# What precision do I need to calculate implied volatility?

I'm developing a software to calculate the implied volatility of an option using the Black & Scholes formula and a trial-and-error method. The implied volatility values I get are correct, but I noticed that they are not the only possible ones.

For example, with a given set of parameters, my trial-and-errors lead me to an implied volatility of 43,21%, which, when used on B&S formula, outputs the price I started with. Great!

But I realized this 43,21% value is just a fraction of a much wider range of possible values (let's say, 32,19% - 54,32%).

Which value should I, then, pick as the 'best' one to show to my user?

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That is strange, what are your parameters? – Bob Jansen Apr 24 '12 at 17:22
My parameters are the ones the B&S formula requires: - Underlying stock current price. - Option strike price. - Time to expire. - Interest rate. - Volatility. The four first ones are known to me, and I want to discover the last one, given an option market price. I start by trying it with the historical volatility and, by trial-and-error, adjust it until the result is the same as option market price. When it comes to out-of-the-money call options, the historical volatility usually is a valid answer, but, as I said before, it is a fraction of a wide range of valid answers. – Mário Marinato Apr 24 '12 at 17:32
And what are your parameter values? – Bob Jansen Apr 24 '12 at 18:32
An example: underlying stock price is 21.19. Option strike price is 35.50. Time to expire is 27 days (which I divide by 365 and use as 0.0740. Interest rate: 9.75. Current option market price: 0.01. Then, as I said on my first comment, the first try is the historical volatility, which is 31.27%, and B&S gives me 0.01 as the fair price. Again: great! But I discovered that, with those same parameter values, the volatilities 10% and 80% also give me the same 0.01 as fair price. – Mário Marinato Apr 24 '12 at 19:47
That's a bit of an edge case of course. This option is very far out of the money. What happens when you try with a more valuable option (and a more useful example)? – Bob Jansen Apr 24 '12 at 20:31

When you compute your trial and errors, you have certainly defined a constant $\epsilon$ which defines when the algorithm has to stop: $(p - \bar{p})^2 < \epsilon$ where $p$ is the market price and $\bar{p}$ is the price you got using the volatility estimate $\bar{\sigma}$ in the BS Formula : $\bar{p}=\text{BS}(\bar{\sigma},\cdot)$. If you want better precision, then you need to reduce $\epsilon$. Consider also that you might be lacking precision from the market price data $p$ that you use.