University of Southern California
Department 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
2 / 3 / 4 / 5 / 6 / 71 / E / O / I / O / - / -
2 / - / E / I / I / -
3 / - / - / O / -
4 / I / - / -
/ 5 / A / I
/ 6 / E
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
2 / 3 / 4 / 5 / 6 / 71 / E / O / I / O / - / -
2 / - / E / I / I / -
3 / - / - / O / -
4 / I / - / -
5 / A / I
6 / E
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
2 / 3 / 4 / 5 / 6 / 71 / 1000 / 10 / 100 / 10 / - / -
2 / - / 1000 / 100 / 100 / -
3 / - / - / 10 / -
4 / 100 / - / -
5 / 10000 / 100
6 / 1000
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 = 1120
TCR (2) = 1000+1000+100+100 =2200
TCR (3) = 10+10 = 20
TCR (4) = 100+1000+100 = 1200
TCR (5) = 10+100+100+10000+100 = 10310
TCR (6) = 100+10+10000+1000 = 11110
TCR (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 6
10
2100
310
4–
510000
71000
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
657241
1-10-10001005th
2100100-3rd
310----10
4-100-10004th
5100001sr
710001002nd
The order of the departments for placement in the layout is
6-5-7-2-4-1-3
Relative locations
Still 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 / C/ H / 6 / D
G / 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.
/ 5000 / 10000 / 500010000 / 6 / 10000
5000 / 10000 / 5000
Place 5 in grid with highest value.
5 / 6Now we want to find the best grid for dept.7
R(7,6) = 1000 R(7,5) = 100
/ 500 / 1000 / 500/ 5 / 6 / 1000
500 / 1000 / 500
50 / 100
500 / 50
1000 / 500
100 / 5 / 6 / 1000
50 / 500
100 / 1000
50 / 500
/ 5 / 6
7
50 / 100
50 / 50
100 / 50
100 / 5 / 6 / 100
50 / 50
100 / 7 / 50
/ 500 / 1000 / 500
50 / 1000
100 / 2 / 1000
100 / 5 / 6 / 500
50 / 100 / 7
Up to now we have not said even one word about size of the department.
All we did was to translate the following representation
2 / 3 / 4 / 5 / 6 / 71 / 1000 / 10 / 100 / 10 / - / -
2 / - / 1000 / 100 / 100 / -
3 / - / - / 10 / -
4 / 100 / - / -
5 / 10000 / 100
6 / 1000
Into the following representation
1 / 34 / 2
5 / 6
7
In the next steps, we will transform the above representation into the following representation.
Prepare Activity Relationship Diagram
/ 1/ 4 / 2 / 3
/ 5 / 6
/ 7
We first apply A relationships
/ 5 / 6Then both A and E relationships
/ 4 / 2/ 5 / 6
Then A, E, and I relationships
/ 1/ 4 / 2
/ 5 / 6
/ 7
Finally all relationships are incorporated
/ 1/ 4 / 2
/ 5 / 6
/ 7
Add Area
Up to this point we have not done even one word on area of the departments.
Now we add area to the above diagram
/ 1/ 4 / 2
/ 5 / 6
Measure of Effectiveness
There are two types of indices to measure the effectiveness of a specific layout.
The first type is Flow times Distance
The second type is Relationship times Adjacency
We 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 Adjacency
We 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 ).
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.
2 / 3 / 4 / 5 / 6 / 71 / 1000 / 10 / 100 / 10 / - / -
2 / - / 1000 / 100 / 100 / -
3 / - / - / 10 / -
4 / 100 / - / -
5 / 10000 / 100
6 / 1000
Since all relationships are positive, therefore if two departments are adjacent ( have an edge in common) their Aij is 0, Otherwise it is 1.
2 / 3 / 4 / 5 / 6 / 71 / 0 / 0 / 0 / 1 / 1 / 1
2 / 0 / 0 / 1 / 0 / 1
3 / 1 / 1 / 0 / 1
4 / 0 / 1 / 1
5 / 0 / 0
6 / 0
Aij table
2 / 3 / 4 / 5 / 6 / 71 / 0 / 0 / 0 / 10 / 0 / 0
2 / 0 / 0 / 100 / 0 / 0
3 / 0 / 0 / 0 / 0
4 / 0 / 0 / 0
5 / 0 / 0
6 / 0
RijAij table
Absolute R×A = 110
Upper Bound for R×A =13520
Lower Bound R×A =0
Relative R×A = 100 (Absolute R×A -LB)/(UB-LB)=0.8%