Homework 22 Solution

# Homework 22 Solution - Homework 22 Solution 6.20 The...

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Unformatted text preview: Homework 22 Solution 6.20 The overall rate of return is just the weighted average of the individual returns: Overall ROR \$60 \$140 20% 12% 14.4% \$200 \$200 6.31 Since these are cost alternatives, do nothing is not an option. If the rate of return for each incremental analysis was less than the MARR, the company should select the first (i.e., cheapest) alternative. 6.32 Since the reverse osmosis process is the more expensive of the two, we need to evaluate (RO–Nano): Capital cost, \$1000 Operating cost, \$1000 per year Revenue, \$1000 per year Nano -3200 -250 1900 RO -4800 -490 2600 (RO–Nano) -1600 -240 700 The RO process brings in \$700K more in revenue but costs \$240K more to operate, so the incremental net revenue is \$440K per year. Setting the present worth of the incremental cash flows equal to zero: PW 1600 440 P | A, i ,10 0 P | A, i ,10 1600 3.6364 440 Looking through the tables at the back of the textbook, we find that (P|A,24%,10) = 3.6819 (P|A,25%,10) = 3.5705 So the incremental ROR is somewhere between 24% and 25%. For all intents and purposes, we have our answer at this point: Choose RO. But the problem statement said to calculate the incremental ROR so we can interpolate the answer from here: 3.6364 3.6819 i 24% 25% 24% 24.4% 3.5705 3.6819 6.35 Since these are cost projects, do nothing is not an option, so our first choice is to select the cheapest alternative (Machine 2). The next-cheapest alternative is Machine 1, which costs \$2000 more but saves \$2700 per year on operating costs: 5.3680 \$2700 F | A, 20%, 4 ERR \$2000 1 4 1 64% Since this is greater than the 20% MARR, Machine 1 is a better choice than Machine 2. Machine 1 is the current champion. The next-cheapest alternative is Machine 3, which costs \$3500 more than Machine 1 and costs more to operate. So Machine 3 is clearly not better than Machine 1. Machine 1 is still the champion. Next cheapest is Machine 5, which costs \$10,000 more than Machine 1 but saves \$500 per year on operating costs. 5.3680 \$500 F | A, 20%, 4 ERR \$10,000 1 4 1 28% So Machine 5 is clearly not better than Machine 1. Machine 1 is still the champion. Finally, Machine 4 costs \$18,000 more than Machine 1 but saves \$3800 per year on operating costs: 5.3680 \$3800 F | A, 20%, 4 ERR \$18,000 1 4 1 3.2% Since this is less than the 20% MARR, Machine 4 is not a better choice than Machine 1, so the final answer is Machine 1. 6.29 The equipment needed to produce the treated vinyl rollers has a service life of just 3 years, so you would have to purchase two of them (in Years 0 and 3) in order to get the same service life as the equipment for producing fiber-impregnated rubber rollers. The cash flows for the two choices are as follows: Year 0 1 2 3 4 5 6 Vinyl -50 -100 -100 -100+5-50=-145 -100 -100 -100+5=-95 Rubber -95 -85 -85 -85 -85 -85 -74 (Rubber-Vinyl) -45 15 15 60 15 15 21 There is only one year with net negative cash flows (Year 0): P 45 The rest have net positive cash flows: 10.1830 F 15 F | A, 21%,6 45 1.21 6 238.5 238.5 ERR 45 3 1 6 1 32% Since this is greater than the 21% MARR, the equipment for making fiber-impregnated rubber rollers is worth the added initial cost. ...
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## This note was uploaded on 12/07/2011 for the course CIVL 4111 taught by Professor Moore,l during the Fall '08 term at U. Memphis.

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