Revision as of 18:44, 19 November 2023 by Admin (Created page with "'''Solution: A''' Book values are linked by BV3(1 + i) – Fr = BV4. Thus BV3(1.04) – 2500(0.035) = BV3 + 8.44. Therefore, BV3 = [2500(0.035) + 8.44]/0.04 = 2398.5. The prospective formula for the book value at time 3 is, where m is the number of six-month periods. <math display = "block"> \begin{array}{l}{{2398.5=2500(0.035){\frac{1-1.04^{-(m-3)}}{0.04}}+2500(1.04)^{-(m-3)}}}\\ {{\mathrm{211=312.5(1.04)^{-(m-3)}}+2500(1.04)^{-(m-3)}}}\\ {{m-3={\frac{\ln(211/312.5)}{...")
Exercise
ABy Admin
Nov 19'23
Answer
Solution: A
Book values are linked by BV3(1 + i) – Fr = BV4. Thus BV3(1.04) – 2500(0.035) = BV3 + 8.44. Therefore, BV3 = [2500(0.035) + 8.44]/0.04 = 2398.5. The prospective formula for the book value at time 3 is, where m is the number of six-month periods.
[[math]]
\begin{array}{l}{{2398.5=2500(0.035){\frac{1-1.04^{-(m-3)}}{0.04}}+2500(1.04)^{-(m-3)}}}\\ {{\mathrm{211=312.5(1.04)^{-(m-3)}}+2500(1.04)^{-(m-3)}}}\\ {{m-3={\frac{\ln(211/312.5)}{-\ln(1.04)}}=10.}}\end{array}
[[/math]]
Thus, m = 13 and n = m/2 = 6.5. Note that the financial calculator can be used to solve for m – 3.
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