Module 5 Computerized Layout Planning

Module 5
1/20/2021 Module 5, Rashmi S, Dept of IE&M 1
Rashmi S
Dept . of IE&M
JSS ATE-B

 Computerized Layout Planning:
 Computerized Relative Allocation of Facility
Techniques (CRAFT)
 Plant layout EvaluationTechniques(PLANET),
 Computerized Relationship Layout Planning
(CORELAP)
 Comparison of computerized layout
techniques
1/20/2021 Module 5, RS, Dept of IE&M 2

 In Computerized methods of Layout Design,
we have two types of Algorithms, i.e:
1. ConstructiveType Algorithm
2. ImprovementType Algorithm

 The key element of Computerized Layout
planning is the representation &
manipulation of the following three types of
information:
1. Numeric information: Space required for an
activity, total flow b/w two activities
2. Logic information: Preferences of the designer,
i.e., the activity relationship chart
3. Graphical information: Drawing of the block
plan

 Constructive Algorithms are of following
types:
1. Automated Layout Design Program (ALDEP)
2. Computerized Relationship Planning
(CORELAP)
 The most famous type in Improvement
types Algorithms is;
1. Computerized Relative Allocation of Facilities
Technique (CRAFT).

 CRAFT is more popular than the other computer
based layout procedures.
 It is improvement algorithm & starts with an
initial layout & proceeds to improve the layout
by interchanging the departments pair wise to
reduce the total material transportation cost
 It does not give the Optimal Layout; but the
results are good & near optimal, which can be
later corrected to suit the need of the layout
planner

1. It attempts to minimize transportation cost,
where Transportation cost = flow x distance x unit cost
2. It Requires assumptions that:
(1) move cost are independent of the
equipment utilization &
(2) move costs are linearly related to the
length of the move

3. Distance matrix used is the rectilinear
distance b/w department centroids
4. CRAFT being a path-oriented method, the
final layout is dependent on the initial layout.
Therefore, a number of initial layouts should
be used as input to the CRAFT
5. CRAFT allows the use of dummy
departments to represent fixed areas in the
layout

6. CRAFT input requirements are as follows:
1. Initial Layout
2. Flow Data
3. Cost per unit distance
4. Total number of departments
5. Fixed departments & their location
6. Area of departments

1. Determine the department centroids
2. Calculate rectilinear distance b/w centroids
3. Calculate transportation cost for the layout
4. Consider department exchanges of either
equal area departments or of departments
sharing common border
5. Determine transportation cost of each
department interchange

6. Select & implement the departmental
interchange that offers the greatest
reduction in transportation cost
7. Repeat the procedure for the new layout
until no interchange is able to reduce the
transportation cost

1. Because the basis is the cost of materials handling, only
production departments are considered.
2. An initial idea of the layout is required. Therefore the
technique only applies to the modification of an existing
layout or new layouts where the outline shape is known
3. The Distance b/w the departments is taken as straight lines
where as in practice movement is usually rectangular along
orthogonal lines

Example 1
Consider the following layout problem with unit cost matrix
(as in table1.2). Use CRAFT algorithm to obtain layout.The
initial layout is shown in table 1.1 & the flow matrix in table 1.3
7 7
7
7
7 7
Table 1.1. Initial Layout
Assume the unit cost perTransfer to be 1
A B
D C

Table: 1.2.Flow Matrix
Department A B C D
A 30 25 45
B 20 15 20
C 10 20 10
D 100 10 5

Solution:
1. Centroids of the department for given initial
layout are as:
(XA,YA) = 3.5, 10.5
(XB,YB) = 10.5, 10.5
(XC,YC) = 10.5, 3.5
(XD,YD) = 3.5, 3.5

Solution:
2. Using the Rectilinear Distance, we draw
the distance matrix as shown in table 1.3
Table. 1.3: Distance Matrix
Departmen
t
A B C D
A 0 7 14 7
B 7 0 7 14
C 14 7 0 7
D 7 14 7 0

Solution:
3. Total material handling cost is calculated as by:
Total cost = Flow x Distance x Unit cost
Fig.1.4.Total Cost Matrix
Department A B C D Cost
A 0 210 350 315 875
B 140 0 105 280 525
C 140 140 0 70 350
D 700 140 35 0 875
Total Cost 2625

Solution:
4. Departmental Interchanges:
 Consider various departmental interchanges
for improvement
 Departmental interchange is possible for
departments having common boundary or
equal area

Solution:
4. Departmental Interchanges:
 Possible Departmental Interchanges are shown in table 1.5
Table 1.5
Departmental pair Reason
A-B Common border & Equal area
A-C Equal area
A-D Common border & Equal area
B-C Common border & Equal area
B-D Equal area
C-D Common border & Equal area

Solution:
5. For the purpose of calculating material
handling cost, interchange would mean
change in the centroid. In the same way as we
calculated the total cost for the initial layout,
we calculate the total cost for each of the
possible interchanges, & select the layout
that gives the least total cost

 ALDEP is basically a construction algorithm,
but it can also be used to evaluate two
layouts
 It uses basic data on facilities & builds a
layout by successively placing the layout
using relationship information b/w the
departments

1. Length & width of facility
2. Area of each department
3. Minimum Closeness Preference (MCP)Value
4. Sweep width
5. Relationship chart showing the closeness
rating
6. Location & size of any restricted area

Procedure Adapted for using ALDEP :
Step#01: Input the following
1. Length & width of facility
2. Area of each department
3. MinimumCloseness Preference (MCP)Value
4. Sweep width
5. Relationship chart showing the closeness rating
6. Location & size of any restricted area

Step#2: One department is selected randomly &
placed in the layout
Step#3: In this step, the algorithm uses minimum
closeness required b/w departments for the
selection of departments to be placed with an
earlier placed department.
Select the department having maximum
closeness rating. If there is no department
having minimum closeness preference then any
dept that remains to be placed is selected

Step#4: If all the departments are placed in the
layout, go to step#5. else go to step#3
Step#5: Compute the total score of the layout
Step#6: If the total score required is
acceptable score, then go to step#7, else go
to step#2
Step#7: Print the current layout & the
corresponding score

Example 2: Develop a layout for the following
problem. Layout & area requirements are shown in
Table2.1
Table.2.1
Department Area (sq.ft) No. of unit squares
1 1200 30
2 800 20
3 600 15
4 1200 30
5 800 20
6 1200 30
7 1200 30
Total 7000 175

Solution: Assume One square in the layout to be equal 40 sq.ft.
No. of unit squares for dept = dept.area in sq.ft/area per square
Let the size of layout be 15x12, & the sweep width be 2 (this
means that we will fill 2 columns simultaneously).
The Relationship chart for the example is follows:
Department 1 2 3 4 5 6 7
1 E O I O U U
2 E U E I I U
3 O U U U O U
4 I E U I U U
5 O I U I A I
6 U I O U A E
7 U U U U I E

Solution: Let department 2 be selected. Number of unit squares
in dept 1 be 20. now 20 square units are filled in 15 x 12 grid as
shown in table2.2
 Since minimum closeness b/w departments required for
selection of department is I=4
 Scan the relationship chart randomly to find the departments
having closeness rating of 4 or greater with department 2

Solution:
 For the above case closeness rating for the pairs
(1-2)=16, (2-5)=4, & (2-6)=4
 Select any dept, say dept 1. Place dept 1 in the layout in
a serpentine pattern as shown in table 2.3
 Repeat the above procedure to get the final layout as
shown in table 2.4

Solution:
 After the final layout is obtained, the score is calculated.
 The score is the sum of the closeness ratings of all the
neighboring departments, see table 2.5
 From the above layout the score is 2 x110 = 220
 A further iteration should be carried out to check if a
better score can be achieved

1. The first department placed in the layout is the one with the
greatest TCR value. If there is a tie, then choose the one with
more A’s (E’s, etc.).
2. If a department has an X relationship with the first one, it is placed
last in the layout and not considered. If a tie exists, choose the one
with the smallest TCR value.
3. The second department is the one with an A (or E, I, etc.).
relationship with the first one. If a tie exists, choose the one with
the greatest TCR value.
CORELAP

4. If a department has an X relationship with the second
one, it is placed next-to-the-last or last in the layout. If
a tie exists, choose the one with the smallest TCR
value.
5. The next department is the one with an A (E, I, etc.)
relationship with the already placed departments. If a
tie exists, choose the one with the greatest TCR value.
6. The procedure continues until all departments have
been placed. ( Placement sequence)

https://users.encs.concordia.ca/~andrea/indu4
21/Presentation%2011%20(Layout%20IV).pdf

 Facilities Planning,Third Edition By JAMESA.Tompkins
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Hiregoudar, B Raghavendra Reddy

Module 5 Computerized Layout Planning

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