initial layout (1)

8/10/2019 Initial Layout (1)

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Copyright 1999. Created by Jose Ventura for the College-Industry Council for Material Handling Education

Initial Layout Construction

Preliminaries

From-To Chart / Flow-Between chart

REL Chart Layout Scores

Traditional Layout Construction

Manual CORELAP Algorithm

Graph-Based Layout Construction

REL Graph, REL Diagram, Planar Graph

Layout Graph, Block Layout

Heuristic Algorithm to Construct a REL Graph

General Procedure


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Relationship (REL) Chart

A Relationship (REL) Chartrepresents

M(M-1)/2 symmetric qualitative

relationships, i.e.,

where

rij{A, E, I, O, U}: Closeness Value

(CV) between activities i and j; rijis an

ordinal value.

A number of factors other than material

handling flow (cost) might be of primary

concern in layout design.

rijvalues when comparing pairs of activities:

A = absolutely necessary 5 %

E = especially important 10 %

I = important 15 %

O = ordinary closeness 20 %

U = unimportant 50 %X = undesirable 5 %

V(rij) = arbitrary cardinal value assigned to rij,

e.g., V(U) = 1, etc.

r12r

13

r23


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Adjacency

Two activities are (fully) adjacentin a layout if they share a common border of positive

lenght, i.e., not just a point.

Two activities are partially adjacentin a layout if they only share one or a finite

number of points, i.e., zero length.

Let aij[0, 1]: adjacency coefficient between activities i and j.

Example: (Fully) adjacent: a12= a13= a24= a34= a45= 1,

Partially adjacent: a14= a23= a25= , and

Non-adjacent: a15= a25= 0.

.adjacentnotaretheyif

and,adjacentpartiallyaretheyif)10(

,adjacentarejandiactivitiesif

0

1

aij

3

1 2

4 5


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Layout Scores

Two ways of computing layout scores:

Layout score based on distance:

where dij= distance between activities i and j.

Layout score based on adjacency:

where aij[0, 1]: adjacency coefficient between activities i and j.

1M

1i

M

1ijijij

d d)r(VLS

1M

1i

M

1ijijij

a a)r(VLS


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Traditional Layout Configuration

An Activity Relationship Diagramis developed from information

in the activity relation chart. Essentially the relationship diagram is a

block diagram of the various areas to be placed into the layout.

The departments are shown linked together by a number of lines. The

total number of lines joining departments reflects the strength of the

relationship between the departments. E.g., four joining lines indicate

a need to have two departments located close together, whereas one

line indicates a low priority on placing the departments adjacent to

each other.

The next step is to combine the relationship diagram with

departmental space requirements to form a Space Relationship

Diagram. Here, the blocks are scaled to reflect space needs while

still maintaining the same relative placement in the layout.

A Block Planrepresents the final layout based on activity

relationship information. If the layout is for an existing facility, the

block plan may have to be modified to fit the building. In the case of

a new facility, the shape of the building will confirm to layout

requirements.

A Rating

E Rating

I Rating

O Rating

U Rating

X Rating

Legend


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Example

Code Reason

1 Flow of material

2 Ease of supervision

3 Common personnel

4 Contact Necessary5 Conveniences

Rating Definition

A Absolutely Necessary

E Especially Important

I Important

O Ordinary Closeness OK

U Unimportant

X Undesirable

1. Offices

2. Foreman

3. Conference Room

4. Parcel Post

5. Parts Shipment

6. Repair and Service Parts

7. Service Areas

8. Receiving

9. Testing

10. General Storage

O

4

I

5

U

U

U

E

3

U

U

E

3

E

5

O

4

U

O

4

U

U

E

3

A

1

O

3

I

2

U

U

U

I

4

U

U

I

2

U

U

U

U

U

I

2

U

U

A

1

U

O

2

U

I

1

U

I

2

U

U

I

2

U

REL chart:


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Example (Cont.)

10

5 8 7

9 6

4 2 3

1Activity Relationship

Diagram


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Example (Cont.)

2

(125)

Space Relationship

Diagram

3

(125)

1(1000)

4

(350)

6

(75)

9(500)

10(1750)

5(500)

8(200)

7(575)


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Manual CORELAP Algorithm

CORELAP is a construction algorithm to create an activity relationship (REL) diagram

or block layout from a REL chart.

Each department (activity) is represented by a unit square.

Numerical values are assigned to CVs:

V(A) = 10,000, V(O) = 10,

V(E) = 1,000, V(U) = 1,

V(I) = 100, V(X) = -10,000.

For each department, the Total Closeness Rating (TCR)is the sum of the absolute

values of the relationships with other departments.


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Procedure to Select Departments

1. The first department placed in the layout is the one with the greatest TCR value. I|f a tie

exists, choose the one with more As.

2. If a department has an X relationship with he first one, it is placed 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. If a tie exists,

choose the one with the greatest TCR value.

4. If a department has an X relationship with he 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 value.

5. The third department is the one with an A relationship with one of the placed departments.

If a tie exists, choose the one with the greatest TCR value.

6. The procedure continues until all departments have been placed.


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Example

1. Receiving

2. Shipping

3. Raw Materials Storage

4. Finished Goods Storage

5. Manufacturing

6. Work-In-Process Storage

7. Assembly

8. Offices

9. Maintenance

AA

E

OU

UA

O

E

E

E

A

A

X

X

AU

U

A

O

O

A

O

A

O

U

E

A

U

E

U

E

AU

O

A

1. Receiving

2. Shipping

3. Raw Materials Storage

4. Finished Goods Storage

5. Manufacturing

6. Work-In-Process Storage

7. Assembly

8. Offices

9. Maintenance

CV values:

V(A) = 125

V(E) = 25

V(I) = 5

V(O) = 1

V(U) = 0

V(X) = -125

Partial adjacency:= 0.5


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Table of TCR Values

Department Summary

Dept.

1 2 3 4 5 6 7 8 9 A E I O U X

TCR Order

1

2

3

4

5

6

7

8

9

- A A E O U U A O

A - E A U O U E A

A E - E A U U E A

E A E - E O A E U

U O A E - A A O A

U O U O A - A O O

U U U A A A - X A

A E E E O O X - X

O U A U A O A X -

3 1 0 2 2 0

2 2 0 1 3 0

3 3 0 0 2 0

2 4 0 1 1 0

4 1 0 2 1 0

2 0 0 4 2 0

4 0 0 0 3 1

1 3 0 2 0 2

3 0 0 2 2 1

402

301

450

351

527

254

625

452

502

(5)

(7)

(4)

(6)

(2)

(8)

(1)

(9)

(3)


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Example (cont.)

7

125

125

125 125

62.5 62.5

62.562.5

7 125

62.5 62.5

62.5187.5

5125

62.5 187.5

187.5 187.5

7 0

62.5 0

5

187.5

187.5

9187.5

62.5 125 62.5

0

62.5125

7 0

125.5 0

5

1.59126.5

0.5 1 0.5

0

163.5

3125

62.5

62.5


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Example (cont.)

7 1255

137.5925 0

100

337.5

37.5

12.5

112.5 12.5

62.5

62.5137.537.5

7

125

5

9125

12.5

387.5

137.5

12.5

162.5 125

62.5

0025

4125 62.5

75

9

1

125

31

0

1

1 1.5

125

188

4

1.5 0.5

21

0.5

0.5

63.5

62.562.5


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Example (cont.)

75

9

75

-60.5

3112.5

1

87.5 -62.5

-112

4

-37.5 12.5

225

12.5

12.5

-37.5

-61.525.5 612.5

0.5 10.5 0.5

75

9

3

1 42

6

8


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Planar Graph

Assumption:

A Planar Graphis a graph that can be drawn in two dimensions with no arc crossing.

.otherwise

,adjacentfullyarejandiactivitiesif

0

1aij

NonplanarPlanar

A graph is nonplanar if it contains either one of the two Kuratowski graphs:


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Relationship (REL) Graph

Given a (block) layout with M activities, a corresponding planar undirected graph,

called the Relationship (REL) Graph, can always be constructed.

REL Graph

1 2

543

6(Exterior)

1 2

543

Block Layout

A REL graph has M+1 nodes(one node for each activity and a node for the exterior ofthe layout. The exterior can be considered as an additional activity. Thearcscorrespond

to the pairs of activities that are adjacent.

A REL graph corresponding to a layout is planar because the arcs connecting two

adjacent activities can always be drawn passing through their common border of

positive length.


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Relationship (REL) Diagram

A Relationship (REL) Diagramis also an undirected graph, generated from the REL

diagram, but it is in general nonplanar.

A REL diagram, including the U closeness values, has M(M-1)/2 arcs. Since a planar

graph can have at most 3M-6 arcs, a REL diagram will be nonplanar if M(M-1)/2 >3M-6.

M(M-1)/2 > 3M-6 M 5.

A REL graph is a subgraph of the REL diagram.

For M 5, at most 3M-6 out of M(M-1)/2 relationships can be satisfied through

adjacency in a REL graph.

An upper bound on LSa, LSaUB, is the sum of the 3M-6 longest V(rij)s.


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Maximally Planar Graph (MPG)

A planar graph with exactly 3M-6 arcs is called Maximally Planar Graph (MPG).

Not MPG since

has only 5 arcs

(5 < 6 = 3M-6)

MPG since

has 6 arcs

The interior facesof a graph are the bounded regions formed by its arcs, and its

exterior faceis the unbounded region formed by its outside arcs.

IF1 IF2

IF3

EF The tetrahedron has three interior faces (IF1, IF2and IF3) and an exterior face (EF)


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Maximally Planar Graph (MPG)

The interior faces and the exterior face of an MPG are triangular, i.e., the faces are

formed by three arcs.

Not triangular

Not an MPG

The REL graph of a given a (block) layout may not be an MPG.

Layout REL Graph

Not an MPG


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Maximally Planar Weighted Graph (MPWG)

An MPG whose sum of arc weights is as large as any other possible MPG is called a

Maximally Planar Weighted Graph (MPWG).

Using the V(rij)sas arc weights, a REL graph that is a MPWG has the maximum

possible LSa, close to LSaUB.

Since it is difficult to find an MPWG, a Heuristic (non-optimal) procedure will be used

to construct a REL graph that is an MPG, but may not be an MPWG (although its LSa

will be close to LSaUB).

The Layout Graphis the dual of the REL graph.

Given a graph G, its dual graph GDhas a node for each face of G and two nodes in GD

are connected with an arc if the two corresponding faces in G share an arc.


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Layout Graph

Example.

The number of nodes in G (primal graph) is the same than the number of faces in GD

(dual graph), and vice versa. In addition,(GD)D= G.

Primal Graph is Planar Dual Graph is planar.

G GD


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Layout Graph (Cont.)

Given a layout, the corresponding layout graphcan always be constructed by placing

the nodes at the corners of the layout where three or more activities meet (including the

exterior of the layout as an activity). The arcs in the graph are the remaining portions of

the layout walls. E.g.,

Layout Graph

1 2

54

3

(Exterior)

Given a REL graph (RG), its corresponding layout graph (LG)is LG = RGD. E.g.,

Layout

c g

a b

d f

e

h

1 2

54

3

6

RG LG

RGD

LGD

Only activity 3 and

exterior meet here

Activities 3, 5, and

exterior meet here


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Layout Graph (Cont.)

If LG is given, then RG = LGD, but for layout construction, the layout is not known

initially, so LG cannot be constructed without RG.

If a planar REL graph (primal graph) exist, the corresponding layout graph (dual graph)

is also planar. Therefore, it is possible theorectically to construct a block layout that willsatisfy all the adjacency requirements. In practice, this is not straightforward because the

space requirements of the activities are difficult to handle.


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Example

Space Requirements:

Dept. Area

A 300

B 200

C 100

D 200

E 100

F (exterior)

REL graph (Primal graph):

A B

C D

F G


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Example (Cont.)

Layout graph (Dual graph):

A B

C D

F G

1

2

3

4

5

7

6

8


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Example (Cont.)

A corner pointis a point where at least three departments meet, including the exterior

department.

Note that each corner point in the block layout corresponds to a node in the layoutgraph. In the first block layout, each corner point is defined by exactly three

departments. In this case, there is a one-to-one correspondence between corner points

and nodes in the layout graph. In the square block layout, there are two corner points

defined by four departments, i.e., (A, B, C, D) and (B, D, E, F). Each of these two corner

points corresponds to two nodes in the layout graph.

Block Layout:Square Block Layout:

(areas are not considered)

A

D

BC

E

8 1 6

7 2 3 4

5

A

D

B

C

E

7

8

8 4

1 5

2 3

F

F


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Heuristic Procedure to Construct a Relationship Graph

1. Rank activities in non-increasing order of TCRk, k = 1, ,M, where

TCRk=

(Note that the negative values of V(rik) and V(rkj) are not included in TCRk).

2. Form a tetrahedron using activities 1 to 4 (i.e., the activities with the four largest TCRks).

3. For k = 5, , M, insert activity k into the face with the maximum sum of weights (V(rij))

of k with the three nodes defining the face (where insert refers to connecting the inserted

node to the three nodes forming the face with arcs).

4. Insert (M+1)thnode into the exterior face of the REL graph.

Max{0, V(r )} Max{0, V(r )}.iki 1

k-1

kjj=k+1

M


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Example

OI

OA

X

U

U

O

UU

E

E

A

B

C

D

E

F

I

E

E

CV values:

V(A) = 81V(E) = 27

V(I) = 9

V(O) = 3

V(U) = 1

V(X) = -243

REL chart:


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Table of TCR Values

Department Summary

Dept.

A B C D E F A E I O U X

TCR Order

A - I O I O A 1 0 2 2 0 0 105 2

B I - X U U E 0 1 1 0 2 1 38 5

C O X - U E E 0 2 0 1 1 1 58 3

D I U U - U E 0 1 1 0 3 0 39 4

E O U E U - O 0 1 0 2 2 0 35 6

F A E E E O - 1 3 0 1 0 0 165 1


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Example (Cont.)

Step 2:

A

C

F D

A

O

E U

E

I = rADV(rAD) = 9


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Example (Cont.)

Step 3: Insert B.

A

C

F D

EF

IF1 IF2

IF3

I

I I

E

E

E

U

U

U

X X

X

Face LSa

EF 9 + 1 + 27 = 37 *

IF1 9 + 27 - 243 = -207

IF2 9 - 243 + 1 = -233

IF3 27 - 243 + 1 = -215

Insert B in EF


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Example (Cont.)

Step 3 (Cont.): Insert B.

A

C

F D

B

IF1

IF2 IF3

IF4

IF5

EF

Face LSa

EF 5

IF1 7

IF2 33 *

IF3 31

IF4 31

IF5 5

Insert E in IF2


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Example (Cont.)

Step 4: Call exterior activity EX.

A

C

F D

B EX

E

Since arcs (AB), (BD),

and (DA) are the outside

arcs, EX connects to

nodes A, B, and D.


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Example (Cont.)

LSaUBis the sum of the 3M - 6 ( 3 6 - 6 = 12), largest V(rij)s.

In the last example,

LSaUB= V(rAF) + V(rBF) + V(rCE) + V(rCF) + V(rDF) + V(rAB) + V(rAD) + V(rAC)

+ V(rAE) + V(rEF) + V(rBD) + V(rBE) = 81 + 27 + 27 + 27 + 27 + 9 + 9 + 3

+ 3 + 3 + 1 + 1 = 218.

For the final REL graph, LSa= 218.

LSaUB= LSa The final REL graph is an MPWG It is optimal.

LSaUB> LSa The final REL graph may not be an MPWG It may not be optimal.

Using the Heuristic procedure, the generated REL graph will always be an MPG since

each face is triangular.


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General Procedure for

Graph Based Layout Construction

1. Given the REL chart, use the Heuristic procedure to construct the REL graph.

2. Construct the layout graph by taking the dual of the REL graph, letting the facility

exterior node of the REL graph be in the exterior face of the layout graph.

3. Convert (by hand) the layout graph into an initial layout taking into consideration the

space requirement of each activity.

REL Chart REL Graph Layout Graph Initial Layout

Space

Requirements


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Example

Step 1: (from before)

A

C

F D

B EX

E

REL Graph


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Example (Cont.)

Step 2: take the dual of RG

C

F

D

EX

E

A

B

Layout Graph


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Example (Cont.)

Step 3:

Initial layout isdrawn as a square,

but could be any

other shape.

Only A and B are

nonrectangular.

B D

A

F

E C

Initial Layout


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Comments

1. If an activity is desired to be adjacent to the exterior of a facility (e.g., a shipping/receiving

department), then the exterior could be included in the REL chart and treated as a normal

activity, making sure that, in step 1 of the general procedure, its node is one of the nodes

forming the exterior face of the REL graph.

2. The area of each interior face of the layout graph constructed in step 2 does not correspond to

the space requirements of its activity.

3. In step 3, the overall shape of the initial layoutshould be usually be rectangularif it

corresponds to an entire building because rectangular buildings are usually cheaperto

build; even if the initial layout corresponds to just a department, a rectangular shape would

still be preferred, if possible.

4. In step 3, the shape of each activity in the initial layout should be rectangular if possible, or at

most L- or T-shaped(e.g., activities A and B), because rectangular shapes require less wall

spaceto enclose and provide more layout possibilities in interiors as compared to other

shapes.


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Comments (Cont.)

5. All shapes should be orthogonal, i.e., all corners are either 90 or 270 (e.g., a triangle is not

an orthogonal shape since its corners could all be 60 ).

6. In step 1, if the LSaof the REL graph is less than LSaUB, then the REL graph may not be

optimal. The following three steps may improve the REC graph for the purpose of increasing

LSa:

a) Edge Replacement: replace an arc in the REL graph with a new arc not previously

in the graph, without losing planarity, if it increases LSa.

b) Vertex Relocation: move a node in the REL graph connected to three arcs to

another triangular face if it increases LSa.

c) Use a different activity to replace one of the four activities of the tetrahedronformed in step 2 of the Heuristic procedure to construct a new REL graph.

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