TRANSPORTATION PROBLEM
SUBMITTED BY
MEENU KOHLI
ASSOCIATE PROF. IN COMMERCE
Introduction
Meaning
Definitions in Transportation Problems
1) Basic Feasible Solution:
A feasible solution to a m-origin, n-destination problem is said to be basic if the number of positive allocations are equal to (m+n-1).
2) Feasible Solution :
A set of positive individual allocations which simultaneously removes deficiencies is called a feasible solution.
3) Optimal Solution :
A feasible solution (not basically basic) is said to be optimal if it minimises the total transportation cost.
Mathematical Formulation of Transportation Problem
Let there be three units, producing scooter, say, A1, A2 and A3 from where the scooters are to be supplied to four depots say B1, B2, B3 and B4.
Let the number of scooters produced at A1, A2 and A3 be a1, a2 and a3 respectively and the demands at the depots be b1, b2, b3 and b4 respectively.
We assume the condition
a1+a2+a3 = b1+b2 + b3 + b4
i.e., all scooters produced are supplied to the different depots.
Let the cost of transportation of one scooter from A1 to B1 be c11. Similarly, the cost of transportations in other casus are also shown in the figure and Table.
Mathematical Formulation of Transportation Problem
Let out of a1 scooters available at A1, x11 be taken at B1 depot, x12 be taken at B2 depot and to other depots as well, as shown in the following figure and table 1.
Total number of scooters to be transported forms A1 to all destination, i.e., B1, B2, B3, and B4 must be equal to a1.
\ x11 + x12 + x13 + x14 = a1 (1)
Similarly, from A2 and A3 the scooters transported be equal to a2 and a3 respectively.
\ x21 + x22 + x23 + x24 = a2 (2)
and x31 + x32 + x33 + x34 = a3 (3)
On the other hand it should be kept in mind that the total number of scooters delivered to B1 from all units must be equal to b1, i.e.,
x11 + x21 + x31 = b1 (4)
Similarly, x12 + x22 + x32 = b2 (5)
x13 + x23 + x33 = b3 (6)
x14 + x24 + x34 = b4 (7)
Mathematical Formulation of Transportation Problem
With the help of the above information we can construct the following table:-
Deposit Time | To B1 | To B2 | To B3 | To B4 | Stock |
From A1 | x11(c11) | x12(c12) | x13(c13) | x14(c14) | a1 |
From A2 | x21(c21) | x22(c22) | x23(c23) | x24(c24) | a2 |
From A3 | x31(c31) | x32(c32) | x33(c33) | x34(c34) | a3 |
Requirement | b1 | b2 | b3 | b4 | |
Mathematical Formulation of Transportation Problem
The cost of transportation from Ai (i=1,2,3) to Bj (j=1,2,3,4) will be equal to
……………………… (8)
where the symbol put before cij xij signifies that the quantities cij xij must be summed over all i = 1,2,3 and all
j = 1,2,3,4.
Thus we come across a linear programming problem given by equations (1) to (7) and a linear function (8).
We have to find the non-negative solutions of the system such that it minimizes the function (8).
Mathematical Formulation of Transportation Problem
Mathematical Formulation of Transportation Problem
Some Definitions:
A) Feasible Solution (F.S.):
A set of non-negative allocations xij > 0 which satisfies the row and column restrictions is known as feasible solution.
B) Basic Feasible Solution (B.F.S.):
A feasible solution to a m-origin and n-destination problem is said to be basic feasible solution if the number of positive allocations are (m + n – 1).
If the number of allocations in a basic feasible solutions are less than (m+n–1), it is called degenerate basic feasible solution (DBFS) (otherwise non-degenerate).
C) Optimal Solution:
A feasible solution (not necessarily basic) is said to be optimal if it minimizes the total transportation cost.
Distributing any commodity from any group of supply centers, called sources, to any group of receiving centers, called destinations, in such a way as to minimize the total distribution cost (shipping cost).
Transportation Problem
TYPES OF TRANSPORTATION PROBLEMS�
Balanced Transportation Problems:
cases where the total supply is equal to the total demand.
Unbalanced Transportation Problems:
cases where the total supply is not equal to the total demand. When the supply is higher than the demand, a dummy destination is introduced in the equation to make it equal to the supply (with shipping costs of $0); the excess supply is assumed to go to inventory. On the other hand, when the demand is higher than the supply, a dummy source is introduced in the equation to make it equal to the demand (in these cases there is usually a penalty cost associated for not fulfilling the demand).
Example 1:
Example 2:
| | Destination | | | Supply |
| D1 | D2 | D3 | D4 | |
S1 | 50 | 75 | 35 | 75 | 12 |
Source S2 | 65 | 80 | 60 | 65 | 17 |
S3 | 40 | 70 | 45 | 55 | 11 |
(D) | 0 | 0 | 0 | 0 | 10 |
Demand | 15 | 10 | 15 | 10 | |
Transportation Tableau:
1. Northwest Corner Method
1. Select from upper left (northwest) corner of matrix and note the supply requirement in the row s, and the demand requirement in the column d.
2. Allocate the minimum of s or d to this variable. If this minimum is s, eliminate all variables in its row from future consideration and reduce the demand in its column by s; if the minimum is d, eliminate all variables in the column from future consideration and reduce the supply in its row by d.
REPEAT THESE STEPS UNTIL ALL SUPPLIES HAVE BEEN ALLOCATED.
Initial Solution Procedure:
Total shipping cost = 2250
LEAST COST ENTRY METHOD OR MATRIX MINIMA METHOD
The least cost method is more economical than north-west corner rule,since it starts with a lower beginning cost. Various steps involved in this method are summarized as under.
Step 1: Find the cell with the least(minimum) cost in the transportation table.
Step 2: Allocate the maximum feasible quantity to this cell.
Step:3: Eliminate the row or column where an allocation is made.
Step:4: Repeat the above steps for the reduced transportation table until all the allocations are made.
Obtain an initial basic feasible solution to the following transportation problem using least cost method.
�
Total Supply = Total Demand = 24
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
Given Transportation Problem is:
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The least cost is 1 corresponds to the cells (O1, D1) and (O3, D4)
Take the Cell (O1, D1) arbitrarily.
Allocate min (6,4) = 4 units to this cell.
THE REDUCED TABLE IS��
The least cost corresponds to the cell (O3, D4). Allocate min (10,6) = 6 units to this cell.
THE REDUCED TABLE IS
The least cost is 2 corresponds to the cells (O1, D2), (O2, D3), (O3, D2), (O3, D3)
Allocate min (2,6) = 2 units to this cell.
Transportation schedule :
O1→ D1, O1→D2, O2→D3, O3→D2, O3→D4
Total transportation cost
= (4×1)+ (2×2)+(8×2)+(4×2)+(6×1)
= 4+4+16+8+6
=Rs. 38.
THANKS