Copperfield Mining Company owns two mines, each of which produces three grades of ore high, medium, and low. The company has a contract to supply a smelting company with at least (and maybe more than) 12 tons of high-grade ore, 8 tons of medium-grade ore, and 24 tons of low-grade ore.

Each mine produces a certain amount of each type of ore during each hour it operates. Mine 1 produces 6 tons of high-grade ore, 2 tons of medium-grade ore, and 4 tons of low-grade ore per hour. Mine 2 produces 2 tons of high-grade ore, 2 tons of medium-grade ore, and 12 tons of low-grade ore per hour.

It costs the company $200 per hour to operate mine 1 and it costs $160 per hour to operate mine 2. The company wants to determine the number of hours it needs to operate each mine so that it can meet its contractual obligations at the lowest cost.

Use Excel to solve the problem and add at least the Answer Report and the Sensitivity Report.

Turn in two files

1) The Excel file you use to answer the questions

2) A Word file with full sentence answers to the questions asked.

A) What are the number of hours it needs to operate each mine so that it can meet its contractual obligations at the lowest cost and what is this lowest cost?

B) Discuss the meaning of the range of optimality for the cost contributions for the two decision variables in the context of the numbers you obtain in Excel.

C) Discuss the meaning of the range of feasibility for the RHS of each constraint, including the limiting numerical values in Excel and also include a discussion of the dual values or shadow prices in Excel involved with each constraint (so discuss high grade, the medium and the low).

M2 (1,3) (0,6) High grade Medium (3,1) grade (0,4) Low grade (0, 2) M1 (2,0) (4,0) (6,0) Slide for hint on range of optimality - change of slope of cost line (the red line) M2 Cost = 200M1 + 160M2 and at best point (from corner point method) (1,3) 200M1 + 160M2 = 680 or (0,6) M2 = 680/160 - (200/160) M1 or (3, 1) M2 = 4.25 -1.25 M1 (0,4) M2 intercept (0,4.25) M1 intercept (3.4,0) (0, 2) M1 (2,0) (4,0) (6,0) Slide for hint on range of feasibility - change of position of high grade ore constraint M2 (1,3) 6M1 + 2M2 > 12 original high grade constraint (0,6) (3, 1) (0,4) The 12 can be smaller or larger by some amount and high grade and medium grade constraints crossing still define best point, albeit at different cost. (0,2) M1 12,0) (4,0) (6,0) M2 (1,3) (0,6) High grade Medium (3,1) grade (0,4) Low grade (0, 2) M1 (2,0) (4,0) (6,0) Slide for hint on range of optimality - change of slope of cost line (the red line) M2 Cost = 200M1 + 160M2 and at best point (from corner point method) (1,3) 200M1 + 160M2 = 680 or (0,6) M2 = 680/160 - (200/160) M1 or (3, 1) M2 = 4.25 -1.25 M1 (0,4) M2 intercept (0,4.25) M1 intercept (3.4,0) (0, 2) M1 (2,0) (4,0) (6,0) Slide for hint on range of feasibility - change of position of high grade ore constraint M2 (1,3) 6M1 + 2M2 > 12 original high grade constraint (0,6) (3, 1) (0,4) The 12 can be smaller or larger by some amount and high grade and medium grade constraints crossing still define best point, albeit at different cost. (0,2) M1 12,0) (4,0) (6,0)