Mathematical Principle of TWAMM

Definition

Assume that TWAMM requires

$N$ blocks to execute a large virtual transaction, and that the pool selling

$X$ sells at a rate of

$x_{rate}$ per block, while the pool selling

$Y$ sells at a rate of

$y_{rate}$ per block. Thus, the total amount of

$X$ sold during the whole period is

$x_{in}=N{x}_{rate}$ and the total amount of

$Y$ sold is

$y_{in}=N{y}_{rate}$.

Also, we note the initial reserves

$x_{reserve}$ and

$y_{reserve}$ for this time period in embeded AMM as

$x_0=x_{ammStart}$ and

$y_0=y_{ammStart}$, respectively.

According to the design of TWAMM, large orders are traded with blocks, each block sells

$x_{rate}$ to get

$y_{out}$, or sells

$y_{rate}$ to get

$x_{out}$, while AMM updates the values of

$x_{reserve}$ and

$y_{reserve}$, the whole process

$N$ transactions in total.

It is worth noting that each block transaction of AMM always follows a constant product market making.

Formula

First, after executing the transaction for the

$n-1$ block, we assume that the

$x_{reserve}$ and

$y_{reserve}$ values of AMM are

$x_{n-1}$ and

$y_{n-1}$, respectively.

Next, the transaction for block

$n$ is executed, with

$X-Pool$ and

$Y-Pool$ feeding

$x_{rate}$ and

$y_{rate}$ to AMM, respectively. Then,

$${\bar{x}}_n=x_{n-1}+x_{rate}$$

$${\bar{y}}_n=y_{n-1}+y_{rate}$$

Since

$x_{rate}$ and

$y_{rate}$ are very small and the exchange follow a constant product market maker, we can obtain that,

$${\displaystyle \frac{x_{out,n}}{y_{rate}}}={\displaystyle \frac{{\bar{x}}_n}{{\bar{y}}_n}}={\displaystyle \frac{x_{n-1}+x_{rate}}{y_{n-1}+y_{rate}}}$$

$${\displaystyle \frac{y_{out,n}}{x_{rate}}}={\displaystyle \frac{{\bar{y}}_n}{{\bar{x}}_n}}={\displaystyle \frac{y_{n-1}+y_{rate}}{x_{n-1}+x_{rate}}}$$

Simplifying,

$$x_{out,n}=y_{rate}\cdot {\displaystyle \frac{{\bar{x}}_n}{{\bar{y}}_n}}=y_{rate}\cdot {\displaystyle \frac{x_{n-1}+x_{rate}}{y_{n-1}+y_{rate}}}$$

$$y_{out,n}=x_{rate}\cdot {\displaystyle \frac{{\bar{y}}_n}{{\bar{x}}_n}}=x_{rate}\cdot {\displaystyle \frac{y_{n-1}+y_{rate}}{x_{n-1}+x_{rate}}}$$

After getting the values of

$x_{out,n}$ and

$y_{out,n}$, we can further obtain the

$x_{reserve}$ and

$y_{reserve}$ values

$x_n$ and

$y_n$ after the transaction of

$n$ blocks.

$$x_n={\bar{x}}_n-x_{out,n}={\bar{x}}_n-y_{rate}\cdot {\displaystyle \frac{{\bar{x}}_n}{{\bar{y}}_n}}=y_{n-1}\cdot {\displaystyle \frac{{\bar{x}}_n}{{\bar{y}}_n}}=y_{n-1}\cdot {\displaystyle \frac{x_{n-1}+x_{rate}}{y_{n-1}+y_{rate}}}$$

$$y_n={\bar{y}}_n-y_{out,n}={\bar{y}}_n-x_{rate}\cdot {\displaystyle \frac{{\bar{y}}_n}{{\bar{x}}_n}}=x_{n-1}\cdot {\displaystyle \frac{{\bar{y}}_n}{{\bar{x}}_n}}=x_{n-1}\cdot {\displaystyle \frac{y_{n-1}+y_{rate}}{x_{n-1}+x_{rate}}}$$

$$AveragePrice{P}_y^x=\sum _{n=1}^N{\displaystyle \frac{x_{out,n}}{N{y}_{rate}}}={\displaystyle \frac{x_{out}}{y_{in}}}$$

$$AveragePrice{P}_x^y=\sum _{n=1}^N{\displaystyle \frac{y_{out,n}}{N{x}_{rate}}}={\displaystyle \frac{y_{out}}{x_{in}}}$$

By observation, we find that

$x_n\cdot y_n=x_{n-1}\cdot y_{n-1}$, which exactly meets the prerequisite requirement of AMM for making a market by following a constant product.

Let

$x_n{y}_n=x_{n-1}{y}_{n-1}=...=x_1{y}_1=x_0{y}_0=k$,

$k$ is a constant.

Fractional Linear Recursion

First find the general formula for

$x_n$ and obtain the value of

$x_{ammEnd}=x_N$, and the same for

$y_n$.

$$x_n=y_{n-1}\cdot {\displaystyle \frac{x_{n-1}+x_{rate}}{y_{n-1}+y_{rate}}}={\displaystyle \frac{k}{x_{n-1}}}\cdot {\displaystyle \frac{x_{n-1}+{\displaystyle \frac{x_{in}}{N}}}{{\displaystyle \frac{k}{x_{n-1}}}+{\displaystyle \frac{y_{in}}{N}}}}=k\cdot {\displaystyle \frac{x_{n-1}+{\displaystyle \frac{x_{in}}{N}}}{{\displaystyle \frac{y_{in}}{N}}\cdot x_{n-1}+k}}$$

Let

$a=\sqrt{{\displaystyle \frac{k{x}_{in}}{y_{in}}}},\bar{a}=\sqrt{{\displaystyle \frac{k{y}_{in}}{x_{in}}}},b=\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}$, then

$$x_n={\displaystyle \frac{x_{n-1}+{\displaystyle \frac{ab}{N}}}{{\displaystyle \frac{b}{aN}}\cdot x_{n-1}+1}}$$

The fractional linear recursive expression for

$y_n$ is as follows:

$$y_n=k\cdot {\displaystyle \frac{y_{n-1}+{\displaystyle \frac{y_{in}}{N}}}{{\displaystyle \frac{x_{in}}{N}}\cdot y_{n-1}+k}}={\displaystyle \frac{y_{n-1}+{\displaystyle \frac{\bar{a}b}{N}}}{{\displaystyle \frac{b}{\bar{a}N}}\cdot y_{n-1}+1}}$$

Solution

First, if

$y_{in}=0$ and

$x_{in}\ne 0$ , then

$$x_n=x_0+{\displaystyle \frac{x_{in}}{N}}\cdot n,x_{ammEnd}=x_N=x_0+x_{in},x_{out}=x_0+x_{in}-x_{ammEnd}=0$$

And,

$$y_n=k\cdot {\displaystyle \frac{y_{n-1}}{{\displaystyle \frac{x_{in}}{N}}\cdot y_{n-1}+k}},{\displaystyle \frac{1}{y_n}}={\displaystyle \frac{1}{y_{n-1}}}+{\displaystyle \frac{x_{in}}{kN}}$$

The calculation yields,

$$y_n={\displaystyle \frac{1}{{\displaystyle \frac{1}{y_0}}+{\displaystyle \frac{x_{in}}{kN}}\cdot n}},y_{ammEnd}=y_N={\displaystyle \frac{k}{x_0+x_{in}}}$$

$$y_{out}=y_0+y_{in}-y_{ammEnd}={\displaystyle \frac{y_0}{x_0+x_{in}}}\cdot x_{in}$$

$$AveragePrice{P}_x^y={\displaystyle \frac{y_{out}}{x_{in}}}={\displaystyle \frac{y_0}{x_0+x_{in}}}$$

---

When

$x_{in}=0$ and

$y_{in}\ne 0$ , the discussion is the same as above.

---

If both

$x_{in}\ne 0$ and

$y_{in}\ne 0$, for general fractional linear recursion, we can use the Fixed-Point Iteration method to solve the problem.

According to

$x_n={\displaystyle \frac{x_{n-1}+{\displaystyle \frac{ab}{N}}}{{\displaystyle \frac{b}{aN}}\cdot x_{n-1}+1}}$, replace

$\lambda ={\displaystyle \frac{\lambda +{\displaystyle \frac{ab}{N}}}{{\displaystyle \frac{b}{aN}}\cdot \lambda +1}}$ and solve for

$\lambda =\pm a$.

$\pm a$ is exactly the two fixed points of the above fractional linear recursive expression.

Further calculations,

$$x_n-a={\displaystyle \frac{a({\displaystyle \frac{N}{b}}-1)(x_{n-1}-a)}{x_{n-1}+{\displaystyle \frac{aN}{b}}}},x_n+a={\displaystyle \frac{a({\displaystyle \frac{N}{b}}+1)(x_{n-1}+a)}{x_{n-1}+{\displaystyle \frac{aN}{b}}}}$$

Dividing the two expressions above,

$${\displaystyle \frac{x_n-a}{x_n+a}}={\displaystyle \frac{N-b}{N+b}}\cdot {\displaystyle \frac{x_{n-1}-a}{x_{n-1}+a}}=(1-{\displaystyle \frac{2b}{N+b}})\cdot {\displaystyle \frac{x_{n-1}-a}{x_{n-1}+a}}$$

According to the geometric progression we get,

$${\displaystyle \frac{x_n-a}{x_n+a}}=(1-{\displaystyle \frac{2b}{N+b}}{)}^n\cdot {\displaystyle \frac{x_0-a}{x_0+a}},{\displaystyle \frac{x_N-a}{x_N+a}}=(1-{\displaystyle \frac{2b}{N+b}}{)}^N\cdot {\displaystyle \frac{x_0-a}{x_0+a}}$$

Similarly,

$${\displaystyle \frac{y_n-\bar{a}}{y_n+\bar{a}}}=(1-{\displaystyle \frac{2b}{N+b}}{)}^n\cdot {\displaystyle \frac{y_0-\bar{a}}{y_0+\bar{a}}},{\displaystyle \frac{y_N-\bar{a}}{y_N+\bar{a}}}=(1-{\displaystyle \frac{2b}{N+b}}{)}^N\cdot {\displaystyle \frac{y_0-\bar{a}}{y_0+\bar{a}}}$$

Limit Value

The basic principle of TWAMM is to decompose long term large orders into infinitely many infinitely small virtual orders, i.e.

$N$ can be taken to infinity, so the limit value of

$x_N$ can be obtained.

$${\displaystyle \frac{x_{ammEnd}-a}{x_{ammEnd}+a}}=\underset{N\to +\mathrm{\infty }}{lim}({\displaystyle \frac{x_N-a}{x_N+a}})=\underset{N\to +\mathrm{\infty }}{lim}(1-{\displaystyle \frac{2b}{N+b}}{)}^N\cdot {\displaystyle \frac{x_0-a}{x_0+a}}=e^{-2b}\cdot {\displaystyle \frac{x_0-a}{x_0+a}}$$

$$x_{ammEnd}=a\cdot {\displaystyle \frac{e^{2b}+c}{e^{2b}-c}}=\sqrt{{\displaystyle \frac{k{x}_{in}}{y_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+c}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-c}}$$

Where，

$$c={\displaystyle \frac{x_0-a}{x_0+a}}={\displaystyle \frac{\sqrt{x_{ammStart}\cdot y_{in}}-\sqrt{y_{ammStart}\cdot x_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}$$

$$x_{out}=x_{ammStart}+x_{in}-x_{ammEnd}$$

$$AveragePrice{P}_y^x={\displaystyle \frac{x_{out}}{y_{in}}}={\displaystyle \frac{x_{ammStart}+x_{in}-x_{ammEnd}}{y_{in}}}$$

Similarly,

$$y_{ammEnd}=\sqrt{{\displaystyle \frac{k{y}_{in}}{x_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+\bar{c}}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-\bar{c}}}$$

$$\bar{c}={\displaystyle \frac{\sqrt{y_{ammStart}\cdot x_{in}}-\sqrt{x_{ammStart}\cdot y_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}=-c$$

$$y_{out}=y_{ammStart}+y_{in}-y_{ammEnd}$$

$$AveragePrice{P}_x^y={\displaystyle \frac{y_{out}}{x_{in}}}={\displaystyle \frac{y_{ammStart}+y_{in}-y_{ammEnd}}{x_{in}}}$$

An important point is that，

$x_{out}$,

$y_{out}$,

$x_{ammEnd}$ and

$y_{ammEnd}$ must larger than 0:

$$e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}>\left|c\right|=\left|{\displaystyle \frac{\sqrt{x_{ammStart}\cdot y_{in}}-\sqrt{y_{ammStart}\cdot x_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}\right|$$

Finally, after a simple verification

$x_{ammEnd}\cdot y_{ammEnd}=x_{ammStart}\cdot y_{ammStart}=k$, the constant product of AMM is still satisfied.

When

$y_{in}\to 0,x_{in}\ne 0$，through L'Hôpital's Rule, we can obtain:

$$y_{ammEnd}=\underset{y_{in}\to 0}{lim}(\sqrt{{\displaystyle \frac{k{y}_{in}}{x_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+\bar{c}}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-\bar{c}}})$$

$$=\underset{y_{in}\to 0}{lim}(\sqrt{{\displaystyle \frac{k{y}_{in}}{x_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+{\displaystyle \frac{\sqrt{y_{ammStart}\cdot x_{in}}-\sqrt{x_{ammStart}\cdot y_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-{\displaystyle \frac{\sqrt{y_{ammStart}\cdot x_{in}}-\sqrt{x_{ammStart}\cdot y_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}}})$$

$$={\displaystyle \frac{1}{{\displaystyle \frac{1}{y_{ammStart}}}+{\displaystyle \frac{x_{in}}{k}}}}={\displaystyle \frac{y_{ammStart}\cdot x_{ammStart}}{x_{ammStart}+x_{in}}}$$

$$x_{ammEnd}=\underset{y_{in}\to 0}{lim}(\sqrt{{\displaystyle \frac{k{x}_{in}}{y_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-\bar{c}}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+\bar{c}}})$$

$$=\underset{y_{in}\to 0}{lim}(\sqrt{{\displaystyle \frac{k{x}_{in}}{y_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-{\displaystyle \frac{\sqrt{y_{ammStart}\cdot x_{in}}-\sqrt{x_{ammStart}\cdot y_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+{\displaystyle \frac{\sqrt{y_{ammStart}\cdot x_{in}}-\sqrt{x_{ammStart}\cdot y_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}}})$$

$$=x_{ammStart}+x_{in}$$

Differential Equation Of TWAMM

Assume the trading rate of

$X-Pool$ and

$Y-Pool$ is uniform according to time，

$${\bar{x}}_{new}=x_{old}+{\displaystyle \frac{x_{in}}{T}}dt$$

$${\bar{y}}_{new}=y_{old}+{\displaystyle \frac{y_{in}}{T}}dt$$

$$x_{new}=y_{old}\cdot {\displaystyle \frac{{\bar{x}}_{new}}{{\bar{y}}_{new}}}=y_{old}\cdot {\displaystyle \frac{x_{old}+{\displaystyle \frac{x_{in}}{T}}dt}{y_{old}+{\displaystyle \frac{y_{in}}{T}}dt}}$$

$$y_{new}=x_{old}\cdot {\displaystyle \frac{{\bar{y}}_{new}}{{\bar{x}}_{new}}}=x_{old}\cdot {\displaystyle \frac{y_{old}+{\displaystyle \frac{y_{in}}{T}}dt}{x_{old}+{\displaystyle \frac{x_{in}}{T}}dt}}$$

$$x_{new}=y_{old}\cdot {\displaystyle \frac{x_{old}+{\displaystyle \frac{x_{in}}{T}}dt}{y_{old}+{\displaystyle \frac{y_{in}}{T}}dt}}={\displaystyle \frac{k}{x_{old}}}\cdot {\displaystyle \frac{x_{old}+{\displaystyle \frac{x_{in}}{T}}dt}{{\displaystyle \frac{k}{x_{old}}}+{\displaystyle \frac{y_{in}}{T}}dt}}=k\cdot {\displaystyle \frac{x_{old}+{\displaystyle \frac{x_{in}}{T}}dt}{{\displaystyle \frac{y_{in}}{T}}dt\cdot x_{old}+k}}$$

Let

$a=\sqrt{{\displaystyle \frac{k{x}_{in}}{y_{in}}}},\bar{a}=\sqrt{{\displaystyle \frac{k{y}_{in}}{x_{in}}}},b=\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}$, then,

$$x_{new}={\displaystyle \frac{x_{old}+{\displaystyle \frac{ab}{T}}dt}{{\displaystyle \frac{b}{aT}}dt\cdot x_{old}+1}}$$

$${\displaystyle \frac{x_{new}-a}{x_{new}+a}}=(1-{\displaystyle \frac{{\displaystyle \frac{2b}{T}}dt}{1+{\displaystyle \frac{b}{T}}dt}})\cdot {\displaystyle \frac{x_{old}-a}{x_{old}+a}}$$

$${\displaystyle \frac{{\displaystyle \frac{x_{new}-a}{x_{new}+a}}-{\displaystyle \frac{x_{old}-a}{x_{old}+a}}}{{\displaystyle \frac{x_{old}-a}{x_{old}+a}}}}=-{\displaystyle \frac{{\displaystyle \frac{2b}{T}}dt}{1+{\displaystyle \frac{b}{T}}dt}}$$

$${\displaystyle \frac{{\displaystyle \frac{x+dx-a}{x+dx+a}}-{\displaystyle \frac{x_{}-a}{x_{}+a}}}{{\displaystyle \frac{x_{}-a}{x_{}+a}}}}=-{\displaystyle \frac{{\displaystyle \frac{2b}{T}}dt}{1+{\displaystyle \frac{b}{T}}dt}}$$

$$\underset{dt\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{x+dx-a}{x+dx+a}}-{\displaystyle \frac{x_{}-a}{x_{}+a}}}{{\displaystyle \frac{x_{}-a}{x_{}+a}}\cdot dt}}=\underset{dt\to 0}{lim}{\displaystyle \frac{-{\displaystyle \frac{2b}{T}}}{1+{\displaystyle \frac{b}{T}}dt}}$$

$${\displaystyle \frac{d\left(ln{\displaystyle \frac{x-a}{x+a}}\right)}{dt}}=-{\displaystyle \frac{2b}{T}}$$

$$ln{\displaystyle \frac{x_{ammEnd}-a}{x_{ammEnd}+a}}-ln{\displaystyle \frac{x_{ammStart}-a}{x_{ammStart}+a}}={\int }_0^T-{\displaystyle \frac{2b}{T}}dt=-2b$$

$${\displaystyle \frac{x_{ammEnd}-a}{x_{ammEnd}+a}}=e^{-2b}\cdot {\displaystyle \frac{x_{ammStart}-a}{x_{ammStart}+a}}$$

$$x_{ammEnd}=a\cdot {\displaystyle \frac{e^{2b}+c}{e^{2b}-c}}=\sqrt{{\displaystyle \frac{k{x}_{in}}{y_{in}}}}\cdot {\displaystyle \frac{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}+c}{e^{2\sqrt{{\displaystyle \frac{x_{in}{y}_{in}}{k}}}}-c}}$$

Where，

$$c={\displaystyle \frac{x_0-a}{x_0+a}}={\displaystyle \frac{\sqrt{x_{ammStart}\cdot y_{in}}-\sqrt{y_{ammStart}\cdot x_{in}}}{\sqrt{x_{ammStart}\cdot y_{in}}+\sqrt{y_{ammStart}\cdot x_{in}}}}$$

---

If we use

$f(t)$ and

$g(t)$ instead of

${\displaystyle \frac{x_{in}}{T}}$ and

${\displaystyle \frac{y_{in}}{T}}$, that is , the trading rate of

$X-Pool$ and

$Y-Pool$ is a function of time.

$$x_{new}=k\cdot {\displaystyle \frac{x_{old}+f(t)dt}{g(t)dt\cdot x_{old}+k}}$$

Let

$a=\sqrt{{\displaystyle \frac{kf(t)}{g(t)}}},\bar{a}=\sqrt{{\displaystyle \frac{kg(t)}{f(t)}}},b=\sqrt{{\displaystyle \frac{f(t)g(t)}{k}}}$, then,

$$x_{new}-x_{old}={\displaystyle \frac{ab\cdot dt-{\displaystyle \frac{b}{a}}dt\cdot x_{old}^2}{{\displaystyle \frac{b}{a}}dt\cdot x_{old}+1}}$$

$${\displaystyle \frac{x_{new}-x_{old}}{dt}}={\displaystyle \frac{ab-{\displaystyle \frac{b}{a}}\cdot x_{old}^2}{{\displaystyle \frac{b}{a}}dt\cdot x_{old}+1}}$$

$$\underset{dt\to 0}{lim}\left({\displaystyle \frac{x_{new}-x_{old}}{dt}}\right)=\underset{dt\to 0}{lim}\left({\displaystyle \frac{ab-{\displaystyle \frac{b}{a}}\cdot x_{old}^2}{{\displaystyle \frac{b}{a}}dt\cdot x_{old}+1}}\right)$$

$${\displaystyle \frac{dx}{dt}}=ab-{\displaystyle \frac{b}{a}}\cdot x^2=f(t)-{\displaystyle \frac{g(t)}{k}}{x}^2={\displaystyle \frac{g(t)}{k}}\cdot ({\displaystyle \frac{kf(t)}{g(t)}}-x^2)$$

Let

$h(t)={\displaystyle \frac{f(t)}{g(t)}}={\displaystyle \frac{a^2}{k}}$,

$${\displaystyle \frac{dx}{dt}}={\displaystyle \frac{g(t)}{k}}\cdot (k\cdot h(t)-x^2)$$

If

$a$ is constant, i.e.the trading rate of

$X-Pool$ and

$Y-Pool$ is synchronized.

$${\displaystyle \frac{d(ln{\displaystyle \frac{x-a}{x+a}})}{dt}}=-2\sqrt{{\displaystyle \frac{f(t)g(t)}{k}}}$$

$$ln{\displaystyle \frac{x_{ammEnd}-a(T)}{x_{ammEnd}+a(T)}}-ln{\displaystyle \frac{x_{ammStart}-a(0)}{x_{ammStart}+a(0)}}={\int }_0^T-2\sqrt{{\displaystyle \frac{f(t)g(t)}{k}}}dt$$

$$x_{ammEnd}=\sqrt{{\displaystyle \frac{kf(T)}{g(T)}}}\cdot {\displaystyle \frac{e^{2{\int }_0^T\sqrt{{\displaystyle \frac{f(t)g(t)}{k}}}dt}+c}{e^{2{\int }_0^T\sqrt{{\displaystyle \frac{f(t)g(t)}{k}}}dt}-c}}$$

Where，

$$c={\displaystyle \frac{x_{ammStart}-a(0)}{x_{ammStart}+a(0)}}={\displaystyle \frac{x_{ammStart}-\sqrt{{\displaystyle \frac{kf(0)}{g(0)}}}}{x_{ammStart}+\sqrt{{\displaystyle \frac{kf(0)}{g(0)}}}}}={\displaystyle \frac{\sqrt{x_{ammStart}\cdot g(0)}-\sqrt{y_{ammStart}\cdot f(0)}}{\sqrt{x_{ammStart}\cdot g(0)}+\sqrt{y_{ammStart}\cdot f(0)}}}$$

At this point, we have completed a rigorous argument and explanation of the mathematical principles of TWAMM and obtained exactly the same conclusion as in the article [The Time-Weighted Average Market Maker - TWAMM].

参考

• [1] The Time-Weighted Average Market Maker - TWAMM