IND 405 PERANCANGAN TATA LETAK FASILITAS
Lecture 7: Computer Aided Aided Layout (Part 1)
References 2
Tompkins, et al. (2010), Facilities Planning , 4th ed., ed., John Wiley & Sons, Sections 6.4.4 – 6.4.8 Tompkins, et al. (2003), Facilities Planning , 3rd ed., ed., John Wiley & Sons, Appendix 6.A
Introduction 3
The use of a computer to facilitate the generation of facility layout. Cannot replace human judgement and experience. Generally do not capture the qualitative characteristics of a layout. Can significantly enchance the productivity of the layout planner and the quality of the final solution. Generate and numerically evaluate a large number of layout alternatives in a very short time.
Recap of algorithm classification 4
1. Type of input data 2. Objective function 3. Department shapes 4. Primary function
Algorithm classification (1) 5
A. Type of Input Data 1.
Qualitative ‘flow’ data (ARC)
2.
Quantitative flow matrix (From-to Chart)
3.
Both qualitative and quantitative data. Receiving (1) E
Milling (2)
O U
Press (3)
E U
Screw machine (4)
I
I
U U
U U
I E
Shipping (7)
I
U
Plating (6)
U
O
A
Stores
O
U
Assembly (5)
From/To
I
St
Mi
Tu
Pr
Pl
-
12
6
9
1
Milling
-
Turning
3
7 -
Press Plate
4
3
1
3 -
Algorithm classification (2) 6
B. Objective Function 1. Minimizing the sum of flows times distances
distance-based objective: m m
min z
f ij .c ij .d ij i 1 j 1
m
= number of derpartments
f ij
= flow from department i to department j (number of unit loads/unit time)
Cij
= cost of moving a unit load one distance unit from department i to department j
dij
= distance from department i to j
Algorithm classification (3) 7
B. Objective Function 2. Maximizing an adjacency score adjacency-based
objective: m m
max z
f ij .x ij i 1 j 1
f ij
= flow values (relationship values) between department i and department j
xij
= 1 if departments i and j are adjacent (share a border) = 0 otherwise
Algorithm classification (4) 8
Normalized adjacency score (efficiency rating): m m
f ij .x ij z
i 1 j 1 m m
f ij i 1 j 1
If there is “negative flow”, normalized adjacency score (efficiency rating):
z
(i, j)
F
f ij .x ij (i, j) F f ij .(1 - x ij )
(i, j)
F
f ij (i, j) F f ij
Where F and F represent the set of departement pairs with positive and negative flow values.
Algorithm classification (5) 9
C. Format used for layout presentation 1. Discrete representation 2. Continuous
representation more flexible than discrete, but more difficult to implement on a computer.
Split and unsplit departments 10
A layout algorithm should not split a department into two or more pieces.
U n s p l i t :
A “dot” that can move only from one grid to an adjacent grid.
Any grid assigned to department i must be reachable from any other such grid.
Algorithm classification (6) 11
D. Primary Function 1.
Improvement starts with an initial layout and seek to improve the objective function through ‘incremental’ changes in the layout.
2.
Construction develops a layout ‘from scratch’, either the building dimensions are assumed to be given or not.
CAL algorithms 12
CRAFT (Computerized Relative Allocation of Facilities Techniques)
MCRAFT (Micro –CRAFT)
BLOCPLAN
MIP (Mixed Integer Programming)
LOGIC (Layout Optimization with Guillotine Induced Cuts)
MULTIPLE (Multi-floor Plant Layout Evaluation)
CORELAP (Computerized Relationship Layout Planning)
ALDEP (Automated Layout Design Program)
Algorithm classification (7) 13
Classification
Techniques
Type of input data
1. 2. 3.
Quantitative: CRAFT – MCRAFT, LOGIC, MULTIPLE Qualitative: ALDEP, CORELAP Both: BLOCPLAN, MIP
Objective functions
1.
Minimizing the sum of flows times distances: CRAFT – MCRAFT, MIP, LOGIC, MULTIPLE Maximizing an adjacency score: ALDEP, CORELAP Both: BLOCPLAN
2. 3. Department shapes Primary function
1. 2.
Discrete representation: CRAFT – MCRAFT, ALDEP, MULTIPLE, CORELAP Continuous representation: BLOCPLAN, LOGIC
1. 2. 3.
Construction: CORELAP Improvement: CRAFT Both: BLOCPLAN, MCRAFT, LOGIC, MULTIPLE, ALDEP
CRAFT (Computerized Relative Allocation of Facilities Technique) 14
Input data: From-to Chart. Layout cost is measured objective function.
by
the
distance-based
Layout representation: discrete.
Shape of departments: not restricted to rectangular.
Type: improvement algorithm (initial layout is needed).
CRAFT – Dummy departments 15
CRAFT can have “dummy” ’ departments. Dummy departments have no flows or interaction with others, but requires a certain area specified by the layout planner.
Dummy departments may be used to: 1.
Fill building irregularities.
2.
Represent obstacles or unstable areas (fixed).
3.
Represent extra space in the facility.
4.
Aid in evaluating aisle locations in the final layout.
CRAFT – Exchangeable departments (1) 16
Can perform two-way or three-way exchanges.
Departments cannot be split.
CRAFT can only exchange those departments that are either adjacent (share a border) or equal in area.
CRAFT – Exchangeable departments (2) 17
Two departments with equal areas, whether adjacent or not, can always be exchanged without shifting the other departments. If two departments are not equal in area, then adjacency is necessary but sometimes not sufficient .
CRAFT algorithm procedure (1) 18
1.
Determine the centroid of each department in the initial layout.
2.
Calculate the rectilinear distance between pairs of department centroids and store the values in a distance matrix.
3.
Determine the initial layout cost by multiplying each entry in the from-to chart with the corresponding entries in the unit cost matrix (c ij) and the distance matrix.
CRAFT algorithm procedure (2) 19
4.
Consider all possible two-way or three-way department exchanges and identifies the best exchange that yields the largest reduction in the layout cost.
5.
Update the layout according to the best exchange and do the next iteration until no further reduction in layout cost can be obtained.
CRAFT example (1) 20
Department Name
FLOW
Number of Grids
A
B
C
D
E
F
G
H
12000
30
0
45
15
25
10
5
0
0
B: Milling
8000
20
0
0
0
30
25
15
0
0
C: Press m/c
6000
15
0
0
0
0
5
10
0
0
12000
30
0
20
0
0
35
0
0
0
8000
20
0
0
0
0
0
65
35
0
F: Plating
12000
30
0
5
0
0
25
0
65
0
G: Shipping
12000
30
0
0
0
0
0
0
0
0
H: Dummy
2000
5
0
0
0
0
0
0
0
0
A: Receiving
D: Screw E: Assembly
Area (ft2)
•
Dummy department H is generated because the total available space (72000 ft 2) exceeds the total required space (70000 ft 2).
•
The locations of Receiving (A) and Shipping (G) are assumed to be
CRAFT example (2) 21
1. Determining centroids from the initial layout.
CRAFT example (3) 22
2. Calculating rectilinear distance, ex. A to B = 6, A to C = 5.5, and so on. 3. Calculating layout cots, ex. A to B = 6 (45), add A to C = 5.5 (15), …, until H to G For initial layout: z = 2974 (equals to 59480 feet). 4. Considering all possible exchanges. 5. Identifying the best exchange 6. Updating the layout.
E
& F.
CRAFT example (4) 23
CRAFT final and massaged layout 24
CRAFT massaged layout 25
Once it is massaged, a layout cannot be generally reevaluate via computer-based layout algorithm, unless the grid size redefine and repeat the process using the new grid size.
CRAFT strenghts and weaknesses 26
(+)
Can have fixed location departments. Each department can be arranged to have rectangular shape (massaging). Can represent initial layout accurately.
(-)
Heuristic (steepest descent procedure)
Final layout is strongly influenced by the initial layout.
locally
optimal.
Exchange can occur if both departments are adjacent or equal in areas.
MCRAFT (Micro CRAFT) 27
Similar to CRAFT, but MCRAFT can exchange any two departments whether they are adjacent or not. MCRAFT can automocatically shift other departments when two unequal-area, non-adjacent departments are exchanged.
Using bands in its algorithms.
The number of bands is specified by the user.
MCRAFT algorithm procedure 28
Input:
From-to chart
The length and width of the building
The number of bands
Layout vector (fill sequence)
Divides the facility into bands, and the grids in each band are then assigned to one or more departments. Starting from the upper-lefthand corner of the building, sweep the bands in a serpentine pattern following a particular sequence of department numbers (layout vector / fill sequence).
MCRAFT example (1) 29
Department Name
Area
FLOW
(ft2)
A
B
C
D
E
F
G
H
12000
0
45
15
25
10
5
0
0
B: Milling
8000
0
0
0
30
25
15
0
0
C: Press
6000
0
0
0
0
5
10
0
0
D: Screw
12000
0
20
0
0
35
0
0
0
8000
0
0
0
0
0
65
35
0
F: Plating
12000
0
5
0
0
25
0
65
0
G: Shipping
12000
0
0
0
0
0
0
0
0
H: Dummy
2000
0
0
0
0
0
0
0
0
A: Receiving
E: Assembly
Building length = 360 ft; width = 200 ft.
Number of bands = 3
Layout vector = 1-7-5-3-2-4-8-6
MCRAFT – Initial layout 30
MCRAFT – Iterations and final layout 31
• Four iterations: exchanging C & E; then C & H; then C & D; and finally B & C. • Final sequence: 1-7-8-5-3-2-4-6
MCRAFT strenghts and weakness 32
(+)
Can exchange any two departments, whether they are adjacent or not.
(-)
Cannot represent the initial layout accurately, except the departments are already arranged in bands. All bands have the same width. Obstacles and fixed departments may change or shift.
CORELAP (Computerized Relationship Layout Planning) 33
Constructing a layout by calculating TCR (Total Closeness Rating ) for each department.
Input: ARC.
Objectives: maximize closeness rating.
Algorithm type: construction.
CORELAP algorithm (1) 34
1.
Calculates the total closeness rating (TCR) for each department. TCR is the sum of the numerical values assigned to the closeness relationships (A=6, E=5, I=4, O=3, U=2, X=1) between a department and all other departments.
2.
The department having the highest TCR is placed in the center of the layout. If there is a tie, apply the tiebreaking rule:
the department having the largest area
the department having the lowest department number.
CORELAP algorithm (2) 35
3.
Bring in the 2nd department that has an “A” relationship with the selected department. If none exists, find an “E” relationship, then “I” and so on. If a tie exists, choose the department with the highest TCR. If there is still a tie, use the tie-breaking rule.
4.
The 3rd department to enter is the department having an “A” relationship with the first dept. If there is a tie, use the TCR rule then the tie-breaking rule. If none exists, find the department having an “A” relationship with the 2nd dept. Then the procedure is repeated considering “E” relationship, then “I”.
CORELAP algorithm (3) 36
5.
Continue with the 4 th department and the rest until all departments have been selected to enter the layout. Each time a department enters the layout, a placement decision must be made based on its placing rating. Placing rating: the sum of the weighted closeness ratings between the department to enter the layout and its neighbors. For example, weights assigned to relationships for placing rating: A = 243, E = 81, I = 27, O = 9, U = 1.
CORELAP algorithm (4) 37
If a tie exists: the boundary lengths of the tied locations are compared (the number of unit square sides that the department to enter has in common with its neighbors).
CORELAP example (1) 38
TCR Calculation for Each Department Dept Name
Receiving (1) E
Milling (2)
O U
Press (3)
I E
U
Screw machine (4)
I U
I
Assembly (5)
I
U
Plating (6)
U
O
A
U U
U
U
Relationships
TCR
A: Receiving
1
E, O, I, O, U, U
19
B: Milling
2
E, U, E, I, I, U
22
C: Press
3
O, U, U, U, U, U
14
D: Screw
4
I, E, U, I, U, U
19
E: Assembly
5
O, I, U, I, A, I
23
F: Plating
6
U, I, O, U, A, E
22
G: Shipping
7
U, U, U, U, I, E
17
I E
Shipping (7)
O
Dept Number
Dept Name A: Receiving
Area Number of Unit (ft2) Area Templates 12000
6
B: Milling
8000
4
C: Press
6000
3
D: Screw
12000
6
8000
4
12000
6
E: Assembly F: Plating
CORELAP example (2) 39
1st department to enter: highest TCR
Dept 5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2nd : “A” with dept 5 Dept 6 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 16 0 0 0
0 16 15 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
Relationship Weight A 243 E 81 I 27 O 9 U 1
Placing Rating = …
CORELAP example (3) 40
A with 5 none A with 6 none E with 5
none
E with 6
Dept
Placing rating?
7
0 0 0 0 0
0 0 0 0 0
0 0 17 0 0
0 16 17 0 0
0 16 15 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
0 12 17 0 0
0 16 17 0 0
0 16 15 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
3rd: Dept 7
I with 5
2
&4
Highest TCR 4th: Dept 2
2
Placing rating?
CORELAP example (4) 41
Placing rating?
0 0 0 0 0
0 0 0 0 0
Placing rating?
0 0 0 0 0
14 14 0 0 0
E with 2 1 & 4 Highest TCR
same
Largest area
same
Lowest number 1 5th: Dept 1
E with 2
4
6th: Dept 4
11 12 17 0 0
11 16 17 0 0
11 12 17 0 0
11 16 17 0 0
0 16 15 0 0 0 16 15 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
0 0 0 0 0
CORELAP example (5) 42
7th: Dept 3
0 0 0 0 0
14 14 0 0 0
11 12 17 0 0
11 16 17 0 0
13 16 15 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
Evaluating the Layout Calculate the layout score: numerical Layout score
all departments
length of
closeness X shortest rating
path
CORELAP Layout Score
43
Relationship Value From To Distance Value x Distance A 6 15 16 0 0 E 5 11 12 0 0 E 5 12 14 0 0 E 5 16 17 0 0 I 4 11 14 0 0 I 4 12 15 2 8 I 4 12 16 0 0 I 4 14 15 3 12 I 4 15 17 0 0 O 3 11 13 0 0 O 3 13 16 0 0 O 3 11 15 2 6 U 2 11 16 0 0 U 2 11 17 1 2 U 2 12 13 2 4 U 2 12 17 0 0 U 2 13 14 2 4 U 2 13 15 1 2 U 2 13 17 2 4 U 2 14 16 1 2 U 2 14 17 1 2
CORELAP weaknesses 44
CORELAP-generated layouts often result in irregular building shapes and will need manual adjustment. The shortest rectilinear path between departments may not always be a realistic measure.
Excercise 45
Using CRAFT (two-way exchange), improve the initial layout and determine the final layout. Assume that department A is fixed.
