0

What deposit made today will provide for a payment of $ \$1000$ in $1$ year and $ \$2000$ in $3$ years, if the effective rate of interest is $0.075$. Answer is $\$2540.15$

I have calculated $d=0.069$ by using $i=\frac{d}{1-d}$

Then by using $A(3)=A(0)(1-d)^{-3}$, I don't get the answer

lulu
  • 70,402
Tosh
  • 1,614
  • You are neglecting the interim $$1000$ payment. For the last two years interest accrues on a reduced amount. – lulu Feb 28 '16 at 12:32
  • I am not understanding. Could you please be more precise @lulu – Tosh Feb 28 '16 at 12:35
  • I'll post a calculation. Also: check your discount factor...for small interest rates $d\sim 1-i$ – lulu Feb 28 '16 at 12:36
  • With $$2540.15$ initial investement and an interest rate of $0.075$ you need to be a magician to receive $$1000$ in just $1$ year. – Jimmy R. Feb 28 '16 at 12:38
  • @JimmyR.,I understand your logic, but that is called the discount function. Please google it, you will be enlightened. – Tosh Feb 28 '16 at 12:41

2 Answers2

1

Let $A_0$ denote the initial amount, and $A_i$ the amount amount $i$ years. We have $$A_1=A_0(1+i)-1000\;\;\;\;\;\;A_2=3000=A_1(1+i)^2$$ Substituting for $A_1$ we see that $$3000=(A_0(1+i)-1000)(1+i)^2=A_0(1+i)^3-1000(1+i)^2\implies A_0=\frac {3000+1000(1+i)^2}{(1+i)^3}$$ Plugging in $i=.075$ yields $$A_0\sim\$3345.1143$$

Note: I see that the problem was revised after I posted. It is, of course, easy to change $\$3000$ to $\$2000$ in the above.

lulu
  • 70,402
1

Hint: if $X$ is the deposit, and after one year you have a payement of $1000$, the residual value after this year is $$ X_1=X(1+i)-1000 $$ that after two years becomes: $$ \left(X(1+i)-1000\right)(1+i)^2=2000 $$ solve for $X$ and you have the result.

Emilio Novati
  • 62,675