ZK + PARTH Bonus: How Many Concurrent Users does Psy need to hit 100,000 TPS

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In our previous blog post, we showed that the TPS of Psy running on a 2-to-1 configuration can be calculated as:

$$TPS(u)=\frac{\text{\# of Transactions}}{\text{Block Time}}=\frac{k_{\text{tx}}*u}{\log_4(u)*(k_r+k_{net})}$$

In reality, Psy uses a 4-to-1 proof configuration, where each decentralized proving job recursively verifiers 4 proofs and 4 delta merkle proofs for performance reasons, so the actual equation we can use to model our TPS is:

$$TPS(u)=\frac{\text{\# of Transactions}}{\text{Block Time}}=\frac{k_{\text{tx}}*u}{\log_4(u)*(k_r+k_{net})}$$

Where:

• $u$ is the number of users participating in a block

• $k_{tx}$ is the average number of transactions each participating user performs during the block

• $k_r$ is the time it takes to recursively verify 2 proofs and associated delta merkle proofs (proving time for each job performed by the decentralized proving network)

• $k_{net}$ is the time it takes to send a proof to a decentralized proof miner (~30kb)

The only downside of using a 4-to-1 configuration is that our proving time is slightly longer, notably ~500ms on consumer hardware instead of 380ms in the 2-to-1 configuration.

Calculating the numbers user required to hit 100,000 TPS

Using our formula, if we wanted to see how many users would need to participate in a block to attain a TPS of 100,000, we would need to solve the equation below for $u$:

$$100000=\frac{k_{\text{tx}}*u}{\log_4(u)*(k_r+k_{net})}$$

To find the inverse, let's first clean up our equation a bit, we will say $x$ represents our desired TPS:

$$t=\frac{k_{\text{tx}}*u}{\log_4(u)*(k_r+k_{net})}$$

Let's clean things up a bit, substituting $\log_4(u)=\frac{\ln(u)}{\ln(4)}$ using the change of base rule for logarithms:

$$t=\frac{k_{\text{tx}}*u}{\log_4(u)*(k_r+k_{net})}=\frac{k_{\text{tx}}*u}{\frac{\ln(u)}{\ln(4)}*(k_r+k_{net})}=\frac{k_{tx}}{\frac{k_r+k_{net}}{\ln(4)}}*\frac{u}{\ln(u)}=\frac{k_{tx}\ln(4)}{k_r+k_{net}}*\frac{u}{\ln{u}}$$

Multiplying by $\frac{k_r+k_{net}}{k_{tx}\ln(4)}$ on both sides we get:

$$\frac{k_r+k_{net}}{k_{tx}\ln(4)}t=\frac{u}{\ln{u}}$$

We want to solve for $u$, so remember our goal is to find some function $f(t)$ such that $u=f(t)$ where $t$ is our desired TPS, so our target equation should start with $u=\text{some function of t}$. To get $u$ isolated we need to find a function $g(x)$ such that $g(\frac{u}{\ln(u)})=u$, that way we can say that:

$$g\left(\frac{k_r+k_{net}}{k_{tx}\ln(4)}t\right)=g\left(\frac{u}{\ln{u}}\right)$$

And since $g(x)$ is defined such that $g(\frac{u}{\ln(u)})=u$:

$$u=g\left(\frac{k_r+k_{net}}{k_{tx}\ln(4)}t\right)$$

Or in other words:

$$\text{Users}(TPS)=g\left(\frac{k_r+k_{net}}{k_{tx}\ln(4)}\cdot TPS\right)$$

Now we just have to find a function $g(x)$ such that $g(\frac{u}{\ln(u)})=u$.

To help us find such a function let's let $y(x)=\frac{x}{\ln{x}}$, therefore it follows that $y^{-1}(x)=g(x)$, so now we have simplified our original problem to finding the inverse of $y(x)$.

$$y=\frac{x}{\ln{x}}$$

Let's first define $z$ to equal $z=\frac{1}{y}=\frac{\ln{x}}{x}$:

$$z=\frac{\ln{x}}{x}$$

We can then multiply both sides by $x$ to get:

$$zx=\ln{x}$$

Raising $e$ to the power of both sides we get:

$$e^{zx}=e^{\ln x}$$

Simplifying, we get:

$$e^{zx}=x$$

Let's now multiply both sides by $-z\cdot e^{-zx}$:

$$e^{zx}\cdot(-z\cdot e^{-zx})=x\cdot (-z\cdot e^{-zx})$$

Simplifying, we get:

$$-z=-zx\cdot e^{-zx}$$

Now let's set $p=-zx$, and replace $-zx$ with $p$ in our equation:

$$-z=p\cdot e^p$$

It turns out that there is a function called the Lambert W function (referred to as $W(x)$) which has the property that:

$$W(x\cdot e^x)=x$$

Let's apply Lambert's W function to both sides of our equation:

$$W(-z)=W(p\cdot e^p)$$

Using the properties we know about Lambert's W function we can further simplify to:

$$W(-z)=p$$

Since $p=-xz$, we can rewrite this as:

$$W(-z)=-xz$$

Since $z=\frac{1}{y}$, we can further rewrite our equation as:

$$W\left(\frac{-1}{y}\right)=-\frac{x}{y}$$

If we multiply both sides by $-y$, we now have an expression for $x$ 🎉:

$$x=-y\cdot W\left(\frac{-1}{y}\right)$$

Hence, we now have a function which meets the desired requirements of $g(x)$:

$$g(x)=-x\cdot W\left(\frac{-1}{x}\right)$$

And since we already showed that:

$$\text{Users}(\text{TPS})=g\left(\frac{k_r+k_{net}}{k_{tx}\ln(4)}\cdot \text{TPS}\right)$$

We now can now write a formula for the number of users that need to submit proofs in a block to attain a certain TPS on Psy:

$$\text{Users}(\text{TPS})=-\text{TPS}\cdot\left(\frac{k_r+k_{net}}{k_{tx}\ln(4)}\right)\cdot W\left(\frac{-1}{\text{TPS}\cdot\left(\frac{k_r+k_{net}}{k_{tx}\ln(4)}\right)}\right)$$

To keep our math simple moving forward, let's set the constant $c$ to equal $\frac{k_r+k_{net}}{k_{tx}\ln(4)}$:

$$c=\frac{k_r+k_{net}}{k_{tx}\ln(4)}$$

Therefore, $\text{Users}(\text{TPS})$ can be rewritten as:

$$\text{Users}(\text{TPS})=-\text{TPS}\cdot c \cdot W\left(\frac{-1}{\text{TPS}\cdot c}\right)$$

Let's use some experimental values from Psy's internal testnet to calculate the number of users we need to attain 100,000 TPS on Psy.

• The proving time required to recursively verify 4 proofs and 4 corresponding delta merkle proofs that link to the nearest common ancestor merkle cap is ~500ms, so $k_r=0.5$

• The proof size is ~30kb, so lets say it takes 100ms to send over the network to any where around the world (it fits in a single TCP packet, so think of this as "ping" time)

• Let's say each user submits an average of 3 transactions per block, so $k_{tx}=3$

To keep our math simple, let's first solve for $c=\frac{k_r+k_{net}}{k_{tx}\ln(4)}$

$$c=\frac{k_r+k_{net}}{k_{tx}\ln(4)}=\frac{0.5+0.1}{3\ln(4)}=\frac{0.6}{3\ln(4)}\approx0.1442695$$

So $\text{Users}(100000)$ can be written as:

$$\text{Users}(100000)=-100000\cdot\frac{0.6}{3\ln(4)}\cdot W\left(\frac{-1}{100000\cdot\frac{0.6}{3\ln(4)}}\right)$$

Approximating for $\frac{0.6}{3\ln(4)}$, we get:

$$\text{Users}(100000)=-14426.95\cdot W(-0.000069314718)$$

To evaluate the Lambert W function, we can use Wolfram Alpha:

Hence, we arrive at the result that a block must have 174,096 users participating in a block to hit 100,000 TPS!

$$\text{Users}(100000)=174096$$

Let's check with our original equation just to make sure:

$$TPS(u)=\frac{\text{\# of Transactions}}{\text{Block Time}}=\frac{k_{\text{tx}}*u}{\log_4(u)*(k_r+k_{net})}$$

Plugging in $u=174096$, we get the correct answer 🎉

$$TPS(174096)=\frac{3*174096}{\log_4(174096)*(0.5+0.1)}\approx 100000.4392383768 \text{ TPS}$$