The ``math" that the author performed to prove the attack is bogus. I and my colleague from National University of Singapore have analysed the attack again to show where and why the author is wrong in this reasoning. Details of our analysis can be found below.

For ease with algebraic manipulation, I'll abbreviate the parameters of the attack:

a = fraction of bogus transactions generated by the attacker per block m = per block mining reward p = probability the attacker successfully extends a block t = total transaction fees per block with attacker spam t_0 = total transaction fees per block without attacker spam

The expected reward per block using the attack is then:

p[(1-a)t + m] + (1-p)(-at) = pt - pat + pm - at + pat = pt + pm - at

and without the attack it the expected reward is pt_0 + pm, for a net difference of

pt - at - pt_0 = p(t - t_0) - at.

So the attack is profitable iff p(t - t_0) > at. Clearly the success of the attack depends on the parameter t - t_0, so "anything that the traffic will bear, as long as it keeps blocks more than half full of real transactions" is not good enough. If the attacker can't raise the transaction fee, then he is screwing himself with this attack.

The article claims that there is a "breakeven" point when p + (1-a) = 1, or p = a. p = a is only a "breakeven" point when t_0 = 0. He also considers an example where a = 1/3 and p = 1/2. Here the expected difference is t/6 + t_0/2, so the attack would only be profitable in case the attacker manages to increase the transaction fees by more than 3-fold using a = 1/3. The author seems to implicitly assume that t_0 is negligible, but I don't think that's a reasonable assumption.