A deposit of 100 is made into a fund at time [math]t=0[/math]. The fund pays interest at a nominal annual rate of discount [math]d[/math] compounded quarterly for the first two years. Beginning at time [math]t=2[/math], interest is credited at a force of interest [math]\delta_t=\frac{1}{t+1}[/math]. At time [math]t=5[/math], the accumulated value of the fund is 260. Calculate [math]d[/math].

• 12.7%
• 12.9%
• 13.1%
• 13.3%
• 13.5%

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

Solution: B

[[math]] 260=\frac{100}{(1-d / 4)^8} e^{\int_2^5 1 /(t+1) d t}=\frac{100}{(1-d / 4)^8} e^{\left.\ln (t+1)\right|_2 ^5}=\frac{100}{(1-d / 4)^8} e^{\ln (6)-\ln (3)}=\frac{100}{(1-d / 4)^8} e^{\ln (6 / 3)}=\frac{100}{(1-d / 4)^8} 2 \text {. } [[/math]]

So [math]1-d / 4=(200 / 260)^{.125}[/math] so [math]d=4 *\left(1-(200 / 260)^{.125}\right)=0.129[/math].

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.