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Author Topic: Alternative coinbase reward formula  (Read 826 times)
cuddlefish
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June 28, 2011, 07:03:54 PM
 #1

Suppose we replace the coinbase reward formula with:

R = 50 - F

R is the reward for a block (NEW COINS ONLY, NOT TRANSACTION FEES, THEY STILL GET THOSE NORMALLY)
F is the sum of all transaction fees in the previous block.

If Bitcoins deflate, (as it recently has) the amount people will pay to miners for their transactions decreases. Thus, the block reward will go up, causing inflation and counteracting this.

OTOH if they inflate, miners will demand higher fees. Thus the block reward will go down, causing deflation.

This system allows Bitcoin to survive theoretically forever.
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Even in the event that an attacker gains more than 50% of the network's computational power, only transactions sent by the attacker could be reversed or double-spent. The network would not be destroyed.
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CydeWeys
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June 28, 2011, 07:11:45 PM
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The proper way to express this formula would be:

Quote
T(n) = 50 - F(n-1) + F(n)

-F(n-1)+F(n) would average to zero over every two blocks (for obvious reasons), so all you're really proposing to do is just reward 50 BTC per block, forever.
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June 28, 2011, 07:15:18 PM
 #3

The proper way to express this formula would be:

Quote
T(n) = 50 - F(n-1) + F(n)

-F(n-1)+F(n) would average to zero over every two blocks (for obvious reasons), so all you're really proposing to do is just reward 50 BTC per block, forever.

Okay then. With difficulty increases, calculate the coinbase reward for the upcoming 2016 block period as the average sum of TX fee in the previous blocks.

ALSO!: My formula calculates the number of NEW coins printed. the REWARD averages to 50, yes. the number of NEW coins does not.
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