ZK + PARTH Bonus: Calculating how many users are needed to achieve 100,000 TPS

In this post, we will derive a formula which calculates the number of users that need to participate in a QED block to attain 100,000 TPS.

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

In reality, QED 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:

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 $u$:

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

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

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

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:

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

Or in other words:

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 $y(x)$.

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

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

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

Simplifying, we get:

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

Simplifying, we get:

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

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

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

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

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

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

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

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

And since we already showed that:

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 QED:

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

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

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

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

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

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

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!

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

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