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Layout design IV.Layout design IV.Ch 6Chapter 6

Layout generationCORELAPALDEPALDEPMULTIPLEMIPMIP

CORELAPCORELAP:: Computerized Relationship Layout PlanningComputerized Relationship Layout PlanningCORELAPCORELAP: : Computerized Relationship Layout PlanningComputerized Relationship Layout Planning

Developed for main frame computersAdjacency‐based method◦ CORELAP uses A=4, E=3, I=2, O=1, U=0 and X=‐1

lvalues

Selection of the departments to enter the layout is based on Total Closeness Ratinglayout is based on Total Closeness Rating◦ Total Closeness Rating (TCR) for a department is the sum of the numerical values assigned to the closeness relationships between the department and all other departments.

m

∑≠=

=m

jijijwTCR

,1

CORELAPCORELAPDepartment selectionDepartment selection

1 The first department placed in the layout is the one with the1. 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 l d l t i th l t If ti i t h th ithplaced last in the layout. If a tie exists, choose the one with

the smallest TCR value.3. The second department is the one with an A relationship with

the first one (or E, I, etc.). If a tie exists, choose the one with ( )the greatest TCR value.

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 valuechoose the one with the smallest TCR value.

5. The third department is the one with most A (E, I, etc.) relationships 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.

CORELAPDepartment placement

• Department neighbors 8 7 6Department neighborso Fully adjacent (in position 1, 3, 5 or 7) with department 0

o Touching (in position 2, 4, 6 or 8) department 0

5

432

1 0

• Placing rating (PR) is the sum of the weighted closeness ratings between the department to enter the layout and its neighbors.

placed}alreadysdepartment{where ==∑ kwPR k

• The placement of departments is based on the following steps:

1. The first department selected is placed in the middle.

placed}already sdepartment{ where∑ kwPRk

ik

p p

2. The placement of a department is determined by evaluating PR for all possible

locations around the current layout in counterclockwise order beginning at the

“ t d ”“western edge”.

3. The new department is located based on the greatest PR value.

CORELAP – Example 1

Given the relationship chart and the departmental dimensionsGiven the relationship chart and the departmental dimensions below determine the sequence of the placement of the departments in the layout based on the CORELAP algorithm. Place the departments in the layout while evaluating each placement.the departments in the layout while evaluating each placement.

Department Sizes Sq.ft. Num of Grids

1. Conf Room 100 2

2. President 200 4

3. Sales 300 6

4. Personnel 500 10

5. Plant Mng. 100 2

6. Plant Eng 500 10

7. P. Supervisor 100 2

8. Controller Office 50 1

9. Purchasing Dept 300 6

A=4 E=3 I=2 O=1 U=0 X= 1

CORELAP – Example 1A=4, E=3, I=2, O=1, U=0, X=‐1

Table of TCR Values

Dept. Department relationships Summary TCR Order

1 2 3 4 5 6 7 8 9 A E I O U X1 ‐ I I U O U U U U 0 0 2 1 5 0 5 712345

I I U O U U U U 0 0 2 1 5 0 5 7I ‐ O U O U U U O 0 0 1 3 4 0 5 8I O ‐ U I O O E U 0 1 2 3 2 0 10 5U U U ‐ O O O O O 0 0 0 5 3 0 5 9O O I O ‐ A A O O 2 0 1 5 0 0 15 15

6789

O O I O A A O O 2 0 1 5 0 0 15 1U U O O A ‐ I O E 1 1 1 3 2 0 12 2U U O O A I ‐ U O 1 0 1 3 3 0 9 3U U E O O O U ‐ I 0 1 1 3 3 0 8 6U O U O O E O I ‐ 0 1 1 4 2 0 9 49 U O U O O E O I ‐ 0 1 1 4 2 0 9 4

The placement sequence: 5‐6‐7‐9‐3‐8‐1‐2‐4


Both options gives the p gsame PR Score

PR = A[5,7]+I[6,7]

4 + 2 6= 4 + 2 = 6

If the location for the department 7 is chosen as pshown, the PR would bePR = A [5,7] =4



PR = E[6,9] = 3

PR = E[6,9]+O[5,9] = 3 + 1= 4



PR I O U (3 9) 2 1 0 3PR = I[3,5] + O[3,7] + U (3,9) = 2 + 1 + 0 = 3

PR = + E + I [8 9] = 3 + 2 = 5PR = + E[3,8] + I [8,9] = 3 + 2 = 5



PR = I[1 3] + U (1,7) = 2 + 0 = 2[1,3] ( , )

TCR = I[1,2] + I[2,3] = 2 +2 = 4

Continue with Department 4.


CORELAP – Example 2Given the relationship chart below and the sequence of theGiven the relationship chart below and the sequence of the

placement of the departments 7‐5‐9‐3‐1‐4‐2‐6‐8, find the best layout with CORELAP algorithm assuming that all the departments have the same size. Use these closeness values: A=125, E=25, I=5, O=1, U=0, X=‐

1. Receiving

125 and consider half weight if the departments are only touching by one point.

g

2. Shipping

3. Raw Materials Storage

AA

EO

EA

4. Finished Goods Storage

5. Manufacturing

UU

AO

E

E

A

U

U

A

A

O

O

UU

EE

U6. Work-In-Process Storage

7. Assembly

A

A

X

A

O

O

A E

A

E

UA

U

O

8. Offices

9. Maintenance

X

X

OA

CORELAPCORELAP –– Example 2Example 2

A=125, E=25, I=5, O=1, U=0, X=‐125

CORELAPCORELAP Example 2Example 2The placement sequence: 7‐5‐9‐3‐1‐4‐2‐6‐8

7

125

125 125

62.5 62.5

7 125

62.5 62.5

5125

187.5 187.5

125 62.562.5 62.5187.562.5 187.5

7 0

62.5 0

5187 5

62.5125

7 0

125.5 0

5

0.563.5

3125

62.5

7 05

187.5

187.5

9187.5

62 5 125 62 5

0

751.59126.5

0 5 1 0 5

0

362.5

62.5 125 62.5 0.5 1 0.5


A=125, E=25, I=5, O=1, U=0, X=‐125


7 1255

100

337.5

12.5 62.5137.537.5

75

12.5

387.5

12.5

62.5

0025

137.5925 0

37.5 112.5 12.5

62.5 1259

125

137.5 1

62.5 125

4

125 62.5

1250 188 62.562.5

75

9

31

1

125

421

0.5

63.5

11 1.5 1.5 0.50.5


A=125, E=25, I=5, O=1, U=0, X=‐125


0.5 10.5 0.5

75

-60.5

3112 5 11212 5

-61.525.5 612.5

753

6

875

9

3112.5

1

-112

4225

12.5

-37.5

75

9

3

1 42

8

7587.5 -62.5 -37.5 12.512.5

CORELAPCORELAP ‐‐ CommentsCommentsCORELAP CORELAP CommentsComments

The final layouts are evaluated by the distance‐based layout score◦ CORELAP uses the shortest rectilinear path pbetween the departments (not always realistic)

The layouts often result in irregular building shapes

ALDEPALDEP –– Automated Layout Design ProgramAutomated Layout Design ProgramALDEP ALDEP Automated Layout Design ProgramAutomated Layout Design Program

Similar to CORELAP (objectives, requirements)The main differences:◦ Randomness◦ Multi‐floor capabilityMulti floor capability◦ CORELAP attempts to produce the best layout, ALDEP produces many layoutsy , p y y

ALDEPALDEP ‐‐ ProcedureProcedureALDEP ALDEP ProcedureProcedure

Department selection◦ Randomly selects the first departmentRandomly selects the first department◦ Out of those departments which have “A” relationship with the first one

(or “E”, “I”, etc. – min level of importance is determined by the user) it selects randomly the second department

◦ If no such department exists it selects the second one completely◦ If no such department exists it selects the second one completely randomly

◦ The selection procedure is repeated until all the departments are selected

D t t l tDepartment placement◦ Starts from upper left corner and extends it downward◦ Vertical sweep pattern◦ Sweep width is determined by the user◦ Sweep width is determined by the userAdjacency‐based evaluation◦ If minimum requirements met, it prints out the layout and the scoresRepeats the procedure (max 20 layouts per run)Repeats the procedure (max 20 layouts per run)User evaluation

ALDEPALDEPVertical sweep pattern

Sweep widthDept size = 8 gridsDept size = 14 grids

1 grid

Dept. size = 8 gridsDept. size = 14 grids

2 grids

Dept. size = 14 grids

2 grids

Dept. size = 8 grids

3 grids

ALDEPALDEP ExampleExampleALDEP ALDEP ExampleExample

Use ALDEP procedure to determine the layout vector construct and evaluate the layout for thevector, construct and evaluate the layout for the facility based on the relationship chart and the departmental dimensions given below. Use the sweep width of 2 and the minimum acceptablesweep width of 2 and the minimum acceptable level of importance “E”. The closeness values: A=64, E=16, I=4, O=1, U=0, X=‐1024

Dept. Area # of unit area templates1 12,000 302 8000 203 6000 154 12,000 305 8000 206 12 000 306 12,000 307 12,000 30


Department selection

Step Department selected Reason for selection1 4 random2 2 “E” i h 42 2 “E” with 43 1 “E” with 24 6 random5 5 “A” with 66 7 random7 3 remaining

Layout vector◦ 4‐2‐1‐6‐5‐7‐3

ALDEP ALDEP ExampleExampleDept. # of unit area

templates1 30

2 20

Layout construction◦ Layout vector: 4‐2‐1‐6‐5‐7‐3S id h 2

3 15

4 30

5 20

6 30◦ Sweep width: 2 6 30

7 30

• Final layout


Adjacency score A=64, E=16, I=4, O=1, U=0, X=‐1024

MULTIPLEMULTIPLE ––Multi‐floor Plant Layout EvaluationMULTIPLE MULTIPLE Multi floor Plant Layout Evaluation

Construction and improvement algorithmDistance‐based algorithmSimilar to CRAFT (departments notSimilar to CRAFT (departments not restricted to rectangular shapes, discrete representation two‐way exchanges)representation, two way exchanges)But MULTIPLE can exchange non‐adjacent departmentsdepartmentsUses spacefilling curves to reconstruct a

l t ft h it tinew layout after each iteration

MULTIPLEMULTIPLE ‐‐ SpacefillingSpacefilling Curves (SFC)Curves (SFC)MULTIPLE MULTIPLE SpacefillingSpacefilling Curves (SFC)Curves (SFC)

Spacefilling curve connects all the grids in a layoutE h id i i it d tl◦ Each grid is visited exactly once

◦ Next grid visited is always adjacentto the current grid (only horizontal orvertical moves)vertical moves)

SFC is generated by the computerSFC allows MULTIPLE to

Map a layout vector into aMap a layout vector into a two‐dimensional layoutProcedure: ◦ The departments are placed based on the layout vector (similar as MCRAFT)

◦ The SFC is followed until the required number of grid for h d h deach department is reached

MULTIPLEMULTIPLE ‐‐ ExampleExampleMULTIPLE MULTIPLE ExampleExample

Create a MULTIPLE layout for the d b l b d h ldepartments below based on the layout vector 1‐2‐3‐4‐5‐6. Then build a new layout by exchanging the departments 1 and 5 Theby exchanging the departments 1 and 5. The facility and SFC are given below.

Department Area (m^2)1 162 82 83 44 165 85 86 12


Layout vector 1‐2‐3‐4‐5‐6Dep. Area1 162 8

2 3 4

2 83 44 165 82 3 46 12

511

6


Layout vector 1‐2‐3‐4‐5‐6Dep. Area1 16

2 3 4

2 83 44 165 82 3 4 5 86 12

511

6


Exchange 1 and 5 ‐ Layout vector 5‐2‐3‐4‐1‐6


Layout vector 5‐2‐3‐4‐1‐6 Dep. Area1 16

4 1

2 83 44 165 84 1

3

5 86 12

3

225 6


Initial layout

L t ftLayout afterthe exchange

MULTIPLEMULTIPLE ‐‐ Conforming CurvesConforming CurvesMULTIPLE MULTIPLE Conforming CurvesConforming Curves

Conforming curves are hand‐generated curvesThey are used:◦ If the building shape is irregular◦ If we want to capture the initial layout exactly◦ If we want to capture the initial layout exactly◦ If there are numerous obstacles (walls)◦ If there are fixed departments

Procedure:◦ May start and end at any grid◦ The curve visits all the grids assigned to a particular department before visiting other department◦ The fixed departments and obstacles are not visitedp

MULTIPLEMULTIPLE ‐‐ Conforming CurvesConforming CurvesMULTIPLE MULTIPLE Conforming CurvesConforming Curves

MULTIPLEMULTIPLEMULTIPLEMULTIPLEFinal MULTIPLE layout for the CRAFT exampleexample

The cost is lower than for the final layout f d b !found by CRAFT!◦ MULTIPLE is very likely to obtain lower‐cost solutions than CRAFT, since it considers a larger so ut o s t a C , s ce t co s de s a a geset of possible solutions at each iteration

MULTIPLEMULTIPLEMULTIPLEMULTIPLE

Final MULTIPLE layout for the CRAFT example may also need massaging to smooth the department borders

MULTIPLEMULTIPLE ‐‐ Construction algorithmConstruction algorithmMULTIPLE MULTIPLE Construction algorithmConstruction algorithm

Any SFC or conforming curves could be used to fill the vacant buildingAny vector can be used as the initial ylayout vectorAlternative layouts can be generated byAlternative layouts can be generated by trying different SFC◦ The cost may not be much different◦ The cost may not be much different

Optimal layout vector: D‐B‐H‐C‐F‐ED B H C F E

Cost z=54,200

Optimal layout vector:D‐E‐F‐B‐C‐H

Cost z=54 900Cost z=54,900

Optimal layout vector: D‐E‐F‐H‐B‐C

Cost z=54,540

MIP‐Mixed Integer Programming

◦ Generally construction type models◦ Could be used to improve a layout as welli b d bj i◦ Distance –based objective

◦ All the departments are assumed to be rectangularrectangular◦ Centroid and the length (width) of the department will fully define its location and p yshape

◦ Continuous representation

Obj i


Objective◦ Minimize the total cost/total distance travelledtravelled

Subject to◦ Rectangular department shapesg p p◦ The minimum and maximum sizes for the width and heights of departments◦ The minimum area requirement of◦ The minimum area requirement of departments◦ Departments are completely located inside the building◦ Departments do not overlap◦ All the departments are located on the floor◦ All the departments are located on the floor


• The notations used in formulations are:• The notations used in formulations are:

Parameters:

Decision variables:Decision variables:

MIP‐Mixed Integer ProgrammingY

),( '''ii yx ),( ''''

ii yx

• Mathematical Program Formulation * (αi, βi)),( ''ii yx ),( '''

ii yxBy

X

(1)

(2)

(3) XBX

Makes the model nonlinear

(4)

(5)

(6)

Make sure that departments do not

( )

(7)

(8)

(9)overlap

(9)

(10)

(11)

(12)(12)

(13)

MIP MIP ‐‐ ExampleExampleFormulate MIP model for the facility with the required dimension and coordinates ofdimension and coordinates of the departments below. Given is also Flow‐between chart and the costs of transfer among the

Flow-between Chart

the costs of transfer among the departments. The available space for the facility is 39 X 43 = 1677m^2.

Total cost• Solution:o Total cost calculation:

MIP MIP Example Example ‐‐ ResultsResults

X Y

A

Upper 35 7.14

Lower 0 0

B

Upper 27.49 12.14

Lower 7.49 7.14

C

Upper 27.49 22.14

Lower 7 49 12 14

22.14

C Lower 7.49 12.14

C

12.14

22.14

B

7.14

A

3527.497.49

MIPMIP‐‐ CommentsCommentsMIPMIP CommentsComments

Benefits of MIP Model◦ Department shapes as well as their areas can be modeled throughDepartment shapes as well as their areas can be modeled through

individually specified lower and upper limits◦ MIP might be able to control length‐width ratio as well

(xi’’ – xi’ ) <= R (yi’’‐yi’) or (yi’’ – yi’ ) <= R (xi’’‐xi’)(yi yi ) < R (xi xi )

◦ However, to use the MIP model effectively, some flexibility in the departmental specifications is required

May be used to improve a given layout as wellSolving the problem exactly (optimal solution) is hard◦ 7‐8 departments is the typical size solvable in a reasonable amount of

time◦ In practice, the number of departments easily exceeds 15‐18In practice, the number of departments easily exceeds 15 18Heuristically, CRAFT could be combined with MIP. ◦ Get an initial layout using CRAFT, use MIP to find the best rectangular

layout designA rectangle is not always an ideal shape for the department

Conclusion Conclusion Layout generation algorithms Layout generation algorithms

There is no commercial package that will suit all the needs, partly due to thesuit all the needs, partly due to the difficulty of the problem, but more due to the fact that facility layout is athe fact that facility layout is a combination of Science and Art.

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