For a certain fund the force of interest δ_t is a linear function of t such that δ₀ = 0 and δ₁ = ln(1.05). Three deposits of $1000 were are made on Jan 1, 2001 (time 0), Jan 1, 2003 and Jan 1, 2005.

What is the total accumulated value on Jan.1, 2006?

• 3875
• 4050
• 4525
• 4575
• 4755

References

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

Solution: E

[[math]] \begin{aligned} \text { Accum.Value } & =1000 e^{\int_0^5 \delta_t d t}+1000 e^{\int_2^5 \delta_t d t}+1000 e^{\int_4^5 \delta_t d t} \\ & =1000 e^{(\ln (1.05))(25 / 2-0)}+1000 e^{(\ln (1.05))(25 / 2-4 / 2)}+1000 e^{(\ln (1.05))(25 / 2-16 / 2)} \\ & =1000\left(1.05^{12.5}+1.05^{10.5}+1.05^{4.5}\right)=1000(1.840205+1.669120+1.245523)=4754.85 \end{aligned} [[/math]]

References

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