Chapter_2_H_W_Solutions

Chapter_2_H_W_Solutions - Problem 2.4 FV = Ccfw(1 r)t 1 r M...

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10/30/2008 1 Problem 2.4 FV = C {[(1 + r) t – 1] / r } + M FV = $700,000[(1+0.062) 8 – 1) / 0.062] + $10M FV = $16,978,160.38 . Problem 2.8 P B = C{1 – [1/(1 + r/2)] 2xn } / r/2 + M / (1+r/2) 2xn P B(A) = $40{1 – [1/(1 + 0.07/2)] 2x9 } / 0.07/2 + $1,000 / (1+0.07/2) 2x9 = $527.59 + $538.36 = $1,065.95 P B(B) = $45{1 – [1/(1 + 0.09/2)] 2x20 } / 0.09/2 + $1,000 / (1+0.09/2) 2x20 = $828.071+ $171.929= $1,000.00 Problem 2.8 (Continued) P B(C) = $30{1 – [1/(1 + 0.10/2)] 2x15 } / 0.10/2 + $1,000 / (1+0.10/2) 2x15 = $461.174+ $231.377= $692.55 P B(D) = $1,000 / (1+0.08/2) 2x14 = $333.48 Problem 2.9 a. P B = $30 A 10 yrs 15% + $1,000 / (1+0.15/2) 2x10 = $541.25 @ r = 16% b P B = $509.09 . % Δ = ($509.09 – $541.25) / $541.25 = –5.94%. c. P B = $30 A 10 yrs 5% + $1,000 / (1+0.5/2) 2x10 = $1,077.95 @ r = 6% b P B = $1,000 % Δ = ($1,000.00 – $1,077.95) / $1,077.95 = –7.23% There is more volatility in a lower interest-rate environment because there was a greater fall –7.23% vs. –5.94%
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Chapter_2_H_W_Solutions - Problem 2.4 FV = Ccfw(1 r)t 1 r M...

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