Answer to Question #222595 in Operations Research for HANNA

Question

Problem 4: John, president of Hardrock Concrete Company, has plants in three

locations and is currently working on three major construction projects, located at

different sites. The shipping cost per truckload of concrete, plant capacities, and

project requirements are provided in the following table.

A. Formulate an initial feasible solution to Hardrock’s transportation problem

using the northwest corner rule.

B. Find the optimal solution using stepping-stone method

Answer 1

Question

John, president of Hardrock Concrete Company, has plants in three

locations and is currently working on three major construction projects, located at

different sites. The shipping cost per truckload of concrete, plant capacities, and

project requirements are provided in the following table.

$\begin{matrix} & project A & project B & project C & plant \space capacities \\ plant 1 & \$10 & \$4 & \$ 11 & 70\\ plant 2& \$12 & \$5 & \$ 8 & 50\\ plant 3 & \$9 & \$7 & \$6 & 30\\ Project \space requirements & 40 & 50 & 60 & 150 \end{matrix}$

A. Formulate an initial feasible solution to Hardrock’s transportation problem

using the northwest corner rule.

B. Find the optimal solution using stepping-stone method

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Soution

A. Formulate an initial feasible solution to Hardrock’s transportation problem

using the northwest corner rule.

$\begin{matrix} & project A & project B & project C & plant \space capacities \\ plant 1 & 10 & 4 & 11 & 70\\ plant 2& 12 & 5 & 8 & 50\\ plant 3 & 9 & 7 & 6 & 30\\ Project \space requirements & 40 & 50 & 60 & 150 \end{matrix}$

(I)The rim values for plant 1=70 and project A=40 are compared.

The smaller of the two i.e. min(70,40) = 40 is assigned to plant 1 project A

This meets the complete demand of project A and leaves 70 - 40=30 units with plant 1.

(II)The rim values for plant1=30 and project B=50 are compared.

The smaller of the two i.e. min(30,50) = 30 is assigned to plant1 project B

This exhausts the capacity of plant1 and leaves 50 - 30=20 units with project B

(III)The rim values for plant2=50 and project B=20 are compared.

The smaller of the two i.e. min(50,20) = 20 is assigned to plant2 project B

This meets the complete demand of project B and leaves 50 - 20=30 units with plant2

(IV)The rim values for plant2=30 and project C=60 are compared.

The smaller of the two i.e. min(30,60) = 30 is assigned to plant2 project C

 This exhausts the capacity of plant2 and leaves 60 - 30=30 units with project C

(V)The rim values for plant3=30 and project C=30 are compared.

The smaller of the two i.e. min(30,30) = 30 is assigned to plant3 project C

Initial feasible solution is

$\begin{matrix} & project A & project B & project C & plant \space capacities \\ plant 1 & 10(40) & 4(30) & 11 & 70\\ plant 2& 12 & 5(20) & 8 (30)& 50\\ plant 3 & 9 & 7 & 6(30) & 30\\ Project \space requirements & 40 & 50 & 60 & 150 \end{matrix}$

The minimum total transportation cost =10×40+4×30+5×20+8×30+6×30=1040

Here, the number of allocated cells = 5 is equal to m + n - 1 = 3 + 3 - 1 = 5

∴ This solution is non-degenerate

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B. Find the optimal solution using stepping-stone method

Iteration-1 of optimality test

1. Create closed loop for unoccupied cells, we get

$\begin{matrix} Unoccupied cell & Closed path & Net cost change\\ C & C→ B→ B→ C & 11 - 4 + 5 - 8=4\\ A & A→ B→ B→ A & 12 - 5 + 4 - 10=1\\ A & A→ C→ C→ B→ B→ A & 9 - 6 + 8 - 5 + 4 - 10=0\\A B & B→ C→ C→ B & 7 - 6 + 8 - 5=4\\\end{matrix}$

Since all net cost change ≥0

So final optimal solution is arrived.

$\begin{matrix} & project A & project B & project C & plant \space capacities \\ plant 1 & 10(40) & 4(30) & 11 & 70\\ plant 2& 12 & 5(20) & 8 (30)& 50\\ plant 3 & 9 & 7 & 6(30) & 30\\ Project \space requirements & 40 & 50 & 60 & 150 \end{matrix}$

The minimum total transportation cost =10×40+4×30+5×20+8×30+6×30=1040

Notice alternate solution is available with unoccupied cell plant 3 project A=0, but with the same optimal value.