Liquidity Operations (L)

LO1 - Withdraw FT from lpFT

Withdraw $\sigma_{\mathscr{L}_x}$ lpFT and $\sigma_{\mathscr{L_y}}$ lpXt to $\lambda_u$ UT

Step1 - Calculate $\sigma_x$ FT according to $\sigma_{\mathscr{L}_x}$ lpFT and $\sigma_y$ XT according to $\sigma_{\mathscr{L_y}}$ lpXT

$$\sigma_x=\sigma_{\mathscr{L}_x} \times \frac{R_x}{S_{\mathscr{L}_x}}$$

$$\sigma_y=\sigma_{\mathscr{L}_y} \times \frac{R_y}{S_{\mathscr{L}_y}}$$

Step2 - Calculate $\sigma_u$ UT to withdraw according to $\sigma_x$ FT and $\sigma_y$ XT

According to $\text{Eq.C2}$, $\varepsilon \sigma_u$ FT and $\sigma_u$ XT can be combined into $\sigma_u$ UT.

If $\varepsilon \sigma_y > \sigma_x$,

Step2 - Calculate new price

Fig. 2

The liquidity impact of withdrawing FT from the pool ($Fig.2$) is equivalent to the liquidity impact of following two operations ($Fig.3$).

1. Sell $-\sigma_x$ FT to the pool and get $-\sigma_y$ XT

2. Combine $(\Delta_x-\sigma_x)$ FT and $\sigma_y$ XT into $\sigma_y$ UT and withdraw

The new price impact of withdrawing FT from the pool will be equivalent to the price impact of the two operations as well. In addition, the price of pool remains the same when UT is provided or withdrawn from the pool by definition. Therefore, the price impact of withdrawing FT from the pool will be equivalent to the price impact of Sell $-\sigma_x$ FT to the pool and get $-\sigma_y$ XT.

LO2 - Withdraw XT

Withdraw $\delta_{\mathscr{L}_y}$ lpXT to $\Delta_y$ XT

Step1 - Calculate $\Delta_y$ XT according to $\delta_{\mathscr{L}_y}$ lpXT and contract state

$$\Delta_y=\delta_{\mathscr{L}_y} \times \frac{R_y}{S_{\mathscr{L}_y}}$$

Step2 - Calculate new price

The liquidity impact of withdrawing XT from the pool is equivalent to the liquidity impact of following two operations.

1. Sell $-\sigma_y$ XT to the pool and get $-\sigma_x$ FT

2. Combine $(\Delta_y-\sigma_y)$ XT and $\sigma_x$ FT into $(\Delta_y-\sigma_y)$ UT and withdraw

The new price impact of withdrawing FT from the pool will be equivalent to the price impact of the two operations as well. In addition, the price of pool remains the same when UT is provided or withdrawn from the pool by definition. Therefore, the price impact of withdrawing FT from the pool will be equivalent to the price impact of Sell $-\sigma_y$ XT to the pool and get $-\sigma_x$ FT.

According to $Eq.5$,

$$x'_{ori} \cdot y'_{ori}=x'_{new} \cdot y'_{new}=(x'_{ori}+\sigma_x)\cdot(y'_{ori}-\sigma_y) = k \text{ (eq.1)}$$

According to $Eq.2$,

$$\sigma_x=\varepsilon (\Delta_y-\sigma_y) \text{ (eq.2)}$$

According to $eq.1$ and $eq.2$ with elimination by substitution,

$$\varepsilon \sigma_y^2 - (x_{ori}'+ \varepsilon \Delta_y+\varepsilon y_{ori}') \cdot \sigma_y+(x_{ori}'y_{ori}'+\varepsilon \Delta_y y_{ori}'-\gamma^{-2}L^2)=0$$

derive

$$\sigma_y=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ (invalid, since } y_{new}' \text{ should be greater than 0)} \lor \sigma_y=\frac{-b-\sqrt{b^2-4ac}}{2a} \\ \sigma_x=\Delta_x-\varepsilon\sigma_y$$

where

$$\begin{cases} a=\varepsilon \\ b=- (x_{ori}'+ \varepsilon \Delta_y+\varepsilon y_{ori}') \\ c= x_{ori}'y_{ori}'+\varepsilon \Delta_y y_{ori}'-\gamma^{-2}L^2 \end{cases}$$

Step3 - Update pool state

$$P_{new}=\frac{y'_{new}}{x'_{new}} = \frac{y_{ori}-\sigma_y+\beta}{x_{ori}+\sigma_x}$$

$$R_{new}=\frac{\frac{1}{P_{new}}+\varepsilon-1}{\theta}$$