6 ms·
Indeed, as I used to tell my ops colleagues when they pointed to RAM utilization graphs, "we paid for all of that RAM, why aren't we using it?"
by wrs 8mo ago
Indeed, as I used to tell my ops colleagues when they pointed to RAM utilization graphs, "we paid for all of that RAM, why aren't we using it?"
- jcelerier 8mo agobecause memory access performance is not O(1) but depends on the size of what's in memory (https://www.ilikebigbits.com/2014_04_21_myth_of_ram_1.html https://www.ilikebigbits.com/2014_04_21_myth_of_ram_1.html). Every byte used makes the whole thing slower.
- SkiFire13 8mo agoMemory access performance depends on the _maximum size of memory you need to address_. You can clearly see it in the graph of that article where L1, L2, L3 and RAM are no longer enough to fit the linked list. However while the working set fits in them the performance scales much better. So as long as you give priority to the working set, you can fill the rest of the biggest memory with whatever you want without affecting performance.
- stevefan1999 8mo agowhy is it not O(1)? It has to service within a deadline time, so it is still constant.
- yunnpp 8mo ago> Every byte used makes the whole thing slower. This is an incorrect conclusion to make from the link you posted in the context of this discussion. That post is a very long-winded way of saying that the average speed of addressing N elements depends on N and the size of the caches, which isn't news to anyone. Key word: addressing.
- rileymat2 8mo agoI am not following, isn't this just a graph that shows that how fast operations happen is largely dependent on the odds that it is in cache at various levels (CPU/Ram/Disk)? The memory operation itself is O(1), around 100 ns, where at a certain point we are doing full ram fetches each time because the odds of it being in CPU cache are low? Typically O notation is an upper bound, and it holds well there. That said, due to cache hits, the lower bound is much lower than that. You see similar performance degradation if you iterate in a double sided array the in the wrong index first.
- adrianN 8mo agoO notation is technically meaningless for systems with bounded resources. That said, yes the performance is depending on the probability of cache hits, notably also the TLB. For large amounts of memory used and random access patterns, assuming logarithmic costs for memory access tends to model reality better.
- johncolanduoni 8mo agoRAM is always storing something, it’s just sometimes zeros or garbage. Nothing in how DRAM timings work is sensitive to what bits are encoded in each cell.
- anvuong 8mo agoHuh? There is nothing called "empty memory". There is always something being stored in the memory, the important thing is whether you care about that specific bits or not. And no, the articles you linked is about caching, not RAM access. Hardware-wise, it doesn't matter what you have in the cells, access latency is the same. There is gonna be some degradation with #read/write cycles, but that is besides the point.
- lelandbatey 8mo agoThe author of that post effectively re-defines "memory"/"RAM" as "data", and uses that to say "accessing data in the limit scales to N x sqrt(N) as N increases". Which, like, yeah? Duh, I can't fit 200PB of data into the physical RAM of my computer and the more data I have to access the slower it'll be to access any part of it without working harder at other abstraction layers to bring the time taken down. That's true. It's also unrelated to what people are talking about when they say "memory access is O(1)". When people say "memory access is O(1)" they are talking about cases where their data fits in memory (RAM). Their experimental results would in fact be a flat line IF they could disable all the CPU caches, even though performance would be slow.
- jasonfarnon 8mo agodo you also say that about hdd space? about money in the bank?
- astafrig 8mo ago> about money in the bank? Yes, generally. That's the entire idea behind the stock market.
- groundzeros2015 8mo agoIt’s counterintuitive but I learned this best by playing RTS games. If you don’t spend money your opponent can outdo you on the map by simply spending their money. But the principle extends, everything you have doing nothing (buildings units etc) is losing. The most efficient process is to have all your resources working for you at all times.
- LaGrange 8mo ago> It’s counterintuitive but I learned this best by playing RTS games. If you don’t spend money your opponent can outdo you on the map by simply spending their money. OK, hear me out over here: We are not in an RTS. Edit: in real-world settings lacking redundancy tends to make systems incredibly fragile, in a way that just rarely matters in an RTS. Which we are _not in_.
- groundzeros2015 8mo agoAgreed. Real life is not an RTS. Optimizing computer or business resources - kind of like one.
- heavyset_go 8mo agoIf you don't have savings to spend for a potential change of tactics, larger players, groups or players with different strategies can easily overtake you as your perfectly efficient economy collapses. Going to also echo the comment that this isn't an RTS
- underlipton 8mo agoBecause OoM errors are oh so fun.
- stouset 8mo agoCaches are automatically released by the OS when demand for memory increases.
- bagels 8mo agoYou eventually run out of caches to evict.
- winter_blue 8mo agoAt that point you can still fall back onto swap on NVME.
- kulahan 8mo agoDoesn’t Apple use pretty damn quick NVME? I wonder how much of a performance drop it actually is. Certainly not as bad as running a swap file on a 5400 rpm HDD…
- gck1 8mo agoIsn't that NVME also very expensive to replace because it's tied to hardware identifiers? If you keep swapping all the time, surely NVME would be the first part to fail
- ua709 8mo agoThis was heavily debated in the 11.4 timeframe because there was risk that this version of the OS could excessively wear NVME. https://appleinsider.com/articles/21/06/04/apple-resolves-m1-mac-ssd-storage-longevity-issue-in-macos-114-beta https://appleinsider.com/articles/21/06/04/apple-resolves-m1... The issue was subsequently resolved but the consensus was with modern wear leveling this isn't so much a thing. I have a 2021 MacBook Pro with the original drive. I use it heavily for development practically every day and just dumped the SMART data. Model Number: APPLE SSD AP1024R === START OF SMART DATA SECTION === SMART overall-health self-assessment test result: PASSED Available Spare: 100% Available Spare Threshold: 99% As always, YMMV
- RunSet 7mo ago"Uneaten food" <CHOMP CHOMP> "is wasted food." <CHOMP CHOMP>