<p>Nowadays, a miner can gain more than another one even when they mine the same amount of blocks due to variance in tx fees. It's expected that, especially once the coinbase reward is gone, this will result in miners competing for "juicy" txs and hence damaging the stability of the system. See <a href="https://dl.acm.org/citation.cfm?id=2978408" rel="nofollow noreferrer">https://dl.acm.org/citation.cfm?id=2978408</a> for future reference.</p>

<p>I've been thinking if it's possible to overcome this simply by changing the rewarding mechanisms of Bitcoin. So let's say we keep a k-length jumping window, take sum of fees in that window and divide it evenly to miners. So if we have k=4 and 1st block has 20 in fees, 2nd has 20 in fees, 3rd has 10 in fees and 4th has 30 in fees each miner will get (20+20+10+30)/4 = 20 in fees.</p>

<p>What would be shortcomings/downsides of such a mechanism ?</p>

CC BY-SA 3.0
2018-01-08T17:50:45.217Z
When using an aggregative method as you have suggested, how would you ensure that the total amount of bitcoin paid in fees, is the same as the total amount of bitcoin collected from fees? The moving averages that are collected wouldn't sum to exactly the sum of the bitcoin paid.
Matthew Stannard
CC BY-SA 3.0
2018-01-08T18:44:00.060Z
`The moving averages that are collected wouldn't sum to exactly the sum of the bitcoin paid.` Why is that? I don't see. Fee of each block is simply Input sum - Output sum. Is there anything more to this than that ?
SpiderRico
CC BY-SA 3.0
2018-01-08T18:54:38.940Z
For each block yes. But not over an aggregation of blocks.
Matthew Stannard
CC BY-SA 3.0
2018-01-08T19:05:07.487Z
But if we add each block individually, what do we miss ? Surely, we can go back k blocks in the chain and aggregate each block, no ?
SpiderRico
CC BY-SA 3.0
2018-01-08T19:21:13.973Z
Hmm, I've misread your question... Yes, what you have described is possible. You may have found a solution to Undercutting attacks.
Matthew Stannard
CC BY-SA 3.0
2018-01-08T19:29:27.710Z
Would you be able to formalize your idea and post it as answer to the following question? https://bitcoin.stackexchange.com/questions/67697/can-undercutting-attacks-be-mitigated
Matthew Stannard
CC BY-SA 3.0
2018-01-08T19:31:53.537Z
@MatthewStannard I'll think a bit more it and possibly try doing some simulations. If I get some nice results, sure :)
SpiderRico
CC BY-SA 3.0
2018-01-08T19:32:08.457Z
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/71362/discussion-between-matthew-stannard-and-spiderrico).
Matthew Stannard
<transaction-fees><mining-theory><mining-profitability>
Dividing miner reward over a k-length jumping window
SpiderRico
# Dividing miner reward over a k-length jumping window

(Asked by: username_0 on 2018-01-08)

Nowadays, a miner can gain more than another one even when they mine the same amount of blocks due to variance in tx fees. It's expected that, especially once the coinbase reward is gone, this will result in miners competing for "juicy" txs and hence damaging the stability of the system. See <https://dl.acm.org/citation.cfm?id=2978408> for future reference.

I've been thinking if it's possible to overcome this simply by changing the rewarding mechanisms of Bitcoin. So let's say we keep a k-length jumping window, take sum of fees in that window and divide it evenly to miners. So if we have k=4 and 1st block has 20 in fees, 2nd has 20 in fees, 3rd has 10 in fees and 4th has 30 in fees each miner will get (20+20+10+30)/4 = 20 in fees.

What would be shortcomings/downsides of such a mechanism ?

**Tags:** transaction-fees, mining-theory, mining-profitability

> **Comments:**
> **username_1**: When using an aggregative method as you have suggested, how would you ensure that the total amount of bitcoin paid in fees, is the same as the total amount of bitcoin collected from fees? The moving averages that are collected wouldn't sum to exactly the sum of the bitcoin paid.  
> **username_0**: `The moving averages that are collected wouldn't sum to exactly the sum of the bitcoin paid.` Why is that? I don't see. Fee of each block is simply Input sum - Output sum. Is there anything more to this than that ?  
> **username_1**: For each block yes. But not over an aggregation of blocks.  
> **username_0**: But if we add each block individually, what do we miss ? Surely, we can go back k blocks in the chain and aggregate each block, no ?  
> **username_1**: Hmm, I've misread your question... Yes, what you have described is possible. You may have found a solution to Undercutting attacks.  
> **username_1**: Would you be able to formalize your idea and post it as answer to the following question? https://bitcoin.stackexchange.com/questions/67697/can-undercutting-attacks-be-mitigated  
> **username_0**: @username_1 I'll think a bit more it and possibly try doing some simulations. If I get some nice results, sure :)  
> **username_1**: Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/71362/discussion-between-username_1-and-username_0).  

---

## 0 answers

