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A mining company owns two mines, each of which produces three grades (high, medium, and low) of ore. The company has a contract to supply

A mining company owns two mines, each of which produces three grades (high, medium, and low) of ore. The company has a contract to supply a smelting company with at least 12 tons of high-grade ore, at least 8 tons of medium-grade ore, and at least 24 tons of low-grade ore. Each hour of operation, mine 1 produces 6 tons of high-grade ore, 2 tons of medium-grade ore, and 4 tons of low-grade ore. Each hour of operation, mine 2 produces 2 tons of high-grade ore, 2 tons of medium-grade ore, and 12 tons of low-grade ore. It costs $200 per hour to operate mine 1 and $160 per hour to operate mine 2.How many hours should each mine be operated so as to meet the contractual obligations at the lowest total operating cost?

Use this sensitivity report to answer the questions below.

If the cost per hour to operate Mine 1 decreased by $10 and the cost per hour to operate Mine 2 increased by $20: Would (1,3) remain optimal? What minimum Cost could be realized?

Question 4 options:

For the first objective function coefficient, a decrease of 10 is 10/40 = 25% of the allowable decrease. For the second objective function coefficient, an increase of 20 is 20/40 = 50% of the allowable increase. The sum of those percentages (75%) is 100%. Thus, (1,3) will remain optimal. The minimum Cost will now be (190)(1) + (180)(3) = $730.

For the first objective function coefficient, an increase of 10 is 10/280 = 3.6 % of the allowable increase. For the second objective function coefficient, a decrease of 20 is 20/93 = 21.5 % of the allowable decrease. The sum of those percentages (25.1%) is 100%. Thus, (1,3) will remain optimal. The minimum Cost will now be (190*) (1) + (180) *(3) = $730.

For the first objective function coefficient, a decrease of 10 is 10/93 = 10% of the allowable decrease. For the second objective function coefficient, an increase of 20 is 20/280 = 7% of the allowable increase. The sum of those percentages (17%) is 100%. Thus, (1,3) will remain optimal. The minimum Cost will now be (190)(1) + (180) (3) = $730.

None of the above

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