## Info

B+

SB

= 7

Here, A and B represent the days of operation per week for each mine; SH, SM, and SL represent excess production of high-, medium-, and low-grade ore, respectively; and SA and SB are days per week that each mine is not operated.

A graphic representation of the linear programming problem was also provided. The graph suggests an optimal solution at point X, where constraints 1 and 2 are binding. Thus,

Substitute A = 2 into the high-grade ore constraint:

A minimum total operating cost per week of \$50,000 is suggested, because

Total Cost = \$10,000A + \$5,000B = \$10,000(2) + \$5,000(6) = \$50,000

Idaho Natural Resources, Ltd., (INR) LP Graph

Days of operation of mine A

Idaho Natural Resources, Ltd., (INR) LP Graph

Days of operation of mine A The consultant's report did not discuss a variety of important long-run planning issues. Specifically, INR wishes to know the following, holding all else equal:

A. How much, if any, excess production would result if the consultant's operating recommendation were followed?

B. What would be the cost effect of increasing low-grade ore sales by 50%?

C. What is INR's minimum acceptable price per ton if it is to renew a current contract to provide one of its customers with 6 tons of high-grade ore per week?

D. With current output requirements, how much would the cost of operating mine A have to rise before INR would change its operating decision?

E. What increase in the cost of operating mine B would cause INR to change its current operating decision?

### ST9.1 Solution

A. If the consultant's operating recommendation of A = 2 and B = 6 were followed, 32 tons of excess low-grade ore production would result. No excess production of high- or mediumgrade ore would occur. This can be shown by solving for SH, SM, and SL at the recommended activity level.

From the constraint equations, we find the following:

(1)

6(2) +

2(6)

- SH = 