Engineering Economy 1 Find the present worth of S4000 the first year, increasing
ID: 1149641 • Letter: E
Question
Engineering Economy
1 Find the present worth of S4000 the first year, increasing by $1000 per year. The interest rate is 8%, and N equals 5 Date: 2. Operating and repair costs for some equipment will be $15,000 per year for the 3 year warranty period. Beginning in year 4-costs will dimb by S2500 per year. The equiprment will be retired after iO years. Find the present worth of these costs if the firm's interest rate is 12%. 3. A car's warranty is 3 years. Upon expiration, annual maintenance starts at $150 and then climbs $25 per year until the car is sold at the end of Year 7. Use a 10% interest rate and find the present worth of these expenses. 4. Demand for a new product will deine as competitors enter the market. If interest is OK, what is an equivalent uniform value? Year Revenue 1 $24,000 2. 18,000 3. 12,000 46.000 5. On a certain piece of machinery, it is estimated that the maintenance expense will be as follows Year Maintenance 1. $100 2.200 3.300 4.400 What is the equivalent uniform annual maintenance cost for the machinery if 6% interest is used?Explanation / Answer
1. Let present worth be denoted by P.W.
P.W = 4000/(1.08) + 5000/(1.08)2 + 6000/(1.08)3+ 7000/(1.08)4+ 8000/(1.08)5
= 3703.70 +4286.69 + 4762.99 + 5145.21 + 5444.67 = 23343.26
So present worth of 4000 increasing 1000 every year is $23343.26.
2.P.W = 15000/(1.12) + 15000/(1.12)2 + 15000/(1.12)3+ (15000+2500)/(1.12)4 + (15000+2*2500)/(1.12)5+ (15000+3*2500)/(1.12)6+ (15000+4*2500)/(1.12)7+ (15000+5*2500)/(1.12)8+ (15000+6*2500)/(1.12)9+ (15000+7*2500)/(1.12)10
=13392.86 + 11957.91+10676.70+27536.59+11348.54+11399.20+11308.73+11106.79+10818.30+10464.13
=$130009.75
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