Tim Denning

5 months ago

SAHIL SAPRU

5 months ago

Discover How We Used Facebook Ads to Grow a New Mobile App from $0 to$15K MRR in Just 6 Months and Our Strategy to Hit $100K a Month. Our client introduced a mobile app for Poshmark resellers in December and wanted as many to experience it and subscribe to the monthly plan. An Error We Committed We initiated a Facebook ad campaign with a "awareness" goal, not "installs." This sent them to a landing page that linked to the iPhone App Store and Android Play Store. Smart, right? We got some installs, but we couldn't tell how many came from the ad versus organic/other channels because the objective we chose only reported landing page clicks, not app installs. We didn't know which interest groups/audiences had the best cost per install (CPI) to optimize and scale our budget. After spending$700 without adequate data (installs and trials report), we stopped the campaign and worked with our client's app developer to set up app events tracking.

This allowed us to create an installs campaign and track installs, trials, and purchases (in some cases).

Finding a Successful Audience

Once we knew what ad sets brought in what installs at what cost, we began optimizing and testing other interest groups and audiences, growing the profitable low CPI ones and eliminating the high CPI ones.

We did all our audience testing using an ABO campaign (Ad Set Budget Optimization), spending $10 to$30 on each ad set for three days and optimizing afterward. All ad sets under $30 were moved to a CBO campaign (Campaign Budget Optimization). We let Facebook's AI decide how much to spend on each ad set, usually the one most likely to convert at the lowest cost. If the CBO campaign maintains a nice CPI, we keep increasing the budget by$50 every few days or duplicating it sometimes in order to double the budget. This is how we've scaled to $400/day profitably. Finding Successful Creatives Per campaign, we tested 2-6 images/videos. Same ad copy and CTA. There was no clear winner because some images did better with some interest groups. The image above with mail packages, for example, got us a cheap CPI of$9.71 from our Goodwill Stores interest group but, a high $48 CPI from our lookalike audience. Once we had statistically significant data, we turned off the high-cost ad. New marketers who are just discovering A/B testing may assume it's black and white — winner and loser. However, Facebook ads' machine learning and reporting has gotten so sophisticated that it's hard to call a creative a flat-out loser, but rather a 'bad fit' for some audiences, and perfect for others. You can see how each creative performs across age groups and optimize. How Many Installs Did It Take Us to Earn$15K Per Month?

Six months after paying $25K, we got 1,940 app installs, 681 free trials, and 522$30 monthly subscriptions. 522 * $30 gives us$15,660 in monthly recurring revenue (MRR).

Reverse Engineering $100K Formula: For$100K/month, we need 3,334 people to pay $30/month. 522 people pay that. We need 2,812 more paid users. 522 paid users from 1,940 installs is a 27% conversion rate. To hit$100K/month, we need 10,415 more installs. Assuming...

With a $400 daily ad spend, we average 40 installs per day. This means that if everything stays the same, it would take us 260 days (around 9 months) to get to$100K a month (MRR).

Conclusion

You must market your goods to reach your income objective (without waiting forever). Paid ads is the way to go if you hate knocking on doors or irritating friends and family (who aren’t scalable anyways).

You must also test and optimize different angles, audiences, interest groups, and creatives.

Nir Zicherman

6 months ago

Only two of the following three options can be achieved: consistency, availability, and partition tolerance

Someone told me that growing from 30 to 60 is the biggest adjustment for a team or business.

I remember thinking, That's random. Each company is unique. I've seen teams of all types confront the same issues during development periods. With new enterprises starting every year, we should be better at navigating growing difficulties.

As a team grows, its processes and systems break down, requiring reorganization or declining results. Why always? Why isn't there a perfect scaling model? Why hasn't that been found?

The Three Things Productive Organizations Must Have

Any company should be efficient and productive. Three items are needed:

First, it must verify that no two team members have conflicting information about the roadmap, strategy, or any input that could affect execution. Teamwork is required.

Second, it must ensure that everyone can receive the information they need from everyone else quickly, especially as teams become more specialized (an inevitability in a developing organization). It requires everyone's accessibility.

Third, it must ensure that the organization can operate efficiently even if a piece is unavailable. It's partition-tolerant.

From my experience with the many teams I've been on, invested in, or advised, achieving all three is nearly impossible. Why a perfect organization model cannot exist is clear after analysis.

The CAP Theorem: What is it?

Eric Brewer of Berkeley discovered the CAP Theorem, which argues that a distributed data storage should have three benefits. One can only have two at once.

The three benefits are consistency, availability, and partition tolerance, which implies that even if part of the system is offline, the remainder continues to work.

This notion is usually applied to computer science, but I've realized it's also true for human organizations. In a post-COVID world, many organizations are hiring non-co-located staff as they grow. CAP Theorem is more important than ever. Growing teams sometimes think they can develop ways to bypass this law, dooming themselves to a less-than-optimal team dynamic. They should adopt CAP to maximize productivity.

Path 1: Consistency and availability equal no tolerance for partitions

Let's imagine you want your team to always be in sync (i.e., for someone to be the source of truth for the latest information) and to be able to share information with each other. Only division into domains will do.

Numerous developing organizations do this, especially after the early stage (say, 30 people) when everyone may wear many hats and be aware of all the moving elements. After a certain point, it's tougher to keep generalists aligned than to divide them into specialized tasks.

In a specialized, segmented team, leaders optimize consistency and availability (i.e. every function is up-to-speed on the latest strategy, no one is out of sync, and everyone is able to unblock and inform everyone else).

Partition tolerance suffers. If any component of the organization breaks down (someone goes on vacation, quits, underperforms, or Gmail or Slack goes down), productivity stops. There's no way to give the team stability, availability, and smooth operation during a hiccup.

Path 2: Partition Tolerance and Availability = No Consistency

Some businesses avoid relying too heavily on any one person or sub-team by maximizing availability and partition tolerance (the organization continues to function as a whole even if particular components fail). Only redundancy can do that. Instead of specializing each member, the team spreads expertise so people can work in parallel. I switched from Path 1 to Path 2 because I realized too much reliance on one person is risky.

What happens after redundancy? Unreliable. The more people may run independently and in parallel, the less anyone can be the truth. Lack of alignment or updated information can lead to people executing slightly different strategies. So, resources are squandered on the wrong work.

Path 3: Partition and Consistency "Tolerance" equates to "absence"

The third, least-used path stresses partition tolerance and consistency (meaning answers are always correct and up-to-date). In this organizational style, it's most critical to maintain the system operating and keep everyone aligned. No one is allowed to read anything without an assurance that it's up-to-date (i.e. there’s no availability).

Always short-lived. In my experience, a business that prioritizes quality and scalability over speedy information transmission can get bogged down in heavy processes that hinder production. Large-scale, this is unsustainable.

Accepting CAP

When two puzzle pieces fit, the third won't. I've watched developing teams try to tackle these difficulties, only to find, as their ancestors did, that they can never be entirely solved. Idealized solutions fail in reality, causing lost effort, confusion, and lower production.

As teams develop and change, they should embrace CAP, acknowledge there is a limit to productivity in a scaling business, and choose the best two-out-of-three path.

Vitalik

1 year ago

An approximate introduction to how zk-SNARKs are possible (part 2)

If tasked with the problem of coming up with a zk-SNARK protocol, many people would make their way to this point and then get stuck and give up. How can a verifier possibly check every single piece of the computation, without looking at each piece of the computation individually? But it turns out that there is a clever solution.

Polynomials

Polynomials are a special class of algebraic expressions of the form:

• x+5
• x^4
• x^3+3x^2+3x+1
• 628x^{271}+318x^{270}+530x^{269}+…+69x+381

i.e. they are a sum of any (finite!) number of terms of the form cx^k

There are many things that are fascinating about polynomials. But here we are going to zoom in on a particular one: polynomials are a single mathematical object that can contain an unbounded amount of information (think of them as a list of integers and this is obvious). The fourth example above contained 816 digits of tau, and one can easily imagine a polynomial that contains far more.

Furthermore, a single equation between polynomials can represent an unbounded number of equations between numbers. For example, consider the equation A(x)+ B(x) = C(x). If this equation is true, then it's also true that:

• A(0)+B(0)=C(0)
• A(1)+B(1)=C(1)
• A(2)+B(2)=C(2)
• A(3)+B(3)=C(3)

And so on for every possible coordinate. You can even construct polynomials to deliberately represent sets of numbers so you can check many equations all at once. For example, suppose that you wanted to check:

• 12+1=13
• 10+8=18
• 15+8=23
• 15+13=28

You can use a procedure called Lagrange interpolation to construct polynomials A(x) that give (12,10,15,15) as outputs at some specific set of coordinates (eg. (0,1,2,3)), B(x) the outputs (1,8,8,13) on thos same coordinates, and so forth. In fact, here are the polynomials:

• A(x)=-2x^3+\frac{19}{2}x^2-\frac{19}{2}x+12
• B(x)=2x^3-\frac{19}{2}x^2+\frac{29}{2}x+1
• C(x)=5x+13

Checking the equation A(x)+B(x)=C(x) with these polynomials checks all four above equations at the same time.

Comparing a polynomial to itself

You can even check relationships between a large number of adjacent evaluations of the same polynomial using a simple polynomial equation. This is slightly more advanced. Suppose that you want to check that, for a given polynomial F, F(x+2)=F(x)+F(x+1) with the integer range {0,1…89} (so if you also check F(0)=F(1)=1, then F(100) would be the 100th Fibonacci number)

As polynomials, F(x+2)-F(x+1)-F(x) would not be exactly zero, as it could give arbitrary answers outside the range x={0,1…98}. But we can do something clever. In general, there is a rule that if a polynomial P is zero across some set S=\{x_1,x_2…x_n\} then it can be expressed as P(x)=Z(x)*H(x), where Z(x)=(x-x_1)*(x-x_2)*…*(x-x_n) and H(x) is also a polynomial. In other words, any polynomial that equals zero across some set is a (polynomial) multiple of the simplest (lowest-degree) polynomial that equals zero across that same set.

Why is this the case? It is a nice corollary of polynomial long division: the factor theorem. We know that, when dividing P(x) by Z(x), we will get a quotient Q(x) and a remainder R(x) is strictly less than that of Z(x). Since we know that P is zero on all of S, it means that R has to be zero on all of S as well. So we can simply compute R(x) via polynomial interpolation, since it's a polynomial of degree at most n-1 and we know n values (the zeros at S). Interpolating a polynomial with all zeroes gives the zero polynomial, thus R(x)=0 and H(x)=Q(x).

Going back to our example, if we have a polynomial F that encodes Fibonacci numbers (so F(x+2)=F(x)+F(x+1) across x=\{0,1…98\}), then I can convince you that F actually satisfies this condition by proving that the polynomial P(x)=F(x+2)-F(x+1)-F(x) is zero over that range, by giving you the quotient:
H(x)=\frac{F(x+2)-F(x+1)-F(x)}{Z(x)}
Where Z(x) = (x-0)*(x-1)*…*(x-98).
You can calculate Z(x) yourself (ideally you would have it precomputed), check the equation, and if the check passes then F(x) satisfies the condition!

Now, step back and notice what we did here. We converted a 100-step-long computation into a single equation with polynomials. Of course, proving the N'th Fibonacci number is not an especially useful task, especially since Fibonacci numbers have a closed form. But you can use exactly the same basic technique, just with some extra polynomials and some more complicated equations, to encode arbitrary computations with an arbitrarily large number of steps.

see part 3

Jeff John Roberts

8 months ago

Jack Dorsey and  Jay-Z Launch 'Bitcoin Academy' in Brooklyn rapper's home

The new Bitcoin Academy will teach Jay-Marcy Z's Houses neighbors "What is Cryptocurrency."
Jay-Z grew up in Brooklyn's Marcy Houses. The rapper and Block CEO Jack Dorsey are giving back to his hometown by creating the Bitcoin Academy.

The Bitcoin Academy will offer online and in-person classes, including "What is Money?" and "What is Blockchain?"
The program will provide participants with a mobile hotspot and a small amount of Bitcoin for hands-on learning.

Students will receive dinner and two evenings of instruction until early September. The Shawn Carter Foundation will help with on-the-ground instruction.

Jay-Z and Dorsey announced the program Thursday morning. It will begin at Marcy Houses but may be expanded.

Crypto Blockchain Plug and Black Bitcoin Billionaire, which has received a grant from Block, will teach the classes.

Jay-Z, Dorsey reunite

Jay-Z and Dorsey have previously worked together to promote a Bitcoin and crypto-based future.

In 2021, Dorsey's Block (then Square) acquired the rapper's streaming music service Tidal, which they propose using for NFT distribution.

Dorsey and Jay-Z launched an endowment in 2021 to fund Bitcoin development in Africa and India.

Dorsey is funding the new Bitcoin Academy out of his own pocket (as is Jay-Z), but he's also pushed crypto-related charitable endeavors at Block, including a \$5 million fund backed by corporate Bitcoin interest.