Facility Design-Week 10 (cont)
Computerized Layout Planning
By
Anastasia L. Maukar
1
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.
2
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:
Moment

d f
ij
(i, j )
ij
where
d
f
ij
 distance
of the (i, j) entry from the diagonal
 the (i, j) entry in the matrix
ij
3
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.
4
Following are some
examples of
questions
addressed by
CRAFT:
• Is this a good
layout?
• If not, can it be
improved?
10 20 30 40 50 60 70 80 90 100
CRAFT
A
D
C
B
10 20 30 40 50 60 70 80 90 100
5
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
6
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
7
CRAFT: Distance Between Two
Departments
• Let
– Centroid of Department A =
xA , yA 
– Centroid of Department B =
 xB , yB 
• Then, the distance between departments A and B, Dist(A,B)
 xA  xB  y A  yB
• Ex: the distance between departments A and C is the rectilinear
distance between their centroids (30,75) and (80,35). Distance
(A,C)
 x A  x C  y A  y C  30  80  75  35  90
8
Centroid of A
= (30,75)
Centroid of C
= (80,35)
Distance (A,C)
= 90
10 20 30 40 50 60 70 80 90 100
CRAFT: Distance Between Two
Departments
A
D
(80,85)
C
B
(30,25)
9
10 20 30 40 50 60 70 80 90 100
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.
10
CRAFT: Total Distance Traveled
(a) Material handling trips(given)
To
From
A
A
B
C
D
2
7
4
5
7
B
3
C
6
7
D
7
7
3
B
C
D
50
90
60
60
110
3
(b)
(b) Distances (given)
To
From
A
A
B
50
C
90
60
D
60
110
50
50
11
To
CRAFT: Total
Distance Traveled
(a) Material handling
trips (given)
(b) Distances (given)
(c) Sample computation:
distance traveled (A,B)
= trips (A,B)  dist (A,B)
=………..
Total distance traveled
= 100+630+240+….
= 4640
From
A
A
B
C
D
2
7
4
5
7
B
3
C
6
7
D
7
7
3
A
B
C
50
90
60
To
From
A
(a)
3
D
60
1 1 0 (b)
B
50
C
90
60
D
60
110
50
A
B
100
C
630
D
240
300
7 7 0 (c)
To
From
A
B
150
C
540
420
D
420
770
50
150
150
12
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.
13
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.
14
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.
15
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.
16
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)
Interchange
• New dist (A,D) = Previous dist (A,C)
C,D
• New dist (B,C) = Previous dist (B,D)
• New dist (B,D) = Previous dist (A,C)
To
From
A
A
To
B
C
D
50
90
60
60
110
A
50
B
50
C
90
60
D
60
110
50
From
B
A
B
C
D
50
60
110
90
60
C
50
60
110
D
90
60
50
50
17
CRAFT: A Sample
Computation of
Savings
(a) Material handling trips
(given)
To
From
A
(c) Sample computation:
distance traveled (A,B)
= trips (A,B)  dist (A,B)
=
Total distance traveled
= 100+420+360+…
= 4480
Savings
=
B
C
D
2
7
4
5
7
B
3
C
6
7
D
7
7
3
A
B
C
D
50
60
110
90
60
To
(b) Distances (rearranged)
A
From
A
B
3
C
50
60
110
D
90
60
50
A
B
C
420
D
360
550
420
To
From
A
100
B
150
C
360
770
D
630
420
(a)
(b)
50
(c)
150
150
18
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.
19
• 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)?
10 20 30 40 50 60 70 80 90 100
CRAFT: Exact Coordinates of Centroids
D
A
C
B
10 20 30 40 50 60 70 80 90 100
20
Let
A1  Area A1 
A 2  Area A2 
 x1 , y 1 
 Centroid of A1

x2 , y2 
 Centroid of A1

50 60 70 80 90 100
CRAFT: Exact Coordinates of Centroids
A
A1
A2
10 20 30 40 50 60 70 80 90 100
Find the centroid of A
21
CRAFT: Exact Coordinates of Centroids
Rectangle
(1)
A1
A2
Total
Area
(2)
X-coordinate
Multiply
of centroid
(2) and (3)
(3)
(4)
X-coordinate of the centroid of A


A1 x1  A 2 x 2
A1  A 2
22
CRAFT: Exact Coordinates of Centroids
Rectangle
(1)
A1
A2
Total
Area
(2)
Y-coordinate
Multiply
of centroid
(2) and (3)
(3)
(4)
Y-coordinate of the centroid of A


A1 y 1  A 2 y 2
A1  A 2
23
50 60 70 80 90 100
CRAFT: Exact Coordinates of Centroids
A
A1
A2
10 20 30 40 50 60 70 80 90 100
Exact coordinate of area A is
24
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
25
Computerized Layout Planning
Graphical Representation
“Points and lines” representation is not convenient for
analysis
26
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
6
Max
s
w
i
Xi
i 1
• 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 27
between the adjacent departments instead of scores assigned to A,
1
Exampl
eReceiving
1
2
E
Milling
U
Press
3
E
U
O
I
Plating
E
1
2
U
3
4
I
Shipping
U
U
U
A
7
I
O
I
6
U
U
Assembly
5
I
U
Screw Machine
4
O
5
6
7
2
1
2
3
4
5
6
7
3
4
5
I
E
O
I
6
O
A
7
4+1
=5
U 16+4+0 =20
=1
U 1+0
---64
E 16
Total Score
U
3
7
U
2
E
=64
=16
106
4
Shipping
Milling
Screw
Machine
E
I
I
Press
O
6
Plating
A
5
Assembly
O
1
Receiving
28
1
2
Receiving
E
Milling
Press
Screw Machine
Assembly
U
3
4
O
5
I
E
U
U
O
I
U
Plating
1
2
U
Exercise: Find the score of the layout
shown below. Use A=8, E=4, I=2, O=1,
U=0 and X=-8.
3
4
I
E
Shipping
U
U
U
A
7
I
O
I
6
U
5
6
7
3
Press
7
Shipping
1
Receiving
6
Plating
2
Milling
4
Screw
Machine
5
Assembly
29
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
m 1
Min
s
m
 c
ij
D ij
i 1 j  i  1
• 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
30
CRAFT - Example 2
Initial Layout
Flow Data
Ma
F ro m /T o
A
B
C
D
A
-
2
4
4
B
1
-
1
3
C
2
1
-
2
D
4
1
0
-
Total Cost
Distance Data
F ro m /T o
A
B
C
D
F ro m /T o
A
B
C
D
A
-
80
A
-
40
25
55
B
40
-
65
75
180
B
40
-
65
25
C
50
65
-
80
195
C
25
65
-
40
D
220
25
0
-
245
D
55
25
40
-
T o ta l
100 220
T o ta l
400
310 170 165 375 1020
31
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
32
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)
33
CRAFT - Example 2
To
From
A
B
C
D
50
25
47.5
35
27.5
40’
A
30’
20’
10’
B
50
20’
C
25
35
D
47.5
27.5
40’
60’
80’
62.5
62.5
34
Layout Evaluation – Distance-Based
Scoring
– Distance-based scoring
– Impact of aisle layout and direction of travel
A
B
C
D
35
Benefits and Problems
• Benefit
– A Computer Program
– Flexibilty
• Problems
– Greedy Algorithm
– Inefficient
– End result may need to be modified
36
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.
37