# Adding Liquidity in BlueLotusDAO

There are some conditions when adding liquidity to the BlueLotusDAO:

1. After LP contributions, the token price is unchanged.
2. $$P\_{min}$$ and $$P\_{max}$$ are also unchanged after LP contributions.

In BlueLotusDAO, the pool for pair X-Y needs to maintain 4 parameters:

1. The initial amount of token $$X$$ that is used for amplification, denoted by $$x\_0$$
2. The initial amount of token $$Y$$ that is used for amplification, denoted by $$y\_0$$
3. The change in token $$X$$ amount after trading activities, denoted by $$\Delta x\_0$$
4. The change in token $$Y$$ amount after trading activities, denoted by $$\Delta y\_0$$

Therefore, the real balances and virtual balances of the reserves are:

*<mark style="background-color:purple;">Real Balances</mark>*

$$
x = x\_0 + \Delta x\_0 \ y = y\_0 + \Delta y\_0
$$

*<mark style="background-color:purple;">Virtual Balances</mark>*

$$
x' = a \cdot x\_0 + \Delta x\_0 \ y' = a \cdot y\_0 + \Delta y\_0
$$

where $$a$$ is the amplification factor. You may find more information about the amplification factor [here](https://genesyblockchain.gitbook.io/genesys-blockchain/bluelotusdao/bluelotusdao-products/bluelotusdao-protocol/programmable-pricing-curve).

The constant product $$x' \cdot y' = (a \cdot x\_0 + \Delta x\_0) \cdot (a \cdot y\_0 + \Delta y\_0) = k'$$.&#x20;

Note that $$P\_{min}$$ and $$P\_{max}$$ at this time are:

$$
\begin{cases} P\_{min} = \cfrac{(y\_0 \cdot a - y\_0)^2}{k'} \ \ P\_{max} = \cfrac{k'}{(x\_0 \cdot a - x\_0)^2} \end{cases}
$$

The current price: $$P = \cfrac{y'}{x'} = \cfrac{a \cdot y\_0 + \Delta y\_0}{a \cdot x\_0 + \Delta x\_0}$$

Liquidity Providers have to contribute in the same proportion for all 4 amount types. We denote the contribution ratio to be $$b$$. LPs have to contribute $$x\_1 + \Delta x\_1$$, $$y\_1 + \Delta y\_1$$ in which:

$$
\begin{cases} x\_1 = b \cdot x\_0 \ \Delta x\_1 = b \cdot \Delta x\_0 \ y\_1 = b \cdot y\_0 \ \Delta y\_1 = b \cdot \Delta y\_0 \end{cases}
$$

The real balances and virtual balances of the reserve after contribution are:

*<mark style="background-color:purple;">Real Balances</mark>*

$$
x = (x\_0 + x\_1) + (\Delta x\_0 + \Delta x\_1) = (b + 1) \cdot (x\_0 + \Delta x\_0) \ y = (y\_0 + y\_1) + (\Delta y\_0 + \Delta y\_1) = (b + 1) \cdot (y\_0 + \Delta y\_0)
$$

*<mark style="background-color:purple;">Virtual Balances</mark>*

$$
x' = a \cdot (x\_0 + x\_1) + (\Delta x\_0 + \Delta x\_1) = (b + 1) \cdot (a \cdot x\_0 + \Delta x\_0) \ y' = a \cdot (y\_0 + y\_1) + (\Delta y\_0 + \Delta y\_1) = (b + 1) \cdot (a \cdot y\_0 + \Delta y\_0)
$$

The constant product, after the LP contribution, becomes:

$$
x' \cdot y' = (b + 1)^2 \cdot (a \cdot x\_0 + \Delta x\_0) \cdot (a \cdot y\_0 + \Delta y\_0) = (b + 1)^2 \cdot k'
$$

$$P\_{min}$$ and $$P\_{max}$$ at this time are:

$$
\begin{cases} P\_{min} = \cfrac{((y\_0 + y\_1) \cdot a - (y\_0 + y\_1))^2}{(b + 1)^2 \cdot k'} = \cfrac{(y\_0 \cdot a - y\_0)^2}{k'} \ P\_{max} = \cfrac{(b + 1)^2 \cdot k'}{((x\_0 + x\_1) \cdot a - (x\_0 + x\_1))^2} = \cfrac{(x\_0 \cdot a - x\_0)^2}{k'} \end{cases}
$$

The current price is updated to be $$P = \cfrac{y'}{x'} = \cfrac{(a \cdot y\_0 + \Delta y\_0) \cdot (b + 1)}{(a \cdot x\_0 + \Delta x\_0) \cdot (b + 1)} = \cfrac{a \cdot y\_0 + \Delta y\_0}{a \cdot x\_0 + \Delta x\_0}$$

We see that after LP contributes, the current price, $$P\_{min}$$ and $$P\_{max}$$ are unchanged. It is similar in the case of LPs withdrawals, where the ratio $$b$$ is negative.

#### Example <a href="#example" id="example"></a>

* Initially, the first LP put 100 $$X$$ and 100 $$Y$$ to the reserve, we have: $$x = 100, y = 100, \Delta x = 0, \Delta y = 0$$.
* A user trades 20 X for 15 Y, so we have the updated parameters: $$x = 100, y = 100, \Delta x = 20, \Delta y = −15$$.
* Suppose an LP wants to contribute 20% of the current token amounts in the pool, so he should deposit:

  $$
  0.2 · 100 + 0.2 · 20 = 24 (X) \ 0.2 · 100 + 0.2 · (−15) = 17 (Y)
  $$

ie. deposit 24X and 17Y tokens.

The parameters are then updated to be: $$x = 120$$, $$y = 120$$, $$\Delta x = 24$$, $$\Delta y = −18$$.