Saving another 100TB of RAM

來源原標題

Saving another 100TB of RAM

作者

f311a

原始標籤

hackernews

來源訊號

HN points 438　HN 留言 97

原始連結

https://blog.cloudflare.com/saving-100-tb-of-ram-with-math/

摘要原文

Cloudflare operates at a scale so big that even after working here for years, it doesn’t seem real. We have thousands of servers all over the world with petabytes of RAM and millions of CPU cores, and all of it is pushed to the max. As vast as those resources feel, they are still finite, and when you need every service to run on every node, it doesn’t leave room for wasted space. At this scale, small improvements are greatly magnified, so even 1%-at-a-time improvements are worth celebrating. And some tweaks add up to a lot more: in this post, we’ll look at how small changes to a single algorithm reduced the memory footprint of one of our Pingora-based services significantly. That allowed us to reclaim more than 100TB of RAM globally, on top of the 100TB of memory the DNS team was able to shed last month . Waste not Maintaining equitable resource sharing between teams is not easy, especially in large organizations. One of the ways Cloudflare ensures the balance is kept is through the tireless efforts of the wonderful Performance team. This story starts with a ticket filed by Ivan who found: Excessive memory usage from pingora-ketama in Pingora Backend Router . The finding was that our internal load-balancing service, Pingora Backend Router (yes, PBR), was using significantly more memory than expected — specifically in structures associated with pingora-ketama, which is our open-source library for handling consistent hashing. In order to talk about how we addressed this seeming overuse of memory, we need to talk about what consistent hashing even is, why we are using it in PBR, and how it became so memory hungry. Along the way, we’ll learn some Rust and even a little math. Consistent hashing Consistent hashing is a widely used method for distributing tasks across multiple servers in a way that does not require large changes when servers are added or removed. Internally we use it to route cacheable requests to servers by URL. This allows us to keep only one copy of a file stored per data center and gives a stable way to find the location of each file. We have mentioned this system before , but let’s take the time to walk through how and why this algorithm is used and how it works. The key concept of consistent hashing is that while hash functions can accept any kind of input, their output is limited to a single unsigned integer (32, 64, or 128-bit integers depending on which hash function). This allows us to relate tasks and servers to each other in a consistent way. Most discussions of consistent hashing have you think of that output space as a continuous, circular ring that wraps around from its max value to zero. This depiction makes for some nice visualizations, but it can also make the simple concept of integer ranges seem more complicated than it needs to be. For our discussion, we’ll represent the 32-bit output of our hash function as a number line. Now, let’s say we have a set of servers, A, B, & C, and a set of tasks t-z. We can map each onto the number line based on the hash of their representative values, so something like IP addresses for servers and cache keys for tasks. Assigning tasks to servers is now just a matter of finding the first server to the left of each task. We can represent this visually by coloring in the region of hashes that will be associated with each server. Notice that the range covered by server C wraps around to the beginning, hence the idea that hashes exist in a ring. And that’s it. At a base level, consistent hashing is this simple — but it doesn’t take long to see that there is room for improvement. Notice that the range covered by server A in our example is significantly larger than that of either B or C. This is a problem because the fraction of the requests a server handles is going to be proportional to the size of its range on the number line. Ideally we would like to guarantee each server will have an equal size, but because hashes are essentially random numbers, we have to talk about the size of the regions in terms of statistics . 😨 Math and consequences First: don’t panic. I promise I'm not about to lie to you and that we will stay safely within the bounds of a day-one probability lesson. When we talk about statistical distributions, there are two big factors that help us quantify uncertainty in helpful ways: expected value and standard deviation . In (over-)simplified terms, expected value gives us a point where measurements based on a distribution will be centered, and standard deviation tells how close to that central point most measurements are likely to be. For consistent hashing, we can calculate these factors for the fractional size of the range associated with one of N servers. (Details on where this formula comes from later). $$m \begin{align*} \text{Exp} &= \frac{1}{N} \\ \text{SD} &= \frac{1}{N}\sqrt{\frac{N-1}{N+1}} \end{align*} m$$ In terms of concrete numbers, let’s say we have 100 servers. The formulas above give: $$m \text{Exp}=1/100 = 1\% \\ \text{SD}= \frac{1}{100}\sqrt{\frac{100-1}{100+1}} \approx 0.99\% m$$ That tells us that we can expect that the range each server handles will be centered around 0.99% of the total and most of the lengths to fall within 1% of what's expected. This sounds good until we realize that that’s 0.99% of the total length . We need to scale the standard deviation by the expected value to see how big the error is as a fraction of the target size. This value is called the coefficient of variation . $$m \text{CV} = \frac{\text{SD}}{\text{Exp}} = \sqrt{\frac{N-1}{N+1}} m$$ At $m N=100, \text{CV} \approx 99\% m$ — meaning some servers will likely be working 99% harder than they should be (handling twice as many requests) while others could be doing practically nothing! Now that we have a way to predict how evenly loaded servers will be using consistent hashing, we can start working on improvements. What if we add hashes? The simplicity of consistent hashing is a double-edged sword. It’s easy to understand and implement because everything is turned into easily-relatable hashes on the same numberline, but any improvements to the system will also need to be relatable to that numberline. That means the solution to any consistent hashing problem can only be more hashes . It’s less like a golden hammer (a tool with which all problems look like nails) and more like a golden nail in that it turns all tools into hammers. To solve the problem of imbalanced workloads, we can add multiple hashes to represent each server instead of just one. We’ll get to the math behind this momentarily, but it should make some intuitive sense that while each individual range has a large standard deviation, adding a bunch together should make their total size even out. If we take our three-server example from the above diagrams and add two more hashes at random for each server, we see that it helps even out each server’s workload. This is an admittedly contrived example. The random nature of the system means there’s no guarantee how much improvement you will get from adding 2 additional hashes per server, but it should make some intuitive sense that combining more of these hash segments together produces a more even distribution. Each segment in the sum has a chance of balancing another. Maybe one is too short; maybe one is too long. This is essentially what the law of large numbers tells us should happen… The obvious problem is it only works for large numbers. In NGINX, the baseline number of hashes per server is hardcoded to 160 , and Pingora uses the same value as the default . I’ll spare you the math for now, but if we go back to our 100-server example, if we use 160 points per server instead of just one, the coefficient of variation (which we can think of like an error margin) drops from about 99% to about 8%, a significant improvement. What if we add more hashes? (438 points, 97 comments on Hacker News)