Initial Layout Recovered]

8/3/2019 Initial Layout Recovered]

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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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From-To and Flow-Between Charts

Given M activities, a From-ToChart

represents M(M-1) asymmetricquantitative

relationships.Example:

where

fij = material flow from activity i to

activity j.

A Flow-Between Chartrepresents

M(M-1)/2 symmetric quantitative

relationships, i.e.,

gij = fij + fji, for all i > j,

where

gij = material flow between

activities i and j.

f12 f13

f23

f32

f21

f31


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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) betweenactivities i and j; rij is anordinal value.

A number of factors other thanmaterial

handling flow (cost) might be ofprimary

concern in layout design.

rij values 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 %

r12r

13

r23


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Adjacency

Two activities are (fully) adjacent in a layout if they share a commonborder of positive lenght, i.e., not just a point.

Two activities are partially adjacent in 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


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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 Diagram is developed frominformation in the activity relation chart. Essentially therelationship diagram is a block diagram of the variousareas to be placed into the layout.

The departments are shown linked together by anumber of lines. The total number of lines joiningdepartments reflects the strength of the relationshipbetween the departments. E.g., four joining linesindicate a need to have two departments located closetogether, whereas one line indicates a low priority onplacing the departments adjacent to each other.

The next step is to combine the relationship diagramwith departmental space requirements to form a SpaceRelationship Diagram. Here, the blocks are scaled toreflect space needs while still maintaining the samerelative placement in the layout.

A Block Plan represents the final layout based onactivity relationship information. If the layout is for an

existing facility, the block plan may have to be modifiedto fit the building. In the case of a new facility, the shape

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 ofthe 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 greatestTCR 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 TCRvalue.

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.


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Procedure to Place Departments

Consider the figure on the right. Assume that adepartment is placed in the middle (position 0). Then, ifanother department is placed in position 1, 3, 5 or 7, it isfully adjacent with the first one. It is placed in

position 2, 4, 6 or 8, it is partially adjacent.

8 7 6

5

432

1 0

For each position, Weighted Placement (WP) is the sum of the numerical values for all

pairs of adjacent departments.

The placement of departments is based on the following steps:

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

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

around the current layout in counterclockwise order beginning at the western edge.

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


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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 XTCR Order

12

3

4

5

6

7

8

9

- A A E O U U A OA - 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 02 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

402301

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 Graph is 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 of thelayout. The exterior can be considered as an additional activity. The arcs correspond 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) Diagram is also an undirected graph,generated from the REL chart, 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 willbe 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 PlanarGraph (MPG).

Not MPG since

has only 5 arcs

(5 < 6 = 3M-6)

MPG since

has 6 arcs

The interior faces of a graph are the bounded regions formed by its arcs, and its exterior

face is the unbounded region formed by its outside arcs.

IF1 IF2

IF3

EF The tetrahedron has three interior faces (IF1, IF2

and 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 possibleMPG is called a Maximally Planar Weighted Graph (MPWG).

Using the V(rij)s as 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, butmay not be an MPWG (although its LSa will be close to LSaUB).

The Layout Graph is the dual of the REL graph.

Given a graph G, its dual graph GD has a node for each face of G andtwo nodes in GD are connected with an arc if the two correspondingfaces 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 PlanarDual Graph is planar.

G GD


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Layout Graph (Cont.)

Given a layout, the corresponding layout graph can 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 isnot known initially, so LG cannot be constructed without RG.

If a planar REL graph (primal graph) exist, the corresponding layoutgraph (dual graph) is also planar. Therefore, it is possibletheorectically to construct a block layout that will satisfy all theadjacency requirements. In practice, this is not straightforwardbecause the space requirements of the activities are difficult tohandle.


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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

E F


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Example (Cont.)

Layout graph (Dual graph):

A B

C D

E F

1

2

3

4

5

7

6

8


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Example (Cont.)

A corner point is 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 layout graph.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 aRelationship 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)th node into the exterior face of the REL graph.

M a x { 0 ,V ( r) } M a x { 0 ,V ( r) } .i ki 1

k - 1

k jj = k + 1

M

=

+


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Example

O

IO

A

X

U

U

O

UU

E

E

A

B

C

D

E

F

I

E

E

CV values:

V(A) = 81

V(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.)

Step2:

A

C

F D

A

O

E U

E

I = rADV(rAD) = 9


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Example (Cont.)

Step 3: InsertB.

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.): InsertE.

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 exterioractivity 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.)

LSaUB is 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 notbe optimal.

Using the Heuristic procedure, the generated REL graph will always bean MPG since each face is triangular.


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General Procedure forGraph 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: (frombefore)

A

C

F D

B EX

E

REL Graph


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Example (Cont.)

Step 2: take the dualof RG

C

F

D

EX

E

A

B

Layout Graph


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Example (Cont.)

Step 3:

Initial layout is drawnas 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 layout should be usually be

rectangular if it corresponds to an entire building because rectangular

buildings are usually cheaper to build; even if the initial layout

corresponds to just a department, a rectangular shape would still bepreferred, 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 space to 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 LSa of 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 thetetrahedron formed in step 2 of the Heuristic procedure to construct a

new REL graph.

Initial Layout Recovered]

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