Suppose the force of interest is:
$$F(t) = \frac{0.02 + 0.01t}{1 + 0.02t + 0.005t^2}$$
, where $t$ is the number of years beginning March 31, 2001.
An investment of $100$ is made on March 31, 2002, and an investment of $200$ is made on March 31, 2006. What is the accumulated value of the two investments on March 31, 2010?
My attempt:
The schedule of events goes as follows:
$t = 0$ - nothing happens
$t = 1$ - $100$ invested
$t = 5$ - $200$ invested
$t = 9$ - Accumulated value?
By definition, the force of interest is $F(t) = \frac{A'(t)}{A(t)} = \frac{a'(t)}{a(t)}$, which tell us that $a(t) = 1 + 0.02t + 0.005t^2$. However, proceeding by formula, we have:
$$a(t) = e^{\int_{0}^{t} F(u) \,du}$$
$$= e^{\int_{0}^{t} \frac{0.02 + 0.01u}{1 + 0.02u + 0.005u^2} \,du}$$
$$= 1 + 0.02t + 0.005t^2$$
We know that $A(1) = 100$ so the accumulated value of this $100$ after $8$ years is:
$$a(8)\cdot A(0) = a(9) \cdot \frac{A(1)}{a(1)} = \frac{100\cdot a(9)}{a(1)} = \frac{100(1 + 0.02(9) + 0.005(9)^2)}{1 + 0.02(1) + 0.005(1)^2} = 154.63$$
We also know that $A(5) = 200$ so the accumulated value of this $200$ after $4$ years is:
$$A(5)\cdot \frac{a(9)}{a(5)} = \frac{200(1 + 0.02(9) + 0.005(9)^2)}{1 + 0.02(5) + 0.005(5)^2} = 258.78$$
So, the total accumulated value is $154.63 + 258.78 = 413.41$
Is this correct? I am not very sure if my approach is correct so any assistance is much appreciated.
Yes, your approach is correct. Basically, we have $F(t)=\frac{a'(t)}{a(t)}=[\ln a(t)]'$. Hence, the growth factor from time $t_1$ to $t_2$ is $$e^{\int_{t_1}^{t_2} F(s)ds}=e^{\ln a(t_2)-\ln a(t_1)}=\frac{a(t_2)}{a(t_1)}.$$
Thus, for a problem like this, you just have to sum all of the cash flows multiplied by their respective growth factors.