1. INITIAL BASIC
FEASIBLE SOLUTION

2. TRANSPORTATION PROBLEM
Transport various
quantities of a single
homogeneous commodity
to different destinations in
such a way that total
transportation cost is
minimum.

3. 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).

4. OPTIMAL SOLUTION OF TRANSPORTATION
PROBLEM
Obtain an
optimal solution
by making
Initial Basic successive
STEP1 Feasible STEP2 improvements in
Solution IBFS until no
further decrease
in transportation
cost is possible

5. INITIAL BASIC FEASIBLE SOLUTION
Methods Available
NORTH WEST LOWEST COST VOGEL’S
CORNER ENTRY APPROXIMATION
METHOD(NWCM) METHOD(LCEM) METHOD(VAM)

6. UNBALANCED TRANSPORTATION
PROBLEM
Demand>Supply
Add dummy column in matrix with zero cost
Supply>Demand
Add dummy row in matrix with zero cost

7. DEMAND>SUPPLY
PRODUCT P Q SUPPLY
OFFICE
A 10 15 20
B 20 30 50
DEMAND 30 60 70
90
PRODUCT P Q SUPPLY
OFFICE
A 10 15 20
B 20 30 50
DUMMY 0 0 20
DEMAND 30 60 90
90

8. SUPPLY>DEMAND
PRODUCT P Q SUPPLY
OFFICE
A 10 15 30
B 20 30 60
DEMAND 20 50 90
70
PRODUCT P Q DUMMY SUPPLY
OFFICE
A 10 15 0 20
B 20 30 0 50
DEMAND 20 50 20 90
90

9. METHOD FOR OPTIMAL SOLUTION
Stepping stone
method
Modified
Distribution
Method(MODI)

10. NORTH WEST CORNER METHOD
Most systematic and easiest
method for obtaining initial
feasible basic solution

11. STEPS TO SOLVE THE PROBLEM
Step1: construct an empty m*n matrix, completed with rows & columns.
Step2: indicate the rows and column totals at the end.
Step3: starting with (1,1)cell at the north west corner of the matrix, allocate maximum possible quantity
keeping in view that allocation can neither be more than the quantity required by the respective warehouses
nor more than quantity available at the each supply centre.
Step 4: adjust the supply and demand nos. in the rows and columns allocations.
Step5: if the supply for the first row is exhausted then move down to the first cell in the second row and first
column and go to the step 4.
Step 6: if the demand for the first column is satisfied, then move to the next cell in the second column and first
row and go to step 4.
Step 7: if for any cell, supply equals demand then the next allocation can be made in cell either in the next
row or column.
Step 8: continue the procedure until the total available quantity is fully allocated to the cells as required.

12. MINIMIZATION USING NWCM
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21 25 19

13. Step1: Check whether problem is
balanced
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21 25 19 65
65

14. Allocate P1W1 max quantity such that it
should not exceed supply/requirement
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21/1 25 19 65
P1 can supply 20 units at max so W1 can be allocated 20
units only by P1

15. Check whether the requirements of W1
fulfilled
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21/1 25 19 65
Now allocate the remaining requirement to next north
west corner element i.e. P2W1

16. Allocate 1 to P2W1
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/27
1
P3 4 5 8 17
DEMAND 21/1/0 25 19 65
In this way repeat it for all columns until demand of
all warehouses is fulfilled

17. No. of allocations=No. of rows + No. of columns-1
(For generating optimal solution)
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/27/0
1 25 2
P3 4 5 8 17/0
17
DEMAND 21/1/0 25/0 19/17/0 65
Here no. of rows=3, no. of columns=3, total allocations=5
5=3+3-1

18. TOTAL TRANSPORTATION COST
PLANT WAREHOUSE COST
P1 W1 7*20=140
P2 W2 5*1=5
P2 W2 7*25=175
P2 W2 3*2=6
P3 W3 8*17=136
TOTAL COST 462

19. LEAST COST ENTRY METHOD
This method takes into
consideration the lowest cost
and therefore takes less time to
solve the problem

20. STEPS TO SOLVE THE PROBLEM
• Step1: select the cell with the lowest transportation cost
among all the rows and columns of the transportation
table. If the minimum cost is not unique then select
arbitrarily any cell with the lowest cost.
• Step2: allocate as many units as possible to the cell
determined in step 1 and eliminate that row in which
either capacity or requirement is exhausted.
• Step3:adjust the capacity and the requirement for the
next allocations.
• Step4: repeat the steps1to3 for the reduced table until
the entire capacities are exhausted to fill the
requirements at the different destinations.

21. MINIMIZATION USING LCEM
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21 25 19 65
Step1 :This problem is balanced problem

22. Find out the cell having least cost
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28/9
19
P3 4 5 8 17
DEMAND 21 25 19/0 65
P2W3 is having least cost i.e. 3 so it is allocated 19 units
as per the requirement of W3

23. In this way allocate next minimum costs
until all requirements are fulfilled
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/9/5/0
4 5 19
P3 4 5 8 17/0
17
DEMAND 21/4/0 25/5/0 19/0 65
Now P2W1 and P3W2 are having same cost, we can
allocate any of them but in this case P2W1 will be
allocated because in P3W2 P3 cant supply anymore units

24. Final matrix with all allocations
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/9/5/0
4 5 19
P3 4 5 8 17/0
17
DEMAND 21/4/0 25/5/0 19/0 65

25. No. of allocations=No. of rows + No. of columns-1
(For generating optimal solution)
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/9/5/0
4 5 19
P3 4 5 8 17/0
17
DEMAND 21/4/0 25/5/0 19/0 65
Here no. of rows=3, no. of columns=3, total allocations=5
5=3+3-1

26. TOTAL TRANSPORTATION COST
PLANT WAREHOUSE COST
P1 W2 6*20=120
P2 W1 5*4=20
P2 W2 7*5=35
P2 W3 3*19=57
P3 W1 4*17=68
TOTAL COST 300

27. MAXIMIZATION USING LCEM
INVESTMENT P Q R S AVAILABLE
1 95 80 70 60 70
2 75 65 60 50 40
3 70 45 50 40 90
4 60 40 40 30 10
DEMAND 40 50 60 60 210
Convert maximization into minimization

28. Subtract all the elements from the largest
element thereby giving minimization case.
INVESTMENT P Q R S AVAILABLE
1 95
95 80 70 60 70
2 75 65 60 50 40
3 70 45 50 40 90
4 60 40 40 30 10
DEMAND 40 50 60 60 210
Here 95 is maximum element so subtract all elements
from 95

29. This is the matrix after subtracting all the
elements from 95
INVESTMENT P Q R S AVAILABLE
1 0 15 25 35 70
2 20 30 35 45 40
3 25 50 45 55 90
4 35 55 55 65 10
DEMAND 40 50 60 60 210
This is now minimization case which can be further
solved as discussed before

30. TRANSPORTATION PROBLEM
Categorized into two types:
Minimization Maximization
problem problem

31. MINIMIZATION PROBLEM
In this transportation
cost is given which is to
be minimized.

32. MAXIMIZATION PROBLEM
In this profit is given which is to be maximized.
To solve this problem we convert the problem into minimization.
Conversion is done by selecting the largest element from Profit
Pay off matrix and then subtracting all elements from largest
element including itself.
Reduced matrix obtain becomes minimization case and then
same steps are taken to solve it as is done in minimization
problem

33. VOGEL’S APPROXIMATION METHOD
• BASIS OF ALLOCATION IS UNIT COST PENALTY
• THE SUBSEQUENT ALLOCATIONS IN CELLS ARE
DONE KEEPING IN VIEW THE HIGHEST UNIT
COST
• IBFS OBATINED BY THIS METHOD IS EITHER
OPTIMAL OR VERY NEAR TO OPTIMAL
SOLUTION
• SO AMOUNT OF TIME REQUIRED TO
CALCULATE THE OPTIMUM SOLUTION IS
REDUCED

34. STEPS TO SOLVE THE PROBLEM
Step1: for each row of the table identify the lowest and the next lowest cost
cell. Find their least cost than the difference shall be zero.
Step 2: similarly find the difference of each column and place it below each
column. These differences found in the steps 1 and 2 are also called
penalties.
Step 3: looking at all the penalties. Identify the highest of them and the row
or column relative to that penalty. Allocate the maximum possible units
to the least cost cell in the selected row or column. Ties should be broken
in this order
Maximum difference least cost cell.
Maximum difference tie least cost cell
Maximum unit’s allocations tie arbitrary
Step 4: adjust the supply and demand and cross the satisfied row or column.
Step 5:Recompute the column and row differences ignoring deleted
rows/columns and go to the step3. Repeat the procedure until the entire
column and row totals are satisfied.

35. MINIMIZATION USING VAM
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21 25 19 65
Step1 :This problem is balanced problem

36. Find out ‘Penalties’(P) for each row & column by
subtracting the smallest element from next
smallest element in the row/column
WAREHOUSE W1 W2 W3 SUPPLY P1
PLANT
P1 7 6 9 20 7-6=1
P2 5 7 3 28 5-3=2 In W3 3 is min cost
19
P3 4 5 8 17 5-4=1
DEMAND 21 25 19 65
P1 5-4=1 6-5=1 9-3=6
6 is max penalty
Find out the maximum penalty and in that row/column
find minimum element and allocate it.

37. The row/column having zero demand/supply left
will be cut by a line
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28/9
19
P3 4 5 8 17
DEMAND 21 25 19/0 65
W3 is having zero demand now so it will be cut and wont
be considered

38. Now find next penalty P2 in same way
WAREHOUSE W1 W2 W3 SUPPLY P2
PLANT
P1 7 6 9 20 7-6=1
P2 5 9 7 3 28/9/0 7-5=2 2 is max penalty
P3 4 5 8 17 5-4=1
19
DEMAND 21/12 25 19/0
P2 5-4=1 6-5=1 ------------
In W1 5 is min cost

39. In this way find out all the penalties until all
allocations are not done
WAREHOUSE W1 W2 W3 SUPPLY P3
PLANT
P1 7 6 9 20 7-6=1
P2 5 9 7 3 28/9/0 ----------
P3 4 12 5 8 17/5 5-4=1
19
DEMAND 21/12/0 25 19/0 In P3 4 is min cost
P3 7-4=3 6-5=1 ------------
3 is max penalty

40. Final matrix with all allocations
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/9/0
9 19
P3 4 5 8 17/5/0
12 5
DEMAND 21/12/0 25/5/0 19/0

41. No. of allocations=No. of rows + No. of columns-1
(For generating optimal solution)
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20/0
20
P2 5 7 3 28/9/0
9 19
P3 4 5 8 17/5/0
12 5
DEMAND 21/12/0 25/5/0 19/0
Here no. of rows=3, no. of columns=3, total allocations=5
5=3+3-1

42. TOTAL TRANSPORTATION COST
PLANT WAREHOUSE COST
P1 W2 6*20=120
P2 W1 5*9=45
P2 W3 3*19=57
P3 W1 12*4=48
P3 W2 5*5=25
TOTAL COST 295

43. COMPARISON OF NWCM,LCEM,VAM
WAREHOUSE W1 W2 W3 SUPPLY
PLANT
P1 7 6 9 20
P2 5 7 3 28
P3 4 5 8 17
DEMAND 21 25 19 65
Minimum Transportation Cost
NWCM=462
LCEM=300
VAM=295

44. MAXIMIZATION CASE
I II III IV CAPACITY
A 40 25 22 33 100
B 44 35 30 30 30
C 38 38 28 30 70
REQUIREMENT 40 20 60 30 200
150
This is an unbalanced maximization problem for
finding out maximum profit involved

45. To solve this first we will convert this in
minimization problem by subtracting the largest
element from all elements
I II III IV CAPACITY
A 4 19 22 11 100
B 0 9 14 14 30
C 6 6 16 14 70
REQUIREMENT 40 20 60 30 200
150
44 is maximum element in the matrix so subtract all the
elements from 44 and re-write in matrix form

46. Now we will balance this matrix
Requirement<Capacity Dummy column will be
added
I II III IV Dummy CAPACITY
A 4 19 22 11 0 100
B 0 9 14 14 0 30
C 6 6 16 14 0 70
REQUIREMENT 40 20 60 30 50 200
200
Now this can be solved by any of the IBFS methods

47. CHOCOLATE
WINING
QUESTION FOR
AUDIENCE

48. Obtain IBFS for the following
Maximization Problem using VAM:
I II III SUPPLY
A 6 4 4 100
B 5 6 7 25
C 3 4 6 75
DEMAND 60 80 85

49. SOLUTION:
I II III SUPPLY
A 6 60 4 40 4 100
B 5 6 15 7 10 25
C 3 4 6 75 75
DUMMY 0 0 25 0 25
DEMAND 60 80 85
MAXIMUM PROFIT= Rs. 1130