Vogel's Approximation Method / Unit Cost Penalty Method

Q1.Find the IBFS for the following TP by VAM.


                D1        D2         D3          D4          Supply
O1            11        13          17           14            250
O2            16        18          14           10            300
O3            21        24          13           10            400
Demand   200       225       275          250

Soln:
STEP I:
The given TP is a balanced one. Since Eai=Ebj=950.Therefore there exists a Feasible Solution.

STEP II:
Calculate Penalty (The difference between the Smallest and next smallest cost in each row and column).

        D1       D2     D3     D4          Supply        P1
        ------------------------
          200
O1  11        13      17        14         250              2(13-11)
O2  16        18       14        10        300              4
O3  21        24      13         10       400               3
      200     225     275        250
P2   5         5         1            0

Among the Penalty calculated Select the highest value and choose the cell which has the lowest cost.

                                                  P1
     50
13           17       14         250     1(14-13)
18           14       10          300    4
24           13       10         400     3
225        275      250
5               1         0   P2


Similarly if we calculate the final one will be      

    125
10       250
125


Solution is given by

   200       50
11        13        17     14
                175               125
16        18        14      10
                             275    125
21        24        13       10

The values are independent ie., the number of closed cycles is m+n-1=3+4-1=6

There are six positive independent allocation which equals to m+n-1=6.This ensures that the solution is non-degenerate Basic Feasible solution.

Transportation Cost=(11 x 200)+(13 x 50)+(18 x 175)+(10 x 125)+(13 x 275)+( 10 X 125)
                              =Rs.12075

Least Cost Method / Matrix Minima Method

Q1.Obtain the initial feasible solution for the following TP using Matrix Minima Method.


                 D1          D2        D3        D4       Supply
O1             1            2            3            4          6
O2             4            3            2            0          8
O3             0            2            2            1         10
Demand     4             6           8             6

Soln:  
STEP I:
Sum up Demand and Supply.
Since Eai=Ebj=24. There exists a feasible solution for TP.

STEP II:
In the Least Cost Method, we have to find out the least value in the given problem and start from there.
In this problem the least value is "0", so start from there,

------------------
1      2      3      4        6
4      3      2      0        8
   4
0      2       2     1       10
------------------
4      6       8      6



-----------------
2        3         4          6
                         6
3         2         0          8
2         2         1          6
------------------
6         8         6

------------
    6
2           3           6
3           2           2
2           2           6
------------
6           8


----------
3        2        2
    0
2         2        6
---------
0         8

Here we have subtracted the demand and supply values , in the previous steps the least values were "0" therefore we did not subtract it .

----
   2
2        2
2        6
---
8

-----
    6
2           6
-----
6

STEP III:
The Solution is given by:

                   6
1             2            3             4
                                 2              6
4              3            2             0
   4               0            6
0               2            2             1


Total Cost = (2 x 6)+(2 x 2)+(0 x 6)+(0 x 4)+(2  x 0)+(2 x 6)
                 = Rs.28

Transportation Problem (TP)

• Feasible Solution: Any set of non-negative allocation which satisfies row and column sum is called a feasible solution.
• Basic Feasible Solution(BFS): A feasible solution is called BFS if the number of non-negative allocation is equal to m+n-1, where m- number of rows , n- number of columns
• Non Degenerated Basic Feasible Solution: Any feasible solution of a TP containing m origins and n destinations are said to be Non Degenerated, if it contains m+n-1 occupied cells and each allocation is in independent positions.
• Allocations are said to be independent if it is impossible to form a closed path.
• Closed Path: It means by allowing horizontal and vertical lines and all the corner cells are occupied.

• Degenerated Basic Feasible Solution:If a BFS contains less than m+n-1 non-negative allocations it is said to be degenerated.

OPTIMAL SOLUTION:

It is a feasible solution which minimizes the total cost.

The solution of a TP can be obtained in two stages namely

1. Initial solution
2. Optimal solution

Initial solution can be obtained by any one of the following methods

• North West Corner Rule(NWCR)
• Least Cost Method (LCM) or Matrix Minima Method
• Vogel's Approximation Method (VAM)