Modified Distribution (MODI) Method
The MODI method is another way of evaluating the initial solution of a transportation problem and finding a more optimal solution with much less iterations compared to the Stepping Stones method.
It allows us to compute improvement indices much quickly for each unused square without drawing all of the closed paths.
MODI provides new means of finding the unused route with the largest negative improvement index. Once the largest index is identified, we are required to trace only one closed path. We can then decide the maximum number of units that can be shipped along that unused route.
We begin with an initial solution obtained by the Northwest Corner Rule.
Step 1 : We must now assign separate values to each row and column as shown below.
In the above table we have three rows for three plants, let us call them (P_{A, }P_{B, }P_{C}) & we have three columns for three Warehouses, let us call them (W_{E, }W_{F, }W_{G}).
Generally, let us assume
Pi = Value assigned to rows.
Wj = Value assigned to columns.
Cij = Cost of square ij (cost of shipping one unit from source i to destination j)
Step 2 : We will now compute values for each occupied square as following.
Pi + Wj = Cij
Let us set up equations for each occupied square in the above table.
Step 3 : After all equations have been written, set P_{A }= 0 and solve the equations to get all the rest of P and W values as follows.
Letting P_{A }= 0, we can easily derive values of W_{E, }P_{B, }W_{F, }P_{C }and W_{G}
P_{A }+ W_{E }= 50
0_{ }+ W_{E }= 50
W_{E }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ 50_{ }= 80
P_{B}_{ }= 30
P_{B}_{ }+ W_{F}_{ }= 40
30_{ }+ W_{F}_{ }= 40
W_{F}_{ }= 10
P_{C}_{ }+ W_{F}_{ }= 70
P_{C}_{ }+ 10_{ }= 70
P_{C}_{ }= 60
P_{C}_{ }+ W_{G}_{ }= 50
60_{ }+ W_{G}_{ }= 50
W_{G}_{ }= 10
Step 4 : Now calculate the improvement index for each unused square using the following formula.
Improvement Index (Iij) = Cij  Pi  Wj
Currently we have four unused squares in the table.
Route AF
Route AG
Route BG
Route CE
Let us calculate the Improvement Indices of all these routes.
Route AF Index
IAF = CAF  PA  WF
IAF = 40  0  10
IAF = 30
Route AG Index
IAG = CAG  PA  WG
IAG = 30 0 + 10
IAG = 40
Route BG Index
IBG = CBG  PB  WG
IBG = 30  30 + 10
IBG = 10
Route CE Index
ICE = CCE  PC  WE
ICE = 90  60  50
ICE = 20
We can see that the above assignment has derived exactly the same results as the Stepping Stones Method.
Now because one of the indices is negative, we can conclude that the current solution is not optimal.
The route that needs to be traced to realize cost savings in the current solution is the Route CE (Plant C to Warehouse E) as this has the largest negative Improvement Index.
Step 5 : After selecting the largest negative improvement index, ship the maximum allowable number of units on that route using the same method used in Stepping stone Method to reduce the total shipping cost.
To find out the maximum number of units that can be shipped to our new money saving route, we will take the following steps.
1. Beginning at the square having the largest negative improvement index, trace a closed path back to it via squares that are currently being used.
2. Beginning with (+) sign at the unused square, place alternating () and (+) signs on each corner square of the closed path just traced as shown below.
3. Select the smallest quantity containing a negative sign in the current closed path. Add this number to all squares of the closed path containing positive signs and subtract it from all squares containing negative signs just as we did in the Steeping Stone Method.
We will arrive at the following solution after carrying out the above exercise.
Step 6 : Compute New Improvement Indices for this new solution. With each new MODI Solution, we must recalculate the values of all Rows and Columns. These values are then used to compute new improvement indices of all unused squares in order to determine whether further reduction in shipping cost is possible.
We will again set up a table and name the rows and columns appropriately, and will compute values for each occupied square.
Equations for each occupied square in the above table.
We have four unused squares in the current solution.
Route AF
Route AG
Route BG
Route CF
Let us calculate the Improvement Indices of all these routes.
Route AF Index
IAF = CAF  PA  WF
IAF = 40  0  10
IAF = 30
Route AG Index
IAG = CAG  PA  WG
IAG = 30 0 + 10
IAG = 40
Route BG Index
IBG = CBG  PB  WG
IBG = 30  30  10
IBG = 10
Route CF Index
ICE = CCF  PC  WF
ICE = 70  40  10
ICE = 20
In the above exercise, we can see that Route BG has a negative improvement index, which means we can realize further cost savings by utilizing this route as compared to the present route selection.
Now we need to decide the maximum number of units that can be shipped to our new money saving route without disturbing the supply and capacity constraints of any of the plants or warehouses.
We will trace a closed path beginning at the square having the largest negative Improvement Index, and back to it via the squares that are currently being utilized.
Beginning with (+) sign at the unused square, place alternating () and (+) signs on each corner square of the closed path just traced as shown below.
We can skip any occupied square falling between our closed path if does not agrees with the supply or the capacity constraints.
Select the smallest quantity containing a negative sign in the current closed path. Add this number to all squares of the closed path containing positive signs and subtract it from all squares containing negative signs as shown in the table below.
Above is the improved solution for Bengal Plumbing problem using Route (BG).
Let us calculate the total shipping cost.
The total computed shipping cost with this iteration has come down to Rs.39000.00 as compared to the previous result of Rs.40000.00
The MODI method is another way of evaluating the initial solution of a transportation problem and finding a more optimal solution with much less iterations compared to the Stepping Stones method.
It allows us to compute improvement indices much quickly for each unused square without drawing all of the closed paths.
MODI provides new means of finding the unused route with the largest negative improvement index. Once the largest index is identified, we are required to trace only one closed path. We can then decide the maximum number of units that can be shipped along that unused route.
We begin with an initial solution obtained by the Northwest Corner Rule.
Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


Plant A

100

Rs.50

Rs.40

Rs.30

100


Plant B

200

Rs.80

100

Rs.40

Rs.30

300


Plant C

Rs.90

100

Rs.70

200

Rs.50

300


Warehouse Requirement

300

200

200

700

Step 1 : We must now assign separate values to each row and column as shown below.
In the above table we have three rows for three plants, let us call them (P_{A, }P_{B, }P_{C}) & we have three columns for three Warehouses, let us call them (W_{E, }W_{F, }W_{G}).
Generally, let us assume
Pi = Value assigned to rows.
Wj = Value assigned to columns.
Cij = Cost of square ij (cost of shipping one unit from source i to destination j)
W_{E}

W_{F}

W_{G}


Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


P_{A}

Plant A

100

Rs.50

Rs.40

Rs.30

100


P_{B}

Plant B

200

Rs.80

100

Rs.40

Rs.30

300


P_{C}

Plant C

Rs.90

100

Rs.70

200

Rs.50

300


Warehouse Requirement

300

200

200

700

Step 2 : We will now compute values for each occupied square as following.
Pi + Wj = Cij
Let us set up equations for each occupied square in the above table.
P_{A }+ W_{E }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ W_{F}_{ }= 40
P_{C}_{ }+ W_{F}_{ }= 70
P_{C}_{ }+ W_{G}_{ }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ W_{F}_{ }= 40
P_{C}_{ }+ W_{F}_{ }= 70
P_{C}_{ }+ W_{G}_{ }= 50
Step 3 : After all equations have been written, set P_{A }= 0 and solve the equations to get all the rest of P and W values as follows.
Letting P_{A }= 0, we can easily derive values of W_{E, }P_{B, }W_{F, }P_{C }and W_{G}
P_{A }+ W_{E }= 50
0_{ }+ W_{E }= 50
W_{E }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ 50_{ }= 80
P_{B}_{ }= 30
P_{B}_{ }+ W_{F}_{ }= 40
30_{ }+ W_{F}_{ }= 40
W_{F}_{ }= 10
P_{C}_{ }+ W_{F}_{ }= 70
P_{C}_{ }+ 10_{ }= 70
P_{C}_{ }= 60
P_{C}_{ }+ W_{G}_{ }= 50
60_{ }+ W_{G}_{ }= 50
W_{G}_{ }= 10
Step 4 : Now calculate the improvement index for each unused square using the following formula.
Improvement Index (Iij) = Cij  Pi  Wj
Currently we have four unused squares in the table.
Route AF
Route AG
Route BG
Route CE
Let us calculate the Improvement Indices of all these routes.
Route AF Index
IAF = CAF  PA  WF
IAF = 40  0  10
IAF = 30
Route AG Index
IAG = CAG  PA  WG
IAG = 30 0 + 10
IAG = 40
Route BG Index
IBG = CBG  PB  WG
IBG = 30  30 + 10
IBG = 10
Route CE Index
ICE = CCE  PC  WE
ICE = 90  60  50
ICE = 20
We can see that the above assignment has derived exactly the same results as the Stepping Stones Method.
Now because one of the indices is negative, we can conclude that the current solution is not optimal.
The route that needs to be traced to realize cost savings in the current solution is the Route CE (Plant C to Warehouse E) as this has the largest negative Improvement Index.
Step 5 : After selecting the largest negative improvement index, ship the maximum allowable number of units on that route using the same method used in Stepping stone Method to reduce the total shipping cost.
To find out the maximum number of units that can be shipped to our new money saving route, we will take the following steps.
1. Beginning at the square having the largest negative improvement index, trace a closed path back to it via squares that are currently being used.
2. Beginning with (+) sign at the unused square, place alternating () and (+) signs on each corner square of the closed path just traced as shown below.
Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


Plant A

100

Rs.50

Rs.40

Rs.30

100


Plant B

() 200

Rs.80

(+) 100

Rs.40

Rs.30

300


Plant C

(+)

Rs.90

() 100

Rs.70

200

Rs.50

300

Warehouse Requirement

300

200

200

700

3. Select the smallest quantity containing a negative sign in the current closed path. Add this number to all squares of the closed path containing positive signs and subtract it from all squares containing negative signs just as we did in the Steeping Stone Method.
We will arrive at the following solution after carrying out the above exercise.
Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


Plant A

100

Rs.50

Rs.40

Rs.30

100


Plant B

100

Rs.80

200

Rs.40

Rs.30

300


Plant C

100

Rs.90

Rs.70

200

Rs.50

300


Warehouse Requirement

300

200

200

700

Step 6 : Compute New Improvement Indices for this new solution. With each new MODI Solution, we must recalculate the values of all Rows and Columns. These values are then used to compute new improvement indices of all unused squares in order to determine whether further reduction in shipping cost is possible.
We will again set up a table and name the rows and columns appropriately, and will compute values for each occupied square.
W_{E}

W_{F}

W_{G}


Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


P_{A}

Plant A

100

Rs.50

Rs.40

Rs.30

100


P_{B}

Plant B

100

Rs.80

200

Rs.40

Rs.30

300


P_{C}

Plant C

100

Rs.90

Rs.70

200

Rs.50

300


Warehouse Requirement

300

200

200

700

Equations for each occupied square in the above table.
P_{A }+ W_{E }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ W_{F}_{ }= 40
P_{C}_{ }+ W_{E}_{ }= 90
P_{C}_{ }+ W_{G}_{ }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ W_{F}_{ }= 40
P_{C}_{ }+ W_{E}_{ }= 90
P_{C}_{ }+ W_{G}_{ }= 50
Set P_{A }= 0 and solve the equations to get all the rest of P and W values as follows.
P_{A }+ W_{E }= 50
0_{ }+ W_{E }= 50
W_{E }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ 50_{ }= 80
P_{B}_{ }= 30
P_{B}_{ }+ W_{F}_{ }= 40
30_{ }+ W_{F}_{ }= 40
W_{F}_{ }= 10
P_{C}_{ }+ W_{E}_{ }= 90
P_{C}_{ }+ 50_{ }= 90
P_{C}_{ }= 40
P_{C}_{ }+ W_{G}_{ }= 50
40_{ }+ W_{G}_{ }= 50
W_{G}_{ }= 10
0_{ }+ W_{E }= 50
W_{E }= 50
P_{B}_{ }+ W_{E }= 80
P_{B}_{ }+ 50_{ }= 80
P_{B}_{ }= 30
P_{B}_{ }+ W_{F}_{ }= 40
30_{ }+ W_{F}_{ }= 40
W_{F}_{ }= 10
P_{C}_{ }+ W_{E}_{ }= 90
P_{C}_{ }+ 50_{ }= 90
P_{C}_{ }= 40
P_{C}_{ }+ W_{G}_{ }= 50
40_{ }+ W_{G}_{ }= 50
W_{G}_{ }= 10
Now we can once again calculate the improvement index for each unused square using the following formula.
Improvement Index (Iij) = Cij  Pi  Wj
Improvement Index (Iij) = Cij  Pi  Wj
We have four unused squares in the current solution.
Route AF
Route AG
Route BG
Route CF
Let us calculate the Improvement Indices of all these routes.
Route AF Index
IAF = CAF  PA  WF
IAF = 40  0  10
IAF = 30
Route AG Index
IAG = CAG  PA  WG
IAG = 30 0 + 10
IAG = 40
Route BG Index
IBG = CBG  PB  WG
IBG = 30  30  10
IBG = 10
Route CF Index
ICE = CCF  PC  WF
ICE = 70  40  10
ICE = 20
In the above exercise, we can see that Route BG has a negative improvement index, which means we can realize further cost savings by utilizing this route as compared to the present route selection.
Now we need to decide the maximum number of units that can be shipped to our new money saving route without disturbing the supply and capacity constraints of any of the plants or warehouses.
We will trace a closed path beginning at the square having the largest negative Improvement Index, and back to it via the squares that are currently being utilized.
Beginning with (+) sign at the unused square, place alternating () and (+) signs on each corner square of the closed path just traced as shown below.
We can skip any occupied square falling between our closed path if does not agrees with the supply or the capacity constraints.
Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


Plant A

100

Rs.50

Rs.40

Rs.30

100


Plant B

() 100

Rs.80

200

Rs.40

(+)

Rs.30

300

Plant C

(+) 100

Rs.90

Rs.70

() 200

Rs.50

300


Warehouse Requirement

300

200

200

700

Select the smallest quantity containing a negative sign in the current closed path. Add this number to all squares of the closed path containing positive signs and subtract it from all squares containing negative signs as shown in the table below.
Transportation
Matrix for Bengal Plumbing


From \ To

Warehouse
E

Warehouse
F

Warehouse
G

Factory
Capacity


Plant A

100

Rs.50


Rs.40


Rs.30

100

Plant B


Rs.80

200

Rs.40

100

Rs.30

300

Plant C

200

Rs.90

Rs.70

100

Rs.50

300


Warehouse Requirement

300

200

200

700

Above is the improved solution for Bengal Plumbing problem using Route (BG).
Let us calculate the total shipping cost.
Improved Solution for Bengal Plumbing
Problem by Stepping Stone Method


From

To

Units Shipped

Cost per unit

Total Cost

Plant A

Warehouse E

100

Rs.50

Rs.5000

Plant B

Warehouse F

200

Rs.40

Rs.8000

Plant B

Warehouse G

100

Rs.30

Rs.3000

Plant C

Warehouse E

200

Rs.90

Rs.18000

Plant C

Warehouse G

100

Rs.50

Rs.5000

Total Shipping Cost

Rs.39000

The total computed shipping cost with this iteration has come down to Rs.39000.00 as compared to the previous result of Rs.40000.00