## Calculation for transaction input-output balling:

1

To roughly calculate transaction send size in bytes we do:

((numberOfInputs*148)+(numberOfOutputs*34)+10)


We want to make a transaction using two inputs of 0.01BTC (Amount to send = 0.015BTC, change left over = 0.005BTC). Number of outputs in this case will be two; as we have two inputs, one of which we are 'tearing-in-two' creating change that will be sent back to us (we are the second output so the change is not lost).

So to calculate the fee

• We calculate the value of each input in BTC multiplied by the age of the input in blocks
• Divide the priority by the size of the transaction in bytes
• If this gives a number less than 0.576 then the transaction requires a fee
• Also if the transaction size exceeds 1000 bytes it requires a fee

If both inputs were 1000 conformations old, a fee would be required:

((0.01*1000) + (0.01*1000)) = 20 //priority
((2*148)+(2*34)+10) = 374 //bytes
20 / 374 = 0.053475935828877004


If both inputs were 11000 conformations old, a fee would not be required:

((0.01*11000) + (0.01*11000)) = 220 //priority
((2*148)+(2*34)+10) = 374 //bytes
220 / 374 = 0.5882352941176471


Is this correct?

If one input was old dust and the other new and large (this looks free to me):

((0.00001*11000) + (1*1000)) = 1000.11 //priority
((2*148)+(2*34)+10) = 374 //bytes
1000.11 / 374 = 2.6740909090909093


Does this explain how you can put a dust-input with a large-input to rescue dust?

Will the threshold of 0.576 change in the future and if so how will we find out?

where does you initial formula come from? Is it form you or from some wiki? – Dennis Kriechel – 2014-06-11T07:04:15.480

http://bitcoinfees.com – BENZ.404 – 2014-06-11T13:50:50.543

The whole section about priority is not relevant anymore; free transaction mining doesn't happen in any significant amount anymore and software support for it has been removed in various places. Every transaction just pays bytes * feerate in fee. – Pieter Wuille – 2017-06-20T07:07:26.637