facilities and design chapter 5

CHAPTER 5MODELS FOR THE LAYOUT PROBLEM

INTRODUCTION There are several objectives in developing a layout, the primary goal is to minimize the cost or time involved in transportation

The focus of this subject is on mathematical models used for the facility layout problem.

For purposes of modeling, we classify the layout problem into two types:

Single – row layout problem

Multi – row layout problem

Single – row layout

Multi – row layout

ABSMODEL 1SINGLE – ROW LAYOUT PROBLEM

ABSMODEL 1 : Assumptions

1. The facilities are square or rectangular and their shapes are known a priori


2. The facilities are arranged along a straight line.


3. the orientation of facilities is known a priori

4. there is no restriction on the shape of the building in which the facilities are to be housed

ABSMODEL 1 : Parameters

Parameters:

n number of departments in the problem

cij cost of moving a unit load by a unit distance between

departments i and j

fij number of unit loads between departments i and j

li length of the horizontal side of department i

dij minimum distance by which departments i and j are to be

separated horizontally

H horizontal dimension of the floor plan

Decision Variable:

xi distance between center of department i and vertical

reference line (VRL)

ABSMODEL 1: Formulation

1

1 1

Minimizen

i

n

ijjiijij xxfc

0.5 1, 2,..., 1i j i j ijx x l l d i n Subject to

. . .

lilj

xi

Facility i Facility j

xj

0.5 0.5 1,2,...,i i iH l x l i n

H

VR

L

ABSMODEL 1: ExampleTVCR repair shop current layout

General repair area

Customer service

Parts display area

15 ft.

75 ft.

ABSMODEL 1: ExampleR o o m

1 2 3 4 5 RoomNumber

Room Name Dimensions(in feet)

R 1 - 12 8 20 0 1 TV/VCR 20 x 10

o 2 12 - 4 6 2 2 Audio 10 x 10

[fij]= o 3 8 4 - 10 0 3 Microwave 10 x 10

m 4 20 6 10 - 3 4 Computer 20 x 10

5 0 2 0 3 - 5 Parts 15 x 10

ABSMODEL 1: Example Solution

DECISION VARIABLES

X1 = Distance between the center of facility 1 and the VRL





FINAL MODEL

Minimize = 12 * ABS( X1 - X2 ) + 8 * ABS( X1 - X3 ) + 20 * ABS ( X1 - X3 ) + 4 * ABS ( X2 - X3 ) + 6 * ABS( X2 - X4 ) + 2 * ABS( X2 - X5 ) + 10 * ABS( X3 - X4 ) + 3 * ABS( X4 - X5 )

ABSMODEL 1: Example Solution

• ABS( X1 - X2 ) > 15 ;

• ABS( X1 - X3 ) > 15 ;

• ABS( X1 - X4 ) > 20 ;

• ABS( X1 - X5 ) > 17.5 ;

• ABS( X2 - X3 ) > 10 ;

• ABS( X2 - X4 ) > 15 ;

• ABS( X2 - X5 ) > 12.5 ;

• ABS( X3 - X4 ) > 15 ;

•ABS( X3 - X5 ) > 12.5 ;

•ABS( X4 - X5 ) > 17.5

•X1 > 10 ;

•X2 > 5 ;

•X3 > 5 ;

•X4 > 10 ;

•X5 > 7.5 ;

•X1 < 65 ;

•X2 < 70 ;

•X3 < 70 ;

•X4 < 65 ;

•X5 < 67.5 ;

SUBJECT TO:

ABSMODEL 1: Example Sol

OPTIMAL SOLUTION USING “GINO OPTIMIZATION SOFTWARE”

MIN = 1657.50

VARIABLES

X1 = 10.00

X2 = 25.00

X3 = 35.00

X4 = 50.00

X5 = 67.50

Verti

cal R

efer

ence

Lin

e (V

RL)

X1 = 10

X2 = 25

X3 = 35

X4 = 50

X5 = 67.5

ABSMODEL 1: Example Sol

TV/VCR Audio Microwave Computer

Verti

cal R

efer

ence

Lin

e (V

RL)

X1 = 10

X2 = 25

X3 = 35

X4 = 50

X5 = 67.5

TV/VCR Audio Microwave Computer Parts

DisplayArea

General repair area

Customer service

Parts display area

Current Layout

Final Layout

MODELS FOR THE MULTIROW LAYOUT PROBLEM WITH FACILITIES OF EQUAL AREA

MODELS FOR THE MULTIROW LAYOUT PROBLEM (with facilities of equal area)

1. Quadratic Assignment Problem (QAP)

2. Linear Mixed-Integer Programming (LMIP)

3. Quadratic set covering model

4. Non-linear model with absolute terms with OF and constraints

Facility 1 Facility 2


Quadratic Assignment Problem (Multi-row Problem)

The QAP formulations requires equal number of facilities and location.

If there are fewer facilities to be assigned to locations, create dummy facilities.

If there are fewer locations than to locations, the problem is infeasible.



Location 1 Location 2

Location 3 Location 4

QAP Parameters

Parameters:n total number of departments and locationsaij net revenue from operating department i at location jfik flow of material from department i to kcjl cost of transporting unit load of material from location j to l

Decision Variable:

1 if department is assigned to location0 otherwiseij

i jx

QAP Formulation

Subject to

1 1 1 1 1 1

Maximize n n n n n n

ij ij ik jl ij kli j i j k l

i k j l

a x f c x x

11

n

jijx

11

n

iijx

1or 0ijx

i=1,2,...,n

j=1,2,...,n

i, j=1,2,...,n

1 1 1 1 1 1

Minimize n n n n n n

ij ij ik jl ij kli j i j k l

i k j l

a x f c x x

1 1 1 1

Minimize n n n n

ik jl ij kli j k l

i k j l

f c x x

Ensures each facility assigned to one location. And guarantee that each location has one facility.

QAP Example

O1 2 3 4 1 2 3 4

f 1 - 17 12 11 S 1 - 1 1 2

[fNij]= f 2 17 - 12 4 [dij]= i 2 1 - 2 1

i 3 12 12 - 4 t 3 1 2 - 1

c 4 11 4 4 - e 4 2 1 -

e

Trip Matrix( trips between each offices) Sites Matrix (dist between each sites)

LonBank Inc,.

QAP Example Solution

Objective Function

Subject to

QAP Example Solution: FINAL MODEL

Objective Function

Subject to

Decision VariablesX22 = 1 if Room 2 assigned to Site 2, 0 if not X23 = 1 if Room 2 assigned to Site 3, 0 if not X24 = 1 if Room 2 assigned to Site 4, 0 if not X32 = 1 if Room 2 assigned to Site 2, 0 if not X33 = 1 if Room 2 assigned to Site 3, 0 if not X34 = 1 if Room 2 assigned to Site 4, 0 if not X42 = 1 if Room 2 assigned to Site 2, 0 if not X43 = 1 if Room 2 assigned to Site 3, 0 if not X44 = 1 if Room 2 assigned to Site 4, 0 if not

QAP Example Solution: OPTIMAL SOLUTION using “GINO OPTIMIZATION SOFTWARE’’

Objective FunctionMIN C = 152.00

Variables Value

X22 = 0X23 = 1X24 = 0X32 = 0X33 = 0X34 = 1X42 = 1 X43 = 0X44 = 0

Decision VariablesX22 = 1 if Room 2 assigned to Site 2, 0 if not X23 = 1 if Room 2 assigned to Site 3, 0 if not X24 = 1 if Room 2 assigned to Site 4, 0 if not X32 = 1 if Room 2 assigned to Site 2, 0 if not X33 = 1 if Room 2 assigned to Site 3, 0 if not X34 = 1 if Room 2 assigned to Site 4, 0 if not X42 = 1 if Room 2 assigned to Site 2, 0 if not X43 = 1 if Room 2 assigned to Site 3, 0 if not X44 = 1 if Room 2 assigned to Site 4, 0 if not

Room 1 Room 2

Room 3 Room 4

Site 1 Site 2

Site 3 Site 4

R1, S1 R4, S2

R2, S3 R3, S4

Final Layout

ABSMODEL 2

Subject to

1

1 1

n

i

n

ijjijiijij yyxxfcMinimize

|xi – xj| + |yi – yj| > 1 i=1,2,...,n–1; j=i+1,...,nxi, yi = integer i=1,...,n

Facility i

Facility j

H

V

HRLVRL

xj

xi

yi

yj

Decision Variablesx – horizontal distance between center of facility I and vertical reference liney - Vertical distance between center of facility I and horizontal reference line

Parameters:Use the ABSMODEL 1 parameters

ABSMODEL 2 Example( using the previous example)

Trip Matrix( trips between each offices)

LonBank Inc,.Dimension: 2’ X 2’


Solution: Model

Objective Function

Subject to

Mangune


OPTIMAL SOLUTION

Objective FunctionMin = 83.00

Variables (ft)X1 = 1Y1 = 1X2 = 2Y2 = 1X3 = 1Y3 = 2X4 = 2Y4 = 2

Verti

cal R

efer

ence

Lin

e (V

RL)

Horizontal Reference Line (HRL)

3 4

1 2

1’

2’

2’

2’

2’ 1’ 1’

1’

MODELS FOR THE MULTIROW LAYOUT PROBLEM WITH FACILITIES OF UNEQUAL AREA

ABSMODEL 3Parameters: n number of departments in the problem cij cost of moving a unit load by a unit distance between

departments i and j fij number of unit loads between departments i and j li length of the horizontal side of department i dij minimum distance by which departments i and j are to be

separated horizontally H horizontal dimension of the floor plan li length of the horizontal side of facility i bi length of the vertical side of facility i dhij horizontal clearance betweenfacilities i and j dvij vertical clearance betweenfacilities i and j

ABSMODEL 3

Subject to

Minimize

|xi – xj| +Mzij> 0.5(li+lj)+dhij i=1,2,...,n–1; j=i+1,...,n|yi – yj| +M(1-zij)> 0.5(bi+bj)+dvij i=1,2,...,n–1; j=i+1,...,nzij(1-zij) = 0 i=1,2,...,n–1; j=i+1,...,nxi, yi > 0 i=1,...,n

1

1 1

n

i

n

ijjijiijij yyxxfc

Facility i

Facility j

HRLVRL

xiyi

yj

xj

bi

li

dvij

dhij lj

bj

ABSMODEL 3 Example

Office Trips Matrix

O 1 2 3 4 5 Office Dimensions (in feet)

f 1 - 10 15 20 0 1 25 x 2

f 2 10 - 30 35 10 2 25 x 20

[fij] = i 3 15 30 - 10 20 3 35 x 30

c 4 20 35 10 - 15 4 30 x 20

e 5 0 10 20 15 - 5 35 x 20

Insurance Company

LINEAR MIXED-INTEGER MODELSFOR THE SINGLE-ROW LAYOUT PROBLEM

LMIP 1

1

1 1

n n

ij ij ij iji j i

c f x x

Minimize

Subject to 0.5 , 1,2,... 1; 1,...,i j ij i j ijx x Mz l l d i n j i n

(1 ) 0.5 1,2,..., 1; 1,...,j i ij i j ijx x M z l l d i n j i n

1,2,..., -1; 1,...i j ij ijx x x x i n j i n

, 1, 2,..., -1; 1,...ij ijx x i n j i n

0 or 1 i=1,2,..., -1; 1,...ijz n j i n

0 i=1,2,...,ix n

LMIP 1 ExampleMachine Dimensions Horizontal Clearance Matrix Flow Matrix

1 2 3 4 5 1 2 3 4 5

1 25x20 1 - 3.5 5.0 5.0 5.0 1 - 25 35 50 0

2 35x20 2 3.5 - 5.0 3.0 5.0 2 25 - 10 15 20

3 30x30 3 5.0 5.0 - 5.0 5.0 3 35 10 - 50 10

4 40x20 4 5.0 3.0 5.0 - 5.0 4 50 15 50 - 15

5 35x35 5 5.0 5.0 5.0 5.0 5.0 5 0 20 10 15 -

Furniture Manufacturer

LMIP 1 Example Solution

Objective FunctionSubject to

LMIP 1 Example Solution

facilities and design chapter 5

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