Homework 27 Solution

# Homework 27 Solution - Homework 27 Solution 8.17 Let X be...

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Homework 27 Solution 8.17 Let X be the number of miles to be treated. If we choose Alternative 1, the EUAC is  0.1030 EUAC \$500,000 A | P,6%,15 \$100X \$51,480 \$100X  If we choose Alternative 2, we will have to apply road oil at a cost of \$130 per barrel. If each barrel covers 1/20 th of a mile, the cost per mile is \$130 × 20 = \$2600 per mile. Note that the road oil must be applied now and annually after that, so the EUAC is 0.1030 EUAC \$50,000 \$2600X A | P,6%,15 \$2600X \$5150 \$2868X Setting these equal and solving for X: \$51,480 + \$100X = \$5150 + \$2868X \$46,330 = \$2768X X = 16.7 miles If the total distance to be surfaced is only 12.5 miles (i.e., less than the breakeven point), the more economical alternative is the one with the higher unit cost, which is Alternative 2. Check: Alternative 1: \$51,480 + \$100 (12.5) = \$52,730 Alternative 2: \$5150 + \$2868 (12.5) = \$41,000 8.18 Let X represent the number of months of use. Since we’re dealing in months, we’ll need a monthly value for the MARR: MARR = r/m = 15% / 12 = 1.25% per month If the engineer leases the computer system, his monthly cost is simply \$800. If he buys the computer

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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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Homework 27 Solution - Homework 27 Solution 8.17 Let X be...

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