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# week 3 intermediate finance(ANSWER KEY) - Solutions Guide...

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Solutions Guide: Please reword the answers to essay type parts so as to guarantee that your answer is an original. Do not submit as your own. Chapter 7-Problem 3 Consider Borden’s 83/4 percent bonds that mature on April 15, 2016. Assume that the interest on these bonds is paid and compounded annually. Determine the value of a \$1,000 denomination Borden bond as of April 15, 2004, to an investor who holds the bond until maturity and whose required rate of return is a. 7 percent b. 9 percent c. 11 percent a. P o = Σ I/(1 + k d ) t + M/(1 + k d ) n t=1 I = .0875 X 1000 = \$87.50 k d = 0.07 M = \$1000 n = 12 years (2012 - 2004) 12 P o = Σ 87.50/(1 + 0.07) t + 1000/(1 + 0.07) 12 t=1 = 87.50(PVIFA .07,12 ) + 1000 (PVIF .07,12 ) = 87.50(7.943) + 1000 (0.444) = \$1139 (tables and calculator) b. I = \$87.50 k d = .09 M = \$1000 n = 12 12 P o = Σ 87.50/(1 + .09) t + 1000/(1 + 0.09) 12 t=1 = 87.50 (PVIFA 0.09,12 ) + 1000(PVIF 0.09,12 ) = 87.50(7.161) + 1000 (0.356) = \$983 (Tables) \$982 (Calculator) c. I = \$87.50 k d = .11 M = \$1000 n = 12 12

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P o = Σ 87.50/(1 + 0.11) t + 1000/(1 + 0.11) 12 t=1 = 87.50 (PVIFA 0.11,12 ) + 1000(PVIF 0.11,12 ) = 87.50(6.492) + 1000 (0.286) = \$854 (tables and calculator) d. I = .0875(1000)/2 = \$43.75 M = \$1000 k d = 0.08/2 = 0.04; n = 12 X 2 = 24 24 P o = Σ 43.75/(1 + 0.04) t + 1000/(1 + 0.04) 24 t=1 = 43.75(PVIFA 0.04,24 ) + 1000(PVIF 0.04,24 ) = 43.75(15.247) + 1000(0.390) = \$1057 (tables and calculator) Chapter 7-P 7 Consider the Leverage Unlimited, Inc., zero coupon bonds of 2008. Th e bonds were issued in 1990 for \$100. Determine the yield to maturity (to the nearest 1/10 of 1 percent) if the a. Issue price in 1990. (Note: To avoid a fractional year holding period, assume that the issue and maturity dates are at the midpoint—July 1—of the respective years.) b. Market price as of July 1, 2004, of \$750. c. Explain why the returns calculated in Parts a and b are different. a. P o = M/(1 + k d ) n = M(PVIF k d ,n ) n = 18 (2008 - 1990); P o = \$100; M = \$1000 \$100 = \$1000(PVIF k d ,18 ) (PVIF k d ,18 ) = 0.100 From Table II, this present value interest factor in the 18-year row is between the values for 13% (0.111) and 14% (0.095). Calculator solution is k d = 13.65 %.
b. P o = \$750; n = 4 (2008 - 2004) \$750 = \$1000(PVIF k d ,4 ) (PVIF k d ,4 ) = 0.750 k d = 7.46% (by calculator) c. Over the period from 1985 to 1999, the general level of interest rates

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