0

You invested $500$ on Jan $1$ $2012$. To save for this amount, you invest $x$ on Jan $1$ $2008$ and $2x$ on July $1$ $2008$. The force of interest is $\delta_t=0.02t$ where $t$ is $0$ on Jan $1$ $2008$. Find $x$.

The accumulated value at $t_{\frac{1}{2}}=A(\frac{1}{2})=x(1+i)^{\frac{1}{2}}+2x$

$=x({e^{(\delta_t)}}^{0.5}+2x)(e^{\int^4_0 0.002t dt})$

This simplifies to

$500=(xe^{\sqrt{0.08}}+2x)(e^{0.16}-1)$

$\therefore x=866.17$

But the answer is $x=142.26$.

Tosh
  • 1,614
  • What interest rate $i$ are you using? Why are you using compound interest? Looking at the exercice I would think that you have to use continuous compounding. That is, your accumulated value $A(t)$ at time $t$ is given by $$A(t) = x e^{0.01t^2},$$ where $x$ is the initial investment. – Cavents May 16 '16 at 08:55
  • @Siron, isn't the force of interest $/delta$ equal to (1+i) in continuous compounding. What I have done I have calculated the future value at $t_{\frac{1}{2}}$. Then I find A(t) from time 0.5 to 4 However, I shall try it your way also to see if it works. Thanks. – Tosh May 16 '16 at 13:13

2 Answers2

2

The accumulation function is $$a(t)=a(t_0)\mathrm e^{\int_{t_0}^t\delta(s)\mathrm d s}=a(t_0)\mathrm e^{0.01(t^2-t_0^2)}$$ So the equation of value is $$ a(t_0)\mathrm e^{0.01(t^2-t_0^2)}+a(t_1)\mathrm e^{0.01(t^2-t_1^2)}=500 $$ For $t_0=0$ we have $a(t_0)=x$ and $t_1=\frac{1}{2}$ we have $a(t_1)=2x$ and then $$ x\left[\mathrm e^{0.01(4^2)}+2\mathrm e^{0.01(4^2-0.5^2)}\right]=500 $$ and then $$x=\frac{500}{3.514672}=142.26$$

alexjo
  • 14,976
1

Hint: The equation is

$$ \large{ \left( x\cdot e^{\int_0^{0.5} 0.02\cdot t \, dt}+2x\right)\cdot e^{\int_{0.5}^4 0.02\cdot t \, dt}=500 }$$

callculus42
  • 30,550