rule based layout planning and its multiple...

Proceedings of the 2014 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland

University, Rochester, MI 48309, USA Copyright © 2014, ISPE and ISAM USA.

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Rule Based Layout Planning and Its Multiple Objectives

Mohammad Komaki, Shaya Sheikh , Behnam Malakooti

Case Western Reserve University

Systems Engineering

Email: [email protected]

Abstract In this paper, we present a novel constructive approach for solving facility layout problems

considering objectives of total flow and qualitative closeness. This approach, called Head-Tail,

utilizes a set of rules to solve single and multiple row layout problems. We apply pair-wise exchange

method to improve the partially constructed layouts. The computational efficiency of Head-Tail

approach is examined by comparing it with Systematic Layout Planning. The results show that

Head-Tail method can find efficient solution with promising quality. A bi-criteria numerical

example that minimizes total flow and total qualitative closeness is also presented.

Key words: Rule-based layout planning, multi-objective optimization, Head-Tail Method

1. Introduction Effective layout designs are necessary in many different businesses such as manufacturing

facilities, and service industries. Diminishing physical space availability, high cost of re-layouts,

safety issues, and substantial long-term investments are good reasons for getting the design right

the first time. Different criteria can be considered in facility layout design including operational

costs, installation costs, utilization of space and equipment, work-force management, materials

handling, inventory work in process, inventory of incoming materials and inventory of outgoing

final products, ease of future expansion, and safety factors. Most of these objectives are difficult to

quantify. In this paper, new methods are introduced for solving multiple objective layout problems.

Researchers have used many criteria to come up with the optimal layout design. Raman et al. [1]

developed three factors of flexibility, productive area utilization and closeness gap to measure the

layout effectiveness. Rosenblatt [2] discussed a facility layout problem with two criteria of

minimizing costs and maximizing closeness rating. Fortenberry and Cox [3] discussed a multi-

criteria facility layout problem that combines quantitative with qualitative criteria and minimizes

total weighted work-flow. Malakooti [4,5,6] presented heuristic methods for solving multi-

objective facility problems.. Malakooti and Tsurushima [7] proposed an expert system for solving

multi-crtieria manufacturing problems. Islier [8] developed a genetic algorithm for solving multi

objective facility layout by minimizing transportation load, maximizing compactness of

departments and minimizing the difference between requested and available area. Pelinescu and

Wang [9] presented a hierarchical multi criteria optimal design of fixture layout. Yang and Kuo

[10] proposed a hierarchical AHP and data envelopment analysis approach for solving layout

design problem. Chen and Sha [11,12] developed a five-phase heuristic method using paired

comparison matrix for solving multi-objective facility layout problem. Sahin and Ramazan [13] and

Singh & Singh [14] have used heuristic method for solving multi-criteria layout problems. In this

paper, we develop a new rule based constructive approach called Head-Tail method. We use this

method for solving bi-criteria layout optimization problems.



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2. Head-Tail Layout Head-Tail approach starts with the pair of departments that have the highest flow as the initial

partial layout. More departments are progressively added to the existing partial layout until all

departments are assigned. The problem is to find the order of departments that minimizes the total

flow as measured in Equation (1). The selection of the location of each department is based on

qualitative or quantitative rules and other pertinent information. We call sp = (i j) as a candidate

pair of departments and sq = ( …) as a partially constructed sequence.

Generic Steps and Rules of Head-Tail Method

1. Consider the ranking of all pairs (i, j). Set the highest ranked pair as sq.

2. Consider the highest ranked unassigned pair, sp = (i , j), such that a. Both i and j are not in sq

b. Either i or j is in sq.

3. Use relevant Head-Tail Rules to combine sp and sq.

4. Repeat steps 2 and 3 until all stations are assigned to sq.

5. Try to improve upon the obtained solution by use of the Pair-Wise Exchange method.

Temporarily Discarding Unassigned Pairs: If a pair sp = (i,j) does not have any common

element with sq = (…...…), then sp = (i,j) is discarded temporarily only in the current iteration, but it

is reconsidered in the next iteration.

Permanently Discarding Assigned Pairs: If a pair sp = (i, j) has two common elements with sq =

(… i.., j…), then sp = (i,j) is discarded permanently (because both i and j are already assigned).

Consider a distance matrix, D, consisting of elements dij where dij is the distance between pairs of

departments i and j. Also consider a flow matrix, W, where each element (wij) represents the flow

from department i to department j. For every pair of departments (i, j) where i<j, create a new wij

equal to (wij + wji). These values are used in calculating the total flow. For a problem with n

departments, there will be n locations. The distances can be presented by a symmetric flow matrix;

that is, the distance from department i to department j, dij, is equal to the distance from department j

to location i, dji. The total flow for a given sequence solution is:

TF =n n

ij iji=1 j=1

w d (1)

Rules for Single Row Layout Order all pairs (i, j) in descending order of wij in a flow Matrix. Label the ordered vectors of wij by

wo. Consider sp = (i j), sq = ( …), and wo = Vector of wij in descending order of wij.

Conflict Resolution Rule: If there is a common department in sequences sp and sq, i.e. sp = (ij)

and sq = (…n i m…), then place the new department, j, next to the common department, i, in

sq on the side that has a higher ranking to its adjacent department, i.e. compare jn and jm flows

and insert j on the side that has higher flow.

Linear Mirror Property: The mirror image of sequence sq, i.e. its reverse sequence, is equivalent

to sq. Both sequences will have the same total flow. In the following, we show how to solve a

single-row layout using Head-Tail.

Example 1: Single-Row Layout Consider the information given below. Suppose the layout should be in the form of a 1x6

rectangle.



3

Distance D

1 2 3 4 5 6

Locations

1 - 1 2 3 4 5

1 2 3 4 5 6

2

- 1 2 3 4

3

- 1 2 3

4

- 1 2

5

- 1

6 -

Flows V

Ranking of Flows V

A B C D E F

A B C D E F

A - 10 20 30 10 100

A - 11 8 7 11 1

B

- 50 60 10 20

B

- 5 3 11 8

C

- 10 20 80

C

- 11 8 2

D

- 50 60

D

- 5 3

E

- 10

E

- 11

F -

F -

Figure 1: Initial Data for Example 1

1. sp = (A↔F); then set sq = (A↔F)

2. sp = (C↔F); The mirror image is sp = (F↔C). Therefore, sq = (A↔F↔C).

3. Discard (B↔D) temporarily as it does not have a common element with sq.

4. sp = (D↔F); there is conflict for location of D, use the Conflict Resolution Rule to find the

location for D. This is shown by the following graphical representation. A F C

? ? D

Since the flow of (D,A) = 30 is more than the flow of (D,C) = 10, locate D between A and

F; therefore, sq = (A↔D↔F↔C). Reconsider (B↔D), the location of B should be decided. (A D F C)

? ? B

The flow of (B, F) = 20 is more than the flow of (B,A) = 10; locate B on the right side of D.

Therefore, sq = (A↔D↔B↔F↔C).

5. Consider (B↔C), discard it because it is a duplicate of elements in the sequence.

6. Consider (D↔E), the location of E should be determined. (A D B F C)

? ? E

The flow of (E,A) = 10 is equal to the flow of (E,B) = 10, therefore E can be put on either side

of D; arbitrarily, locate it on the right side of D. Therefore, sq = (A↔D↔E↔B↔F↔C).

The final layout is (A↔D↔E↔B↔F↔C). Using the above given distances and Equation 1, the

total flow of this layout is Total Flow = (10*3) + (20*5) + (30*1) + (10*2) + (100*4) + (50*2) + (60*2)

+ (10*1) + (20*1) + (10*4) + (20*3) + (80*1) + (50*1) + (60*3) + (10*2) = 1260

Rules for Multi-Row Layout We can extend the single row Head-Tail method to solve multiple row layout problems. The single

row approach expands horizontally. The multi-row layout approach uses all rules of the single row



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approach, but allows both horizontal and vertical expansions.

The following rules show how to progressively augment sq by combining existing sq = (… i

…), the partially built layout, with a new sp = (ij); i.e. find a location for department j. By use

of the following rules, all departments will be assigned and a complete solution will be obtained.

Expanding Phase Rule 1 (Attaching Sequences, Horizontal and Vertical) a) Horizontal: If there is one common department in the pair of sequences sp & sq, such that the

common department is either at the head or the tail of sequence sp and sq, then form a new sequence

by attaching sp to sq by their common department.

For example, given sq = (CBE) and sp = (AC), new sequence is sq = (ACBE).

b) Vertical: A sequence can be attached in the middle of an existing sequence at a 90° angle if the

new sequence has one common department with the existing sequence.

For example, adding the sequence (BF) to the sequence (ABCD) in Figure 2a results in

the sequence in Figure 2b.

Rule 2 (Attaching Sequences, Diagonal): If there is a common department in sequences sp and sq,

e.g. sp = (ij) and sq = (… i …), and Rule 1 fails, i.e. department j cannot be placed next to i

in sq, then insert it at a 45° angle on the side which is available and has the highest ranking priority

with respect to its adjacent departments.

Rule 3 (Extending): Keep the existing sequence as open (extended) as possible by stretching all

connected lines as straight as possible.

That is, add the new sequence to the old sequence in the form of a straight line if possible. For

example, adding (EF) to the sequence in Figure 2b results in the sequence in Figure 2c.

Reshaping Phase Rule 4 (Rotation): To fit the partial constructed layout tree into the given layout locations, if

possible, use vertical, horizontal, or diagonal rotation of branches while satisfying closeness

preferences. In Example 2, in Figure 5c, E is rotated around D resulting in Figure 5d to allow B to

fold down to C in Figure 5e.

Rule 5 (Folding): To fit the partially constructed layout tree into the given layout locations, if

possible, bend the branches of the layout tree while satisfying closeness preferences.

For example, suppose the given layout is 2x3 where the given constructed sequence is shown in

Figure 2c. Consider folding branch B-F-E. It can be folded to the right or left. Because the ranking

of (E,C) is higher than (E,A), fold the branch to the right connecting E to C as shown in Figure 2d.

Figure 2d uses the folding rule to fit into the given layout space. To do this, bend A and D upwards

and connect them as shown in Figure 2e. This is the final layout which now fits in the 2x3 space.

A B C D

A B C

F

D

A B C

F

D

E

A B C

F

D

E

B C

F E

A D

Figure 2a: Expanding

Figure 2b: Expanding

Figure 2c: Expanding

Figure 2d:

Folding Figure 2e:

Folding

Fitting Phase Rule 6 (Conflict Resolution for Multi-Row): If the generated solution by using the reshaping

rules does not fit into the given layout space, cut low priority (ranking) departments and paste them



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into the closest empty spaces based on closeness preferences.

Suppose that Figure 3a shows the sequence obtained from Phase I where the layout space is 2x3. In

Phase II folding results in Figure 3b. The sequence given in Figure 3b cannot be folded to fit the

layout’s shape of 2x3. Suppose that the departments (A,B) have the lowest flow of all attached

departments in the layout; then A is the best choice to be cut. Apply rule 6 by cutting department A

and pasting it to the open space. The result is shown in Figure 3c. A B C

F

D E

A B C

F

D

E

A

B C

F

D

E Figure 3a:Given Solution Figure 3b:Folding Figure 3c:Cut and Paste A

Example 2: Head-Tail Multi-Row Layout Consider the information given below. Suppose the layout should be in the form of a 2x3 rectangle.

Distance D

1 2 3 4 5 6

Locations

1 - 1 2 1 2 3

1 2 3

2

- 1 2 1 2

4 5 6

3

- 3 2 1

4

- 1 2

5

- 1

6 -

Flows V

Ranking of Flows V

A B C D E F

A B C D E F

A - 10 20 30 10 100

A - 11 8 7 11 1

B

- 50 60 10 20

B

- 5 3 11 8

C

- 10 20 80

C

- 11 8 2

D

- 50 60

D

- 5 3

E

- 10

E

- 11

F -

F -

Figure 4: Initial Data for Example 2

Expanding Steps

1. sp = (AF). sq = (AF) 2.sp = (CF) or sp = (FC). sq = (AFC)

3. sp = (BD). Ignore sp temporarily. 4. sp = (DF). sp can be attached to sq, see Figure 5a.

5. sp = (BD). B can be attached to D, see Figure 5b.

6. sp = (DE). E can be attached to D. Since (E,C) has a higher flow than (E,A), E is attached to

the right side of D. See Figure 5c

All departments are assigned; reshape to fit the layout to the 2x3 block requirement.

Reshaping Steps

7. Consider Figure 5c. (E,C) has a flow of 20 compared to a flow of 10 for (E,A). (B,C) has a

flow of 50 compared to a flow of 10 for (B,A). Therefore, the advantage of having B next to C

(an improvement of 40) outweighs the advantage from having E next to C (an improvement of

10). Therefore, in Figure 5c, E can be rotated from left to the right side. This rotation will allow

B to fold towards (closer) to C instead of A. The result of this rotation rule is shown in Figure

5d.

8. In Figure 5e, B is folded down toward C.

The final solution layout is shown in Figure 5f. The total flow for this solution is Total Flow =



6

700.

A F C

D

A F C

D

B

A F C

D

B

E

Figure 5a: Expanding Figure 5b: Expanding Figure 5c: Expanding

A F C

D

B

E

A F C

D

B

E

A F C

D B E

Figure 5d: Rotation of E Figure 5e: Folding Figure 5f: Solution

Example 3: Comparing Head-Tail Layout and SLP Consider a facility consisting of seven departments, each with individual space requirements as

shown in the table below. Each department can be broken up into units or blocks of 100 square

feet. The total allotted area for the new facility is 2400 square feet and must be laid within a

rectangular space with a dimension of 4x6 blocks. Each department can fit into a square or a

rectangle (e.g. a department can be presented by a 1x6 or 2x3 block shape). Department 1 2 3 4 5 6 7 Sum

Square feet 600 200 400 200 500 100 400 2400

No. of Blocks 6 2 4 2 5 1 4 24

The matrix below shows the given desired proximity or closeness for each pair of departments. We

first use SLP to find a layout solution to this problem for a. 400x600 facility, and b. 200x1200

facility.

1 2 3 4 5 6 7

1 A O A I A A

2 U E O E X

3 X U U E

4 U O O

5 I I

6 X

7

Figure 6 show the initial solution where each department is presented by a node, and the desired

closeness for each pair of departments is presented by lines that connect the two departments.

Figure 7 shows the first iteration of the rearrangement of the relationship diagram to minimize the

strain on these bands.

1

3

2 7

5

6

4 Figure 6: Initial graph to present the problem Figure 7: First iteration to improve the total closeness

Figure 8 shows the next iteration. Re-arrangement of the network should be repeated until no more

34

7

6 5

2

13

4

7

6 5

2

13

4

7

6 5

2

13

4

7

6 5

2

1



7

improvement can be achieved. Once the final graphical relationship is achieved, the space

requirements for each department are considered. Figure 9 shows the area of each department.

2

5

4

6

3

7

1

1

1

1 1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

6

6

6

2

2

2

2

2

2 3

3

3 3

3

3

3

3

3

3

3

3

4

4

4 4

4

4

7

7

7

7

7

7

7

7

7

7

7

5

5 5

5

5

5

5 5

5

5

5

Figure 8: Further rearrangement to improve the total

closeness

Figure 9: Relationship diagram showing departmental

space requirements

Now we can collapse and rotate the spatial relationship diagram shown in Figure 9 to fit into the

4x6 space. This is shown in Figure 10. For the 2x12 block facility, the solution is shown in Figure

11.

2 2 1 1 4 4

5 6 1 1 7 7 5 5 5 4 2 1 1 1 7 7 3 3

5 5 1 1 7 7 5 5 6 4 2 1 1 1 7 7 3 3

5 5 3 3 3 3

Figure 10: Relationship diagram for a

4x6 area

Figure 11: Layout solution for a 2x12 area

The above solution generated by SLP is based on space requirements, availability, closeness

preferences, and other relative factors. The A, E, I, O, U, and X ratings can be converted to the

numerical values, 5, 4, 3, 2, 1, and 0, respectively. Then the ranking of the pairs of departments

will be performed. Note that pairs that tie have same ranking.

Rankings of Pairs (i,j)

1 2 3 4 5 6 7

1 - 1 11 1 8 1 1

2 - 15 5 11 5 X

3 - X 15 15 5

4 - 15 11 11

5 - 8 8

6 - X

7 -

The construction of the layout by the Head-Tail method is as follows.

Expanding Steps

Step 1: sp = (12). sq = (12) Step 2: sp = (14). sq = (412).

Steps 3, 4, and 5: Departments 6, 7, and 3 are added; see Figure 12.

Step 6: Next consider sp = (15). Department 5 can be located in four possible locations

diagonally from department 1 as shown in Figure 13, Step 6. Because pairs (5,2) and (5,6) have the

highest ranking, it will be located next to 2 and 6.

6 1

2

4

6 1

2

4

7

6 1

2

4

7 3

6 1

2

4

7 3

5



8

Step 3: Connect 6 to 1 Step 4: Connect 7 to 1 Step 5: Connect 3 to 7 Step 6:

Connect 5 to 6 and 2

Figure 12: Graphical Representation of layout tree Steps 3, 4, 5, and 6.

Since all departments are assigned, the expansion phase is complete.

Reshaping Steps

Step 7: Consider the next pair, (4↔6). Departments 4 and 6 cannot be folded. Consider the next

pair, (4↔7). Departments 4 and 7 cannot be folded. Consider the next pair, (2↔3). Fold 3 closer

to 2 (see Step 7 in Figure 13).

Fitting Steps

Step 8: Present the actual size and shape of the departments.

Step 9: Move the departments to fit in the layout size of 4x6 units.

6 1

2

4

7

3 5

5

5

5

6

5

5

1

1

1

1

1

1

2 2

4 4

3

3

3

3

7

7

7

7

5 5 2 2 7 3

5 6 1 1 7 3

5 4 1 1 7 3

5 4 1 1 7 3

Step 7: Connect 3 to 2 Step 8: Insert Spaces Step 9: Final Layout

Figure 13: Graphical Representation of Steps 7, 8, and 9

The final layout is shown in Step 9 of Figure 13. The center of each department is presented by a

“●”. The distances between each pair of departments are shown in the following table.

1 2 3 4 5 6 7

1 200 300 200 310 250 200

2 400 400 350 250 300

3 500 510 450 100

4 210 150 400

5 80 410

6 350

7

The total preferential closeness is: Total Closeness = n n

i=1 j=1

Closeness(i,j)*Distance(i,j) = 13,620

Comparison of Head-Tail and SLP Solutions: The above example was solved by SLP method.

The layout and the distances for the SLP solution are shown in Figure 14. The total closeness for

the SLP solution is 15,500. The Rule-based Planning solution (Example 3) was 13,620 which is

considerably better than the solution found by the SLP method.

1 2 3 4 5 6 7

1 300 300 300 330 150 250

2 600 400 230 150 550

3 400 390 450 250

4 630 450 150

5 180 480

6 400

7



9

2 2 1 1 4 4

5 6 1 1 7 7

5 5 1 1 7 7

5 5 3 3 3 3

Figure 14: SLP Layout Solution

3. BI-CRITERIA LAYOUT PLANNING In bi-criteria facility layout problem, the two objectives are:

Minimize Total Flow f1 = n n

ij iji=1 j=1

w d

Minimize Total Closeness f2 = n n

ij iji=1 j=1

w p

For example, suppose that the flow between departments A and B is very high and from the total

flow point of view should be located very close to each other. However, from a safety point of

view, they should be very far from each other. The above bi-criteria problem can be used to handle

such conflicting objectives.

Suppose that an additive utility function can be used to rank multi-criteria layout alternatives. Let

w1 and w2 be the weights of importance of the two objective functions, respectively. First find the

normalized values of the two objective functions. To normalize the flow information, find the

largest flow (dij,max) and smallest flow (dij,min), and then use the formula shown below to normalize

all flow values presented by dij: they will be between 0 and 1. Similarly, closeness information can

be normalized between 0 and 1 using the following formula where pij,max is the largest and pij,min is

the smallest pij values.

vij’ = ij ij,min

ij,max ij,min

(w - w )

(w - w ) pij’ =

ij ij,min

ij,max ij,min

(p -p )

(p -p )

Then for each pair of departments (i,j), calculate

zij = w1dij’ + w2pij’

The matrix of zij values is called the composite matrix, Z. The additive utility function (U) is

defined as the weighted sum of the flow and the closeness values using weights of importance of

w1 and w2 for the flow and closeness, respectively. In order to minimize the additive utility

function U = w1f1’ + w2f2

’,

Minimize U = n n

ij 1 ij 2 iji-1 j=1

d (w d ' + w p ') =n n

ij iji=1 j=1

d z (2)

The value of zij = w1dij’ + w2pij’ is constant for i, j, and given weights. Therefore, they only need to

be calculated once for each pair of departments. Therefore, for the generated Z matrix, the bi-

criteria layout problem can be solved by Head-Tail method. The alternative with the smallest U

value is the best layout.

Example 4: Bi-Criteria Layout

Distances for the SLP Solution



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For a four department facility layout problem presented in Figure 14, find the best bi-criteria layout

solution. Suppose that the decision maker’s weights of importance for flow and closeness

objectives are w1 = 0.3 and w2 = 0.7, respectively. Consider using 0 to 5 numerical ratings for X to

A qualitative ratings. Location Flow Matrix

1 2 3 4 A B C D

A - 100 40 10

Assignment B - 60 30

A B C D C - 90

D -

Qualitative Matrix Converted Qualitative Matrix

A B C D A B C D

A - X/0 E/4 A/5 A - 0 4 5

B - U/1 E/4 B - 1 4

C - X/0 C - 0

D - D -

Figure 15: Initial data for the bi-criteria layout problem

First convert qualitative closeness data to quantitative data as presented in Figure 15. Then

normalize flow matrix and closeness matrix values as shown in Figure 16. Normalized Flow

Matrix

Normalized Qualitative

Matrix

(vij’) = (vij-10)/(100-10) (pij’) = (pij-0)/(5-0)

A B C D A B C D A B C D

A - 1 0.333 0 A - 0 0.8 1 A - 0.3 0.66 0.7

B - 0.556 0.222 B - 0.2 0.8 B - 0.307 0.627

C - 0.889 C - 0 C - 0.267

D - D - D -

Figure 16: Normalized flow and qualitative closeness values

A composite matrix, Z, can be generated by using zij = w1vij’ + w2pij’= 0.3vij’ + 0.7pij’.

Solving the Bi-Criteria Problem by the HEAD-TAIL Method: First rank all pairs in descending

order of zij values: they are (A,D), (A,C), (B,D), (B,C), (A,B), and (C,D).

The steps of HEAD-TAIL for assigning departments are:

1) (AD); 2) (CAD); 3) (CADB)

This is the final solution, and its mirror is (BDAC). The objective values for this layout are:

f1 = 2*100 + 1*40 + 1*10 + 3*60 + 1*30 + 2*90 = 640 and f2 =2*0+1*4+1*5+3*1+1*4+2*0 =16

No improved solution can be found by applying the Pairwise Exchange Method to the final solution

of this example.

Using the additive utility as the performance measure for each facility, the first iteration of the Pair-

Wise Exchange method produces the results shown below.

Coefficient Facility

Pairs

Initial

distance

Pair-Wise Exchange new distance

AB AC AD BC BD CD

0.3 AB 1 1 1 2 2 3 1

0.66 AC 2 1 2 1 1 2 3

0.7 AD 3 2 1 3 3 1 2



11

0.307 BC 1 2 1 1 1 1 2

0.627 BD 2 3 2 1 1 2 1

0.267 CD 1 1 3 2 2 1 1

U = ij iji j

zd 5.548 1722 4.682 4.828 4.828 4.748 1788

The lowest score is 4.682 for swapping A and C.

Improved Layout: C B A D

The swap is made and the initial layout for the second iteration is as follows:


pairs

Initial

distance


AB AC AD BC BD CD

0.3 AB 1 1 1 2 2 1 1

0.66 AC 2 1 2 3 1 2 1

0.7 AD 1 2 3 1 1 1 2

0.307 BC 1 2 1 1 1 3 2

0.627 BD 2 1 2 1 3 2 1

0.267 CD 3 3 1 2 2 1 3

U= ij iji j

zd 4.682 4.402 5.548 4.748 4.682 4.762 4.402

The lowest score is 4.402 for swapping A and B. The new layout is.

Improved Layout: C A B D

Next iteration is:


pairs

Initial

distance


AB AC AD BC BD CD

0.3 AB 1 1 2 1 1 2 1

0.66 AC 1 2 1 3 1 1 2

0.7 AD 2 1 3 2 2 1 1

0.307 BC 2 1 1 2 2 3 1

0.627 BD 1 2 1 1 3 1 2

0.267 CD 3 3 2 1 1 2 3

U= ij iji j

zd 4.402 4.682 4.828 1788 1722 4.042 4.682

The lowest score is 4.042 for swapping B and D. The new layout is shown below.

Improved Layout: C A D B

Continuing with the next iteration:


pairs

Initial

distance


AB AC AD BC BD CD

0.3 AB 2 2 3 1 1 1 2

0.66 AC 1 3 1 2 2 1 1

0.7 AD 1 1 2 1 1 2 1

0.307 BC 3 1 2 3 3 2 1

0.627 BD 1 1 1 2 2 1 3

0.267 CD 2 2 1 1 1 3 2

U= ij iji j

zd 4.042 4.748 4.468 4.762 4.762 4.402 4.682



12

No more improvement is possible, stop. The layout remains with f1 (total flow) = 640 and f2 (total

closeness) = 32. Final Layout: C A D B

Efficient Frontier By systematically varying weights (w1, w2) in Equation (2), a set of efficient bi-criteria alternatives

can be generated. To do this, first generate the Z matrix for each given set of weights and then use

head-Tail method to solve the problem. Note that for a given set of weights, when PWE method is

used, alternatives generated in different iterations of PWE method may also be efficient and should

be recorded. That is, each Pairwise Exchange solution in each iteration of PWE method can be

efficient. Consider the set of four weights given in the first row of the following table. For each

given set of weights, solve the problem by Head-Tail method. The generated solution is recorded

under each set of weights. The set of generated four alternatives are all efficient. (w1, w2) (0.99, 0.01) (0.6, 0.4) (0.5, 0.5) (0.01, 0.99)

Layout ABCD BACD DBAC CADB

Min. f1 420 460 580 640

Min. f2 32 28 20 16

Efficient? Yes Yes Yes Yes

Conclusion

In SLP, the process of improving the initial graph of the layout becomes very difficult as the

number of departments increases. Also the evaluation process in SLP is subjective and depends on

the layout designer. A different approach for solving layout problem is by use of Head-Tail Layout

method which is a powerful tool for solving both structured and ill-structured layout problems.

Head-Tail can solve very large layout problems with minimal computational efforts. Because this

method is Head-Tail, it can incorporate complicated constraints while developing the layout

solution. Furthermore, it can be used in conjunction with Expert systems approaches to solve

layout problems.

References: [1] Raman D., S. V. Nagalingam, G. C.I. Lin, “Towards measuring the effectiveness of a facilities layout”, Robotics

and Computer-Integrated Manufacturing, doi:10.1016/j.rcim.2007.06.003, 2008.

[2] Rosenblatt, M. J., "The facility layout problem: a multi-goal approach." International Journal of Production

Research, 17, 1979, pp. 323-332.

[3] Fortenberry, J. C., J. F. Cox, “Multiple criteria approach to the facilities layout problem”, International Journal of

Production Research, Vol. 23, Issue 4, 1985, pp.773-782.

[4] Malakooti, B., "Multi-Objective Facility Layout: A Heuristic Method to Generate All Efficient Alternatives",

International Journal of Production Research, Vol. 27, No. 7, 1989, pp. 1225-1238.

[5] Malakooti, B., “Computer Aided Facility Layout Selection with Applications to Multiple Criteria Manufacturing

Planning Problems.” Large Scale Systems: Theory and Applications, Special Issue on Complex Systems Issues in

Manufacturing, Vol. 12, 1987, pp. 109-123.

[6] Malakooti, B., A. Tsurushima, "An Expert System Using Priorities for Solving Multiple Criteria Facility Layout

Problems." Inter. Journal of Production Research, Vol. 27, No. 5, 1989, pp. 793-808.

[7] Yang, Z., B. Malakooti, “A Multiple Criteria Neural Network Approach for Layout of Machine -Part Group

Formation in Cellular Manufacturing.” Third IE Research Conference, 1994, pp. 249-254.

[8] Islier A. A., “A genetic algorithm approach for multiple criteria facility layout design”, International Journal of

Production Research, 1998, Vol. 36, No. 6, pp. 1549-1569.

[9] Pelinescu D. M., M. Y. Wang, "Multi-objective optimal fixture layout design", Robotics and Computer-Integrated

Manufacturing, Vol. 18, 2002, pp. 365-372.

http://www.sciencedirect.com/science/journal/07365845

http://www.sciencedirect.com/science/journal/07365845

http://dx.doi.org/10.1016/j.rcim.2007.06.003

http://vorlon.cwru.edu/~bxm4/papers/21-1989.pdf





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[10] Yang T., Ch. Kuo, “A hierarchical AHP/DEA methodology for the facilities layout design problem”, European

Journal of Operational Research, 147, 2003, pp. 128–136.

[11] Chen C.-W., D. Y. Sha, “Heuristic approach for solving the multi-objective facility layout problem”, International

Journal of Production Research, Vol. 43, No. 21, 1, November 2005, pp. 4493–4507.

[12] Sha, D., C. Chen, "A New Approach to the Multiple Objective Facility Layout Problem", Integrated

Manufacturing Systems, Vol. 12, No.1, 2001, pp. 59-66.

[13] Şahin, Ramazan. “A Simulated Annealing Algorithm for Solving the Bi-objective Facility Layout Problem.”

Expert Systems with Applications, Vol. 38, Issue 4, 2011, pp. 4460-4465.

[14] Singh S. P., V. K. Singh, “Three-Level AHP-Based Heuristic Approach for a Multi-Objective Facility Layout

Problem.” International Journal of Production Research, Vol. 49, Issue 4, 2011, pp. 1105-1125.

http://www.informaworld.com/smpp/title~db=all~content=t713696255

http://www.tandfonline.com/loi/tprs20?open=49#vol_49

http://www.tandfonline.com/toc/tprs20/49/4

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