The United Aluminum Company of Cincinnati produces three grades (high, medium, a

Question

The 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:

                             Table 1: Capacity at each Mill by Grade of Aluminum

Aluminum Grade

Capacity (tons per day)

Mill 1

Mill 2

High

6

2

Medium

2

2

Low

4

10

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 mill1 and $7,000 to operate mill 2. The company wants to know

3. What would be the optimal solution if the company needs to supply at least 10 tons of high-grade aluminum?

A.        $28000

B.        $24000

C.         $18000

D.         cannot be determined

Aluminum Grade

Capacity (tons per day)

Mill 1

Mill 2

High

6

2

Medium

2

2

Low

4

10

Answer Report Objective Cell (Min) Ce Name SB$16 Min mize Total Cost Objective Function Variable Cells Ce Name $B$190 X1 days on mill 1 $C$19 2 days on mill 2 Constraints Ce Name DS23 High grade aluminum re red Ihs $D$24 Medium rade aluminum required Ihs SD$25 Low grade aluminum required Ihs Sensitivity Report Variable Cells Ce Name $B$19 x1 days on mill 1 $C$19 x2 days on mill 2 Constraints Ce Name DS23 High grade aluminum required Ihs $D$24 Medium grade aluminum required Ihs SD$25 Low grade aluminum required lhs Original Value Final Value 24000 Original Value Final Value nteger Contin 4 0 Contin Cell Value Formula Status 24, $D$23 -$E$23 Not Bind 8 $D$24> $E$24. Bindin 16 SD$25 $E$25 Not Binding Reduced Objective Fina Value Coefficient Cost 6000 1000 7000 Fina Shadow Constraint Value R.H. Side Price 24 12 3000 16 Slack 12 11 Allowable Increase 1000 1E+30 Allowable Increase 12 1E+30 11 Allowable Decrease 000 1000 Allowable Decrease 1E+30 1E+30

Explanation / Answer

Let the Mill 1 be operated for x days and Mill 2 for y days.

Model Formulation:

Minimize C = 6000x + 7000y

Subject to

6x + 2y 10, that is 3x + y 5 (Note here we took 10 (atleast 10 tons of high-grade aluminum) instead of 12 to solve the problem)

2x + 2y 8, that is x + y 4

4x + 10y 5

x, y 0

So pt(x,y) C

3x + y = 5; x + y = 4 (0.5,3.5)............................... 27500

3x + y = 5; x=0 (0,5) .................................35000

x + y = 4; y = 0 (4,0).................................24000

By drawing the graphs of the above three equation and checking for optimal solution we get

The optimal Solution is x=4, y = 0, c = 24000

The company must operate Mill1 for 4 days and not operate Mill2.

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