Question

17. United Aluminum Company of Cincinnati produces three grades (high, medium, and low) of aluminum at two mills. Each mill has a different production capacity (in tons per day) for each grade, as follows:

                                                Mill

Aluminum Grade                    1          2         

High                                        6          2      12

Medium                                   2          2      8

Low                                         4          10     5

The company has contracted with a manufacturing firm to supply at least 12 tons of high-grade aluminum, 8 tons of medium-grade aluminum, and 5 tons of low-grade aluminum. It costs United $6,000 per day to operate mill 1 and $7,000 per day to operate mill 2. The company wants to know the number of days to operate each mill in order to meet the contract at the minimum cost.

Solve the linear programming model formulated in Problem 16 for United Aluminum Company graphically.

a. How much Extra (i.e., surplus) high- , medium- , and low-grade Aluminum does the company produce at the optimal solution?

b. What would be the effect on the optimal solution if the cost of operating mill 1 increased from $6000 to $7500 per day ?

c. What would be the effect on the optimal solution if the company could supply only 10 tons of high-grade Aluminum. ?

Answer 1

17.

Linear programming model is formulated as below:

Let X1 and X2 be the number of days Mill 1 and Mill 2 to be operated

Minimize 6000X1+7000X2

s.t.

6X1+2X2 >= 12

2X1+2X2 >= 8

4X1+10X2 >= 5

X1, X2 >= 0

Solution of the LP model using graphical method is as follows:

+ ro 6x + 2y212 2x + 2y28 4x + 10 > 5 (0.6) 6000x + 7000y = 24000 (1,3) (4,0)

X1 is plotted along x-axis and X2 is plotted along y-axis

Feasible region is unbounded, with three corner points at (4,0), (1,3), and (0,6)

Value of objective function is calculated at each of the corner points:

X1	X2	Objective value
4	0	24000
1	3	27000
0	6	42000

Minimum value of objective function is 24,000 at corner point (4,0)

Therefore, optimal solution is:

X1 = 4

X2 = 0

Objective value = 24,000

The solution implies that mill 1 should be operated for 4 days and mill 2 should be operated for 0 days

Total cost of this plan = $ 24,000

-----------------------------

a)

Surplus high-grade Aluminum produced = (6*4+2*0)-12 = 12 tons

Surplus medium-grade Aluminum produced = (2*4+2*0)-8 = 0 tons

Surplus low-grade Aluminum produced = (4*4+10*0)-5 = 11 tons

-----------------------------

b)

If the cost of operating mill 1 is increased from $ 6000 to $ 7500 per day, then the point of optimality will shift to (1,3)

+ rn 6x + 2y212 2x + 2y28 4x + 10y > 5 7500x + 7000y = 28500 (1,3)

New optimal solution is:

X1 = 1 days

X2 = 3 days

Total cost of this plan = 7500*1+3*7000

= $ 28,500

-----------------------------

c)

if the company could supply only 10 tons of high-grade Aluminum, the optimal will NOT change

because production of high grade aluminum is already in the optimal solution, So, the solution will not change unless the requirement of high-grade aluminum is increased to 24 tons or more.

6x + 2y210 2x + 2y8 4x + 10y > 5 6000x + 7000y = 24000 X (4,0)