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...")
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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.

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