Chapter 3 Solutions

# Chapter 3 Solutions - MIME 310 ENGINEERING ECONOMY...

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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 #3 – THE TIME VALUE OF MONEY 1. Monetary flow diagrams are used to illustrate the occurrence of inflows and outflows of funds over a project's life. Downward-pointing arrows represent disbursements, and upward- pointing arrows, receipts. These diagrams are very useful in solving time value problems. i) \$5000 \$4000 \$3000 0 1 2 3 4 5 ii) \$500 1 2 3 4 5 6 7 8 0 9 10 \$2000 17

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iii) \$1000 \$300 \$400 0 1 2 3 4 \$500 \$600 5 \$500 iv) \$1500 0 1 2 3 4 5 6 \$1650 \$1815 \$1996 \$2196 \$2416 \$10 000 2. The compound amount (FV) after five years is determined by applying the compound amount factor. i) With annual compounding FV = 500 (F/P,8%,5) = 500 (1.4693) = \$734.65 ii) With quarterly compounding Rate per quarter: 8% / 4 = 2% Number of quarters: 5 (4) = 20 FV = 500 (F/P,2%,20) = 500 (1.4859) = \$742.95 18
iii) With monthly compounding Rate per month: 8% / 12 = 0.667% Number of months: 5 (12) = 60 FV = 500 (F/P,0.667%,60) = 500 (1.4901) = \$745.05 iv) With continuous compounding FV = 500 e (r•n) = 500 (2.7183) (0.08)(5) = 500 (1.4918) = \$745.90 Note : Alternatively, parts ii, iii and iv can be solved by applying the compound amount factors associated with the relevant effective annual interest rate and a period of five years. EIR = (1 + r / m) m - 1 ii) EIR = (1 + 0.02) 4 - 1 = 0.0824 or 8.24% FV = 500 (F/P,8.24%,5) = 500 (1.4857) = \$742.85 iii) EIR = (1 + 0.00667) 12 - 1 = 0.0830 or 8.30% FV = 500 (F/P,8.30%,5) = 500 (1.4898) = \$744.90 iv) EIR = e r - 1 = 2.7183 (0.08) - 1 = 0.0833 or 8.33% FV = 500 (F/P,8.33%,5) = 500 (1.4919) = \$745.95 The small discrepancies are due to round-off error. 3. The compound amount (FV) is determined by applying the compound amount factor. Monthly periods are used. The total interest earned is the difference between the compound amount and the principal. Rate per month: 6% / 12 = 0.5% Number of months: 5 (12) + 9 = 69 FV = 600 (F/P,0.5%,69) = 600 (1.4108) = \$846.46 Interest earned: 846.46 - 600 = \$246.46 4. The effective annual interest rate is determined by applying the following relationship: EIR = (1 + r / m) m - 1 19

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i) EIR = (1 + 0.12 / 2) 2 - 1 = 0.1236 or 12.36% ii) EIR = (1 + 0.12 / 4) 4 - 1 = 0.1255 or 12.55% iii) EIR = (1 + 0.12 / 12) 12 - 1 = 0.1268 or 12.68% 5. i) Nominal interest rate: r = 12 (0.0075) = 0.09 or 9% Effective interest rate: EIR = (1 + r / m) m - 1 = (1 + 0.0075) 12 - 1 = 0.0938 or 9.38% ii) As interest is charged on the unpaid balance, the situation described is that of compound interest. The total interest paid on a loan of P is:
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Chapter 3 Solutions - MIME 310 ENGINEERING ECONOMY...

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