Nature of Location Decisions
Location decisions are strategic decisions.
• The reasons for location decisions
• Growth
– Expand existing facilities
– Add new facilities
• Production Cost
• Depletion of Resources
Factors Affecting Plant location
Regional Factors
Site Factors
Community Factors
Multiple Plant Strategies
Regional; Raw Material
1-Location of raw material
Raw material oriented factories;
weight of input >>> weight of output
Iron ore
Coal
Steel
plant
Iron & Steel
Alumina
Coke
Electricity
Aluminum
plant
Aluminum
Regional; Raw Material
These types of plants tends to be closer to
the raw material resources.
Indeed row material or any other important
input.
Regional; Market
2-Location of market
Market oriented plants;
Space required for output >>>
space required for input.
Car manufacturing, Appliances
Regional; Labor, Water, Electricity
3-Labor, water, Electricity
Availability of skilled labor, productivity and
wages, union practices
Availability of water; Blast furnace requires a
high flow of water
Availability of electricity; Aluminum plant strongly
depends on availability and cost of electricity,
it dominates all other inputs.
Community
1-Quality of Life;
Cost of living, housing, schools, health care,
entertainment, church
2- Financial support;
Tax regulations, low rate loans for new industrial
and service plants
Site
1-Land;
Cost of land, development of infrastructure.
2-Transportation;
Availability and cost of rail road, highways, and
air transportation.
3-Environment;
Environmental and legal regulations and
restrictions
Decentralization
Small is beautiful; Instead of a single huge
plant in one location, several smaller plants in
different locations
Decentralization based on product
Decentralization based on geographical area
Decentralization based on process
Decentralization based on Product
Each product or sub-set of products is made in one
plant
Each plant is specialized in a narrow sub-set of
products.
Lower operating costs due to specialization.
Decentralization Based on Geographical Area
Each plant is responsible for a geographical region,
Specially for heavy or large products.
Lower transportation costs.
Decentralization Based on Process
Car industry is an example.
Different plants for engine, transmission, body
stamping, radiator.
Specialization in a process results in lower costs and
higher quality.
Since volume is also high, they also take advantage of
economy of scale.
However, coordination of production of all plants
becomes an important issue and requires central
planning and control
Trends in Global Locations
• Foreign producers locating in U.S.
– “Made in USA”
– Currency fluctuations
• Just-in-time manufacturing techniques
• Focused factories
• Information highway
BEP in Location Analysis
• Cost-volume Analysis
– Determine fixed and variable costs
– Plot total costs
– Determine lowest total costs
Example
Fixed and variable costs for
four potential locations
Location
A
B
C
D
Fixed
Cost
$250,000
100,000
150,000
200,000
Variable
Cost
$11
30
20
35
Solution
Fixed
Costs
A
B
C
D
$250,000
100,000
150,000
200,000
Variable
Costs
$11(10,000)
30(10,000)
20(10,000)
35(10,000)
Total
Costs
$360,000
400,000
350,000
550,000
Graphical Solution
$(000)
800
700
600
500
400
300
200
100
0
0
D
B
C
A
A Superior
C Superior
B Superior
2
4
6
8
10
Annual Output (000)
12
14
16
Center of Gravity ; Single Facility Location
Center of gravity is a method to find
the optimal location of a single facility
The single facility is serving a set of demand centers
or
It is being served by a set of supply centers
The objective is to minimize the total transportation
Transportation is Flow ×Distance
Examples of Single Facility Location Problem
There are a set of demand centers in different
locations and we want to find the optimal location for
a Manufacturing Plant or
a Distribution Center (DC) or
a Warehouse
to satisfy the demand of the demand centers
or
There are a set of suppliers for our manufacturing
plant in different locations and we want to find the
optimal location for our Plant to get its required
inputs
The objective is to minimize total Flow × Distance
Center of Gravity ; Single Facility Location
Suppose we have a set of demand points.
Suppose demand of all demand points are equal.
Suppose they are located at locations Xi, Yi
Where is the best position for a DC to satisfy
demand of these points
Distances are calculated as straight line not rectilinear.
There is another optimal solution for the case when
distances are rectilinear.
Optimal Single Facility Location
The coordinates of the optimal location of the DC is
X

X
n
Y

Y
i
n
i
Example
We have 4 demand points.
Demand of all demand points are equal.
Demand points are located at the following locations
Example
Where is the optimal location for the center serving
theses demand points
(3,5)
(8,5)
(5,4)
(2,2)
Solution
Where is the optimal location for the center serving
theses demand points (2,2) (3,5) (5,4) (8,5)
X

X
n
Y

Y
i
n
i
2358

 4.5
4
25 45

4
4
Solution
The optimal location for the center serving
theses demand points
(3,5)
(8,5)
(5,4)
(2,2)
Center of Gravity ; Single Facility Location
Suppose we have a set of demand points.
Suppose they are located at locations Xi, Yi
Demand of demand point i is Qi.
Now where is the best position for a DC to satisfy
demand of these points
Again; the objective is to minimize transportation.
Optimal Single Facility Location
The coordinates of the optimal location of the DC is
QX

X
Q
i
i
QY

Y
Q
i i
i
i
Example
Where is the optimal location for the center serving
theses demand points
900 (3,5)
100 (8,5)
200
(5,4)
800 (2,2)
Solution
Where is the optimal location for the center serving
theses demand points 800 : (2,2) 900 : (3,5) 200 : (5,4)
QX

X
Q
i
100 : (8,5)
i

i
(800)2  (900)3  (200)5  (100)8

800  900  200  100
X  3.05
Solution
Where is the optimal Y location for the center serving
theses demand points 800 : (2,2) 900 : (3,5) 200 : (5,4)
QY

Y
Q
i i

100 : (8,5)
i
(800)2  (900)5  (200)4  (100)5

800  900  200  100
Y  3.7
Solution
The optimal location for the center serving
theses demand points
(900)
(100)
(200)
(800)
Example
Where is the optimal location for the center serving
theses demand points
900 (1,3)
100 (6,3)
200
(3,2)
800 (0,0)
Solution
Where is the optimal location for the center serving
theses demand points
800 : (0,0) 900 : (1,3) 200 : (3,2) 100 : (6,3)
QX

X
Q
i
i

i
( 800 )( 0 )  ( 900 )( 1 )  ( 200 )( 3 )  ( 100 )( 6 )

800  900  200  100
X  1.05
Solution
Where is the optimal location for the center serving
theses demand points
800 : (0,0) 900 : (1,3) 200 : (3,2) 100 : (6,3)
QY

Y 
Q
i
i

i
( 800 )( 0 )  ( 900 )( 3 )  ( 200 )( 2 )  ( 100 )( 3 )

800  900  200  100
Y  1.7
Solution
The optimal location for the center serving
theses demand points is at the same location
(900)
(100)
(200)
(800)
The Transportation Problem
D
(demand)
S
(supply)
S
(supply)
D
(demand)
D
(demand)
S
(supply)
Transportation problem : Narrative representation
There are 3 plants, 3 warehouses.
Production of Plants 1, 2, and 3 are 300, 200, 200 respectively.
Demand of warehouses 1, 2 and 3 are 250, 250, and 200 units
respectively.
Transportation costs for each unit of product is given below
1
Plant 2
3
1
16
14
13
Warehouse
2
18
12
15
3
11
13
17
Formulate this problem as an LP to satisfy demand at minimum
transportation costs.
Transportation problem I : decision variables
300
1
x11
1
x12
250
x13
x21
200
2
x23
x31
200
x22
x32
3
x33
2
3
250
200
Transportation problem I : decision variables
x11 = Volume of product sent from P1 to W1
x12 = Volume of product sent from P1 to W2
x13 = Volume of product sent from P1 to W3
x21 = Volume of product sent from P2 to W1
x22 = Volume of product sent from P2 to W2
x23 = Volume of product sent from P2 to W3
x31 = Volume of product sent from P3 to W1
x32 = Volume of product sent from P3 to W2
x33 = Volume of product sent from P3 to W3
We want to minimize
Z = 16 x11 + 18 x12 +11 x13 + 14 x21 + 12 x22 +13 x23 +
13 x31 + 15 x32 +17 x33
Transportation problem I : supply and demand constraints
x11 + x12 + x13 = 300
x21 + x22 + x23 =200
x31 + x32 + x33 = 200
x11 + x21 + x31 = 250
x12 + x22 + x32 = 250
x13 + x23 + x33 = 200
x11, x12, x13, x21, x22, x23, x31, x32, x33  0
Origins
s1
s2
si
1
2
i
We have a set of ORIGINs
Origin Definition: A source of material
- A set of Manufacturing Plants
- A set of Suppliers
- A set of Warehouses
- A set of Distribution Centers (DC)
In general we refer to them as Origins
There are m origins i=1,2, ………., m
sm
m
Each origin i has a supply of si
Destinations
We have a set of DESTINATIONs
Destination Definition: A location with
a demand for material
- A set of Markets
- A set of Retailers
- A set of Warehouses
- A set of Manufacturing plants
In general we refer to them as Destinations
1
2
j
d1
d2
di
There are n destinations j=1,2, ………., n
Each origin j has a supply of dj
n
dn
Transportation Model Assumptions
• Total supply is equal to total demand.
• There is only one route between each pair of
origin and destination
• Items to be shipped are all the same
• for each and all units sent from origin i to
destination j there is a shipping cost of Cij per
unit
Cij : cost of sending one unit of product from origin i to destination j
C11
C12
1
C21
2
C2n
i
C2
1
2
2
C1
j
n
m
n
Xij : Units of product sent from origin i to destination j
X11
1
X12
X21
2
1
2
X22
X2n
i
m
X1n
j
n
The Problem
1
2
The problem is to determine how
much material is sent from each
origin to each destination, such
that all demand is satisfied at the
minimum transportation cost
1
2
i
j
m
n
The Objective Function
1
1
If we send Xij units
2
from origin i to destination j,
2
its cost is Cij Xij
We want to minimize
i
m
Z   Cij X ij
j
n
Transportation problem I : decision variables
200
1
x11
1
x12
150
x13
x21
200
2
x23
x31
200
x22
x32
3
x33
2
3
250
200
Transportation problem I : supply and demand constraints
x11 + x12 + x13
=200
+x21 + x22 + x23
x11
+ x21
x12
+ x22
x13
=200
+x31 + x32 + x33
=200
+ x31
=150
+ x32
+ x23
=250
+ x33
= 200
Transportation Problem Solution Algorithms
Transportation Problem is a special case of LP models.
Each variable xij appears only in rows i and m+j. Furthermore,
The coefficients of all variables are equal to 1 in all constraints.
Based on these properties, special algorithms have been
developed. They solve the transportation problem much faster
than general LP Algorithms. They only apply addition and
subtraction
If all supply and demand values are integer, then the optimal
values for the decision variable will also come out integer. In
other words, we use linear programming based algorithms to
solve an instance of integer programming problems.
Data for the Transportation Model
Supply
Supply
Supply
Demand
Demand
Demand
• Quantity demanded at each destination
• Quantity supplied from each origin
• Cost between origin and destination
Data for the Transportation Model
20
Supply Locations
40
50
Waxdale
Brampton
Seaford
$300
$800
$600
$400
$700
$100
$700
Min.
$200
Milw.
Demand Locations
$900
Chicago
Our Task
Our main task is to formulate the problem.
By problem formulation we mean to prepare a tabular
representation for this problem.
Then we can simply pass our formulation ( tabular
representation) to EXCEL.
EXCEL will return the optimal solution.
What do we mean by formulation?
D -1
D -2
D -3
Supply
O -1
600
400
300
20
O -2
700
200
900
40
O -3
800
700
100
50
Demand
30
20
60
110
Excel
Excel
Excel
Excel
Excel
Excel
Assignment; Solve it using excel
We have 3 factories and 4 warehouses.
Production of factories are 100, 200, 150 respectively.
Demand of warehouses are 80, 90, 120, 160 respectively.
Transportation cost for each unit of material from each origin to
each destination is given below.
1
Origin 2
3
1
4
12
8
Destination
2
3
7
7
3
8
10
16
4
1
8
5
Formulate this problem as a transportation problem
Excel : Data