1. Facility Design-Week 10 (cont) Computerized Layout PlanningBy
Anastasia L. Maukar
Created by Lorna StJohn
For Operations Management 345
Professor Tom Foster
Wednesday, May 08, 2002

2. CRAFT- Computerized Relative Allocation of Facilities Technique
Created in 1964 by Buffa and Armour
Process layout approach
A heuristic computer program
Compares process departments
CRAFT requires an initial layout, which is improved by CRAFT.
CRAFT is a tool used to help improve the existing layout of the facilities. The facility is improved by switching two or three departments to help arrange the facility in an optimal floor plan. This procedure requires the following inputs: From-To Chart, Cost Matrix, Distances (determined for a given layout) and an Initial layout. Craft is used when the number of departments is so large that the computation by hand would be very intensive and make the improvement not worth the time for many companies.
Process Layout – also known as a functional layout is usually used in job shops or a batch production facility. In this layout, activities are grouped together in work centers or departments according to the process or function they perform. The goal of process layout is to minimize the material handling costs in manufacturing. The departments that incur the most interdepartmental movement should be located the closest to on another.
Advantages – Flexibility
Disadvantages –
Inefficient – Jobs do not flow through in an orderly fashion, therefore backtracking occurs often.
Idle time – More idle time may be experienced while workers are waiting for more work to arrive from different departments
Heuristics – An algorithm that returns good results but not necessarily the optimal result for the layout. CRAFT uses a pairwise exchange algorithm that may not return the optimal result because the final solution depends on the initial layout of the plant.

3. From-To Chart
Determines which of the two departments has a better from-to matrix, we can calculate the moment of the matrix as follows:
After creating the computations you will compare each combination of matrixes to determine which has the lowest number. the lowest number determine the most optimal layout. Once you have determined the optimal matrixes you will lay out the departments in a block diagram to get the best fit.

4. CRAFT
Input for CRAFT:
initial spatial array/layout
flow data
cost data
Number and location of fixed department
Secara umum, dapat ditambahkan dummy yg berfungsi untuk:
Fill building irregularities.
Represent obstacles or unusable areas in facility
Represent extra space in the facility
Aid in evaluating aisle locations in the final layout.

5. CRAFT Following are some examples of questions addressed by CRAFT:
Is this a good layout?
If not, can it be improved?

6. CRAFT The possible interchange: only pairwise interchage
only three-way interchange
pairwise interchage followed by three-way interchange
three-way interchage followed by pairwise interchange
best of two-way and three-way interchage

7. CRAFT: Distance Between Two Departments
Consider the problem of finding the distance between two adjacent departments, separated by a line only.
People needs walking to move from one department to another, even when the departments are adjacent.
An estimate of average walking required is obtained from the distance between centroids of two departments.
The distance between two departments is taken from the distance between their centroids.
People walks along some rectilinear paths. An Euclidean distance between two centroids is not a true representative of the walking required. The rectilinear distance is a better approximation.
So, Distance (A,B) = rectilinear distance between centroids of departments A and B

8. CRAFT: Distance Between Two Departments
Let
Centroid of Department A =
Centroid of Department B =
Then, the distance between departments A and B, Dist(A,B)
Ex: the distance between departments A and C is the rectilinear distance between their centroids (30,75) and (80,35). Distance (A,C)

9. CRAFT: Distance Between Two Departments
100
Centroid of A
= (30,75)
Centroid of C
= (80,35)
Distance (A,C)
= 90 A D 90 80
(80,85) 70 60 C 50 40 B 30 20
(30,25) 10 10 20 30 40 50 60 70 80 90
100

10. CRAFT: Total Distance Traveled
If the number of trips between two departments are very high, then such departments should be placed near to each other in order to minimize the total distance traveled.
Distance traveled from department A to B = Distance (A,B)  Number of trips from department A to B
Total distance traveled is obtained by computing distance traveled between every pair of departments, and then summing up the results.
Given a layout, CRAFT first finds the total distance traveled.

11. CRAFT: Total Distance Traveled
Material handling trips(given)
(b)
(b) Distances (given)

12. CRAFT: Total Distance Traveled
Material handling trips (given)
(b) Distances (given)
(b)
(c) Sample computation:
distance traveled (A,B)
= trips (A,B)  dist (A,B)
=………..
Total distance traveled
= ….
= 4640
(c)

13. CRAFT: Savings
CRAFT then attempts to improve the layout by pair-wise interchanges.
If some interchange results some savings in the total distance traveled, the interchange that saves the most (total distance traveled) is selected.
While searching for the most savings, exact savings are not computed. At the search stage, savings are computed assuming when departments are interchanged, centroids are interchanged too. This assumption does not give the exact savings, but approximate savings only.
Exact centroids are computed later.

14. CRAFT: Savings
Savings are computed for all feasible pairwise interchanges. Savings are not computed for the infeasible interchanges.
An interchange between two departments is feasible only if the departments have the same area or they share a common boundary.
Feasible pairs are {A,B}, {A,C}, {A,D}, {B,C}, {C,D}
and an infeasible pair is {B,D}
In this example savings are not computed for interchanging B and D. Savings are computed for each of the 5 other pair-wise interchanges and the best one chosen.
After the departments are interchanged, every exact centroid is found. This may require more computation if one or more shape is composed of rectangular pieces.

15. CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange
To illustrate the computation of savings, we shall compute the savings from interchanging Departments C and D
New centroids:
A (30,75) Unchanged
B (30,25) Unchanged
C (80,85) Previous centroid of Department D
D (80,35) Previous centroid of Department C
Note: If C and D are interchanged, exact centroids are C(80,65) and D(80,15). So, the centroids C(80,85) and D(80,35) are not exact, but approximate.

16. CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange
The first job in the computation of savings is to reconstruct the distance matrix that would result if the interchange was made.
The purpose of using approximate centroids will be clearer now.
If the exact centroids were used, we would have to recompute distances between every pair of departments that would include one or both of C and D.
However, since we assume that centroids of C and D will be interchanged, the new distance matrix can be obtained just by rearranging some rows and columns of the original distance matrix.

17. CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange
The matrix on the left is the previous matrix, before interchange. The matrix on the right is after.
Dist (A,B) and (C,D) does not change.
New dist (A,C) = Previous dist (A,D)
New dist (A,D) = Previous dist (A,C)
New dist (B,C) = Previous dist (B,D)
New dist (B,D) = Previous dist (A,C)
Interchange
C,D

18. CRAFT: A Sample Computation of Savings
Material handling trips
(given)
(b) Distances (rearranged)
(b)
(c) Sample computation:
distance traveled (A,B)
= trips (A,B)  dist (A,B) =
Total distance traveled
= …
= 4480
Savings =
(c)

19. CRAFT: Improvement Procedure
To complete the exercise
1. Compute savings from all the feasible interchanges. If there is no (positive) savings, stop.
2. If any interchange gives some (positive) savings, choose the interchange that gives the maximum savings
3. If an interchange is chosen, then for every department find an exact centroid after the interchange is implemented
4. Repeat the above 3 steps as longs as Step 1 finds an interchange with some (positive) savings.

20. CRAFT: Exact Coordinates of Centroids
Sometimes, an interchange may result in a peculiar shape of a department; a shape that is composed of some rectangular pieces
For example, consider the layout (from example) and interchange departments A and D. The resulting picture is shown on the right.
How to compute the exact coordinate of the centroid (of a shape like A)?

21. CRAFT: Exact Coordinates of Centroids
100 A A1 90 80 70 60 A2 50 10 20 30 40 50 60 70 80 90
100
Find the centroid of A

22. CRAFT: Exact Coordinates of Centroids
X-coordinate Multiply
Rectangle Area of centroid (2) and (3)
(1) (2) (3) (4) A1 A2
Total
X-coordinate of the centroid of A

23. CRAFT: Exact Coordinates of Centroids
Y-coordinate Multiply
Rectangle Area of centroid (2) and (3)
(1) (2) (3) (4) A1 A2
Total
Y-coordinate of the centroid of A

24. CRAFT: Exact Coordinates of Centroids
100 A A1 90 80 70 60 A2 50 10 20 30 40 50 60 70 80 90
100
Exact coordinate of area A is

25. CRAFT: Some Comments
An improvement procedure, not a construction procedure
At every stage some pairwise interchanges are considered and the best one is chosen
Interchanges are only feasible if departments have the same area; or they share a common boundary
Departments of unequal size that are not adjacent are not considered for interchange
Estimated cost reduction may not be obtained after interchange (because the savings are based on approximate centroids)
Strangely shaped departments may be formed

26. Computerized Layout Planning
Graphical Representation
“Points and lines” representation is not convenient for analysis

27. Layout Evaluation
An Algorithm needs to distinguish between “good” layouts and “bad” ones
Develop scoring model, s = g (X )
Adjacency-based scoring (Komsuluk Bazli Skorlama)
Based on the relationship chart and diagram
Xi is the number of times an adjacency i is satisfied, i=A, E, I, O, U, X
Aldep uses (wi values) A=64, E=16, I=4, O=1, U=0, and X=-1024
Scoring model has intuitive appeal; the ranking of layouts is sensitive to the weight values. Layout “B” may be preferred to “C” with certain weights but not with others.
Therefore, the specification of the weights is very important.
The weights wi can also be represented by the flow amounts between the adjacent departments instead of scores assigned to A, E, I, O, U, X.
wi is the weight for class i.
Xi is the number of adjacencies in class i.

28. Example 1 2 3 4 5 6 7
I O
E I U
O U A E
4+1 =5
=20
1+0 =1
----
64 =64
16 =16
Total Score 106 1 2 3 4 5 6 7 U I E O A
Receiving
Milling
Press
Screw Machine
Assembly
Plating
Shipping 3
Press 7
Shipping 2
Milling 4
Screw Machine U U E E I I 6
Plating 5
Assembly 1
Receiving O A O

29. 1 2 3 4 5 6 7 U I E O A
Receiving
Milling
Press
Screw Machine
Assembly
Plating
Shipping
Exercise: Find the score of the layout shown below. Use A=8, E=4, I=2, O=1, U=0 and X=-8. 3
Press 7
Shipping 6
Plating 4
Screw Machine 1
Receiving 2
Milling 5
Assembly

30. Layout Evaluation (cont’d)
Distance-based scoring (Mesafe Bazli Skorlama)
Approximate the cost of flow between activities
Requires explicit evaluation of the flow volumes and costs
cij covers both the i to j and the j to i material flows
Dij can be determined with any appropriate distance metric
Often the rectilinear distance between department centroids
Assumes that the material flow system has already been specified (cij=flow required* cost /flow-distance)
Assumes that the variable flow cost is proportional to distance
Distance often depends on the aisle layout and material handling equipment

31. CRAFT - Example 2
Initial Layout Ma
Flow Data
Total Cost
Distance Data

32. CRAFT - Example 2 A & D  interchange  Total cost = 1.095
A (60, 10) dan D (25, 30)
A & C  interchange  Total cost = 99
C & D  interchange  Total cost = 1.040
B & D  interchange  Total cost = 945  estimated total cost 
B & C  tidak dapat dipertukarkan

33. CRAFT - Example 2
Yang dipilih adalah pertukaran B & D  menghasilkan layout baru dg department centroid, sesuai dengan luas yang diinginkan pada layout awal
(XA, YA) = (25, 30)
(XB, YB) = (55, 10)
(XC, YC) = (20, 10)
(XD, YD) = (67.5, 25)

34. CRAFT - Example 2 To From A B C D 50 25 47.5 35 27.5 62.5 20’ 40’ 60’
80’
30’
10’

35. Layout Evaluation – Distance-Based Scoring
Impact of aisle layout and direction of travel

36. Benefits and Problems Benefit Problems A Computer Program Flexibilty
Greedy Algorithm
Inefficient
End result may need to be modified
A greedy algorithm is one that always takes the best immediate solution
In a job shop the flow through the system is not always constant and causes fluctuations in the process
Once CRAFT has determined a solution it may need to be massaged to create a layout that fits in the plant

37. Summary
It is beneficial to use CRAFT but you should also realize that the program is not flawless.
The user must understand how the program works of the end product is not as efficient as you had hope.