Revision as of 13:32, 26 November 2023 by Admin (Created page with "An investment of 1000 accumulates to 1200 at the end of 4 years. If the force of interest is 1.25δ during the first two years and δ during the next two years, find the equivalent effective annual interest rate i for the first year. <ul class="mw-excansopts"> <li>0.036</li> <li>0.04</li> <li>0.048</li> <li>0.052</li> <li>0.062</li> </ul> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwin...")
ABy Admin
Nov 26'23
Exercise
An investment of 1000 accumulates to 1200 at the end of 4 years. If the force of interest is 1.25δ during the first two years and δ during the next two years, find the equivalent effective annual interest rate i for the first year.
- 0.036
- 0.04
- 0.048
- 0.052
- 0.062
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.
ABy Admin
Nov 26'23
Solution: B
[[math]]
1200=1000 \exp \left(\int_0^4 \delta_r d r\right)=1000 \exp \left(\int_0^2 1.25 \delta d t+\int_2^4 \delta d t\right)=1000 \exp (2.5 \delta+2 \delta)=1000 \exp (4.5 \delta)
[[/math]]
For the first year [math]1+i=e^{\int_0^1 1.25 \delta}=(1200 / 1000)^{1.25 / 4.5}[/math] so [math]i=1.2^{1.25 / 4.5}-1=0.05195[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.