Answer to Question #243601 in Operations Research for mafi

Question

A transportation problem involves the following costs, supply, and demand.

To

From 1 2 3 4 Supply

1 $500 750 300 450 12

2 650 800 400 600 17

3 400 700 500 550 11

Demand 10 10 10 10

Required:

i. Find the initial solution using the northwest corner method, the minimum cell cost

method, and Vogel's Approximation Method. Compute total cost for each.

ii. Using the VAM initial solution, find the optimal solution using the modified distribution

method (MODI).

Answer 1

Solution:

Northwest Corner Method

To

From 1 2 3 4 Supply

1 $500 750 300 450 12

10 2

2 650 800 400 600 17

8 9

3 400 700 500 550 11

1 10

Demand 10 10 10 10 40

Z = 300X₁₃ + 450X₁₄ + 800X₂₂ + 600X₂₄ + 400X₃₁ + 550X₃₄

Z = 300 (10)

+ 450 (2) + 800 (10) + 600 (7) + 400 (10) + 550 (1)

Z = 3,000 + 900 + 8,000 + 4,200 + 4,000 + 550

Answer: Total Cost: Z = $ 20,650

Minimum cell cost method:

Initial feasible solution is

The minimum total transportation cost =300×10+450×2+800×10+600×7+400×10+550×1=20650

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6

∴ This solution is non-degenerate

Vogel’s Approximation Method

To

From 1 2 3 4 Supply

1 $500 750 300 450 12

2 10 300

2 650 800 400 600 17

7 10 200

3 400 700 500 550 11

10 1 150

Demand 10 10 10 10 40

Z = 750X₁₂ + 450X₁₄ + 800X₂₂ + 400X₂₃ + 400X₃₁ + 700X₃₂

Z = 750 (2) + 450 (10) + 800 (7) + 400 (10) + 400 (10) + 700 (1)

Z = 1,500 + 4,500 + 5,600 + 4,000 + 4,000 + 700

Answer: Total Cost: Z = $ 20,300

ii.

X₁₂  : u₁ + v₂ = 750 Let u₁ = 0

X₁₄ : u₁ + v₄ = 450 Let v₂ = 750

X₂₂ : u₂ + v₂ = 800 Let v₄ = 450

X₂₃ : u₂ + v₃ = 400 Let u₂ = 50

X₃₁ : u₃ + v₁ = 400 Let v₃ = 350

X₃₂ : u₃ + v_2 = 700 Let u₃ = –50

v₁ = 450

v_i v₁ = 450 v₂ = 750 v₃ = 350 v₄ = 450

To

u_i From 1 2 3 4 Supply

u₁ = 0 1 $500 750 300 450 12

2 10

u₂ = 50 2 650 800 400 600 17

7 10

u₃ = –50 3 400 700 500 550 11

10 1

Demand 10 10 10 10 40

X₁₁ : k₁₁ = 500 – 0 – 450 = 50

X₁₃ : k₁₃ = 300 – 0 – 350 = –50

X_21 : k₂₁ = 650 – 50 – 450 = 150

X₂₄ : k₂₄ = 600 – 50 – 450 = 50

X₃₃ : k₃₃ = 500 – (–50) – 350 = 200

X₃₄ : k₃₄ = 550 – (–50) – 450 = 150

To

From 1 2 3 4 Supply

1 $500 - 750 + 300 450 12

2 10

2 650 + 800 - 400 600 17

7 10

3 400 700 500 550 11

10 1

Demand 10 10 10 10 40

To

From 1 2 3 4 Supply

1 $500 750 300 450 12

2 10

2 650 800 400 600 17

7 10

3 400 700 500 550 11

10 1

Demand 10 10 10 10 40

X₁₃ : u₁ + v_3 = 300 Let u₁ = 0

X₁₄   : u₁ + v₄ = 450 Let v₃ = 300

X₂₂   : u₂ + v₂ = 800 Let v₄ = 450

X₂₃   : u₂ + v₃ = 400 Let u₂ = 100

X₃₁   : u₃ + v₁ = 400 Let v₂ = 700

X₃₂   : u₃ + v₂ = 700 Let u₃ = 0

v₁ = 400

v_i v₁ = 450 v₂ = 750 v₃ = 350 v₄ = 450

To

u_i From 1 2 3 4 Supply

u₁ = 0 1 $500 750 300 450 12

2 10

u₂ = 100 2 650 800 400 600 17

9 8

u₃ = 0 3 400 700 500 550 11

10 1

Demand 10 10 10 10 40

X₁₁ : k₁₁ = 500 – 0 – 400 = 100

X₁₂ : k₁₂ = 750 – 0 – 700 = 50

X₂₁ : k₂₁ = 650 – 100 – 400 = 150

X₂₄ : k₂₄ = 600 – 100 – 450 = 100

X₃₃ : k₃₃ = 500 – 0 – 300 = 200

X₃₄ : k₃₄ = 500 – 0 – 450 = 50

Z = 300X₁₃ + 450X₁₄ + 800X₂₂ + 400X₂₃ + 400X₃₁ + 700X₃₂ 

Z = 300 (2) + 450 (10) + 800 (9) + 400 (8) + 400 (10) + 700 (1)

Z = 600 + 4,500 + 7,200 + 3,200 + 4,000 + 700

Answer: Z = $ 20,200