Question

1. A steel company has two mills. Mill 1 costs $70,000 per day to operate, and it can produce 400 tons of high-grade steel,500 tons of medium-grade steel, and 450 tons of low-grade steel each day. Mill 2 costs $60,000 per day to operate, and it can produce 350 tons of high-grade steel, 600 tons of medium-grade steel, and 400 tons of low-grade steel each day. The company has orders totaling 100,000 tons of high-grade steel, 150,000 tons of medium-grade steel, and 124,500 tons of low-grade steel. How many days should the company run each mill to minimize its costs and still fill the orders?

1). Please set up the LP model for the above situation.

2). How many days should we run Mill 1 (M1) and how many days should we run Mill 2 (M2) at optimal?

3). Whether we have slack or surplus variable applicable in this question? If we do, what is the value of slack (or/and surplus) variable? What is the meaning behind it?

4). What is the cost at the optimal solution?

Answer 1

x = number of days to operate mill 1.
y = number of days to operate mill 2.

The objective function is 70,000 * x + 60,000 * y.

The constraint functions are:

400x + 350y >= 100,000 (hi grade steel requirement)
500x + 600y >= 150,000 (medium grade steel requirement)
450x + 400y >= 124,500 (lo grade steel requirement)
x,y >= 0 (number of days can't be negative)

The corner points are (0,311.25), (210,75), (300,0)

the cost at (0,311.25) is 0 * 70,000 + 311.25 * 60,000 = 18,675,000.
the cost at (310,75) is 210 * 70,000 + 75 * 60,000 = 19,200,000.
the cost at (300,0) is 300 * 70,000 + 0 * 60,000 = 21,000,000.

the minimum cost is when you run mill2 for 311.25 daya.

all the constraint functions need to be met when x = 0 and y = 311.25

400x + 350y = 108937.5 which is > 100,000.
500x + 600y = 186,750 which is > 150,000.
450x + 400y = 124,500 which is e3qual to 124,500.

both x and y are > 0.

all the constraint are met and the cost is minimum when x = 0 and y = 311.25.

that means mill 2 gets a lot of work and mill 1 doesn't get any.