Facility Layout 3 Noncomputerized Block Layouts Graph Based

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: 1. Build a graph that represents adjacency 2. Convert to block layout 3. Evaluate block layout (usually A-based)

Step 1: Construct the graph. 1. 2. 3. 4. 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. REL 1 2 3 4 5 1 - 9 8 10 0 2 - - 12 13 7 3 - - - 20 0 4 - - 2 5 - - - 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 3 rd department to enter the graph. REL 1 2 3 4 5 1 - 9 8 10 0 2 - - 12 13 7 3 - - - 20 0 4 - - 2 5 - - - Add department 2 to graph Select the next department with the largest sum of weights with 3 & 4.

Select the 4 th department to enter the graph. REL 1 2 3 4 5 1 - 9 8 10 0 2 - - 12 13 7 3 - - - 20 0 4 - - 2 5 - - - Add department 1 to graph Select the next department with the largest sum of weights with 2, 3 & 4.

Select the 5 th department to enter the graph. REL 1 2 3 4 5 1 - 9 8 10 0 2 - - 12 13 7 3 - - - 20 0 4 - - 2 5 - - - 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

SLP Construction Approach 1. Determine Department Entry Order (DEO) from REL chart 2. Add blocks to layout based on DEO 3. Maximize Adjacency score

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

Generate DEO

Create Block Layout from DEO and Relationships A = 64 3 4 E = 16 I =4 2 1 5 O=1 U=0 Adjacency Score = Efficiency = 168/173 = 97%

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

Pairwise Exchange - Greedy Algorithm Material Flow Matrix To/ From 1 2 3 4 1 - 10 15 20 2 - - 10 5 3 - - - 5 4 - - Select “Best” Exchange Layout 2, TC = 95 3 2 1 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 Exchange 2 & 3 4

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