Mathematical Models for Facility Location
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Transcript of Mathematical Models for Facility Location
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)