CH3 Solutions

# CH3 Solutions - 306-310 ENGINEERING ECONOMY SOLUTIONS TO...

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1 3 0 6 - 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. The monetary flow diagrams are illustrated below. i) ii) 1 2 3 4 5 0 6 7 8 \$10 000 \$10 000 \$1000 1 2 3 4 5 0 6 7 8 \$1000 \$750

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2 iii) 2. With simple interest, the principal amount and the interest are due at the end of the loan period of 24 months. Therefore: F = P + P i n = 1500 + 1500 (0.02) (24) = 1500 + 720 = \$2220 3. Donald gave Minnie two notes of \$1000 carrying simple interest. At the end of the first year, Donald paid Minnie an amount of: F = P + P i n = 1000 + 1000 (0.08) (1) = 1000 + 80 = \$1080 At the end of the second year, Donald paid Minnie an amount of: F = P + P i n = 1000 + 1000 (0.08) (2) = 1000 + 160 = \$1160 The total amount of interest paid by Donald was \$240 (80 + 160). 4. The effective interest rate (EIR) per period is found by applying the following relationship: EIR = (1 + r / m) m - 1 in which r is the nominal interest rate per period, i.e., one year in this problem, and m is the number of times per period that interest is paid, i.e., compounded. For annual compounding: EIR = r = 12% For monthly compounding: EIR = (1 + 0.12 / 12) 12 - 1 = 0.1268 or 12.68% For weekly compounding: EIR = (1 + 0.12 / 52) 52 - 1 = 0.1273 or 12.73% For daily compounding: EIR = (1 + 0.12 / 365) 365 - 1 = 0.1275 or 12.75% For continuous compounding: EIR = e 0.12 - 1 = 0.1275 or 12.75% 1 2 3 4 5 0 6 7 8 \$7500 \$2500 \$2000 \$1600 \$1280 \$1024 \$819.20
3 5. Let n represent the number of periods necessary to double an amount P, with annual compounding at a rate of 12 percent, and r, the nominal annual interest rate associated with continuous compounding. With compounding once per period: F = P (1 + 0.12) n = 2P With continuous compounding: F = P (e r [n/2] ) = 2P Therefore, we have: P (1.12) n = P (e r [n/2] ) Solving for r: r = 2 ln(1.12) = 0.2267 or 22.67% 6. Let r represent the nominal annual interest rate and assume an amount P that triples in value over a 20-year period. With simple interest: P + P (r) (20) = 3P r = (3 - 1) / 20 = 0.1 or 10% With annual compounding: P (1 + r) 20 = 3P r = 3 1/20 - 1 = 0.0565 or 5.65% With quarterly compounding: P (1 + r / 4) 4 [20] = 3P r = 4 (3 [1/80] - 1) = 0.0553 or 5.53% With continuous compounding: P • e r [20] = 3P r = (ln 3) / 20 = 0.0549 or 5.49% 7. Let P represent the amount borrowed by the king. Therefore, his quarterly payment was 0.05 P. The interest rate charged per quarter (i) is found by solving the following relationship: P = 0.05 P (P/A,i,40) Therefore, (P/A,i,40) = 20 Using the series present value factor table (or a financial calculator), we find that: For i=3%, (P/A,i,40) = 23.1148, and for i=4%, (P/A,i,40) = 19.7928 Using linear interpolation:

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4 i = 3% + (23.1148 - 20.0000) / (23.1148 - 19.7928) = 3.94% Therefore, the effective annual interest rate is: (1 + 0.0394) 4 - 1 = 0.1672 or 16.72% 8. Considering that the first instalment is akin to a down-payment, Clarabelle's monthly instalment (A) is determined as follows:
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## This note was uploaded on 10/06/2009 for the course MIME 310 taught by Professor Bilido during the Summer '08 term at McGill.

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CH3 Solutions - 306-310 ENGINEERING ECONOMY SOLUTIONS TO...

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