Fund A accumulates at rate [math]12 \%[/math] convertible monthly. Fund [math]\mathrm{B}[/math] accumulates with force of interest [math]\delta_t=t / 6[/math], in years. At time zero, 1 is deposited into each fund. Let [math]T[/math] be the time when the two funds are equal.

Determine [math]T[/math], in years.

• [math]12 \ln (1.01)[/math]
• [math]12 \ln (1.12)-\ln (1.01)[/math]
• [math] 12 \ln (1.12)[/math]
• [math]144 \ln (1.01)[/math]
• [math]144\ln(1.12)[/math]

References

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

Solution: D

Fund A has value [math](1.01)^m[/math] after [math]m[/math] months. Fund [math]\mathrm{B}[/math] has value [math]e^{\int_0^t r / 6 d r}[/math] after [math]t[/math] years or [math]e^{\int_0^{m / 12} r / 6 d r}[/math] after [math]m[/math] months. Thus

[[math]] 1.01^m=e^{r^2 /\left.12\right|_0 ^{m / 12}}=e^{m^2 / 12^3} \text { so } [[/math]]

[math]m \ln (1.01)=m^2 / 12^3[/math] so [math]m=12^3 \ln (1.01)[/math] months. Thus [math]T=m / 12=12^2 \ln (1.01)[/math].

References

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