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

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)

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.

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

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

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

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

Construct Block Layout from graph Total Weight = (9+8+10+0) + (12+13+7) + 20 + 2 = 81 To get total weight, sum all weights – they are all adjacent, so it is the adjacency-based score.

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

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

Generate DEO

Create Block Layout from DEO and Relationships 3 168 Adjacency Score = Efficiency = Score = (4 + 0 +16 + 0) + (16 + 64 + 4) + (64 + 0) + 0 = 168 Effecicency = score / [(4 + 4 + 16 + 0) + (16+64+4) + (64+0) + 0] = 172 4 2 168/173 = 97% 1 5

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)

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

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

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