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Adding Liquidity in Blue Lotus DAO

There are some conditions when adding liquidity to the Blue Lotus DAO:
  1. 1.
    After LP contributions, the token price is unchanged.
  2. 2.
    ​
    PminP_{min}
     and
    PmaxP_{max}
     are also unchanged after LP contributions.
In Blue Lotus DAO, the pool for pair X-Y needs to maintain 4 parameters:
  1. 1.
    The initial amount of token
    XX
     that is used for amplification, denoted by
    x0x_0
    ​
  2. 2.
    The initial amount of token
    YY
     that is used for amplification, denoted by
    y0y_0
    ​
  3. 3.
    The change in token
    XX
     amount after trading activities, denoted by
    Δx0\Delta x_0
    ​
  4. 4.
    The change in token
    YY
     amount after trading activities, denoted by
    Δy0\Delta y_0
    ​
Therefore, the real balances and virtual balances of the reserves are:
Real Balances
x=x0+Δx0y=y0+Δy0x = x_0 + \Delta x_0 \\ y = y_0 + \Delta y_0
Virtual Balances
x′=a⋅x0+Δx0y′=a⋅y0+Δy0x' = a \cdot x_0 + \Delta x_0 \\ y' = a \cdot y_0 + \Delta y_0
where
aa
 is the amplification factor. You may find more information about the amplification factor here.
The constant product
x′⋅y′=(a⋅x0+Δx0)⋅(a⋅y0+Δy0)=k′x' \cdot y' = (a \cdot x_0 + \Delta x_0) \cdot (a \cdot y_0 + \Delta y_0) = k'
.
Note that
PminP_{min}
 and
PmaxP_{max}
 at this time are:
{Pmin=(y0⋅a−y0)2k′Pmax=k′(x0⋅a−x0)2\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=y′x′=a⋅y0+Δy0a⋅x0+Δx0P = \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
bb
. LPs have to contribute
x1+Δx1x_1 + \Delta x_1
,
y1+Δy1y_1 + \Delta y_1
 in which:
{x1=b⋅x0Δx1=b⋅Δx0y1=b⋅y0Δy1=b⋅Δy0\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:
Real Balances
x=(x0+x1)+(Δx0+Δx1)=(b+1)⋅(x0+Δx0)y=(y0+y1)+(Δy0+Δy1)=(b+1)⋅(y0+Δy0)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)
Virtual Balances
x′=a⋅(x0+x1)+(Δx0+Δx1)=(b+1)⋅(a⋅x0+Δx0)y′=a⋅(y0+y1)+(Δy0+Δy1)=(b+1)⋅(a⋅y0+Δy0)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′⋅y′=(b+1)2⋅(a⋅x0+Δx0)⋅(a⋅y0+Δy0)=(b+1)2⋅k′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'
​
PminP_{min}
 and
PmaxP_{max}
 at this time are:
{Pmin=((y0+y1)⋅a−(y0+y1))2(b+1)2⋅k′=(y0⋅a−y0)2k′Pmax=(b+1)2⋅k′((x0+x1)⋅a−(x0+x1))2=(x0⋅a−x0)2k′\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=y′x′=(a⋅y0+Δy0)⋅(b+1)(a⋅x0+Δx0)⋅(b+1)=a⋅y0+Δy0a⋅x0+Δx0P = \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,
PminP_{min}
 and
PmaxP_{max}
 are unchanged. It is similar in the case of LPs withdrawals, where the ratio
bb
 is negative.

Example

  • Initially, the first LP put 100
    XX
     and 100
    YY
     to the reserve, we have:
    x=100,y=100,Δx=0,Δy=0x = 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,Δx=20,Δy=−15x = 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)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=120x = 120
,
y=120y = 120
,
Δx=24\Delta x = 24
,
Δy=−18\Delta y = −18
.
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Last modified 5mo ago