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Licho, 2024-02-18

This article's goal is to investigate the Ergon's fee burning mechanism in the face of the recent realization that it doesn't do the job as it should. I would like to explain the math that made me realize this mechanism is not enough and make it the part of the ongoing community discussion on that topic.

It is possible to find the value of the Ergon equilibrium. Let us construct the expression describing the dynamics of hashrate, the same way it was constructed in the Ergon whitepaper. We start with a simple observation, that the miners follow profitability. If their rewards from mining are higher than their costs, they join the network. We symbolically express the rate of hashrate changes as follows:

$h'=\alpha(pR-\varepsilon h)$,

Which means that the rate of hashrate changes are driven by the difference between the reward value $R$ and the cost $\varepsilon$ per unit of hashrate $h$. Alpha is the proportionality factor. It measures how fast the miners respond to profitability changes.

Proportional reward is the proposition to put `R=h`

. Then the equation reads:

$h'=\alpha(ph-\varepsilon h)$.

Now we will skip to the price dynamics, derived in the Ergon paper,

$p'=\frac{p*D(t)-pN'}{N}$.

To recap - $p$ is the price, $D(t)$ is an arbitrary demand function for new coins, $N$ is the supply and $N'$ is the reward changing rate - the supply expansion. In the simple case $N'=R=h$.

We have a system of equations:

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

To find the equilibrium, we are looking for a constant solution: $p(t)= p_0$ and $h(t)= h_0$. Then $N'$. Derivatives of $p$ and $h$ are just zero. We will learn that in order to have such a solution, $D$ also must be constant.We have to solve the following system of equations:

$\left\{\begin{array}{l} 0=\frac{p_0*D-p_0h_0}{N}\\ 0=\alpha(p_0h_0-\varepsilon h_0)\\ N'=h_0 \end{array}\right.\,.$

The solution is very straight forward. We get:$p_0=\varepsilon$,

$N=h_0t+C$ and

$h_0=D$.

We can now notice that demand $D$ must not dependend on time for this solution to work.

In the paper, we took for granted that $R$ is nothing more than the issuance, $h$, but in fact it is also the transaction fees rate $f$, like fees paid per day or per block. They are denominated in the coin units because the fees are indeed paid in those units,

$R=h+f$.

In the protocol, we burn half of all fees to correct for the additional inflation they would introduce. We will solve it for the general ratio denote it as $\beta$.

We modify the equations:

$N'=h-\beta f$,

$R=h+(1-\beta)f$.

Now the system reads:

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

Once again we assume the constant solution $p(t)= p^f_0$ and $h(t)= h^f_0$.The system then reads:

$\left\{\begin{array}{l} 0=\frac{p^f_0*D-p^f_0(h^f_0-\beta f)}{N}\\ 0=\alpha(p^f_0(h^f_0+(1-\beta )f)-\varepsilon h^f_0)\\ N'=h^f_0-\beta f \end{array}\right.\,.$

From the first equation follows that $h^f_0=D+\beta f$. We insert that to the second one and we get the dreaded information that:$p^f_0=\varepsilon \frac{D+\beta f}{D+f}$.

Let us call the equilibrium modifying factor $\frac{p^f_0}{p_0}$ as $\gamma= \frac{D+\beta f}{D+f}$.

With no burning ($\beta=0$) the fees would modify the equilibrium by a factor of $\frac{D}{D+f}$ which meand that in the limit of fees infinitely larger than the demand the equilibrium would be at zero, but with burning in place:

$\lim_{\frac{D}{f}\rightarrow 0}\gamma=\beta.$

It means that if we burn 50%, when the equilibrium can be pushed down by 50% due to the fees subsidizing hashrate.The topic continues in part 2