Chapter 3, Problem 8RP

Explanation of Solution

Linear Programming (LP) formulation:

• Given table:

	Steel 1		Steel 2
Mill	Cost	Time (Minute)	Cost	Time (Minute)
1	$10	20	$11	22
2	$12	24	$9	18
3	$14	28	$10	30

• Consider that x_ij be the tons of steel “i” manufactures at steel mill “j”, for (i=1,2 and j=1,2,3).
• The objective is to minimize the cost of manufacturing desired steel.
  • z= ((cost of manufacturing) (tons of steel 1 manufactured in mill 1,2 and 3) + (cost of manufacturing) (tons of steel 2 manufactured in mill 1,2 and 3)
• Thus the objective function is,
  • Min z= $10x_{11}+12x_{12}+14x_{13}+11x_{21}+9x_{22}+10x_{23}$

Constraint 1:

• At least, 500 tons of steel 1 must be produced each month
• [tons of steel 1 produced in mill 1+tons of steel 1 produced in mill 2+ tons of steel 1 produced in mill 3]≥500
• That is, $x_{11}+x_{12}+x_{13}\ge 500$

Constraint 2:

• At least, 600 tons of steel 2 must be produced each month
• [tons of steel 2 produced in mill 1+tons of steel 2 produced in mill 2+ tons of steel 2 produced in mill 3]≥600
• That is, $x_{21}+x_{22}+x_{23}\ge 600$

Constraint 3:

• At most, 200 hours of blast furnace time are available in each mill 1.
• [(time required in mill 1) (tons of steel 1 produced in mill 1)+(time required in mill 1) (tons of steel 2 produced in mill1)]≤200
• That is,

$\begin{array}{l}20x_{11}+22x_{21}\le 200\\ \Rightarrow 20x_{11}+22x_{21}\le 200(60)\\ \Rightarrow 20x_{11}+22x_{21}\le 12000\end{array}$

Constraint 4:

• At most, 200 hours of blast furnace time are available in each mill 2.
• [(time required in mill 2) (tons of steel 1 produced in mill 2)+(time required in mill 2) (tons of steel 2 produced in mill2)]≤200
• That is,

$\begin{array}{l}24x_{12}+18x_{22}\le 200\\ \Rightarrow 24x_{12}+18x_{22}\le 200(60)\\ \Rightarrow 24x_{12}+22x_{22}\le 12000\end{array}$

Constraint 5:

• At most, 200 hours of blast furnace time are available in each mill 3