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.

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.