Answer to Question #299230 in Operations Research for true teach

Question

1.     The Hardrock Concrete Company has plants in three locations and is currently working on three major construction projects, each located at a different site. The shipping cost per truckload of concrete, daily plant capacities, and daily project requirements are provided in the table below.

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.

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