Mathematical Models for Facility Location

Prof Arun Kanda

Department of Mech Engg

Indian Institute of Technology, Delhi

A Case Study

A Decision Model for a Multiple Objective

Plant Location Problem

Prem Vrat And Arun Kanda

INTEGRATED MANAGEMENT, July 1976, Page 27-33

Objective of Location

• To set up a straw board plant (Packaging material) from industrial waste

Plant

Sources of Industrial waste

Industries needing packaging material

Relevant Factors for Plant LocationNotation Factor

A Nearness to raw material source B Availability and dependability

of power

C Transport facilities D Labour supply E Employee facilities F Competition for the market

G Nearness to market H Govt. Incentives

I Cost of land

Objectives

Weightages to various objectives

O1 (W1)

O2 (W2)

…………. …………. On

(Wn)

Measures of effectiveness of various alternatives

A1 P11 P12 …………. P1n j

n

jjWPE

111

A2 P21 P22 ………… P2n

n

jjWjPE

122

AL

TE

RN

AT

IVE

S . . . .

. . . .

………… ………… ………… …………

. . . .

Am Pm1 Pm2 ………… pmn

n

jPmjWjEm

1

Triangular Matrix

O2 O3 …….. On Scores

O1 O1 - 2 S1

O2 S2

O3 On Sn

Applying Pareto PrincipleB C D E F G H I

A A-2 A-1 A-3 A-3 F-1 A-2 A-2 A-3

B C-1 B-1 B-3 F-2 G-2 H-2 I-1

C C-1 C-3 F-2 G-1 H-1 C-1

D D-3 F-3 G-2 H-2 I-2

E F-3 G-3 H-3 I-3

Major difference = 3 F F-1 F-1 F-1

Medium difference=2 G H-2 I-1

Minor difference = 1 H H-2

SUMMARYNotation Factor Total

Points weightage factor(%)

A Nearness to raw material source 16 23.0

B Availability and dependability of power

4 5.7

C Transport facilities 6 8.6 D Labour supply 3 4.3

E Employee facilities 0 0.0

F Competition for the market 14 20.0

G Nearness to market 8 11.4 H Govt. Incentives 12 17.0

I Cost of land 7 10.0

Total 10 100

Decision Matrix for Alternative Locations

A B C D F G H I Total Points

Alternative Location

.230 .057 .086 .043 .200 .114 .170 .100

Panipat 90 80 100 50 100 50 90 90 86.01 Sonepat 80 100 80 70 100 85 80 85 85.98 Rohtak 100 80 90 70 100 60 100 100 *91.16 Meerut 90 50 80 90 80 60 70 60 75.05 Faridabad 50 60 90 100 50 100 50 50 61.87 Gurgaon 55 65 50 60 100 95 60 70 71.26 Ghaziabad 60 50 80 100 60 90 50 60 64.60

* Optimal Location.

Normalization I

80

P

20

Poi

nts

Capital Cost

L C H

Normalization II

80

20

Poi

nts

Capital Cost

L L’ H

D A

B

C1

C2

Normalization III

80

20

Poi

nts

Labour Attitudes

| Restive | Satisfactory Cooperative |

60

Normalization IV

On

.

.

.O2

O1

Poi

nts

X1 X2 - - - - - - Xn

Mathematical Models for Facility Location

Single Facility Location

• New lathe in a job shop

• Tool crib in a factory• New warehouse• Hospital, fire station,

police station• New classroom

building on a college campus

• New airfield for a number of bases

• Component in an electrical network

• New appliance in a kitchen

• Copying machine in a library

• New component on a control panel

Problem Statement

• m existing facilities at locations • P1(a1,b1), P2(a2,b2) … Pm(am,bm)• New facility is to be located at point X (x,y)• d(X,Pi) = appropriately defined distance between X

and Pi– Euclidean, Rectilinear, Squared Euclidean– Generalized distance, Network

• The objective is to determine the location X so as to minimize transportation related costs

Sum (i=1,n) wi d(X,Pi), where wi is the weight associated with the ith existing facility (product of Cost/distance & the expected number of annual trips between X and Pi)

Single Facility Location

P1 (w1)

P3 (w3)

Pn (wn)

X

d(X,P1)

d(X,P2) d(X,P n-1)

d(X,Pn)

P2 (w2)

Pn-1 (wn-1)d(X,P3)

Commonly Used Distances

Rectilinear: | (x-ai) | +| (y-bi)|

Euclidean : [ (x-ai)2 + (y-bi)2]1/2

Squared Euclidean: [(x-ai)2 +(y-ai)2 ]

Other , Network

X (x,y)

Pi (ai,bi)

X (x,y)

Pi (ai,bi)

X (x,y)

Pi(ai,bi)

Rectilinear Distances

• Z = Total cost • = Sum (i =1,n) [ wi | (x-ai) + (y-bi)|]• = Sum (i=1,n) [wi |(x-ai)| + wi |(y-bi)| ]• = Sum (i=1,n) wi |(x-ai)| + Sum (i=1,n) wi |(y-bi)| • = f1(x) + f2(y)• Thus to minimize Z we need to minimize

f1(x) and f2(y) independently.

Example 1(Rectilinear Distance Case)

• A service facility to serve five offices located at (0,0), (3,16),(18,2) (8,18) and (20,2) is to be set up. The number of cars transported per day between the new service facility and the offices equal 5, 22, 41, 60 and 34 respectively.

• What location for the service facility will minimize the distance cars are transported per day?

Solution (x-coordinate) Existing facility

x-coordinate value

Weight Cumulative weight

1 0 5 5

2 3 22 27< 81

4 8 60 87> 81

3 18 41 128

5 20 34 162

x* = 8

Solution (y-coordinate)

Existingfacility

y-coordinatevalue

Weight Cumulativeweight

1 0 5 5

3,5 2 41+34 80< 81

2 16 22 102>81

4 18 60 162

y* = 16

Example 2Squared Euclidean Case

CENTROID LOCATION

x* = Σ wi ai /Σ wi =( 0 x5 + 3x22 + 18x41 + 8x60 + 20x34)/162

= 12.12

y* = Σ wibi/Σ wi = (0x5 + 16x22 + 2x41 + 18x60 + 2x34)/162

= 9.77

(Compare with the median location of (8,16)

R2

R1

Rm

M1

M2

Mn

1

2

m

m+1

m+2

m+n

P

Minimax Problems

*

For the location ofemergency facilitiesour objective wouldbe to minimize the maximum distance

Cost Contours

Increasing Cost

Cost Contourshelp identify alternativefeasible locations

Summary

• Decision Matrix approach to handle multiple objectives in Plant Location

(problem of choosing the best from options)

• Single Facility Location Models– Rectilinear distance– Squared Euclidean– Euclidean distance– (to generate the best from infinite options)

Summary (Contd)

• Notion of Minisum and Minimax problem

(Objective depending on the context)

• Use of Cost Contours to accommodate practical constraints

(Moving from ideal to a feasible solution)