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Author Topic: Crediting transaction fees back to original miners after split  (Read 347 times)
Krellan
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April 09, 2013, 05:18:26 PM
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I'm curious: A Bitcoin block, over time, gets split up into multiple smaller blocks as it is spent in small transactions.  If a large transaction is required again, those smaller blocks can be joined up again, in any order with any other blocks.  So, how then does the Bitcoin transaction fee get divided up?

Does each transaction get traced back to the original mined block, across all joins/splits over time, to properly account for each miner's contribution to this particular transaction, so the payment can be divided up fairly?  Or would this be too much overhead, and a simpler technique is used?  I tried to read up how the Bitcoin protocol rewards miners over time, but couldn't find an answer for how it handles transactions that are a mixture of value from various different mined blocks over time.  Am I not getting something here?

Thanks!

Josh

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April 09, 2013, 05:26:18 PM
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None of that is correct.

A transaction is a transaction.
A block is a collection of transactions.
All blocks reference the prior block creating a blockchain.
The collectively forms the public ledger. A record that can be used to authenticate future transactions.

Unconfirmed transactions are relayed between nodes.
The pool of unconfirmed txs is called the memory pool.
When miners begin to work on a new block they take txs from the memory pool, put them into the block (a simplification) and search for a solution to the block.
Transaction fee (plus block subsidy - currently 25 BTC) is paid to the miner who solves the block.
When a block is solved it is relayed to all peers who verify it and add it to the blockchain.

If the block you solve has 230 txs with fees totally 1.1 BTC you (as the miner who solved the block) would receive 26.1 BTC (1.1 BTC tx fees + 25 BTC subsidy).  Most miners today work in pools, in this case the pool is actually the "miner".
Krellan
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April 09, 2013, 05:39:52 PM
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Interesting, thanks.

So, when a new block is discovered by mining, it has to be attached to a previous transaction?  I thought it began something new.

Also, I thought transaction fees were something that continuously kept being generated over time, as value is spent and sent around the network, eventually trickling back to the original miner who found the block, essentially paying them a dividend over time.  Or, is it a one-time reward, as you mentioned below in your example?

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Gerald Davis


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April 09, 2013, 05:51:31 PM
 #4

You keep interchanging the words blocks and transactions.

They are distinct and different objects.

A transaction is a transaction. 
A block is a collection of transactions.

A block references the block hash of the previous block.   This cryptographically links the blocks together to securely forms the blockchain.

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Also, I thought transaction fees were something that continuously kept being generated over time, as value is spent and sent around the network, eventually trickling back to the original miner who found the block, essentially paying them a dividend over time.  Or, is it a one-time reward, as you mentioned below in your example?

Weirdly nobody has ever mentioned this for like three years and then suddenly this week you are like the third or fourth person who had that misconception.

No a transaction fee is paid on a single tx.  It becomes part of the single block reward for the block where the tx is added.  The block reward is paid to a single miner (or in the case of a pool the single pool) which solved the block.  There is no perpetuate revenue stream from a single transaction.  Now new transactions (and thus new fees) are continually being created but each tx and its fee is separate.
Krellan
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April 09, 2013, 08:31:07 PM
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Nice, thanks for clearing that up!

I think what's happened is a lot of content has been written about Bitcoin in the last few weeks (obviously), some good, some bad, and a lot of people are reading it and getting misconceptions.

Is this all in a FAQ somewhere?  I looked around and read various FAQ's, but didn't see this mentioned.  I rather like your explanation, it clears things up.

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