Transportation problem using vogel\'s approximation method calculator PDF

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Home > Operation Research calculators > Vogel's approximation method calculator

Algorithm and examples Solve transportation problem using vogel's approximation method Type your data, for seperator you can use space or tab for sample click random button

3 2 4 1 70

2 4 3 1 50

3 1 3 2 80

2 2 4 1 60

60 70 80 90

OR

Minimize Transportation Cost Supply Constraints : 4

Maximize Profit

, Demand Constraints : 4

Generate

D1

D2

D3

D4

Supply

S1

3

2

3

2

60

S2

2

4

1

2

70

S3

4

3

3

4

80

S4

1

1

2

1

90

Demand

70

50

80

60

Initial basic feasible solution by Method : 4. Row minima method MODI method (Optimality test) Stepping stone method (Optimality test) Find

Solution

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Solution will be displayed step by step (In 3 parts) Solution Find Solution using Row minima method D1 D2 D3 D4 Supply S1

3

2

3

2

60

S2

2

4

1

2

70

S3

4

3

3

4

80

S4

1

1

2

1

90

Demand 70 50 80 60

Solution: TOTAL number of supply constraints : 4 TOTAL number of demand constraints : 4 Problem Table is D1 D2 D3 D4

Supply

Calculus

About us Geometry

Pre-Algebra

S1

3

Demand 70

2

3

2

60

50

80

60

Here Total Demand = 260 is less than Total Supply = 300. So We add a dummy demand constraint with 0 unit cost and with allocation 40. Now, The modified table is D 1 D 2 D 3 D 4 D dummy

Supply

S1

3

2

3

2

0

60

S2

2

4

1

2

0

70

S3

4

3

3

4

0

80

S4

1

1

2

1

0

90

50

80

60

40

Demand 70

Problem is Maximization, so convert it to minimization by subtracting all the elements from max element (4) D 1 D 2 D 3 D 4 D dummy

Supply

S1

1

2

1

2

4

60

S2

2

0

3

2

4

70

S3

0

1

1

0

4

80

S4

3

3

2

3

4

90

50

80

60

40

Demand 70

In 1 st row, The smallest transportation cost is 1 in cell S 1D 1. The allocation to this cell is min(60,70) = 60. This exhausts the capacity of S 1 and leaves 70 - 60=10 units with D 1 Table-1 D1

D2 D3 D4

D dummy

Supply

S1

1(60)

2

1

2

4

0

S2

2

0

3

2

4

70

S3

0

1

1

0

4

80

S4

3

3

2

3

4

90

Demand

10

50

80

60

40

In 2 nd row, The smallest transportation cost is 0 in cell S 2D 2. The allocation to this cell is min(70,50) = 50. This satisfies the entire demand of D 2 and leaves 70 - 50=20 units with S 2 Table-2 D1

D2

D3 D4

D dummy

Supply

S1

1(60)

2

1

2

4

0

S2

2

0(50)

S3

3

2

4

20

0

S4

3

1

1

0

4

80

3

2

3

4

90

Demand

10

0

80

60

40

In 2 nd row, The smallest transportation cost is 2 in cell S 2D 4. The allocation to this cell is min(20,60) = 20. This exhausts the capacity of S 2 and leaves 60 - 20=40 units with D 4 Table-3

S1

D1

D2

D3

D4

D dummy

Supply

1(60)

2

1

2

4

0

2

0(50)

3

2(20)

4

0

S2

In 3 rd row, The smallest transportation cost is 0 in cell S 3D 4. The allocation to this cell is min(80,40) = 40. This satisfies the entire demand of D 4 and leaves 80 - 40=40 units with S 3 Table-4 D1

D2

D3

D4

D dummy

Supply

S1

1(60)

2

1

2

4

0

S2

2

0(50)

3

2(20)

4

0

S3

0

1

1

0(40)

4

40

S4

3

3

2

3

4

90

Demand

10

0

80

0

40

In 3 rd row, The smallest transportation cost is 0 in cell S 3D 1. The allocation to this cell is min(40,10) = 10. This satisfies the entire demand of D 1 and leaves 40 - 10=30 units with S 3 Table-5 D2

D3

1(60)

2

1

2

0(50)

3

S3

0(10)

1

1

S4

3

3

Demand

0

0

D1 S1 S2

D dummy

Supply

2

4

0

2(20)

4

0

0(40)

4

30

2

3

4

90

80

0

40

D4

In 3 rd row, The smallest transportation cost is 1 in cell S 3D 3. The allocation to this cell is min(30,80) = 30. This exhausts the capacity of S 3 and leaves 80 - 30=50 units with D 3 Table-6

S1

D1

D2

D3

D4

D dummy

Supply

1

2

4

0

3

2(20)

4

0

1(30) 0(40)

4

0 90

1(60)

2

S2

2

0(50)

S3

0(10)

1

S4

3

3

2

3

4

Demand

0

0

50

0

40

In 4 th row, The smallest transportation cost is 2 in cell S 4D 3. The allocation to this cell is min(90,50) = 50. This satisfies the entire demand of D 3 and leaves 90 - 50=40 units with S 4 Table-7 D1

D2

D3

D4

D dummy

Supply

S1

1(60)

2

1

2

4

0

S2

2

0(50)

3

2(20)

4

0

S3

0(10)

1

1(30) 0(40)

4

0

S4

3

3

2(50)

3

4

40

Demand

0

0

0

0

40

In 4 th row, The smallest transportation cost is 4 in cell S 4D dummy. The allocation to this cell is min(40,40) = 40. Table-8

D1

D2

D3

D4

D dummy

Supply

3

3

2(50)

3

4(40)

0

0

0

0

0

0

D3

D4

D dummy

Supply

1

2

4

60

2 (20) 4

70

4

Demand

Initial feasible solution is D1

D2

S1

1 (60) 2

S2

2

S3

0 (10) 1

1 (30) 0 (40) 4

80

S4

3

3

2 (50) 3

4 (40)

90

50

80

60

40

D4

D dummy

2

Demand 70

0 (50) 3

Allocations in the original problem D1

D2

S1

3 (60) 2

S2

2

S3 S4

D3

0

60

2 (20) 0

70

4 (10) 3

3 (30) 4 (40) 0

80

1

1

2 (50) 1

0 (40)

90

50

80

40

Demand 70

3

Supply

4 (50) 1

60

The maximum profit = 3 × 60 + 4 × 50 + 2 × 20 + 4 × 10 + 3 × 30 + 4 × 40 + 2 × 50 + 0 × 40 = 810 Here, the number of allocated cells = 8 is equal to m + n - 1 = 4 + 5 - 1 = 8 ∴ This solution is non-degenerate

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