Not that I have given the formula much thought of exactly the best method to price. But I’d consider something like:
msg\_cost = (p*k^2) + (2^{log(l)})
Where l is the last msg\_cost of the last block.
Not that I have given the formula much thought of exactly the best method to price. But I’d consider something like:
Where l is the last msg\_cost of the last block.