 1 5

13. Pairwise Exchange – Improvement Method
Material Flow Matrix
To/
From 1 2 3 4 - 10 15 20 5
Objective: Minimize Distance-based score
Layout 1: (equal-sized departments) 1 2 3 4
Total Cost =
10(1) + 15(2) + 20(3) + 10(1) + 5(2) + 5(1) = 125
(1,2)
(1,3)
(1,4)
(2,3)
(2,4)
(3,4)

14. Pairwise Exchange - Greedy Algorithm
Layout 1
Material Flow Matrix 1 2 3 4
To/
From 1 2 3 4 - 10 15 20 5
TC (1-2) = {2,1,3,4}
10(1) + 15(1) + 20(2) + 10(2) + 5(3) + 5(1) = 105
TC (1-3) = {3,2,1,4}
10(1) + 15(2) + 20(1) + 10(1) + 5(2) + 5(3) = 95
TC (1-4) = {4,2,3,1} = 120
TC (2-3) = {1,3,2,4} = 120
TC (2-4) = {1,4,3,2} = 105
TC (3-4) = {1,2,4,3} = 125
Select “Best” Exchange
Exchange 1 & 3

15. Pairwise Exchange - Greedy Algorithm
Layout 2, TC = 95
Material Flow Matrix 1 2 3 4
To/
From 1 2 3 4 - 10 15 20 5
TC (1-2) = {3,1,2,4} = 95
TC (1-3) = {1,2,3,4} = 125
TC (1-4) = {3,2,4,1} = 110
TC (2-3) = {2,3,1,4} = 90
TC (2-4) = {3,4,1,2} = 105
TC (3-4) = {4,2,1,3} = 105
Select “Best” Exchange
Exchange 2 & 3

16. Pairwise Exchange - Greedy Algorithm
Layout 3, TC = 90
Material Flow Matrix 2 3 1 4
To/
From 1 2 3 4 - 10 15 20 5
TC (1-2) = 95
TC (1-3) = 120
TC (1-4) = 125
TC (2-3) = 105
TC (2-4) = 100
TC (3-4) = 95
No improvement possible
Finished at “BEST” layout with TC = 90