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Licho, 2024-02-19

In the previous article The problem with fees I showed that the current mechanism of burning fees is insufficient to give us a stable equilibrium independent on the external conditions and allows for up to 50% changes in the equilibrium provided the transaction fees heavily dominate the block subsidy. I'd like to present a solution I came up with, along with benefits and drawbacks. As before, it's also not a statement, but a voice in an ongoing discussion.

The set of equations from the previous article that describe the behavior of the system in the presence of the fees reads:

$\left\{\begin{array}{l} p'=\frac{p*D(t)-p(h-\beta f)}{N}\\ h'=\alpha(p(h+(1-\beta )f)-\varepsilon h)\\ N'=h-\beta f \end{array}\right.\,.$

The equilibrium is then modified by the factor $\gamma=\frac{D+\beta f}{D+f}$ Let us propose an expression for $\beta$ that will cancel out the terms related to the fees in the crucial places, so we can trade the uncertainty for certainty.$\beta = 1+(1-\frac{1}{\gamma})\frac{h}{f}$

We are trying to force some constant $gamma$ between 1 and 0, to always be specific and known. We put this expression in the equations and we see, that most of the nasty terms cancel out. We get:

$\left\{\begin{array}{l} p'=\frac{p*\~D-p\frac{h}{\gamma}}{N}\\ h'=\alpha(p(\frac{h}{\gamma}-\varepsilon h)\\ N'=\frac{h}{\gamma}-f \end{array}\right.\,.$

We can now call $\frac{h}{\gamma}\equiv \~h$, $\gamma \varepsilon\equiv\~\varepsilon$ which are just the old functions, but rescaled and $D+f\equiv \~D$ denoting the fees as part of the demand function. We recover the exactly the same set of equations as the ones without the fees, but with $\~h$, $\~\varepsilon$ and $\~D$.$\left\{\begin{array}{l} p'=\frac{p*(D+f)-p\~h}{N}\\ h'=\alpha(p(\~h-\~\varepsilon \~h)\\ N'=\~h-f \end{array}\right.\,.$

The equilibrium solution of this set is the same as before, but also rescaled by the constant $\~p_0=\~\varepsilon=\gamma \varepsilon$.

We will take a closer look at the $\beta$ parameter:

$\beta = 1+(1-\frac{1}{\gamma})\frac{h}{f}$

We see that the limit of $\frac{h}{f}$ being very high, the burning ratio approaches 1, meaning 100% of the fees burned. It contrasts with the constant burning rate from the part 1, where in the limit of constant burning rate, the equilibrium factor approached the rate.

Now let us take the lower limit. The expression $(1-\frac{1}{\gamma})$ is negative for $0<\gamma<1$, so if the ratio of the fees to ussuance gets very big - tiny fees - it would go negative as fees don't subsidize the mining enough to have it at the set $\gamma \varepsilon$. Beta was defined as a number between 1 and 0 so the result is nonsensical. The mechanism would have to kick in once beta becomes a positive number. Before that, the original dynamics from part 1 would apply. We would have two regimes of operating, where equilibrium is floating between $\varepsilon$ and $\gamma \varepsilon$ and a regime of full $\gamma \varepsilon$.

The fees that a miner actually gets becomes fully determined by the current hashrate. The term for that value reads:

$(1-\beta)f=(1-\frac{1}{\gamma})\frac{h}{\cancel{f}}\cancel{f}=(1-\frac{1}{\gamma})h$

It means that the amount of the fees a miner is allowed to receive is determined by the hashrate. This dynamics makes me think that this mechanism doesn't really improve censorship resistance in the hard burining regime. Any additional fee that would be meant to motivate a miner to include the transaction would have to be burned. Would a miner pick a controversial, potentially censored transaction that burns coins over one that doesn't when both pay him the same amount - it's unclear to me.

This mechanism is a limit to how much fees can there be, nothing more.

Setting the burn ratio to something big, like 90%, essentially limits the equilibrium movements to a 10% range but it's not as abrupt and binary as the smart beta. It's not that elegant but it might be more desirable than the flat limit on the fees. If it is done, the fees would probably need to go up.