Bitcoin Forum

I just recently sent some BTC from a wallet to cold storage, and it pulled the BTC from 3 different wallet address in order to get the amount I wanted to send.  The total transaction size showed 668 bytes, roughly 3 times the typical size.  Because of this, the fee I paid was over $4.

Is it normal that each wallet address that BTC is pulled from adds approx. 220 bytes to the transaction size?

I know that new wallet addresses are often generated for each deposit for privacy/security reasons, but from a fee standpoint wouldn't it cost a lot less to reuse a wallet address more than once?  This way, when money is sent from the wallet it's less likely to need to pull from so many addresses?

Transaction size in bytes is calculated more or less as:

(#Inputs * 148) + (#outputs * 34) + 10 bytes.

It isn't the number of addresses per se, but the number of inputs... For example, I've seen transactions that have 10 inputs... all from 1 address... but it is still 10 inputs... so 10 * 148 = 1480 bytes, + the output bytes + the 10 fixed bytes...

Also, when you send multiple inputs to an address... it doesn't create one big amount of BTC... they are all still separate chunks...

ie. if you send 1.0 BTC, 0.1 BTC and 0.01 BTC... you don't have a single amount of 1.11 BTC in your wallet... you still have 3 separate UTXOs that you can then spend.

Thanks for the response.  So if I understand correctly: If I received say 5 separate deposits, regardless of whether they all came into one address or into multiple addresses, it would still be the same amount of inputs (and therefore same transaction size) if I send all of these deposits to another wallet in a single transaction?

Correct