Revision as of 22:42, 20 November 2023 by Admin (Created page with "'''Solution: A''' Let <math>a, b</math>, and <math>c</math> represent the face values of the three bonds. One, two, three, and four years from now, respectively: the 1-year bond provides payments of <math>1.01 a, 0,0,0</math>; the 3-year bond provides payments of <math>0.05 \mathrm{~b}, 0.05 \mathrm{~b}, 1.05 \mathrm{~b}, 0</math>; and the 4-year bond provides payments of <math>0.07 c, 0.07 c, 0.07 c, 1.07 c</math>. The total payments one, two, three, and four years fro...")
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Exercise

AAdmin
Nov 20'23

Answer

Solution: A

Let [math]a, b[/math], and [math]c[/math] represent the face values of the three bonds. One, two, three, and four years from now, respectively: the 1-year bond provides payments of [math]1.01 a, 0,0,0[/math]; the 3-year bond provides payments of [math]0.05 \mathrm{~b}, 0.05 \mathrm{~b}, 1.05 \mathrm{~b}, 0[/math]; and the 4-year bond provides payments of [math]0.07 c, 0.07 c, 0.07 c, 1.07 c[/math]. The total payments one, two, three, and four years from now must match the liabilities. Therefore, we have

[[math]] \begin{aligned} & 1.01 a+0.05 b+0.07 c=5766 \\ & 0.05 b+0.07 c=X \\ & 1.05 b+0.07 c=15421 \\ & 1.07 c=7811 \end{aligned} [[/math]]


Note that to find [math]X[/math], we do not need the first equation. Solving the fourth equation for [math]c[/math] yields [math]c=\frac{7811}{1.07}=7300[/math]. Substituting this value of [math]c[/math] into the third equation and solving for [math]b[/math] yields

[[math]] b=\frac{15421-0.07(7300)}{1.05}=14200 \text {. } [[/math]]


Finally, substituting these values of [math]b[/math] and [math]c[/math] into the second equation yields [math]X=0.05(14200)+0.07(7300)=1221[/math]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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