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Here's the full problem:

You have two options to repay a loan. You can repay $\mathbb{$}6000$ now and $\mathbb{$}5940$ in one year, or you can repay $\mathbb{$}12000$ in $6$ months. Find the annual effective interest rate(s) i at which both options have the same present value.

In the book I am using, they define the following:

Discount function $\to v(t)=\frac{1}{a(t)}$ where a(t) represents accumulation function

Discount factor $\to v=\frac{1}{1+i} \implies a(t)=(1+i)^t=v^{-t}$ (recalling compound interest) where $i$ is interest

Present value $\to PV_{a(t)}(\mathbb{$}L \mathbb{\;at\;} t_0)=\mathbb{$}Lv(t_0)$ where $\mathbb{$}L$ is the loan

My attempt at work is as follows, but I'm under the impression that it is not correct:

Equating the two payment options using the present value equation, we obtain $$(6000+5940)\frac{1}{1+i}=12,000\frac{1}{\sqrt{1+i}}$$ $$\implies \frac{11940}{12,000}=\sqrt{1+i}$$ $$\implies (0.995)^2-1=i$$ $$\implies =i-.0009975$$

Was I thinking correctly at the approach to this problem or was I completely off? I'm fully confident that I should not have a negative interest rate. Or for that matter, an interest rate that small. If anyone could give me aid, I'd be very much grateful!

user146925
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1 Answers1

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Outline: The PV in the first option is $6000+\frac{5940}{1+i}$. After all, the PV of the first payment of $6000$ is exactly $6000$, since it is being paid now.

The equation you end up with, after multiplying through by $1+i$, is a quadratic in $\sqrt{1+i}$. Let $x=\sqrt{1+i}$ and solve.

André Nicolas
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  • Okay yeah that's where I was thinking I went wrong. Wait, so I then end up with

    $$6000 + \frac{5940}{1+i} = \frac{12000}{\sqrt{1+i}}$$ ???

     – user146925 Aug 31 '14 at 16:41
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    Yes, and then you multiply both sides by $1+i$. Letting $x=\sqrt{1+i}$ you get the equation $6000x^2-12000x+5940=0$. Solve, using the Quadratic Formula. Knock off a $0$ everywhere first. (Actually, they made the numbers nice, and you could even factor.) – André Nicolas Aug 31 '14 at 16:51
  • Yeah I divided everything by two and then applied the quadratic equation. It's nice to have to use a calculator. I've now found the answer. Thank you for the help, Andre. :) – user146925 Aug 31 '14 at 16:55
  • You are welcome. The little $\sqrt{1+i}$ trick, and relatives, is useful elsewhere. – André Nicolas Aug 31 '14 at 16:57
  • LOL this is what I meant to say before "Yeah I divided everything by 6000 and then applied the quadratic equation. It's nice to not have to use a calculator. I've now found the answer. Thank you for the help, Andre. :)" Oh and yeah I've known about that type of substitution for years now. Just needed a little refresher :P – user146925 Aug 31 '14 at 17:04