Types of layout Algorithms
Construction algorithms
Improvement algorithms
Graph-based method
Pairwise exchange method
ALDEP
CRAFT
CORELAP
MCCRAFT
PLANET
MULTIPLE
1
AUTOMATED LAYOUT DESIGN PROGRAM (ALDEP)
• 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 between the departments
• Data requirements
• Area of Departments
• REL Chart with A,E,I,O,U,X rankings
• Sweep width
• Minimum Department Preferences (MDP)
• Ranking values:
• A
= 64,
Absolutely necessary
• E
= 16,
Especially important
• I
= 4,
Important
• O
= 1,
Ordinarily important
• U
= 0,
Unimportant
• X
= -1024,
Undesirable
2
ALDEP Facility Representation
Discrete Representation
Facility Width
Facility Length
Sweep Method
Facility Width
Facility Length
ALDEP Facility Representation
Facility Width
Facility Length
A
A
A
A
A
A
Dept. size = 8 squares
Facility width = 6 squares
Sweep width = 1 square
A
A
Facility Width
Facility Length
C
C
C
C
C
C
C
C
C
C
C
C
Dept. size = 14 squares
Facility width = 6 squares
Sweep width = 2 squares
C
C
Example
Dept.
Area(sq.m)
1
12,000
2
8000
3
6000
4
12,000
5
8000
6
12,000
7
12,000
A Timber Processing facility need to be located in a plot who’s Length is 360 meters
Width is 200 meters. Create a layout using ALDEP to get a optimal facility Layout
with sweep value is 2 & MDP=> E(16)
Example
Dept.
Area
Number of Squares
1
12,000
30
2
8000
20
3
6000
15
4
12,000
30
5
8000
20
6
12,000
30
7
12,000
30
From the above table we try to scale it down by considering common
area value of 400 sq.m. We get number of squares.
Departments 1,4,6,7 have same number of squares. Select any one
Randomly
Dept.
Area
Number of Squares
1
12,000
30
2
8000
20
3
6000
15
4
12,000
30
5
8000
20
6
12,000
30
7
12,000
30
Length of layout matrix = 360/20=18
Facility Length=18
Facility Width=10
width of layout matrix
200/20=10
• One square is 400 sq.m
• Length of square =20m
• Width of square= 20m
• Length of plot is 360m
• Width of plot is 200m
175
• We randomly select Dept. 4 as first dept. in the layout matrix.
• If a dept. is said to be fixed then start with that dept. or office.
• Now sweep Value is 2. So fill the layout matrix accordingly.
• Now compare Dept. 4 with our dept. to find which is nearer.
• Dept. 4
->1,2,3,5,6,7
• Dept. Rank ->I,E,U,I,U,U
So E (Dept. 2) is selected
• Now dept. 2 is entered then its now compared with other Depts.
• To find which is nearer to the Dept. 2
• Now compare Dept. 2 with other Depts.
• Dept. 2 ->1,3,5,6,7
• Dept. Rank->E,U,I,I,U So E (Dept. 1) is selected
The sequence of selection is 4-2-1-6-5-7-3
Scoring the layout
Neighbouring
Departments
Rel Grade
Rel Value
4-2
E
16
4-1
I
4
2-1
E
16
1-6
U
0
6-5
A
64
6-7
E
16
5-7
I
4
7-3
U
0
120
Total scoring = 120*2=240 Since both the ways we can have neighbour pairs
In class Exercise
Dept.
Area(sq.m)
1
12,000
2
8000
3
6000
4
12,000
5
8000
6
12,000
7
12,000
A Timber Processing facility need to be located in a plot who’s Length is 360 meters
Width is 200 meters. Create a layout using ALDEP with Dept. 2 is fixed to get a
optimal facility Layout with sweep value is 2 & MDP=> I(4)
In class Exercise-Solution
Steps
Depts.
Area(sq.m)
1
2
2
1
Fixed
“E” with 2
3
4
5
6
7
4
5
6
7
3
“I” with 1
“I” with 1
“A” with 5
“E” with 6
Remaining
Total scoring = 104*2=208 Since both the ways we can have neighbour pairs
CORELAP
CORELAP
• Computerized relationship layout planning.
• Computer algorithm was Developed by R.C.Lee.
• Scale of layout is limited to a maximum dimension 40 X 40.
• Input requirements of CORELAP:
•
•
•
no. of departments and their area
closeness relationship as given by REL chart.
weighted ratings for REL chart entries.
• Optional input formation :
•
•
•
scale of output
building length to width ratio
department preassigned.
Problem:
• Create a layout using CORELAP.
Step 1: Area of Department and Scale
Number of Squares is calculated using scale information
If input for scale is not given corelap assumes 1 Squares
= 6000 Sq. ft. For the first run usually scale is to be
assumed by corelap. For improvement one can specify
the scale suitably.
Number of squares for each department is fixed by
dividing the area by scale value and result is kept as
integer.
Care Should be taken to see that the total number of
squares multiplied by the scale value equals
approximately the total area.
Contd…
Step 2: Determination of Placement
order of departments
For the purpose of finding the order of placing
departments in the layout the following data is needed.
1. REL Chart
2. Closeness values for entries in the REL chart
Closeness values:
A-
6
EIOUXS–
5
4
3
2
1
Refuse to diagonal
Contd…
REL CHART:
1 S
2 E
U
3 O
U
S
4 I
E
U
S
5 O
I
U
I
S
6 U
I
O
U
A
S
7 U
U
U
U
I
E
S
1
2
3
4
5
6
7
Contd…
• Using the REL Chart information total closeness
rating is computed for all the departments

Total closeness rating (TCR) is the sum of the numerical values
assigned to the closeness relationship between a department
and all other department.
Contd…
Contd…
• The following steps are carried out to select the
placement order of departments
• Step 1: Department with largest TCR is selected for
first placement. In case of ties the department with
the largest area is selected. For the example,
department 5 is selected. Dept 5 has the largest TCR
value 23
Contd…
Step 2:
Review the closeness relationship of dept.5 with
other departments to find the department that has
high closeness relationship with dept5.
closeness of dept. 5 with :
1, 2, 3, 4, 6, 7
O, I, U, I, A, I
Dept 6 has the highest closeness relationship with the
first selected dept 5. Dept 6 is selected as second in
placement order.
Contd…
• Step 3: Scan the REL chart for the closeness values for the unsigned
departments with department 5.
closeness of dept. 5 with : 1, 2, 3, 4, 6, 7
O, I, U, I, A, I
None of the unassigned has an “A” closeness with dept 5
Check the closeness of unassigned department with dept 6 (next in the
placement order)
closeness of dept. 6 with :1, 2, 3, 4, 7
U, I, O, U, E
None of the unassigned has an “A” closeness with dept 6
Check whether any of the unassigned dept has an “E” relationship with the
departments already included in the placement order. None of the
unassigned has “E” closeness with dept 5. Dept 7 has an “E” closeness with
the dept 6. Dept 7 is included third in the order.
Contd…
Step 4: Repeat the procedure of step 3 with the
unassigned departments.
1, 2, 3, 4
With Dept 5 :
O, I, U, I
Dept 6:
U, I, O, U
Dept 7:
U, U, U, U
Dept 2 and 4 have an “I” closeness with dept 5. In the case
of ties, select based on highest TCR. Dept 2 (TCR=22) is
selected and included forth in the placement order
 Repeat this until all departments are included.
Placement order: 5,6,7,2,1,4,3
Step 3: Placement of department in
the layout
Step 1: the first department in the
placement order (dept 5) is placed at the
centre of the (40 X 40) layout.
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Contd…
• Step 2: the second department from the list is placed
adjacent to the first department.
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
6
0
0
0
0
0
6
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Contd…
• Step 3: Department 7 is to be placed next in the
layout. It can be seen that there are many
arrangements in which dept 7 can be placed as an
adjacent department to the layout.
• To aid evaluation of different placing alternatives
placing Rating (PR) is calculated.
• Placing rating is the sum of the weighted closeness
ratings between the entering department and its new
neighbours.
• CORELAP uses the following weighted closeness
rating values:
Contd…
• Dept 7 has 2 squares. The following placements are
possible
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
6
0
0
0
0
7
6
5
0
0
Dept 7 with 6 = E= 81
PR =81
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
• Option 2:
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
6
0
0
0
0
0
6
5
0
0
Dept 7 with 6 = E= 81
with 5= I=27
PR= 108
0
0
7
7
0
0
0
0
0
0
0
0
0
0
0
0
0
0
• Option 3:
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
0
0
0
0
6
7
0
0
0
0
6
5
0
0
Dept 7 with 6 = E= 81
with 5= I=27
PR= 108
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Contd…
• Select the Option 3 ( random between 2 and 3 ).
Department 7 is placed as shown in figure below.
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
0
0
0
0
6
7
0
0
0
0
6
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Contd…
Step 4: Now Dept 2 is placed. Dept 2 has 1 square.
the best location is to place dept 2 adjacent to two
depts (dept 6 and dept 7) so that the PR can be
maximized (PR =28). All other arrangement will have
PR less than 28.
0
0
0
0
0
0

0
0
0
0
0
0
0
0
2
7
0
0
0
0
6
7
0
0
0
0
6
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Repeat this procedure until all the departments are
selected
Final Arrangement
0
0
0
0
0
0
0
4
4
0
0
0
0
1
2
7
0
0
0
1
6
7
0
0
0
3
6
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Step 4: Total Score for the layout
Total score for a layout is the sum of the products
found by multiplying length of the shortest path
between all pairs of department times the numerical
closeness for the pair
Total score = Summation over all pairs (closeness
rating X length of the shortest path)
Path is calculated by the rectilinear distance from
the border of one department to the border of
another department. Therefore if 2 departments are
adjacent they will have common border. The path
will be zero