Revision as of 14:29, 19 November 2023 by Admin (Created page with "A loan of 100 is to have all principal and accrued interest paid at the end of five years. Interest accrues at an annual effective rate of 5\% for the first two years and at a force of interest at time <math>t</math> in years <math>(t>2)</math> of <math>\delta_t=\frac{1}{1+t}</math>. Calculate the equivalent annual effective rate of discount for the five-year period. <ul class="mw-excansopts"><li>10.9%</li><li>13.0%</li><li>14.6%</li><li>15.4%</li><li>17.1%</li></ul>...")
ABy Admin
Nov 19'23
Exercise
A loan of 100 is to have all principal and accrued interest paid at the end of five years. Interest accrues at an annual effective rate of 5\% for the first two years and at a force of interest at time [math]t[/math] in years [math](t\gt2)[/math] of [math]\delta_t=\frac{1}{1+t}[/math].
Calculate the equivalent annual effective rate of discount for the five-year period.
- 10.9%
- 13.0%
- 14.6%
- 15.4%
- 17.1%
ABy Admin
Nov 19'23
Solution: C
[[math]]\begin{aligned} & 100(1+0.05)^2 \exp \left(\int_2^5 \frac{1}{1+t}\right)=100(1.1025) \exp \left[\left.\ln (1+t)\right|_2 ^5\right] \\ & =110.25\left(\frac{6}{3}\right)=220.50 \\ & \frac{100}{(1-d)^5}=220.50 \\ & (1-d)^5=\frac{100}{220.50}=0.453515 \\ & 1-d=0.853726 \\ & d=0.1463\end{aligned}[[/math]]