University of Southern CaliforniaDepartment of Industrial and Systems Engineering

ISE 310L Production I; Facilities and Logistics Spring 2000

March Lecture Handouts (Part 1) A Prototype Algorithm For Facilities Layout Design

Dr. Ardavan Asef-Vaziri

Given a From-To Chart ( or an Activity Relationship Chart) and the size of departments, we want to generate a Block Layout. Shape of the departments are restricted to right angle polygons.Indeed we want to move from

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The algorithm presented in this handout is a combination of a set of 3 algorithms in chapter 7 of your book.These algorithms are Systematic Layout Planning, Relationship Diagramming, and Pairwise interchange. The algorithm is composed of 3 parts

Analyze the Relationships (No area involved) Quantify the qualitative relationships Define a measure for ordering Identify the order for placement Identify the relative location of each department Prepare the relationship diagram

Include Area Prepare the Space Relationship Diagram Prepare the Block Layout

Improve the Layout Evaluate The layout Improve the layout

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Quantify the qualitative terms

We may make any reasonable convention to transform qualitative characters into quantitative terms.

Two examples are given below

Example 1

U : 0 O : 20 I : 21 E : 22 A : 24 X : -24

Example 2

A : 10,000 E : 1000 I : 100 O : 10 U : 0 X : -10000

We use the second example to quantify the above activity relationship chart

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Define the measure of ordering

The Total Closeness Rating (TCR) of a department with respect to a set of departments is equal to the sum of the relationships of this department with the other departments.

Calculate TCR of each department with respect to a set containing all e other departments.

TCR (1) = 1000+10+100+10 = 1120TCR (2) = 1000+1000+100+100 =2200TCR (3) = 10+10 = 20TCR (4) = 100+1000+100 = 1200TCR (5) = 10+100+100+10000+100 = 10310TCR (6) = 100+10+10000+1000 = 11110TCR (7) = 100+1000 = 1100

Define the order of entrance

The department with the highest TCR is selected first.Department 6 goes from set of unselected departments to the set of selected departments.

Given e department i belong to the set of selected departments, i.e. i S, and department J belong to the set of unselected departments, i.e. J S’

Then TCR of each unselected department with respect to selected departments is equal to sum of the relationships of the unselected department with all selected departments.

TCR (i ) = R(i,j) iS jS’

Department 6 is the only department in the set of selected department. Calculate TCR of each unselected department with respect to 6.

Unselected department TCR with respect to the selected department 61 02 1003 104 –5 100007 1000

The department with the highest TCR is selected next. Department 5 has the highest TCR.

Repeat- For each unselected department.- Calculate its TCR with respect to the selected departments- Select the unselected department with highest TCR

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The order of the departments for placement in the layout is

6-5-7-2-4-1-3

Relative locationsStill no area is involved. All departments are of the same size.

Two departments could have an edge in common, a node in common, or nothing in common. An edge full relationship is added to the value of in common the candidate location.

Two departments a node in half of relationships is added to the Common value of the candidate location.

Nothing in nothing is added common

Suppose department 6 is placed . the next department is department 5. Candidate locations are those grids with at least a node in common with department 6.

A B CH 6 DG F E

The relationship between 6 and 5 is 10,000. If we locate department 5 at location B or D or F or H. the relationship is fully satisfied. Value of the grid locations for department 5 are shown below.

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Place 5 in grid with highest value.

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Now we want to find the best grid for dept.7 R(7,6) = 1000 R(7,5) = 100

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Up to now we have not said even one word about size of the department.All we did was to translate the following representation

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Into the following representation

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In the next steps, we will transform the above representation into the following representation.

Prepare Activity Relationship Diagram

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We first apply A relationships

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Then both A and E relationships

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Then A, E, and I relationships

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Finally all relationships are incorporated

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Add AreaUp to this point we have not done even one word on area of the departments.

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Now we add area to the above diagram

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Measure of EffectivenessThere are two types of indices to measure the effectiveness of a specific layout.The first type is Flow times DistanceThe second type is Relationship times AdjacencyWe have already discussed Flow times Distance measure of effectiveness/It was Flow times Distance Straight Rectilinear Aisle Restricted Aisle (for example unidirectional loop)

The second type of measure of effectiveness id Relationship times AdjacencyWe discuss it in the following sections,.Both F×D and R×A can be measured in both Absolute and Relative weights. We have already defined Absolute F×D, Now we first define absolute R×A, and then relative R×D, finally we introduce relative F×D.

Let’s forget partial adjacency. In other words, if two departments have one edge in common, they are adjacent, otherwise they are not, even if hey have one node in common. If two department have positive relationship ( Rij > 0 )

If they are adjacent, then their relationship is satisfied and ( Aij = 0 ), Otherwise their relationship is not satisfied and ( Aij = 1 ).

If two department have negative relationship ( Rij < 0 )If they are not adjacent, then their relationship is satisfied and ( Aij = 0 ), Otherwise their relationship is not satisfied and ( Aij = -1 ).

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Department 1 is adjacent to 2, 3, 4. Their relationships R(1,3) = 10, R(1,2) =1000, R(1,4) = 100. For all of these departments A(1,j)=0. Nothing is added to the score of the layout.

10+1000+100= 1110 is added to the score of the layout.

To make it algorithmic, we prepare the Aij matrix.

We have the quantitative version of Relationship chart.

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Since all relationships are positive, therefore if two departments are adjacent ( have an edge in common) their Aij is 0, Otherwise it is 1.

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Aij table

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Rij Aij table

Absolute R×A = 110Upper Bound for R×A =13520Lower Bound R×A =0

Relative R×A = 100 (Absolute R×A -LB)/(UB-LB)=0.8%