Liquidity provision

Given a liquidity pool consisting of two assets, A and B, and the invariant $I(A,B)$, where $A$ and $B$ are the amounts of the two assets in the pool (the "pool reserve"). For simplicity, we denote this as $I$.

Suppose a user provides liquidity with amounts $a$ and $b$. After the liquidity is added, the invariant value is $I(A+a,B+b)$. For simplicity, we denote this as $I^{\prime }$.

Suppose before adding the liquidity, the supply of LP token is $L$. We mint user new LP tokens of the following amount:

$$l=(1-f)\left(\frac{I^{\prime }}{I}-1\right)L$$

Here, $f$ is a fee rate we charge on the amount of LP tokens mint. Without this fee, the following exploit would be possible: provide unbalanced liquidity, then immediately withdraw balanced liquidity. This effectively achieves a fee-less swap.

The fee rate should be a function over $(A,B,a,b)$, reflecting how unbalance the user liquidity is:

• If user liquidity is prefectly balanced, that is, $\frac{a}{A}=\frac{b}{B}$, fee rate should be zero: $f(A,B,a,b)=0$.
• If user liquidity is prefertly unbalanced, that is, one-sided (e.g. $a\ne 0$ but $b=0$), then the fee rate should be a value such that if the attack is carried out, the output is equal to doing a swap normally.

Our objective for the rest of this article, is to work out the expression of the fee function $f(A,B,a,b)$.

Fee rate

Consider the case where the user liquidity is unbalanced. Without losing generality, let's suppose $\frac{a}{A}>\frac{b}{B}$. That is, the user provides a more than abundant amount of A, and a less than sufficient amount of B.

Scenario 1

In the first scenario, the user withdraws liquidity immediately after provision. He would get:

$$\frac{l}{L+l}(A+a)$$

$$\frac{l}{L+l}(B+b)$$

Here, $\frac{l}{L+l}$ is the portion of the pool's liquidity owned by the user. We can work out its expression as:

$$\frac{l}{L+l}=\frac{\frac{I^{\prime }}{I}-1}{\frac{I^{\prime }}{I}-f}=\frac{r-1}{r-f}$$

where $r=\frac{I^{\prime }}{I}$, which represents how much the invariant increases as a result of the added liquidity.

Scenario 2

In the second scenario, the user does a swap of $\Delta a$ amount of A into $(1-s)\Delta b$ amount of B, where $s$ is the swap fee rate, which is a constant. The swap must satisfy the invariant:

$$I(A,B)=I(A+\Delta a,B-\Delta b)$$

The user now has $A-\Delta a$ amount of A and $B+(1-s)\Delta b$ amount of B.

As discussed in the previous section, we must choose a fee rate $f$ such that the two scenarios are equivalent. This means the user ends up with the same amount of A and B in both scenarios:

$$\frac{r-1}{r-f}(A+a)=A-\Delta a$$

$$\frac{r-1}{r-f}(B+b)=B+(1-s)\Delta b$$

We can rearrange these into a cleaner form:

$$\frac{A+a}{A-\Delta a}=\frac{B+b}{B+(1-s)\Delta b}$$

$$f=r-(r-1)\frac{A+a}{A-\Delta a}=r-(r-1)\frac{B+b}{B+(1-s)\Delta b}$$

We can use the first equation to work out either $\Delta a$ or $\Delta b$, and put it into the second equation to get $f$.

Xyk pool

The xyk pool has the invariant:

$$I(A,B)=AB$$

Our previous system of equations takes the form:

$$AB=(A+\Delta a)(B-\Delta b)$$

$$\frac{A+a}{A-\Delta a}=\frac{B+b}{B+(1-s)\Delta b}$$

$$f=r-(r-1)\frac{A+a}{A-\Delta a}$$

TODO...