This document discusses transportation problems and their solution. It begins by stating the aim is to find an optimal transportation schedule that minimizes transportation costs. It then provides an example transportation problem table and defines key terms. The remainder of the document explains assumptions of transportation models, their applications, and steps to solve them. It covers obtaining an initial basic feasible solution using methods like the Northwest Corner Rule, Least Cost Method, and Vogel's Approximation Method. It also discusses obtaining the optimal basic solution using the Stepping Stone Method.

1. Transportation Problem
By :Deepika Bansal(14MBA5012)

2. Aim of Transportation Model
To find out optimum transportation
schedule keeping in mind cost of
transportation to be minimized.

3. Transportation Problem
Albuquerque
(300 units
required)
Des Moines
(100 units
capacity)
Evansville
(300 units
capacity)
Fort Lauderdale
(300 units
capacity)
Cleveland
(200 units
required)
Boston
(200 units
required)
Figure C.1

4. 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.

5. CONT……
• 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

6. IN SIMPLE LANG……
• “The transportation problem is to transport
various amounts of a single homogenous
commodity, which are initially stored at
various origins , to different destinations in
such a way that the total transportation cost is
a minimum”.

7. Assumptions of the Model
• Availability of the quantity.
• Transportation of items.
• Cost per unit.
• Independent cost.
• Objective.

8. Application of Transportation Problem
 Minimize shipping costs
 Determine low cost location
 Find minimum cost production schedule

9. TERMINOLOGY USED IN
TRANSPORTATIONAL MODEL
 Feasible solution: Non negative values of xij where i=1, 2……….m
and j=1, 2,…n which satisfy the constraints of supply and demand is
called feasible solution.
 Basic feasible solution: If the no of positive allocations are (m+n-1).
 Optimal solution: A feasible solution is said to be optimal solution if
it minimizes the total transportation cost.
 Balanced transportation problem: A transportation problem in
which the total supply from all sources is equal to the total demand
in all the destinations.
 Unbalanced transportation problem: Problems which are not
balanced are called unbalanced.
 Matrix terminology: In the matrix, the squares are called cells and
form columns vertically and rows horizontally.
 Degenerate basic feasible solution: If the no. of allocation in basic
feasible solutions is less than (m+n-1).

10. 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

11. Steps to solve a Transportation Model
• Formulate the problem and setup in the
matrix form.
• Obtain the Initial Basic Feasible solution.
• Test the initial solution for optimality.
• Updating the solution.

12. GENERAL CONEPTS
ROW
COLUMN

13. CONT…………
• To solve the transportation problem, it is required
that the sum of the supplies at the sources equal
the sum of the demands at the destinations. If
the total supply is greater than the total demand,
a dummy destination is added with demand
equal to the excess supply, and shipping costs
from all sources are zero. Similarly, if total supply
is less than total demand, a dummy source is
added.

14. Phases of Solution of Transportation
Problem
• Phase I- obtains the initial basic feasible
solution
• Phase II-obtains the optimal basic solution

15. Initial Basic Feasible Solution
North West Corner Rule (NWCR)
Least Cost Method
Vogle Approximation Method (VAM)

16. North- West Corner Method (NWCM)
• The simplest of the procedures, used to
generate an initial feasible solution is, NWCM.
It is so called because we begin with the North
West or upper left corner cell of our
transportation table.

17. North-West Corner Method
Step1: Select the upper left (north-west) cell of the
transportation matrix and allocate the maximum
possible value to X11 which is equal to min(a1,b1).
Step2:
• If allocation made is equal to the supply available at
the first source (a1 in first row), then move vertically
down to the cell (2,1).
• If allocation made is equal to demand of the first
destination (b1 in first column), then move horizontally
to the cell (1,2).
• If a1=b1 , then allocate X11= a1 or b1 and move to cell
(2,2).
Step3: Continue the process until an allocation is made
in the south-east corner cell of the transportation table.

18. Example: Solve the Transportation Table to find Initial Basic
Feasible Solution using North-West Corner Method.
Total Cost =19*5+30*2+30*6+40*3+70*4+20*14
= Rs. 1015
Supply
19 30 50 10
5 2
70 30 40 60
6 3
40 8 70 20
4 14
Demand 34
S1
S2
S3
7
9
18
5 8 7 14
D1 D2 D3 D4

19. Least-Cost Method
• Least-Cost Method consist in allocating as
much as possible in the lowest cost cell and
then further allocation is done in th cell with
second lowest cost cell and so on.

20. Least-Cost Method
Step1: Select the cell having lowest unit cost in the entire
table and allocate the minimum of supply or demand values
in that cell.
Step2: Then eliminate the row or column in which supply or
demand is exhausted. If both the supply and demand
values are same, either of the row or column can be
eliminated.
In case, the smallest unit cost is not unique, then select the
cell where maximum allocation can be made.
Step3: Repeat the process with next lowest unit cost and
continue until the entire available supply at various sources
and demand at various destinations is satisfied.

21. Supply
19 30 50 10
70 30 40 60
40 8 70 20
8
Demand 34
S3 18
5 8 7 14
S1 7
S2 9
D1 D2 D3 D4

22. Supply
19 30 50 10
70 30 40 60
40 8 70 20
8
Demand 34
S3 18
5 8 7 14
S1 7
S2 9
D1 D2 D3 D4
Supply
19 50 10
7
70 40 60
40 70 20
Demand 34
D1 D3 D4
7 14
7
9
S1
S2
S3 10
5
Supply
70 40 60
40 70 20
7
Demand 34
S2 9
D1 D3 D4
S3 10
5 7 7
Supply
70 40
7
40 70
Demand 34
9
3S3
5 7
S2
D1 D3 Supply
70
2
40
3
Demand 345
S2 2
S3 3
D1

23. • The total transportation cost obtained by
this method
= 8*8+10*7+20*7+40*7+70*2+40*3
= Rs.814
Here, we can see that the Least Cost
Method involves a lower cost than the
North-West Corner Method.

24. Vogel’s Approximation Method (VAM)
• In this method, each allocation is made on the
basis of the opportunity (or penalty or extra)
cost that would have been incurred if
allocations in certain cells with minimum unit
transportation cost were missed. In this
method allocations are made so thet the
penalty cost is minimized.

25. Vogel’s Approximation Method (VAM)
Step1: Calculate penalty for each row and column by taking the
difference between the two smallest unit costs. This penalty or extra
cost has to be paid if one fails to allocate the minimum unit
transportation cost.
Step2: Select the row or column with the highest penalty and select the
minimum unit cost of that row or column. Then, allocate the
minimum of supply or demand values in that cell. If there is a tie,
then select the cell where maximum allocation could be made.
Step3: Adjust the supply and demand and eliminate the satisfied row or
column. If a row and column are satisfied simultaneously, only of
them is eliminated and the other one is assigned a zero value.Any
row or column having zero supply or demand, can not be used in
calculating future penalties.
Step4: Repeat the process until all the supply sources and demand
destinations are satisfied.

26. Supply Row Diff.
19 30 50 10
70 30 40 60
40 8 70 20
8
Demand 34
Col.Diff.
9
10
12
21 22 10 10
D1 D2 D3
S1
S2
S3
5 8 7
D4
14
7
9
18
Supply Row Diff.
19 50 10
5
70 40 60
40 70 20
Demand 34
Col.Diff.
D1 D3 D4
S1 7
S2 9
S3 10
5 7 14
21 10 10
9
20
20
Supply Row Diff.
50 10
40 60
70 20
10
Demand 34
Col.Diff. 10 10
40
20
50
D3 D4
S1 2
S2 9
S3 10
7 14
Supply Row Diff.
50 10
2
40 60
Demand 34
Col.Diff.
20
10 50
40
7 4
S1 2
S2 9
D3 D4
Supply Row Diff.
40 60
7 2
Demand 34
Col.Diff.
20
7 2
S2 9
D3 D4

27. The total transportation cost obtained by this method
= 8*8+19*5+20*10+10*2+40*7+60*2
= Rs.779
Here, we can see that Vogel’s Approximation Method
involves the lowest cost than North-West Corner Method
and Least Cost Method and hence is the most preferred
method of finding initial basic feasible solution.

28. Optimum Basic Solution
Stepping Stone Method
Modified Distribution Method a.k.a. MODI Method

29. 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

30. 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.

31. Problem Illustration
FROM
TO A.
ALBUQUERQUE
B.
BOSTON
C.
CLEVELAND
FACTORY
CAPACITY
D. DES MOINES 5 4 3
100
E. EVANSVILLE 8 4 3
300
F. FORT
LAUDERDALE
9 7 5
300
WAREHOUSE
DEMAND 300 200 200 700

32. Initial Feasible Solution using
Northwest Corner Rule
FROM
TO A.
ALBUQUERQUE
B.
BOSTON
C.
CLEVELAND
FACTORY
CAPACITY
D. DES MOINES 5 4 3
100
E. EVANSVILLE 8 4 3
300
F. FORT
LAUDERDALE
9 7 5
300
WAREHOUSE
DEMAND 300 200 200 700
100
200 100
100 200
IFS= DA + EA +EB + FB + FC = 100(5) + 200(8) + 100(4) + 100(7) + 200(5)
= 500 + 1600 + 400 + 700 + 1000 = 4200

33. $5
$8 $4
$4
+ -
+-
Optimizing Solution using
Stepping-Stone Method
To (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
100
200
200
+-
-+
1
100
201 99
99
100200Figure C.5
Des Moines-
Boston index
= $4 - $5 + $8 - $4
= +$3

34. Stepping-Stone Method
To (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
100
200
200
Figure C.6
Start
+-
+
-+
-
Des Moines-Cleveland index
= $3 - $5 + $8 - $4 + $7 - $5 = +$4

35. Stepping-Stone Method
To (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
100
200
200
Evansville-Cleveland index
= $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)

36. 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

37. Stepping-Stone Method
To (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
100
200
200
Figure C.7
+
+-
-
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

38. Stepping-Stone Method
To (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
200
100
100
200
Figure C.8
Total Cost = $5(100) + $8(100) + $4(200) + $9(100) + $5(200)
= $4,000

39. 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

40. Special Issues in Modeling
Figure C.9
New
Des Moines
capacity
To (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
250
850
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
50200
250
50
150
Dummy
150
0
0
0
150
Total Cost = 250($5) + 50($8) + 200($4) + 50($3) + 150($5) + 150(0)
= $3,350

41. 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

42. To Customer
1
Customer
2
Customer
3
Warehouse 1
Warehouse 2
Warehouse 3
Customer
demand 100 100 100
Warehouse
supply
120
80
100
300
$8
$7
$2
$9
$6
$9
$7
$10
$10
From
Special Issues in Modeling
0 100
100
80
20
Figure C.10
Initial solution is degenerate
Place a zero quantity in an unused square and
proceed computing improvement indices

45. STEPS
1.Construct a transportation table with the given cost
of transportation and rim requirement.
2.Determine IBFS.
3.For current basic feasible solution check degeneracy
and non-degeneracy.
rim requirement=stone square(non-degeneracy)
rim requirement != stone square(degeneracy)
4.Find occupied matrix.
5.Find unoccupied matrix.

46. Steps (contd…)
6.Find opportunity cost of unoccupied cells using
formula:
opportunity cost =actual cost-implied cost
dij= cij - (ri+kj)
7.Unoccupied cell evaluation:
(a) if dij>0 then cost of transportation unchanged.
(b) if dij=0 then cost of transportation unchanged.
(c) if dij<0 then improved solution can be obtain and
go to next step.

47. STEPS(contd…)
8.Select an unoccupied cell with largest –ve
opportunity cost among all unoccupied cell.
9.Construct closed path for the occupied cells
determined in step 8.
10.Assign as many as units as possible to the
unoccupied cell satisfying rim conditions.
11.Go to step 4 and repeat procedure until
All dij>=0 i.e reached to the optimal solution.

48. SPECIAL CASES
• Balanced problem
• Unbalanced problem
• Non -degeneracy
• Degeneracy :occurs in two cases
1. Degeneracy occurs in initial basic solution.
2. Degeneracy occurs in during the test of optimality.
• Profit maximization

49. .
PROBLEM1:Shipping costs are Rs. 10 per kilometer.. What shipping schedule should be used. if
the matrix given below the kilometers from source to destination.
destination
a b c availability
x 50 30 220 1
y 90 45 170 3
z 50 200 50 4
Requirement 3 3 2 8

50. A B C
X
Y
Z
30 220
50
90 45 170
50 200 50
30 220
1
2
3
2
2
2 2
3
P1 P2 P3 P4
20 20 20 20
45 45 45 _
150 150 _ _
P1 40 15 120
P2 40 15 _
P3 40 15 _
IBFS BY USING VAM METHOD

51. A B C
X
Y
Z
30 220
50
90 45 170
50 200 50
30 220
1
2
3
2
2
2 2
3
P1 P2 P3 P4
20 20 20 20
45 45 45 _
150 150 _ _
P1 40 15 120
P2 40 15 _
P3 40 15 _
IBFS BY USING VAM METHOD

52. A B C
X
Y
Z
30 220
50
90 45 170
50 200 50
30 220
1
2
3
2
2
2 2
3
P1 P2 P3 P4
20 20 20 20
45 45 45 _
150 150 _ _
P1 40 15 120
P2 40 15 _
P3 40 15 _
IBFS BY USING VAM METHOD

53. 3030 220
90 45 170
50 200 50
1 E
3
2 2
STONE SEQUARE =4
RIM REQUIREMENT =M+N-1=3+3-1=5
DEGENERACY
NOW STONE SEQUARE =5
RIM REQUIREMENT =5
NON –DEGENERACY

54. 50 30
45
50 50
50 30 50
0
15
0
OCCUPIED MATRIX UNOCCUPIED MATRIX
10525
170
50
65
30
65
170

55. OPTIMUL SOLUTION
X A 50*1=50
X B 30*E=_
Y B 45*3=135
Z A 50*2=100
Z C 50*2=100
385 *10=3850

56. PROBLEM2:DETERMINE THE OPTIMUM SOLUTION FOR THE COMPANY OF
TRASPOTATION PROBLEM(USING NWCM AND MODI METHOD)
8 8 15
15 10 17
3 9 10
REQUIREMENT 150 80 50
120
80
80
CAPACITY
F1
F2
F3
W1 W2 W3
WAREHOUSE
FACTORY

57. W1 W2 W3
F1
F2
F3
8 8 15
15
3
10 17
9 10
120
30 50
30 50
150 80 50
120
80
80
IBFS WITH NWCM

58. OCCUPIED MATRIX UNOCCUPIED MATRIX
8
15 10
9 10
15 10 11
-7
0
-1
15 10 11
-7
0
-1
5 11
6
-11
3 4
11
14

59. 1588
15 10 17
3 9 1
0
10
120
30 50
30 50
+
+_
_
8
15
3
8 15
10 17
9 10
120
E
80
30 50
STONE SEQUARE=RIM REQUIREMENT
DEGENERACY OCCUAR
LOOP CONSTRUCT

60. OCCUPIED MATRIX UNOCCUPIED MATRIX
8 8
10
3 10
3 3 10 3 3 10
5
7
0
5
7
0
5
6
0
0
15
1710
3

61. OPTIMUM SOLUTION
F1 W1 8*120 =960
F1 W2 8*E = _
F2 W2 10*80 =800
F3 W1 3*30= 90
F3 W3 10*50 =500
2,350 RS