Transportation Problem

TRANSPORTATION PROBLEMS

TRANSPORTATION PROBLEM

A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points. While trying to find the best way, generally a variable cost of shipping the product from one supply point to a demand point or a similar constraint should be taken into consideration.

General Description of a Transportation Problem

EXAMPLEA company has three production facilities S1,S2, S3 with production capacity of 7, 9 and 18 units (in 100s) per week of a product respectively. These units are to be shipped to four warehouses D1, D2, D3, D4 with requirement of 5, 6,7 and 14 units (in 100s) per week, respectively. The transportation costs (in Rupees) per unit between factories to warehouses are given in the table on next slide.

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 7 9

S2 70 30 40 60 9 10

S3 40 8 70 20 18 12

Demand 5 8 7 14 34

Column Penalty

21 22 10 10

TRANSPORTATION COST MATRIX

OPTIMIZATION OF TRANSPORTATION PROBLEM

Phase 1Phase 1 Calculating Basic Solution Calculating Basic Solution

(Using VAM)(Using VAM)

D1 D2 D3 D4 Supply RowPenalty

S119 30 50 10 7 9

S270 30 40 60 9 10

S340 8 70 20 18 12

Demand5 8 7 14 34

Column Penalty 21 22 10 10

Step 1: From the Table we determine the penalty (element – smallest in its row (column) ) for each row and column.

Step 2: Identify the row or column with the largest

penalty among all the rows and columns. If the penalties corresponding to two or more rows or

columns are equal we select the topmost row and the extreme left column.

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 7 9

S2 70 30 40 60 9 10

S3 40 8 70 20 10 12

Demand 5 8 7 14 34

Column Penalty

21 22 10 10

Step 3: Select Xij as a basic variable if Cij is the minimum cost in the row or column with largest penalty. Choose Xij as high as possible. Eliminate ith row or jth column depending upon whether ai or bj is the smaller of the two.

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 7 9

S2 70 30 40 60 9 20

S3 40 8 70 20 10 20

Demand 5 8 7 14 34

Column Penalty

21 22 10 10

Step 4: The step (ii) is now performed on the reduced matrix until all the basic variables have been identified.

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 2 9

S2 70 30 40 60 9 20

S3 40 8 70 20 10 20

Demand 5 8 7 14 34

Column Penalty

21 22 10 10

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 2 40

S2 70 30 40 60 9 20

S3 40 8 70 20 10 50

Demand 5 8 7 14 34

Column Penalty

21 22 10 10

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 2 40

S2 70 30 40 60 9 20

S3 40 8 70 20 0 50

Demand 5 8 7 4 34

Column Penalty

21 22 10 10

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 2 40

S2 70 30 40 60 9 20

S3 40 8 70 20 0 50

Demand 5 8 7 4 34

Column Penalty

21 22 10 50

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 0 40

S2 70 30 40 60 9 20

S3 40 8 70 20 0 50

Demand 5 8 7 2 34

Column Penalty

21 22 10 50

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 0 40

S2 70 30 40 60 7 20

S3 40 8 70 20 0 50

Demand 5 8 7 0 34

Column Penalty

21 22 40 50

D1 D2 D3 D4 Supply RowPenalty

S1 19 30 50 10 0 40

S2 70 30 40 60 0 20

S3 40 8 70 20 0 50

Demand 5 8 7 0 34

Column Penalty

21 22 40 50

Phase 2 Calculating Optimal Solution

MODIFIED DISTRIBUTION (MODI or u-v) METHOD

Step 1 Introduce dual variables corresponding to the row and column

constraints.

Eg. u1+v1=c11 = 19

D1 D2 D3 D4 Supply Row Constraints

S1 19 30 50 10 7 U1

S2 70 30 40 60 9 U2

S3 40 8 70 20 18 U3

Demand 5 8 7 14 34 U4

Column Constraints

v1 v2 v3 v4

Equations

Step 2: Any basic feasible solution of the transportation problem has m + n – 1; Xjj >0. Thus there will be m + n - 1 equations to determine m + n dual variables. One of the dual variables can be chosen arbitrarily. It is to be also noted that as the primal constraints are equations, the dual variables are unrestricted in sign.

Putting v4=0 (as it appears maximum number of times)u3=20 v1=9u2=60 v3=-20u1=10 v2=-12

Step 3: If X ij = 0, the dual variables computed in 3 are compared with the Cij values of this allocation as: Cij – Ui - Vj . If all Cij – Ui - Vj ≥ 0, then by an application of complementary slackness theorem it can be shown that the corresponding solution of the transportation problem is optimum. If one or more of Cij – Ui - Vj < 0, we choose the cell with least value of Cij – Ui - Vj and allocate as much as possible subject to the row and the column constraints. The allocation of a number of adjacent cell are adjusted so that a basic variable becomes non basic.

Basic Allocation

D1 D2 D3 D4 Supply Row Constraints

S1 5 2 7

S2 7 2 9

S3 8 10 18

Demand 5 8 7 14 34

Column Constraints

Negative Value indicates that there is a possibility of cost reduction by assigning to this cell.

D1 D2 D3 D4 Supply Row Constraints

S1 +32 +60 7 10

S2 +1 -18 9 60

S3 +11 +70 18 20

Demand 5 8 7 14 34

Column Constraints

9 -12 -20 0

Re-Allocation

D1 D2 D3 D4 Supply Row Constraints

S1 5 2 7

S2 (+) 7

2 (-)

9

S3 8(-)10 (+)

18

Demand 5 8 7 14 34

Column Constraints

RE-ALLOCATION

D1 D2 D3 D4 Supply Row Constraints

S1 5 2 7 u1

S2 2 7

9 u2

S3 6 12 18 u3

Demand 5 8 7 14 34

Column Constraints

v1 v2 v3 v4

Step 4A fresh set of dual variables are

computed and entire procedure is repeated.

D1 D2 D3 D4 Supply Row Constraints

S1 19 30 50 10 7 U1

S2 70 30 40 60 9 U2

S3 40 8 70 20 18 U3

Demand 5 8 7 14 34 U4

Column Constraints

v1 v2 v3 v4

U1+v1=19 u1+v4=10u2+v2=30 u2+v3=40u3+v2=8 u3+v4=20Putting u1=0 : v1=19,v2=-2,v3=8, v4=10, u2=32, u3=10

All positive values indicate that no further reduction in cost is possible. Hence it is optimal solution.

D1 D2 D3 D4 Supply Row Constraints

S1 +32 +42 7 0

S2 +19 +14 9 32

S3 +11 +52 18 10

Demand 5 8 7 14 34

Column Constraints

19 -2 8 10

D1 D2 D3 D4 Supply Row Constraints

S1 19 30 50 10 7 U1

S2 70 30 40 60 9 U2

S3 40 8 70 20 18 U3

Demand 5 8 7 14 34 U4

Column Constraints

v1 v2 v3 v4

Optimal Solution

Unbalanced Problem

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