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An investor deposits USD 300 in a bank account at time 0, reinvests all interest payments and continuously invests USD 300 per annum, until the total value of the deposits reaches USD 3312. At that point the investor stops making additional deposits, but still lets the interest payments accumulate in the account.

The ODE for the value of the deposits, $V$, over time is then, $$\frac{dV}{dt}=r(t)V(t)+I(t),$$ where $I(t)=300$ until $V(t)$ reaches $V$=3312, at which point $I(t)$ instantaneously switched to $I(t)=0.$ Also, $r(t)=\frac{1}{20+\frac{t}{2}}$.

Derive an expression for the value of the asset as a function of time, $V(t)$, $t\geq 0.$

$\textbf{My Approach: }$

I tried the problem by using the formula, $$V(t)=e^{-P(T)}\Bigg{(}\int_{0}^{T}e^{P(t)}q(t)dt + c\Bigg{)}, \text{ } c\in \mathbb{R},$$ where $P(t)=\int_{0}^{t}r(t)dt$. I don't know how to progress any further. Any help will be greatly appreciated. Thanks in advance

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Let be $V_0=300$ the deposit at $t=0$, $V(T)=3312$ the value at $t=T$, and $r(t)=\frac{1}{20+t/2}=\frac{b}{a+t}$ the interest rate with $a=40$ and $b=2$. Denote $$I(t)=\rho\left(1-H(t-T)\right)$$ the rate of payment at any time in the range $[0,T]$ where $\rho=300$ and $H(t)$ is the Heaviside function or step function defined as $$H(t)=\begin{cases} 1 & t> 0\\ 0 & t\le 0 \end{cases}$$ so that $$I(t)=\begin{cases} 0 & t> T\\ \rho & 0\le t\le T \end{cases}$$ So you have to solve $$ \frac{\mathrm d V}{\mathrm d t}=r(t) V(t)+I(t)\tag 1 $$ that is a first order non-linear ODE. It's better to solve the equation (1) separately in the time intervals $t\le T$ and $t>T$ considering the following two linear ODE $$ \begin{align} \frac{\mathrm d V}{\mathrm d t}&=r(t) V(t)+\rho & 0\le t\le T\tag 2\\ \frac{\mathrm d V}{\mathrm d t}&=r(t) V(t) & t> T\tag 3 \end{align} $$ The general solution of $\dot x(t)+p(t)x(t)=q(t)$ is $$ x(t)=\frac{1}{u(t)}\left(\int u(t)q(t)\mathrm{d}t+C\right) \qquad \text{where } u(t)=\mathrm{e}^{\int p(t)\mathrm{d}t} $$ Then we have for equation (2) $$ V(t)=\mathrm{e}^{P(t)}\left(\int_0^t \rho \mathrm{e}^{-P(s)}\mathrm{d}s+C\right)\tag 4 $$ where $$P(t)=\int_0^t r(s)\mathrm{d}s=b\log\left(1+\frac{t}{a}\right)\tag 5$$ and $\mathrm{e}^{P(t)}=\left(1+\frac{t}{a}\right)^b$.

The condition $V(0)=V_0$ leads to $C=V_0\mathrm{e}^{-P(0)}=V_0$ because $P(0)=0$. Then the (4) becomes $$ \begin{align} V(t)&=V_0\mathrm{e}^{P(t)}+ \mathrm{e}^{P(t)}\int_0^t \rho \mathrm{e}^{-P(s)}\mathrm{d}s\\ &=V_0\left(1+\frac{t}{a}\right)^b+\rho\mathrm{e}^{P(t)}\int_0^t \left(1+\frac{s}{a}\right)^{-b}\mathrm{d}s\\ &=V_0\left(1+\frac{t}{a}\right)^b+\rho\left(1+\frac{t}{a}\right)^b\left[\frac{(a+t)^{1-b}a^b}{(1-b)}-\frac{a}{1-b}\right]\\ &=V_0\left(1+\frac{t}{a}\right)^b+\rho\left[\frac{a+t}{1-b}-\frac{(a+t)^{b}a^{1-b}}{1-b}\right]\\ &=V_0\left(1+\frac{t}{a}\right)^b+\frac{\rho a}{1-b}\left[\left(1+\frac{t}{a}\right)-\left(1+\frac{t}{a}\right)^b\right] \tag 6 \end{align} $$ The solution for (3) is $$ V(t)=V(T)\mathrm{e}^{P(t)-P(T)}\tag 7 $$ where from (6) $$V(T)=V_0\left(1+\frac{T}{a}\right)^b+\frac{\rho a}{1-b}\left[\left(1+\frac{T}{a}\right)-\left(1+\frac{T}{a}\right)^b\right]\tag 8$$ Finally the solution is

$$ V(t)= \begin{cases} V_0\left(1+\frac{t}{a}\right)^b+\frac{\rho a}{1-b}\left[\left(1+\frac{t}{a}\right)-\left(1+\frac{t}{a}\right)^b\right] & 0\le t\le T\\ V(T)\mathrm{e}^{P(t)-P(T)} & t> T \end{cases}\tag 9 $$ or $$ V(t)= \begin{cases} V_0\mathrm{e}^{P(t)}+\frac{\rho a}{1-b}\left[\mathrm{e}^{P(t)/b}-\mathrm{e}^{P(t)}\right] & 0\le t\le T\\ V(T)\mathrm{e}^{P(t)-P(T)} & t> T \end{cases}\tag {10} $$

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  • $\begingroup$ Your math is incorrect! $\endgroup$ – jxjxjx Mar 8 '16 at 2:30
  • $\begingroup$ @tx123 may be, but where? $\endgroup$ – alexjo Mar 8 '16 at 5:45
  • $\begingroup$ @tx123 I've checked and it seems to be right. $\endgroup$ – alexjo Mar 8 '16 at 14:35
  • $\begingroup$ Yeah, now it is correct. Initially, your integration was incorrect. $\endgroup$ – jxjxjx Mar 8 '16 at 19:53
  • $\begingroup$ @tx123 I forgot to multiply in the first line of (6) by the factor $\mathrm{e}^{P(t)}$. I fixed it. $\endgroup$ – alexjo Mar 8 '16 at 20:00
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Let's suppose $P$ is total annual deposits made continuously, then the change in value of total deposits $dV_t$ is (assuming no condition on additional deposits) $$dV_t= V_t r dt + P dt $$ where we assumed $r$ is constant. Solving above differential equation, we have: $$V_T = V_0 e^{rT} + \frac{P}{r} (e^{rT} -1)$$

Assuming $t_1$ is the time period at which $V_T$ reach the limit $(K)$ (which is 3312 in the question), thereafter investor stop making deposits. So, the value of deposits at time $t_1$ is: $$V_{t_1} = V_0 e^{r t_1} + \frac{P}{r} (e^{rt_1} -1) = K$$ You can solve the above equation numerically to derive $t_1$. After $t_1$, there is no fresh deposits, so your account will grow normally and value of portfolio at $T \{T > t_1 \}$ is $$V_T = V_{t_1} e^{r(T-t_1)} = K e^{r(T-t_1)} \quad \quad \,T > t_1$$

So, your expression for the vale of deposits is:

\begin{equation} V_T = \begin{cases} V_0 e^{rT} + \frac{P}{r} (e^{rT} -1) \quad & T \leq t_1\\ Ke^{r(T-t_1)} \quad \quad \quad \quad \quad & T> t_1 \end{cases} \end{equation}


Edit: Assuming interest rate $r$ as a function of time.

In this case, the value of deposits at time $T$ is: $$V_T = V_0 e^{\int_{0}^{T}r(t)dt} + \int_{0}^{T} Pe^{\int_{0}^{t}r(s)ds}dt$$

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  • $\begingroup$ That's true for $r(t)=r=$constant, but $r(t)=\frac{1}{20+t/2}$ $\endgroup$ – alexjo Mar 7 '16 at 17:38
  • $\begingroup$ @alexjo Since $r(t)$ is deterministic function of time, you can easily replace $e^{rT}$ with $e^{\int_{0}^{T} r(t) dt }$. It won't lead to any complexities. $\endgroup$ – Neeraj Mar 7 '16 at 19:03
  • $\begingroup$ @alexjo I considered r(t) as constant to keep the things simple and to provide the idea how it can be solved. $\endgroup$ – Neeraj Mar 7 '16 at 19:07
  • $\begingroup$ @alexjo I have added the answer for r(t) = $\frac{1}{20 + \frac{t}{2}}$. $\endgroup$ – Neeraj Mar 7 '16 at 19:57
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    $\begingroup$ You can simplify it to $2ln(1+\frac{t}{40})$ $\endgroup$ – jxjxjx Mar 8 '16 at 19:57

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