I am to the blockchain space. I was reading the Ethereum whitepaper, and I couldn't understand this example:

Suppose Alice wants to send 11.7 BTC to Bob. First, Alice will look for a set of available UTXO that she owns that totals up to at least 11.7 BTC. Realistically, Alice will not be able to get exactly 11.7 BTC;


  1. Why Alice needs to get 11.7 BTC if she already owns those UTXO?
  2. Why she can't have exactly 11.7 BTC?
  3. Why does she need to get those UTXO from somewhere if she already owns that amount?

say that the smallest she can get is 6+4+2=12.


  1. Where did these 3 figures come from?

She then creates a transaction with those three inputs and two outputs. The first output will be 11.7 BTC with Bob's address as its owner, and the second output will be the remaining 0.3 BTC "change", with the owner being Alice herself.


  1. Why does she need to pay this extra 0.3 BTC change?

1 Answer 1


UTXO or Unspent Transaction Output is simply a way to break cryptocurrency into fractional pieces. As such, in answer to your questions:

  1. The process of transfer will firstly have to evaluate wether or not Alice has enough UTXO to undertake the transfer. Of course, if this check fails, she will not be able to send that amount.
  2. She may have either less than or more than 11.7BTC in UTXO, but in order to combine her tokens in such a way as to be able to send 11.7BTC, there will be a number of aspects which require an amount != 11.7BTC in any form; in this example, her assets don't exactly equal the amount to send. But in others, you have to consider gas etc... which makes the net figure != 11.7BTC.
  3. She has to get that amount in order to send it.
  4. These figures are for the sake of example.
  5. Alice needs to add her UTXO together in order to make the tx. So! Since her denominations used to create the transaction are >11.7BTC, the 'change' will remain with Alice post trsnaction. She doesn't have to 'pay' it so much as it has to be provided in order to push the transaction through.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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