MODI Method Examples, Transportation Problem

8/11/2019 MODI Method Examples, Transportation Problem

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MODI Method Examples:

Transportation ProblemIn the previous section, we provided the steps in MODI method

(modified distribution method)to solve a t ransportation

problem. In this section, we provide an example. Let's solve the

following example:

This example is the largest and the most involvedyou have read so far. So you must read t he stepsand the explanation mindfully.

Example: MODI Method (Modified

Distribution Method)

Consider the transportation problempresented in the following

table.

Distribution centre

D1 D2 D3 D4 Supply

Plant

P1 19 30 50 12 7

P2 70 30 40 60 10P3 40 10 60 20 18

Requirement 5 8 7 15

Determine the optimal solution of the above problem.

Solution.

An initial basic feasible solution is obtained by Matrix Minimum

Method and is shown in table 1.

Table 1

Distribution centre

D1 D2 D3 D4 Supply

Plant

P1 19 30 50 7

P2 30 60 10

P3

60 18


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Initial basic feasible solution

12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.

Calculating uiand vjusing ui + vj= cij

Substituting u1= 0, we get

u1 + v4= c140 + v4= 12 or v4= 12

u3 + v4= c34u3+ 12 = 20 or u3= 8

u3 + v2= c328 + v2= 10 or v2= 2

u3 + v1= c318 + v1= 40 or v1= 32

u2 + v1= c21 u2+ 32 = 70 or u2= 38

u2 + v3= c2338 + v3= 40 or v3= 2

Table 2

Distribution centre

D1 D2 D3 D4 Supply ui

Plant

P1 19 30 50 7 0

P2 30 60 10 38

P360 18 8


vj 32 2 2 12

Calculating opportunity costusing c ij ( ui + vj )

Unoccupied cells Opportunity cost

(P1, D1) c11 ( u1 + v1 ) = 19 (0 + 32) = 13

(P1, D2) c12 ( u1 + v2 ) = 30 (0 + 2) = 28

(P1, D3) c13 ( u1 + v3 ) = 50 (0 + 2) = 48

(P2, D2) c22 ( u2 + v2 ) = 30 (38 + 2) = 10

(P2, D4) c14 ( u2 + v4 ) = 60 (38 + 12) = 10

(P3, D3) c33 ( u3 + v3 ) = 60 (8 + 2) = 50

Table 3

Distribution centre


Plant

P1 7 0

P210 38

P318 8



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vj 32 2 2 12

Now choose the smallest (most) negative value from opportunity

cost (i.e., 13) and draw a closed path fromP1D1. The following

table shows the closed path.

Table 4

Choose the smallest value with a negative position on the closed

path(i.e., 2), it indicates the number of units that can be shipped

to the entering cell. Now add this quantity to all the cells on thecorner points of the closed path marked with plus signs and

subtract it from those cells marked with minus signs. In this way,

an unoccupied cell becomes an occupied cell.

Now again calculate the values for ui& vjand opportunity cost.

The resulting matrix is shown below.

Table 5

Distribution centre


Plant

P1 7 0

P2 10 51

P3 18 8


vj 19 2 11 12

Choose the smallest (most) negative value from opportunity cost

(i.e., 23). Now draw a closed path fromP2D2.

Table 6


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Examples Solving Equations Solve Equations Linear Equations

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MODI Method Examples, Transportation Problem

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