Revision as of 23:32, 18 November 2023 by Admin (Created page with "A bank agrees to lend 10,000 now and X three years from now in exchange for a single repayment of 75,000 at the end of 10 years. The bank charges interest at an annual effective rate of 6% for the first 5 years and at a force of interest <math display = "block"> \delta_t = \frac{1}{t+1}, \quad \textrm{for} \, t \geq 5.</math> Calculate X <ul class="mw-excansopts"><li>23,500</li><li>24,000</li><li>24,500</li><li>25,000</li><li>25,500</li></ul> {{soacopyright | 2023 }}")
ABy Admin
Nov 18'23
Exercise
A bank agrees to lend 10,000 now and X three years from now in exchange for a single repayment of 75,000 at the end of 10 years. The bank charges interest at an annual effective rate of 6% for the first 5 years and at a force of interest
[[math]] \delta_t = \frac{1}{t+1}, \quad \textrm{for} \, t \geq 5.[[/math]]
Calculate X
- 23,500
- 24,000
- 24,500
- 25,000
- 25,500
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
ABy Admin
Nov 18'23
Solution: D
The effective annual interest rate is
[[math]]\dot{l}=(1-d)^{-1}-1=(1-0.055)^{-1}-1=5.829 \%.
[[/math]]
The balance on the loan at time 2 is 15, 000, 000(1.0582)2 =16, 796,809.
The number of payments is given by
[[math]]1,200,000 a_{\overline{n}|} = 16, 796,809[[/math]]
which gives [math]n = 29.795 \Rightarrow 29[/math]
payments of 1,200,000. The final equation of value is
[[math]]
\begin{align*}
1.200,000a_{\overline{{{29}}}|}+X(1.0582)^{-30}=16,796,809 \\
X=(16,796,809-16,621,012)(5.45799)=959,490.
\end{align*}
[[/math]]
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.