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= 10x11+12x12+14x13+11x21+9x22+10x23

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, x11+x12+x13≥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, x21+x22+x23≥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,

20x11+22x21≤200⇒20x11+22x21≤200(60)⇒20x11+22x21≤12000

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,

24x12+18x22≤200⇒24x12+18x22≤200(60)⇒24x12+22x22≤12000

Constraint 5:

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

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Knowledge Booster

	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

Solution Summary: The author explains how the objective function is to minimize the cost of manufacturing desired steel.

Explanation of Solution

Chapter 3 Solutions