facility design-week 10 ( cont ) computerized layout planning

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Facility Design-Week 10 (cont) Computerized Layout Planning

By Anastasia L. Maukar

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CRAFT- Computerized Relative Allocation of Facilities Technique

• Created in 1964 by Buffa and Armour• Process layout approach• A heuristic computer program

– Compares process departments– CRAFT requires an initial layout, which is improved

by CRAFT.

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From-To Chart

• Determines which of the two departments has a better from-to matrix, we can calculate the moment of the matrix as follows:

fd ijji

ijMoment

),(

matrix in theentry j) (i, the

diagonal thefromentry j) (i, theof distance where

fd

ij

ij

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CRAFT

• Input for CRAFT:– initial spatial array/layout– flow data– cost data– Number and location of fixed department

• Secara umum, dapat ditambahkan dummy yg berfungsi untuk:– Fill building irregularities.– Represent obstacles or unusable areas in facility– Represent extra space in the facility– Aid in evaluating aisle locations in the final layout.

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CRAFT Following are some

examples of questions addressed by CRAFT:

• Is this a good layout?

• If not, can it be improved?

10 20 30 40 50 60 70 80 90 100

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3040

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A

B

C

D

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CRAFT

The possible interchange:– only pairwise interchage– only three-way interchange– pairwise interchage followed by three-way

interchange– three-way interchage followed by pairwise

interchange– best of two-way and three-way interchage

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• Consider the problem of finding the distance between two adjacent departments, separated by a line only.

• People needs walking to move from one department to another, even when the departments are adjacent.

• An estimate of average walking required is obtained from the distance between centroids of two departments.

• The distance between two departments is taken from the distance between their centroids.

• People walks along some rectilinear paths. An Euclidean distance between two centroids is not a true representative of the walking required. The rectilinear distance is a better approximation.

• So, Distance (A,B) = rectilinear distance between centroids of departments A and B

CRAFT: Distance Between Two Departments

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• Let – Centroid of Department A =– Centroid of Department B =

• Then, the distance between departments A and B, Dist(A,B)

• Ex: the distance between departments A and C is the rectilinear distance between their centroids (30,75) and (80,35). Distance (A,C)

9035758030 CACA yyxx

BABA yyxx

AA yx ,

BB yx ,


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Centroid of A= (30,75)

Centroid of C= (80,35)

Distance (A,C)= 90

10 20 30 40 50 60 70 80 90 100

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A

B

C

D

(80,85)

(30,25)


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• If the number of trips between two departments are very high, then such departments should be placed near to each other in order to minimize the total distance traveled.

• Distance traveled from department A to B = Distance (A,B) Number of trips from department A to B

• Total distance traveled is obtained by computing distance traveled between every pair of departments, and then summing up the results.

• Given a layout, CRAFT first finds the total distance traveled.

CRAFT: Total Distance Traveled

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A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo

(a) Material handling trips(given)

(b)

(b) Distances (given)A B C D

A 50 90 60B 50 60 110C 90 60 50D 60 110 50

FromTo


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A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo

(a) Material handling trips (given)

(b) Distances (given)

(a)

(b)

(c)Total distance traveled = 100+630+240+…. = 4640

A B C DA 50 90 60B 50 60 110C 90 60 50D 60 110 50

FromTo

A B C DA 100 630 240B 150 300 770C 540 420 150D 420 770 150

FromTo

(c) Sample computation: distance traveled (A,B) = trips (A,B) dist (A,B) =………..


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• CRAFT then attempts to improve the layout by pair-wise interchanges.

• If some interchange results some savings in the total distance traveled, the interchange that saves the most (total distance traveled) is selected.

• While searching for the most savings, exact savings are not computed. At the search stage, savings are computed assuming when departments are interchanged, centroids are interchanged too. This assumption does not give the exact savings, but approximate savings only.

• Exact centroids are computed later.

CRAFT: Savings

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• Savings are computed for all feasible pairwise interchanges. Savings are not computed for the infeasible interchanges.

• An interchange between two departments is feasible only if the departments have the same area or they share a common boundary. – Feasible pairs are {A,B}, {A,C}, {A,D}, {B,C}, {C,D}– and an infeasible pair is {B,D}

• In this example savings are not computed for interchanging B and D. Savings are computed for each of the 5 other pair-wise interchanges and the best one chosen.

• After the departments are interchanged, every exact centroid is found. This may require more computation if one or more shape is composed of rectangular pieces.

CRAFT: Savings

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CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange

• To illustrate the computation of savings, we shall compute the savings from interchanging Departments C and D

• New centroids:A (30,75) UnchangedB (30,25) UnchangedC (80,85) Previous centroid of Department DD (80,35) Previous centroid of Department C

• Note: If C and D are interchanged, exact centroids are C(80,65) and D(80,15). So, the centroids C(80,85) and D(80,35) are not exact, but approximate.

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• The first job in the computation of savings is to reconstruct the distance matrix that would result if the interchange was made.

• The purpose of using approximate centroids will be clearer now. • If the exact centroids were used, we would have to recompute

distances between every pair of departments that would include one or both of C and D.

• However, since we assume that centroids of C and D will be interchanged, the new distance matrix can be obtained just by rearranging some rows and columns of the original distance matrix.

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• The matrix on the left is the previous matrix, before interchange. The matrix on the right is after.

• Dist (A,B) and (C,D) does not change. • New dist (A,C) = Previous dist (A,D)• New dist (A,D) = Previous dist (A,C)• New dist (B,C) = Previous dist (B,D)• New dist (B,D) = Previous dist (A,C)

A B C DA 50 60 90B 50 110 60C 60 110 50D 90 60 50

FromToA B C D

A 50 90 60B 50 60 110C 90 60 50D 60 110 50

FromTo

InterchangeC,D

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CRAFT: A Sample Computation of Savings

A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo

(a)

(b)

(c)

A B C DA 50 60 90B 50 110 60C 60 110 50D 90 60 50

FromTo

A B C DA 100 420 360B 150 550 420C 360 770 150D 630 420 150

FromTo

(a) Material handling trips (given)

(b) Distances (rearranged)

Total distance traveled = 100+420+360+… = 4480Savings =

(c) Sample computation: distance traveled (A,B) = trips (A,B) dist (A,B) =

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CRAFT: Improvement Procedure

• To complete the exercise1. Compute savings from all the feasible interchanges. If there is

no (positive) savings, stop.2. If any interchange gives some (positive) savings, choose the

interchange that gives the maximum savings3. If an interchange is chosen, then for every department find an

exact centroid after the interchange is implemented 4. Repeat the above 3 steps as longs as Step 1 finds an

interchange with some (positive) savings.

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CRAFT: Exact Coordinates of Centroids

• Sometimes, an interchange may result in a peculiar shape of a department; a shape that is composed of some rectangular pieces

• For example, consider the layout (from example) and interchange departments A and D. The resulting picture is shown on the right.

• How to compute the exact coordinate of the centroid (of a shape like A)?

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A

B

C

D

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Find the centroid of A

10 20 30 40 50 60 70 80 90 100

5060

7080

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A A1

A2

A1of Centroid

A1of Centroid

A2Area A1Area

Let

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2

1

yx

yxAA

,

,

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X-coordinate Multiply Rectangle Area of centroid (2) and (3) (1) (2) (3) (4) A1

A2 Total

X-coordinate of the centroid of A


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2211

AAxAxA

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Y-coordinate Multiply Rectangle Area of centroid (2) and (3) (1) (2) (3) (4) A1

A2Total

Y-coordinate of the centroid of A


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2211

AAyAyA

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Exact coordinate of area A is

10 20 30 40 50 60 70 80 90 100

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A A1

A2

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CRAFT: Some Comments

• An improvement procedure, not a construction procedure• At every stage some pairwise interchanges are considered and the

best one is chosen• Interchanges are only feasible if departments have the same area;

or they share a common boundary• Departments of unequal size that are not adjacent are not

considered for interchange • Estimated cost reduction may not be obtained after interchange

(because the savings are based on approximate centroids)• Strangely shaped departments may be formed

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Computerized Layout PlanningGraphical Representation

“Points and lines” representation is not convenient for analysis

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Layout Evaluation– An Algorithm needs to distinguish between “good” layouts

and “bad” ones– Develop scoring model, s = g (X )– Adjacency-based scoring (Komsuluk Bazli Skorlama)

• Based on the relationship chart and diagram

• Xi is the number of times an adjacency i is satisfied, i=A, E, I, O, U, X

• Aldep uses (wi values) A=64, E=16, I=4, O=1, U=0, and X=-1024• Scoring model has intuitive appeal; the ranking of layouts is

sensitive to the weight values. Layout “B” may be preferred to “C” with certain weights but not with others.

• Therefore, the specification of the weights is very important.• The weights wi can also be represented by the flow amounts

between the adjacent departments instead of scores assigned to A, E, I, O, U, X.

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1

i

iiXwsMax

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12

34

56

7

12

34

56

7

UU

UU

IE

UI

OU

A

OI

UI

IE

U

OU

EReceiving

Milling

Press

Screw Machine

Assembly

Plating

Shipping

3

Press

7

Shipping

6Plating

2

Milling

4

Screw Machine

5

Assembly

1

Receiving

A

E I

E

O

I

O

1234567

1 2 3 4 5 6 7

I OE I U

O U

AE

4+1 =516+4+0=201+0 =1----64 =6416 =16

Total Score 106

Example

U U

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Exercise: Find the score of the layout shown below. Use A=8, E=4, I=2, O=1, U=0 and X=-8.

12

34

56

7

12

34

56

7

UU

UU

IE

UI

OU

A

OI

UI

IE

U

OU

EReceiving

Milling

Press

Screw Machine

Assembly

Plating

Shipping

3Press

7Shipping

6Plating

4Screw

Machine

1Receiving

2Milling

5Assembly

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Layout Evaluation (cont’d)– Distance-based scoring (Mesafe Bazli Skorlama)

• Approximate the cost of flow between activities• Requires explicit evaluation of the flow volumes and costs

• cij covers both the i to j and the j to i material flows• Dij can be determined with any appropriate distance metric

–Often the rectilinear distance between department centroids• Assumes that the material flow system has already been

specified (cij=flow required* cost /flow-distance)• Assumes that the variable flow cost is proportional to

distance• Distance often depends on the aisle layout and material

handling equipment

1

1 1

m

i

m

ijijijDcsMin

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CRAFT - Example 2

A B

C D

Initial Layout

From/To A B C DA - 2 4 4B 1 - 1 3C 2 1 - 2D 4 1 0 -

Flow Data

From/To A B C DA - 40 25 55B 40 - 65 25C 25 65 - 40D 55 25 40 -

Distance Data From/To A B C D TotalA - 80 100 220 400B 40 - 65 75 180C 50 65 - 80 195D 220 25 0 - 245

Total 310 170 165 375 1020

Total Cost

Ma

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CRAFT - Example 2

• A & D interchange Total cost = 1.095• A (60, 10) dan D (25, 30)• A & C interchange Total cost = 99• C & D interchange Total cost = 1.040• B & D interchange Total cost = 945

estimated total cost • B & C tidak dapat dipertukarkan

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CRAFT - Example 2

• Yang dipilih adalah pertukaran B & D menghasilkan layout baru dg department centroid, sesuai dengan luas yang diinginkan pada layout awal– (XA, YA) = (25, 30)– (XB, YB) = (55, 10)– (XC, YC) = (20, 10)– (XD, YD) = (67.5, 25)

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CRAFT - Example 2

ToFrom A B C D

A 50 25 47.5

B 50 35 27.5

C 25 35 62.5

D 47.5 27.5 62.5

20’ 40’ 60’ 80’

40’

30’

20’

10’

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Layout Evaluation – Distance-Based Scoring– Distance-based scoring– Impact of aisle layout and direction of travel

A B

C D

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Benefits and Problems

• Benefit– A Computer Program– Flexibilty

• Problems– Greedy Algorithm– Inefficient– End result may need to be modified

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Summary

It is beneficial to use CRAFT but you should also realize that the program is not flawless.

The user must understand how the program works of the end product is not as efficient as you had hope.

facility design-week 10 ( cont ) computerized layout planning

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