Solutions+Problem+Set+3-1

# Solutions+Problem+Set+3-1 - E120 Principles of Engineering...

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E120 Principles of Engineering Economics Fall 2010 Problem Set #3 Solutions 1. a. Since 6 months is 6/24 = 1/4 of 2 years, using our rule (1 + 0.2) 1/4 = 1.0466 So the equivalent 6 month rate is 4.66% b. Since one year is half of 2 years, (1.2) 1/2 = 1.0954 So the equivalent 1 year rate is 9.54% c. Since 1 month is 1/24 of 2 years, using our rule (1 + 0.2) 1/24 = 1.00763 So the equivalent 1 month rate is 0.763%. 2. Dropped. 3. Using the formula for converting from an EAR to an APR quote (1 + APR/k) k = 1.05 Solving for the APR APR = k((1.05) 1/k - 1) With semiannual payments k = 2, so APR = 4.939% With monthly payments k = 12, so APR = 4.889% 4. Using the PV of an annuity formula with N = 10 payments and C = \$100 with r = 4.067% per 6 month interval, since there is an 8% APR with monthly compounding: 8% / 12 = 0.6667% per month, or (1.006667) 6 – 1 = 4.067% per 6 months. PV = (\$100/0.04067) × (1 – 1.04067 -10 ) = \$808.39 5. To find out what is owed compute the PV of the remaining payments using the loan

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## This note was uploaded on 11/22/2010 for the course ENGIN 120 taught by Professor Ilan during the Fall '08 term at Berkeley.

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Solutions+Problem+Set+3-1 - E120 Principles of Engineering...

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