Revision as of 12:05, 18 November 2023 by Admin (Created page with "You are given the following information about a perpetuity with annual payments: #The first 15 payments are each 2500, with the first payment to be made three years from now. #Beginning with the 16th payment, each payment is k% larger than the previous payment. #Using an annual effective interest rate of 3.5%, the present value of the perpetuity is 115,000. Calculate k <ul class="mw-excansopts"><li>1.66</li><li>1.74</li><li>1.78</li><li>1.83</li><li>1.89</li></ul> {{...")
ABy Admin
Nov 18'23
Exercise
You are given the following information about a perpetuity with annual payments:
- The first 15 payments are each 2500, with the first payment to be made three years from now.
- Beginning with the 16th payment, each payment is k% larger than the previous payment.
- Using an annual effective interest rate of 3.5%, the present value of the perpetuity is 115,000.
Calculate k
- 1.66
- 1.74
- 1.78
- 1.83
- 1.89
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
ABy Admin
Nov 18'23
Solution: E
The value at time 17 of the payments beginning at time 18 is
[[math]]
2500\left(\frac{1+k}{1.035}+\frac{\left(1+k\right)^{2}}{0.035^{2}}+\cdots\right)=2500\frac{\frac{1+k}{1.035}}{1-\frac{1+k}{1.035}}=2500\frac{1+k}{0.035-k}
[[/math]]
The total present value is
[[math]]
\begin{array}{c}{{1\,15,000=2500(1.035^{-2})a_{\overline{15}|0.035}+2500v^{17}\,\frac{1+k}{0.035-k}}}\\ {{46(0.035-k)=10.751\,6(0.035-k)+0.55720(1+k)}} \\
k=\frac{1.61-0.3763\,1-0.55720}{46-10.751\,6+0.5720}=0.01889
\end{array}
[[/math]]
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.