Transportation Problem

Transportation Problem

By : Alvin G. Niere
Misamis University

Aim of Transportation Model

To find out optimum transportation
schedule keeping in mind cost of
transportation to be minimized.

What is a Transportation Problem?
• The transportation problem is a special type of
LPP where the objective is to minimize the cost of
distributing a product from a number of sources
or origins to a number of destinations.

• Because of its special structure the usual simplex
method is not suitable for solving transportation
problems. These problems require special
method of solution.

The Transportation Problem
• The problem of finding the minimum-cost
distribution of a given commodity
from a group of supply centers (sources) i=1,…,m
to a group of receiving centers (destinations)
j=1,…,n
• Each source has a certain supply (si)
• Each destination has a certain demand (dj)
• The cost of shipping from a source to a
destination is directly proportional to the number
of units shipped

Simple Network Representation
Sources Destinations

Supply s1 1 Demand d1
1

Supply s2 2
2 Demand d2
…

…
xij

n Demand dn
Supply sm m

Costs cij

Transportation-5

Application of Transportation Problem

 Minimize shipping costs

 Determine low cost location

 Find minimum cost production schedule

 Military distribution system

Two Types of Transportation Problem

• Balanced Transportation Problem
where the total supply equals total demand

• Unbalanced Transportation Problem
where the total supply is not equal to the
total demand

Phases of Solution of Transportation
Problem

• Phase I- obtains the initial basic feasible
solution

• Phase II-obtains the optimal basic solution

Initial Basic Feasible Solution

North West Corner Rule (NWCR)

Row Minima Method

Column Minima Method

Least Cost Method

Vogle Approximation Method (VAM)

Optimum Basic Solution

Stepping Stone Method

Modified Distribution Method a.k.a. MODI Method

Optimum Basic Solution:
Stepping-Stone Method
1. Select any unused square to evaluate
2. Beginning at this square, trace a closed path
back to the original square via squares that
are currently being used
3. Beginning with a plus (+) sign at the unused
corner, place alternate minus and plus signs at
each corner of the path just traced

4. Calculate an improvement index by first
adding the unit-cost figures found in each
square containing a plus sign and subtracting
the unit costs in each square containing a
minus sign
5. Repeat steps 1 though 4 until you have
calculated an improvement index for all
unused squares. If all indices are ≥ 0, you have
reached an optimal solution.

Problem Illustration

TO A. B. C. FACTORY
FROM ALBUQUERQUE BOSTON CLEVELAND CAPACITY

D. DES MOINES 5 4 3
100

E. EVANSVILLE 8 4 3
300

F. FORT 9 7 5
LAUDERDALE 300

WAREHOUSE
DEMAND 300 200 200 700

Initial Feasible Solution using
Northwest Corner Rule
TO A. B. C. FACTORY
FROM ALBUQUERQUE BOSTON CLEVELAND CAPACITY

D. DES MOINES 5 4 3
100 100

E. EVANSVILLE 8 4 3
200 100 300

F. FORT 9 7 5
LAUDERDALE 100 200 300

WAREHOUSE
DEMAND 300 200 200 700

IFS= DA + EA +EB + FB + FC = 100(5) + 200(8) + 100(4) + 100(7) + 200(5)
= 500 + 1600 + 400 + 700 + 1000 = 4200

Optimizing Solution using
To (A) (B) (C) Factory
From Albuquerque Boston Cleveland capacity

$5 $4 $3
(D) Des Moines 100
- 100 Des Moines-
+
$8 $4 $3
Boston index
(E) Evansville 200 100 300
+ - = $4 - $5 + $8 - $4
$9 $7 $5
(F) Fort Lauderdale 100 200 300 = +$3
Warehouse
requirement 300 200 200 700

99 $5 1 $4
100
- +

+ -
201 $8 99 $4
Figure C.5 200 100

$5 $4 Start $3
(D) Des Moines 100 100
- +
$8 $4 $3
+ -
$9 $7 $5
(F) Fort Lauderdale 100 200 300
+ -
Warehouse

Des Moines-Cleveland index
Figure C.6 = $3 - $5 + $8 - $4 + $7 - $5 = +$4

$5 $4 $3

$8 $4 $3

$9 $7 $5
Evansville-Cleveland index
Warehouse = $3 - $4 + $7 - $5 = +$1
(Closed path = EC - EB + FB - FC)
Fort Lauderdale-Albuquerque index
= $9 - $7 + $4 - $8 = -$1
(Closed path = FA - FB + EB - EA)

1. If an improvement is possible, choose the
route (unused square) with the largest
negative improvement index
2. On the closed path for that route, select the
smallest number found in the squares
containing minus signs
3. Add this number to all squares on the closed
path with plus signs and subtract it from all
squares with a minus sign

$5 $4 $3

$8 $4 $3
- +
$9 $7 $5
+ -
Warehouse
1. Add 100 units on route FA
2. Subtract 100 from routes FB
3. Add 100 to route EB
4. Subtract 100 from route EA
Figure C.7

$5 $4 $3

$8 $4 $3

$9 $7 $5

Warehouse

Total Cost = $5(100) + $8(100) + $4(200) + $9(100) + $5(200)
= $4,000
Figure C.8

Special Issues in Modeling

 Demand not equal to supply
 Called an unbalanced problem
 Common situation in the real world
 Resolved by introducing dummy sources
or dummy destinations as necessary with
cost coefficients of zero

Total Cost
Special50($8) + 200($4)in50($3) + 150($5) + 150(0)
= 250($5) +
Issues + Modeling
= $3,350
Dummy capacity
From Albuquerque Boston Cleveland

$5 $4 $3 0

$8 $4 $3 0
(E) Evansville 50 200 50 300

$9 $7 $5 0

Warehouse
requirement 300 200 200 150 850

New
Figure C.9 Des Moines
capacity


 Degeneracy
 To use the stepping-stone
methodology, the number of occupied
squares in any solution must be equal to
the number of rows in the table plus the
number of columns minus 1
 If a solution does not satisfy this rule it is
called degenerate

To Customer Customer Customer Warehouse
From 1 2 3 supply
$8 $2 $6
Warehouse 1 100 100

$10 $9 $9
Warehouse 2 0 100 20 120

$7 $10 $7
Warehouse 3 80 80

Customer
demand 100 100 100 300

Initial solution is degenerate
Place a zero quantity in an unused square and
Figure C.10 proceed computing improvement indices

Transportation Problem

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Transportation Problem