Revision as of 22:15, 17 November 2023 by Admin (Created page with "'''Solution: A''' The accumulation function is: <math display = "block"> a(t)=e^{\int_{0}^{t}{\frac{1}{8+r}}d r}=e^{\ln(8+r)_{0}^{\iota}}={\frac{8+t}{8}}. </math> Using the equation of value at end of 10 years: <math display = "block"> \begin{align*} 20,000=\textstyle{\int_{0}^{10}}(8k+t k){\frac{a(10)}{a(t)}}d t=k\int_{0}^{10}(8+t){\frac{18/8}{(8+t)/8}}d t=k\int_{0}^{10}18d t \\ ={180k\Rightarrow k}={\frac{20,000}{180}}=111. \end{align*} </math> {{soacopyright | 2...")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: A
The accumulation function is:
[[math]]
a(t)=e^{\int_{0}^{t}{\frac{1}{8+r}}d r}=e^{\ln(8+r)_{0}^{\iota}}={\frac{8+t}{8}}.
[[/math]]
Using the equation of value at end of 10 years:
[[math]]
\begin{align*}
20,000=\textstyle{\int_{0}^{10}}(8k+t k){\frac{a(10)}{a(t)}}d t=k\int_{0}^{10}(8+t){\frac{18/8}{(8+t)/8}}d t=k\int_{0}^{10}18d t \\
={180k\Rightarrow k}={\frac{20,000}{180}}=111.
\end{align*}
[[/math]]
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