0

Found the formula in the following link. I can't understand with my knowledge. Can anybody explain me with an example?

profit = ExchangeA_sell * (1 - ExchangeA_fee) - ExchangeB_buy * (1 + ExchangeB_fee).

One more thing, Is this formula good for all exchanges? Cryptsy and Bter have calculated differently their net total. Bter sell offer is different from cryptsy sell offer. In Bter the trade fee is deducted just opposite to cryptsy in terms of buy and sell currency?

  • I explained it in more detail here. Maybe it is more clarifying – Mathias711 Jul 1 '14 at 13:48
  • I'm not quite sure what you mean in the other questions. The net total does not matter at all for arbitrage. And what in the sell offer is different? The offers itself are of course different, because it are different exchanges. I tried to explain something similiar as your last question here. But please rephrase – Mathias711 Jul 1 '14 at 14:02
  • @Mathias711 - bitcoin.stackexchange.com/questions/28263/…. This is my problem. I am stuck at here. – Jsd Jul 2 '14 at 9:11
  • I saw the question, I tried to read it, but important stuff came through. I will comment my first idea of it – Mathias711 Jul 2 '14 at 9:12
1

As explained by http://www.investopedia.com/terms/p/profit.asp:

profit = total revenue - total expenses

or in plain English:

profit = (what you had received after selling all your coins) - (what you had paid to buy your coins)

so, for example:

If you had spent $1000 for Bitcoin priced at $500/BTC on exchange A, and say the fee is 1%, you will received

$1000/$500*(1-0.01) = 1.98 BTC

If you move and sell your 1.98 BTC at for $600/BTC on exchange B, which also charges a 1% fee, you will receive

1.98*$600*(1-0.01) = $1176.12

So your profit will be

$1176.12 - $1000 = $176.12

Remember to also subtract any withdrawal/deposit fees if applicable.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.