5 ms·

Making Highrise faster with memcached

15 points by bdotdub 18y ago

soundsop 18y ago
People are not going to feel the difference between a page rendered in 50ms and one rendered in 100ms. I don't think that every 50ms difference is unimportant, but the difference between a 50ms render time and 100ms render time may be unimportant. Anyone have a feel for what the maximum render time is for a page to feel fast?
- river_styx 18y ago
  One thing they're missing here is that optimization is also about optimizing resources, not just the speed of an individual request/response cycle. If each request takes 50% less time, the app has 100% more resources to handle other requests. So even though the individual users don't feel a difference, the app as a whole is much more efficient.
mdasen 18y ago
A lot of the time, memcached won't make things faster: it's meant to make things more scalable. On small sites, memcached won't increase your speed, but when you start getting lots of users, it's wonderful! Memcached allows for O(1) lookups. Your database server probably uses b-tree indexes which are great, but they're log(n) lookups. If you aren't dealing with a large number of hits per second, your database will do very well. Once you start getting a lot of lookups per second n * log(n) starts looking a good bit slower than n * 1. Memcached also allows you to scale effortlessly simply by adding servers. Databases can do replication, but only to a point. Remember, even in multi-master replication schemes, every write must be done on every server. Memcached shards your data so that the writes are only done on one server and the load is distributed and knows where to read based on the hash. And every time you're able to read from memcached, you're lowering the load on the piece that is less easily scaled - your database. So, even non-cached loads should get better simply because some of the lookups that would have gone to the database are now hitting memcached.
- palish 18y ago
  Memcached allows for O(1) lookups. Your database server probably uses b-tree indexes which are great, but they're log(n) lookups. A B-tree lookup also requires reading from the harddrive, which can require a lot more time than reading from memory.
  - mdasen 18y ago
    Why? There's no reason that a b-tree must be stored on a hard drive. In fact, there's no reason why MySQL can't exist entirely in memory for read purposes (clearly you want writes written to the disk). I'm not going to comment on configuring specific databases, but there's nothing about b-tree data structures that means you couldn't create an entirely in-memory data store that used b-tree indexes. The difference between disk and memory storage is MemcacheDB and memcached. RDBMS love to leave stuff in memory if they can and that makes the reads skip the harddrive.
    - palish 18y ago
      You are technically correct. My words were imprecise -- indexing into a B-tree does not require accessing the harddrive. But it is a rather large possibility, and it's something to be aware of. From Wikipedia ( http://en.wikipedia.org/wiki/B-tree http://en.wikipedia.org/wiki/B-tree ): "B-trees have substantial advantages over alternative implementations when node access times far exceed access times within nodes. This usually occurs when most nodes are in secondary storage such as hard drives. By maximizing the number of child nodes within each internal node, the height of the tree decreases, balancing occurs less often, and efficiency increases. Usually this value is set such that each node takes up a full disk block or an analogous size in secondary storage. While 2-3 B-trees might be useful in main memory, and are certainly easier to explain, if the node sizes are tuned to the size of a disk block, the result might be a 257-513 B-tree (where the sizes are related to larger powers of 2)." So yes, "there's no reason that a b-tree must be stored on a hard drive". There's also no reason why a car couldn't also have a built-in toaster. :) As far as in-memory data structures go, there are better choices than B-trees. Why use a suboptimal data structure?

15 points by bdotdub 18y ago

People are not going to feel the difference between a page rendered in 50ms and one rendered in 100ms. I don't think that every 50ms difference is unimportant, but the difference between a 50ms render time and 100ms render time may be unimportant. Anyone have a feel for what the maximum render time is for a page to feel fast?

river_styx 18y ago
One thing they're missing here is that optimization is also about optimizing resources, not just the speed of an individual request/response cycle. If each request takes 50% less time, the app has 100% more resources to handle other requests. So even though the individual users don't feel a difference, the app as a whole is much more efficient.

mdasen 18y ago

A lot of the time, memcached won't make things faster: it's meant to make things more scalable. On small sites, memcached won't increase your speed, but when you start getting lots of users, it's wonderful! Memcached allows for O(1) lookups. Your database server probably uses b-tree indexes which are great, but they're log(n) lookups. If you aren't dealing with a large number of hits per second, your database will do very well. Once you start getting a lot of lookups per second n * log(n) starts looking a good bit slower than n * 1. Memcached also allows you to scale effortlessly simply by adding servers. Databases can do replication, but only to a point. Remember, even in multi-master replication schemes, every write must be done on every server. Memcached shards your data so that the writes are only done on one server and the load is distributed and knows where to read based on the hash. And every time you're able to read from memcached, you're lowering the load on the piece that is less easily scaled - your database. So, even non-cached loads should get better simply because some of the lookups that would have gone to the database are now hitting memcached.

palish 18y ago
Memcached allows for O(1) lookups. Your database server probably uses b-tree indexes which are great, but they're log(n) lookups. A B-tree lookup also requires reading from the harddrive, which can require a lot more time than reading from memory.
- mdasen 18y ago
  Why? There's no reason that a b-tree must be stored on a hard drive. In fact, there's no reason why MySQL can't exist entirely in memory for read purposes (clearly you want writes written to the disk). I'm not going to comment on configuring specific databases, but there's nothing about b-tree data structures that means you couldn't create an entirely in-memory data store that used b-tree indexes. The difference between disk and memory storage is MemcacheDB and memcached. RDBMS love to leave stuff in memory if they can and that makes the reads skip the harddrive.
  - palish 18y ago
    You are technically correct. My words were imprecise -- indexing into a B-tree does not require accessing the harddrive. But it is a rather large possibility, and it's something to be aware of. From Wikipedia ( http://en.wikipedia.org/wiki/B-tree http://en.wikipedia.org/wiki/B-tree ): "B-trees have substantial advantages over alternative implementations when node access times far exceed access times within nodes. This usually occurs when most nodes are in secondary storage such as hard drives. By maximizing the number of child nodes within each internal node, the height of the tree decreases, balancing occurs less often, and efficiency increases. Usually this value is set such that each node takes up a full disk block or an analogous size in secondary storage. While 2-3 B-trees might be useful in main memory, and are certainly easier to explain, if the node sizes are tuned to the size of a disk block, the result might be a 257-513 B-tree (where the sizes are related to larger powers of 2)." So yes, "there's no reason that a b-tree must be stored on a hard drive". There's also no reason why a car couldn't also have a built-in toaster. :) As far as in-memory data structures go, there are better choices than B-trees. Why use a suboptimal data structure?