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I am presented with an investment opportunity where I am given #481,000 on day 1. Thereafter, every 10 days, I am required to give back #50,000 every for 100 days (10 * 50000 = 500000).

How do I calculate the interest rate I am paying?

I am guessing I have to use the present value of annuity problem to find out the interest rate.

Am I wrong in my reasoning below?

Solving for i(interest rate) using the present value formula in wolfram alpha, I get 0.7107%. Since the period is 10 days, I divide by 10 and multiply by 365 to get 25.94%.

I know that it has to be greater than 14% but I am a little surprised that it is 26%.

Thanks guys.

newbie
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  • Because of the calculation by using Wolfram Alpha you can post the link to the calculation. – callculus42 Aug 02 '15 at 15:28
  • Could you make an edit of your question and post the link there? The comment hasn´t catched the link in a right way. – callculus42 Aug 02 '15 at 17:11
  • After some manipulations a think I have got the link. But it shows me neither the result nor the calculation. You should calculate it by your own and post it. With it I can proof, if I would have calculated in the same way. – callculus42 Aug 02 '15 at 17:30
  • Since you repay the cash from the start to the end of a 100 days period in roughly uniform manner, the effective duration of you payment is about a half of the whole period. As a rough estimate, imagine you repay all the $500,000$ at the middle ( day 55 ), the interest rate becomes $(\frac{500}{481} - 1)/(\frac{55}{365}) \approx 26%$... – achille hui Aug 02 '15 at 22:15

1 Answers1

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Let the interest rate for using some money for ten days be $p$. Then after the $i^{th}$ back payment due every tenth day you still owe

$i=1\ \ $: $481,000\cdot(1+p)-50,000$

$i=2\ \ $: $481,000\cdot(1+p)^2-50,000\cdot(1+p)-50 000$

$i=3\ \ $: $481,000\cdot(1+p)^3-50,000\cdot(1+p)^2-50 000\cdot(1+p)-50 000$

$\vdots$

$i=10\ $:$481,000\cdot(1+p)^{10}-50,000\cdot\sum_{j=0}^9(1+p)^j=0$

since you have paid back by the $10^{th}$ payment all the capital and the accumulated interest.

Now, let

$$Q=\sum_{j=0}^9(1+p)^j.$$

So, $$(1+p)Q=\sum_{j=0}^9(1+p)^{j+1}=\sum_{j=1}^{10}(1+p)^{j}=\sum_{j=1}^{9}(1+p)^{j}+(1+p)^{10}.$$

With this

$$(1+p)Q-Q=pQ=(1+p)^{10}-1,$$ hence $$Q=\frac{(1+p)^{10}-1}p$$

That is, we have to solve the following equation for $p$.

$$481,000\cdot(1+p)^{10}-50,000\cdot\frac{(1+p)^{10}-1}p=0$$

or

$$\frac p{1-\frac1{\ (1+p)^{10}}}=\frac{50}{481}$$

then

$$p\approx 0.00711\%.$$ So, we pay an interest rate of $0.711\%$ for ten days. The yearly counterpart is $$\sim 26\%.$$

zoli
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