Question

Solve the given problem by NWCM (north-west corner method), LCM (the least cost method) and VAM (Vogel's approximation method) and find optimal solution by UV method.
11 13 17 14 Supply
16 18 14 10 250
21 24 13 10 300
Demand 200 225 275 400

Answer 1

Answer to Question #72474, Math / Discrete Mathematics

Solve the given problem by NWCM (north-west corner method), LCM (the least cost method) and VAM (Vogel's approximation method) and find optimal solution by UV method.

Solution.

North-west corner method:

$$Total\ Cost = 200 \cdot 11 + 50 \cdot 13 + 175 \cdot 18 + 125 \cdot 10 + 125 \cdot 13 + 275 \cdot 10 = 11625$$

UV method:

$$c_{ij} = u_i + v_j \text{ for occupied cells}$$

The reduced costs for unoccupied cells:

$$reduced\ cost = c_{ij} - u_i - v_j$$ $$c_{13} = 17 - 5 = 12; \quad c_{14} = 14 - 2 = 12; \quad c_{21} = 16 - 5 - 11 = 0$$ $$c_{24} = 60 - 5 - 2 = 53; \quad c_{31} = 21 - 8 - 11 = 2; \quad c_{32} = 24 - 8 - 13 = 3$$

Since all the current reduced costs are non-negative, this is the optimal solution.

$$The\ minimum\ cost = Total\ Cost = 11625$$

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Answer to Question #72474, Math / Discrete Mathematics

Least cost method:

Total Cost = 200 · 11 + 50 · 13 + 50 · 18 + 125 · 24 + 250 · 10 + 275 · 10 = 12000

Vogel's approximation method:

Penalties:

$$13 - 11 = 2 \text{ for row 1}$$ $$16 - 10 = 6 \text{ for row 2}$$ $$13 - 10 = 3 \text{ for row 3}$$ $$16 - 11 = 5 \text{ for column 1}$$ $$18 - 13 = 5 \text{ for column 2}$$ $$13 - 10 = 3 \text{ for column 3}$$ $$14 - 10 = 4 \text{ for column 4}$$

The highest penalty occurs in the second row. The minimum $c_{ij}$ in this row is $c_{23} = 10$. So $x_{23} = 300$ and second row is eliminated.

Penalties:

$$13 - 11 = 2 \text{ for row 1}$$

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Answer to Question #72474, Math / Discrete Mathematics

$$13 - 10 = 3 \text{ for row 3}$$ $$21 - 11 = 10 \text{ for column 1}$$ $$24 - 13 = 11 \text{ for column 2}$$ $$17 - 13 = 4 \text{ for column 3}$$ $$14 - 10 = 4 \text{ for column 4}$$

The highest penalty occurs in the second column. The minimum $c_{ij}$ in this column is $c_{12} = 13$. So $x_{12} = 300$ and second column is eliminated.

Penalties:

$$14 - 11 = 3 \text{ for row 1}$$ $$13 - 10 = 3 \text{ for row 3}$$ $$21 - 11 = 10 \text{ for column 1}$$ $$17 - 13 = 4 \text{ for column 3}$$ $$14 - 10 = 4 \text{ for column 4}$$

The highest penalty occurs in the first column. The minimum $c_{ij}$ in this column is $c_{11} = 11$. So $x_{11} = 150$ and first column is eliminated.

Penalties:

$$17 - 14 = 3 \text{ for row 1}$$ $$13 - 10 = 3 \text{ for row 3}$$

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Answer to Question #72474, Math / Discrete Mathematics

$$17 - 13 = 4 \text{ for column 3}$$ $$14 - 10 = 4 \text{ for column 4}$$

The highest penalty occurs in the $4^{\text{th}}$ column. The minimum $c_{ij}$ in this column is $c_{34} = 10$. So $x_{34} = 275$ and $4^{\text{th}}$ column is eliminated.

$$\text{Total Cost} = 16 \cdot 50 + 10 \cdot 250 + 13 \cdot 225 + 11 \cdot 25 + 21 \cdot 125 + 10 \cdot 275 = 11875$$

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