MODI Method / Modification Distribution

Q1.Solve the TP.

                 D1       D2      D3       D4       Supply
O1            11        20       7           8           50
O2            21        16      20         12          40
O3             8         12       8            9          70
Demand    30        25      35          40


Soln:
STEP I:
Eai=160 Ebj=130.The given TP is unbalnced Therefore add a Destination Column.

               D1     D2       D3      D4     D5      Supply
O1          11      20         7         8       0           50
O2          21      16        20       12      0           40
O3            8       12         8         9      0           70
Demand   30      25        35       40    30

Eai=Ebj=160.Now the TP is Balanced.

STEP II:
Now calculate the IBFS using VAM .
The Solution will be,

                      35
11     20      7        8     0
                               10    30
21     16     20      12     0
   30      25   15        15
8       12     8         9       0


In the above table we have 7 independent non negative allocation and m=n-1=7. non degenerate Basic feasible solution is obtained.

Total Cost= Rs.1160

STEP III:
Now select the row or column which has the maximum number of allocations.
Here in the above table we have the maximum number of allocations in the 3rd row.
Therefore assign u3=0
-----------------------------------------
4            9                  35      0         4
  11          20            7            8           0            u1=-1
------------------------------------------
10          1             9                 10        30
   21         16           20          12          0           u2=3
------------------------------------------
   30            25           15          15     3
8              12           8             9           0             u3=0
--------------------------------------------
v1=8       v2=12     v3=8      v4=9     v5=-3


c31=u3+v1=8      c32=u3+v2=12      c33=u3+v3=8      c34=u3+v4=9

c24=u2+v4=12    c25=u2+v5=0        c13=u1+v3=7
         u2=3               v5=-3                         u1=-1


Now calculate

This should be calculated for all the cells which where not allocated.(This should be calculated for the cells in the table which was obtained as a result of applying VAM)

Now (Delta)ij >=0 .Therefore Optimal Solution with Total Cost= Rs.1160.

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

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)