A pull request on github for dynamic fee calculation was just submitted. How does this new dynamic fee calculation work?

2 Answers 2


According to the pull request from moneromooo-monero on behalf of ArticMine:

The dynamic fee is based on the block size in a past window, and the current block reward. It is expressed by kB. This function calculates an estimate for a dynamic fee which will be valid for the next grace_blocks

According to dEBRUYNE_1 this change will take effect with v4 in January 2017 and can be represented with the following formula:

The fee will vary based on the base reward and the current

block size limit:

fee = (R/R0) * (M0/M) * F0

R: base reward

R0: reference base reward (10 monero)

M: block size limit

M0: minimum block size limit (60000)

F0: 0.002 monero

According to this recent study, there are some theoretical (but serious) security risks in situations where transaction fees exceed block rewards. (R/R0) will help keep fees in check by reducing them as block reward falls to buffer against this risk.

(M0/M) acts as a sanity check. When blocks are mostly empty it does not make sense to increase the cap. If the network (that previously had relatively empty blocks) is temporarily spammed in order to increase the block size limit, the network does not benefit so the (M0/M) will punish miners by reducing transaction fees. However when legitimate network usage increases over time the amount of increased transactions will compensate for the reduction in transaction fees as the block size cap scales naturally.

With regard to your spam attack tag and how this will change things:

Transaction fees will scale downwards on a per tx basis in the case of an increasing block size limit as the result of the "M" being larger in the (M0/M) part of the equation. However total transaction fees will increase overall as the number of transactions more than cancels out the reduced fee per tx. Spamming the network to increase the block size will remain expensive after this pull request is activated. In addition, for the reason described below such an impact on tx fees would only be temporary.

It is important to note that the (M0/M) portion of new dynamic fee calculation scales in both directions. Appropriate transaction fees differ based on network utilization which is capable of both increasing and decreasing due to countless unpredictable market factors. This new dynamic tx fee formula allows for both upwards and downwards tx fee scaling.

  • You say that the fee will scale upwards if the block size limit "M" increases, but as "M" is in the denominator of the formula, if "M" increases, "M0 / M" decreases, therefore the fee decreases.
    – glv
    Oct 30, 2016 at 18:09
  • @glv You are right about my numerator and denominator mixup. My description was based on the way I thought it should work which apparently conflicts with that formula. Hopefully someone else can explain why the fee scales in the opposite manner I expected Oct 30, 2016 at 19:02
  • @glv Maybe my explanation is correct and the formula accidentally swapped the M with M0. I will come back and edit my answer after I know for sure. Oct 30, 2016 at 19:08
  • 3
    (R/R0) - fee goes down with the reward, but the price should go up long-term. This seems like a good idea considering a recent study showing bad incentives if fee>reward. (M0/M) - fee goes down with the block size cap. So, if the blocks are empty, you lose the fee by increasing the cap. If they're full, you're supposed to reduce the fee/kb, but win more total fees if you increase the block size cap and fill them up until the top.
    – JollyMort
    Oct 30, 2016 at 21:33
  • @JollyMort that makes sense. I will correct my answer based on that explanation Oct 30, 2016 at 23:58

In the meantime, some more information has been presented here:

How does the dynamic blocksize and the dynamic fees work together in Monero?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.