Revision as of 18:43, 19 November 2023 by Admin (Created page with "'''Solution: E''' Book values are linked by BV3(1 + i) – Fr = BV4. Thus 1254.87(1.06) – Fr = 1277.38. Therefore, the coupon is Fr = 52.7822. The prospective formula for the book value at time 3 is <math display = "block"> \begin{array}{l}{{1254.87=52.7822{\frac{1-1.06^{-(n-3)}}{0.06}}+1890(1.06)^{-(n-3)}}}\\ {{375.1667=1010.297(1.06)^{-(n-3)}}}\\ {{n-3={\frac{1}{-1}}{\frac{1}{75.1667/1010.297)}}=17.}}\end{array} </math> Thus, n = 20. Note that the financial calcul...")
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Exercise


ABy Admin
Nov 19'23

Answer

Solution: E

Book values are linked by BV3(1 + i) – Fr = BV4. Thus 1254.87(1.06) – Fr = 1277.38. Therefore, the coupon is Fr = 52.7822. The prospective formula for the book value at time 3 is

[[math]] \begin{array}{l}{{1254.87=52.7822{\frac{1-1.06^{-(n-3)}}{0.06}}+1890(1.06)^{-(n-3)}}}\\ {{375.1667=1010.297(1.06)^{-(n-3)}}}\\ {{n-3={\frac{1}{-1}}{\frac{1}{75.1667/1010.297)}}=17.}}\end{array} [[/math]]

Thus, n = 20. Note that the financial calculator can be used to solve for n – 3.

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