We study backrunning strategies and the effect of TimeBoost policy on them. The modeling and analysis allows to answer the following questions:
- how much of backrunning opportunity value goes to the time advantaged party?
- how much of the value goes to the party that does not have time advantage and instead tries probabilistic strategies?
- how much both parties spend on gas fees?
- what are the effects of gas fees on strategies and value distribution?
In the following, we make modeling assumptions. Backrunning opportunities are arriving according to a Poisson process, with rate \lambda. Each opportunity is worth v, a random variable distributed according to a probability distribution function F.
There is a player, Alice, that has A time advantage to get her transactions scheduled on the chain. Alice has latency L to reach the sequencer. The only property about A and L that needs to hold is that A>>L, which is easily satisfied in practice since A=200ms.
There is another player, Bob, who does not have a time advantage, but can employ a probabilistic strategy of sending transactions to the sequencer. We assume he sends transactions according to a Poisson process, with rate \eta. We can treat Bob as an aggregate player of all players other than Alice.
The price of one transaction in gas fees is denoted by g. Here we assume that both successful and reverted transactions cost the same. That is, Bob’s expected cost per time unit from employing the probabilistic strategy is -g\eta.
Once a backrunning opportunity arrives, Alice can decide between extracting it and not extracting it. Extracting it costs g – a transaction fee.
The probability to win the opportunity in case of trying to extract is equal to p=e^{-\eta L}, which is the same as the probability that Bob has not submitted a transaction in the past between times t-A+L and t-A where t is the current time. Therefore, the expected utility of Alice in case of trying is equal to pv.
As long as expected utility of trying is higher than the cost of trying, i.e., pv-g>0, equivalent to v>\frac{g}{p}, she tries to extract the opportunity.
When pv-g<0, equivalent to v<\frac{g}{p}, she does not try to extract the opportunity.
Let v^*:=\frac{g}{p} be the threshold above which Alice tries to extract and below which she does not. Then, the expected payoff of Bob in one time unit is:
\mathbb{E}_{F}[v|v<v^*]F(v^*)\lambda + (1-p)E_F[v|v>v^*](1-F(v^*))\lambda- g\eta.
The expected payoff of Alice in one time unit is:
p\lambda\mathbb{E}[v|v>v^*](1-F(v^*)) - \lambda (1-F(v^*))g.
Expected fees spent by both players in one time unit are equal to g(\eta+\lambda(1-F(v^*))).
Decision Problem
Bob decides on \eta, this determines which opportunities will Alice take. Assuming mild smoothness conditions on F, there will be one optimal strategy \eta.
The first order condition allows to solve for optimal \eta, assuming suitable conditions on F. It is feasible to get estimates for the parameters F, g, \lambda, and L, and solve for the optimal \eta.
There is an easy observation in place: There are parameter settings where we get \eta=0 is optimal, meaning that Bob doesn’t want to try. There are also examples where \eta>0. Namely, for fixed F: \eta=0 when either L is too small, g is too large or \lambda is too small. This results into Alice capturing almost all the backrunning opportunity value and paying low transaction fees.
We can easily specify reasonable sufficient conditions for which we get \eta>0. Finding sufficient and necessary conditions may turn out to be harder.
Comparative statics:
- \eta is decreasing in g, in other words, increasing the transaction cost g discourages Bob, which can be viewed as a positive outcome for the chain.
- \eta is increasing in \lambda, i.e., the more value there is, the more frequently Bob tries with probabilistic strategies.
Numerical example:
We consider one numerical example in which \lambda = 2, F is a uniform random variable over [0,1], g is 0.01 and L is 0.02. Out of average of 1 total unit of value that becomes available in a unit of time, we have that Alice profits approximately 0.48, Bob profits 0.155, Bob pays in fees 0.345 (\eta is 34.5) and Alice pays in fees 0.02. In contrast, under the First Come First Serve ordering policy, with a sufficiently large number of competitors, all backrunning opportunity value would be spent on fees, resulting into a higher network resource consumption.
The Game
In this section, we allow Alice to also introduce her own probabilistic strategy of submitting transactions. This results into a game with two players choosing their strategies of how often to send probabilistic transactions to exploit backruns.
Suppose Alice sends her transactions according to a Poisson process with frequency \phi. This costs her \phi g per time unit. Assume than an opportunity happens at time t=0. The only case when Alice wins using her own probabilistic strategy is when she sends the last transaction to the sequencer at time t=-T, where T\leq L and Bob did not send a transaction at t=-T_B, where 200 <T_B\leq 200+T_A. The probability of the second is equal to \frac{\phi}{\phi+\eta}. Let the probability of this event be denoted by q. Let’s calculate q. The probability that Alice’s transaction is happening until time t is equal to (1-e^{-\phi t}). This is a distribution function, therefore, the density function is \phi e^{-\phi t}. Similarly, the probability that Bob’s transaction is happening until time \tau is equal to (1-e^{-\eta\tau}) and the density function it \eta e^{-\eta \tau}. Then, we have q=\int_{0}^{L}\phi e^{-\phi t} \int_{t}^{\infty} \eta e^{-\eta \tau} d\tau dt = \frac{\phi}{\phi+\eta}(1-e^{-(\phi+\eta)L}). This allows us to express expected utilities of both players as only functions of \eta and \phi. Assuming mild smoothness properties on F allows to solve equilibrium values of \eta and \phi, by first solving best responses. There are sets of parameters where equilibrium solution matches the solution from the previous section where \phi=0, i.e., Alice does not try probabilistic strategies in the equilibrium. In particular, this is the case when the expected backrunning value is not too high.