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MIME 310 - Solutions Chapter 4

# MIME 310 - Solutions Chapter 4 - MIME 310 ENGINEERING...

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M I M E 3 1 0 E N G I N E E R I N G E C O N O M Y SOLUTIONS TO PROBLEM SET #4 – PRODUCTION AND COST ANALYSES 1. The recovery method with the highest unit profit (revenue less cost) is preferred. Process B has a higher recovery than process A, but it also has a higher cost. Thus, a tradeoff exists between the two processes. The economic value of the higher recovery is a function of the value of recovered metal. At low values of the metal, process A, with the lower cost, is preferred. However, for high values of the metal, process B is preferred, because the value of recovering more metal outweighs the higher cost of the process. For one tonne of mine product containing an amount of metal M, the profit is given by: M • Recovery (Unit revenue - Unit cost) Unit profit for process A: 0.75 M (6.00 - 3.00) = 2.25 M Unit profit for process B: 0.8 M (6.00 - 3.25) = 2.2 M At a unit metal value of \$6.00 per tonne, process A has the highest unit profit, and it is therefore preferred. Let v be the unit metal value. Process B is more economical when: (Profit B - Profit A ) > 0 Profit B - Profit A = [0.8 M (v - 3.25)] - [0.75 M (v - 3)] = M [0.05 v - 0.35] Therefore, process B is preferred when M [0.05 v - 0.35] > 0 or when v > 7 The recovery method should be changed from A to B when the value of the metal exceeds \$7/tonne. 2. The total cost function is linear because the unit variable costs are constant within the plant's range of operating capacities. i) Let Q represent the production rate. At the break-even rate (Q * ), total revenue equals total costs. TR = p • Q = 500 Q TC = FC + VC = FC + vc • Q = 15 000 + 320 Q 26

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