What is Transportation Model?

What is Transportation Model?
· It is a linear programming problem in which a product is to be transported from a number of sources to a
number of destinations at the minimum cost or maximum profit.
Historical Perspective
· The two (2) contribution below are mainly responsible for the development of Transportation Model
o “The Distribution of a Product from Several sources to numerous Localities” – Study by: F. L.
Hitchcock (1941
o “Optimum Utilization of the Transportation System” – Study by: T. C. Koopman (1947)
· 1953 – A. Charnes and W. W. Cooper developed the Stepping Stone Method (SSM)
· 1955 – The Modified Distribution Method (MODI) was developed
Sample Problem:
The M Tech Sources Company needs to make a plan on the least costly shipments to meet demand

Steps in Solving the Transportation Problem
1. Set-up the Transportation Simplex Tableau

Transportation Simplex Tableau – It is a table that summarizes all the relevant data in a concise manner and
facilitates the search for progressively better solutions
2. Develop an Initial Solution
3. Test the Solution for Improvement
4. Develop the Improved Solution

North West Corner Rule

example:

The Amulya Milk Company has three plants located throughout a state with production capacity 50, 75 and 25 gallons. Each day the firm must furnish its four retail shops R₁, R₂, R₃, & R₄ with at least 20, 20 , 50, and 60 gallons respectively. The transportation costs (in Rs.) are given below.

The economic problem is to distribute the available product to different retail shops in such a way so that the total transportation cost is minimum

Solution.

Starting from the North west corner, we allocate min (50, 20) to P₁R₁, i.e., 20 units to cell P₁R₁. The demand for the first column is satisfied. The allocation is shown in the following table.

Table 1

Now we move horizontally to the second column in the first row and allocate 20 units to cell P₁R₂. The demand for the second column is also satisfied.

Table 2

Proceeding in this way, we observe that P₁R₃ = 10, P₂R₃ = 40, P₂R₄ = 35, P₃R₄ = 25. The resulting feasible solution is shown in the following table.

Final Table

Here, number of retail shops(n) = 4, and
Number of plants (m) = 3

Number of basic variables = m + n – 1 = 3 + 4 – 1 = 6.

The total transportation cost is calculated by multiplying each x_ij in an occupied cell with the corresponding c_ij and adding as follows:

20 X 3 + 20 X 5 + 10 X 7 + 40 X 8 + 35 X 2 + 25 X 2 = 670

MINIMUM COST / GREEDY METHOD

In the given matrix, the supply of each source A, B, C is given Viz. 50units, 40 units, and 60 units respectively. The weekly demand for three retailers D, E, F i.e. 20 units, 95 units and 35 units is given respectively. The shipping cost is given for all the routes.

The minimum transportation cost can be obtained by following the steps:

The minimum cost in the matrix is Rs 3, but there is a tie in the cell BF, and CD, now the question arises in which cell we shall allocate. Generally, the cost where maximum quantity can be assigned should be chosen to obtain the better initial solution. Therefore, 35 units shall be assigned to the cell BF. With this, the demand for retailer F gets fulfilled, and only 5 units are left with the source B.

Again the minimum cost in the matrix is Rs 3. Therefore, 20 units shall be assigned to the cell CD. With this, the demand of retailer D gets fulfilled. Only 40 units are left with the source C.

The next minimum cost is Rs 4, but however, the demand for F is completed, we will move to the next minimum cost which is 5. Again, the demand of D is completed. The next minimum cost is 6, and there is a tie between three cells. But however, no units can be assigned to the cells BD and CF as the demand for both the retailers D and F are saturated. So, we shall assign 5 units to Cell BE. With this, the supply of source B gets saturated.

The next minimum cost is 8, assign 50 units to the cell AE. The supply of source A gets saturated.

The next minimum cost is Rs 9; we shall assign 40 units to the cell CE. With his both the demand and supply of all the sources and origins gets saturated.

The total cost can be calculated by multiplying the assigned quantity with the concerned cost of the cell. Therefore,

Total Cost = 50*8 + 5*6 + 35*3 +20*3 +40*9 = Rs 955.

Vogel’s Approximation Method

The concept of Vogel’s Approximation Method can be well understood through an illustration :

First of all the difference between two least cost cells are calculated for each row and column, which can be seen in the iteration given for each row and column. Then the largest difference is selected, which is 4 in this case. So, allocate 20 units to cell BD, since the minimum cost is to be chosen for the allocation. Now, only 20 units are left with the source B

2. Column D is deleted, again the difference between the least cost cells is calculated for each row and column, as seen in the iteration below. The largest difference value comes to be 3, so allocate 35 units to cell AF and 15 units to the cell AE. With this, the Supply and demand of source A and origin F gets saturated, so delete both the row A and Column F.

3. Now, single column E is left, since no difference can be found out, so allocate 60 units to the cell CE and 20 units to cell BE, as only 20 units are left with source B. Hence the demand and supply are completely met.

Now the total cost can be computed, by multiplying the units assigned to each cell with the cost concerned. Therefore,

Total Cost = 20*3 + 35*1 + 15*4 + 60*4 + 20*8 = Rs 555