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TRANSPORTATION PROBLEMTRANSPORTATION PROBLEM

Finding Initial Basic Feasible SolutionFinding Initial Basic Feasible Solution

Shubhagata Roy

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NorthNorth- -West Corner MethodWest Corner Method

Step1: Select the upper left (north-west) cell of thetransportation matrix and allocate the maximum possiblevalue to X 11 which is equal to min(a 1,b 1).

Step2: If allocation made is equal to the supply available at the

first source (a 1 in first row), then move vertically down tothe cell (2,1). If allocation made is equal to demand of the first

destination (b 1 in first column), then move horizontally tothe cell (1,2).

If a 1=b 1 , then allocate X 11 = a 1 or b 1 and move to cell (2,2).Step3: Continue the process until an allocation is made in

the south-east corner cell of the transportation table.

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ExampleExample: : Solve the Transportation Table to find InitialSolve the Transportation Table to find InitialBasic Feasible Solution using NorthBasic Feasible Solution using North- -West Corner Method.West Corner Method.

Total Cost =19*5+30*2+30*6+40*3+70*4+20*14Total Cost =19*5+30*2+30*6+40*3+70*4+20*14= Rs. 1015= Rs. 1015

Suppl19 30 50 10

5 270 30 40 60

6 340 8 70 204 14

De and 34

S1

S2

S3

7

9

18

5 8 7 14

D1 D2 D3 D4

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Least Cost MethodLeast Cost Method

Step1: Select the cell having lowest unit cost in the entiretable and allocate the minimum of supply or demand valuesin that cell.

Step2: Then eliminate the row or column in which supply or demand is exhausted. If both the supply and demandvalues are same, either of the row or column can beeliminated.In case, the smallest unit cost is not unique, then select thecell where maximum allocation can be made.

Step3: Repeat the process with next lowest unit cost andcontinue until the entire available supply at various sourcesand demand at various destinations is satisfied.

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Supply70

240

3Demand 345

S 2 2

S 3 3

D1

Supply

70 40 60

40 70 207

Demand 34

S3 10

5 7 7

S2 9

D1 D3 D4

Supply70 40

740 70

Demand 34

9

3S3

5 7

S2

D1 D3

Supply

19 30 50 10

70 30 40 6 0

40 8 70 208

Demand 34

S 3 18

5 8 7 14

S 1 7

S 2 9

D1 D2 D3 D4

Supply

19 50 10

770 40 6 0

40 70 20

Demand 34

7

9

S1

S2

S3 10

5

D1 D3 D4

7 14

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The total transportation cost obtained by this methodThe total transportation cost obtained by this method

= * +10*7+20*7+40*7+70*2+40*3= * +10*7+20*7+40*7+70*2+40*3= Rs. 14= Rs. 14Here, we can see that theHere, we can see that the Least Cost Method Least Cost Method involves ainvolves alower cost than thelower cost than the NorthNorth- -West Corner Method West Corner Method ..

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V ogels Approximation MethodV ogels Approximation MethodStep1: Calculate penalty for each row and column by taking the

difference between the two smallest unit costs. This penalty or extra cost has to be paid if one fails to allocate the minimumunit transportation cost.

Step2: Select the row or column with the highest penalty andselect the minimum unit cost of that row or column. Then,

allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocationcould be made.

Step3: Adjust the supply and demand and eliminate the satisfiedrow or column. If a row and column are satisfiedsimultaneously, only of them is eliminated and the other one isassigned a z ero value.Any row or column having z ero supplyor demand, can not be used in calculating future penalties.

Step4: Repeat the process until all the supply sources and

demand destinations are satisfied.

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Supply Row Diff.19 30 50 10

70 30 40 60

40 8 70 208

Demand 34Col.Diff.

9

10

12

21 22 10 10

D1 D2 D3

S 1

S 2

S 3

5 8 7

D4

14

7

9

18

Supply Row Diff.19 50 10

570 40 60

40 70 20

Demand 34Col.Diff.

D1 D3 D4

S1 7

S2 9

S3 10

5 7 1421 10 10

9

20

20

Supply Row Diff.50 10

40 60

70 2010

Demand 34Col.Diff. 10 10

40

20

50

D3 D4

S 1 2

S 2 9

S 3 10

7 14

Supply Row Diff.40 60

7 2Demand 34Col.Diff.

20

7 2

S2 9

D3 D4Supply Row Diff.

50 102

40 60

Demand 34

Col.Diff.

20

10 50

40

7 4

S 1 2

S 2 9

D3 D4

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The total transportation cost obtained by this methodThe total transportation cost obtained by this method

= * +19*5+20*10+10*2+40*7+60*2= * +19*5+20*10+10*2+40*7+60*2= Rs.779= Rs.779Here, we can see thatHere, we can see that V ogels Approximation Method V ogels Approximation Method involves the lowest cost thaninvolves the lowest cost than NorthNorth- -West Corner Method West Corner Method andand Least Cost Method Least Cost Method and hence is the most preferredand hence is the most preferredmethod of finding initial basic feasible solution.method of finding initial basic feasible solution.