2. Aim of Transportation Model
To find out optimum transportation
schedule keeping in mind cost of
transportation to be minimized.
3. 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.
4. 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
5. 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
6. Application of Transportation Problem
Minimize shipping costs
Determine low cost location
Find minimum cost production schedule
Military distribution system
7. 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
8. Phases of Solution of Transportation
Problem
• Phase I- obtains the initial basic feasible
solution
• Phase II-obtains the optimal basic solution
9. Initial Basic Feasible Solution
North West Corner Rule (NWCR)
Row Minima Method
Column Minima Method
Least Cost Method
Vogle Approximation Method (VAM)
11. 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
12. Stepping-Stone Method
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.
13. 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
14. 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
15. Optimizing Solution using
Stepping-Stone Method
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
16. Stepping-Stone Method
To (A) (B) (C) Factory
From Albuquerque Boston Cleveland capacity
$5 $4 Start $3
(D) Des Moines 100 100
- +
$8 $4 $3
(E) Evansville 200 100 300
+ -
$9 $7 $5
(F) Fort Lauderdale 100 200 300
+ -
Warehouse
requirement 300 200 200 700
Des Moines-Cleveland index
Figure C.6 = $3 - $5 + $8 - $4 + $7 - $5 = +$4
17. Stepping-Stone Method
To (A) (B) (C) Factory
From Albuquerque Boston Cleveland capacity
$5 $4 $3
(D) Des Moines 100 100
$8 $4 $3
(E) Evansville 200 100 300
$9 $7 $5
(F) Fort Lauderdale 100 200 300
Evansville-Cleveland index
Warehouse = $3 - $4 + $7 - $5 = +$1
requirement 300 200 200 700
(Closed path = EC - EB + FB - FC)
Fort Lauderdale-Albuquerque index
= $9 - $7 + $4 - $8 = -$1
(Closed path = FA - FB + EB - EA)
18. Stepping-Stone Method
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
19. Stepping-Stone Method
To (A) (B) (C) Factory
From Albuquerque Boston Cleveland capacity
$5 $4 $3
(D) Des Moines 100 100
$8 $4 $3
(E) Evansville 200 100 300
- +
$9 $7 $5
(F) Fort Lauderdale 100 200 300
+ -
Warehouse
requirement 300 200 200 700
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
20. Stepping-Stone Method
To (A) (B) (C) Factory
From Albuquerque Boston Cleveland capacity
$5 $4 $3
(D) Des Moines 100 100
$8 $4 $3
(E) Evansville 100 200 300
$9 $7 $5
(F) Fort Lauderdale 100 200 300
Warehouse
requirement 300 200 200 700
Total Cost = $5(100) + $8(100) + $4(200) + $9(100) + $5(200)
= $4,000
Figure C.8
21. 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
22. Total Cost
Special50($8) + 200($4)in50($3) + 150($5) + 150(0)
= 250($5) +
Issues + Modeling
= $3,350
To (A) (B) (C) Factory
Dummy capacity
From Albuquerque Boston Cleveland
$5 $4 $3 0
(D) Des Moines 250 250
$8 $4 $3 0
(E) Evansville 50 200 50 300
$9 $7 $5 0
(F) Fort Lauderdale 150 150 300
Warehouse
requirement 300 200 200 150 850
New
Figure C.9 Des Moines
capacity
23. Special Issues in Modeling
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
24. Special Issues in Modeling
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