When I run my WebSocket test, I found the following interesting memory usage results:
Server stated, no connection
[{total,573263528},
{processes,17375688},
{processes_used,17360240},
{system,555887840},
{atom,472297},
{atom_used,451576},
{binary,28944},
{code,3774097},
{ets,271016}]
44 processes,
System:705M,
Erlang Residence:519M
100K Connections
[{total,762564512},
{processes,130105104},
{processes_used,130089656},
{system,632459408},
{atom,476337},
{atom_used,456484},
{binary,50160},
{code,3925064},
{ets,7589160}]
100044 processes,
System: 1814M,
Erlang Residence: 950M
200K Connections
( restart server and create from 0 connection, not continue from case 2)
[{total,952040232},
{processes,243161192},
{processes_used,243139984},
{system,708879040},
{atom,476337},
{atom_used,456484},
{binary,70856},
{code,3925064},
{ets,14904760}]
200044 processes,
System:3383M,
Erlang: 1837M
The figures with "System:" and "Erlang:" are provided htop, others are output of memory() call from erlang shell. Please look at the total and erlang residence memory. When there is no connection, these two are roughly same, with 100K connections, residence memory is a little larger than total, with 200K connections, residence memory is almost double the total.
Can anybody explain?
The most probable answer for your quersion is memory fragmentation.
Allocating OS memory is expensive, so Erlang tries to manage memory for you.
When Erlang allocates memory, it creates an entity called "carrier", which consists of many "blocks". Erlang memory(total) reports the sum of all the block sizes (memory actually used). OS reports the sum of all carriers sizes (sum of memory used and preallocated). Both sum of blocks sizes and carrier sizes can be read from Erlang VM. If (block sizes)/(carrier sizes) << 1, than VM has hard time with freeing the carriers. There might be many big carriers with only couple of blocks used. You can read it with: erlang:system_info({allocator,Type}). but there is an easier way. You can check it using Recon library:
http://ferd.github.io/recon/recon_alloc.html
Firstly check:
recon_alloc:fragmentation(current).
and next:
recon_alloc:fragmentation(max).
This should explain the difference between total memory reported by Erlang VM and OS. If you are sending many small messages over websockets, you can decrease the fragmentation by running Erlang with 2 options:
erl +MBas aobf +MBlmbcs 512
First option will change the block allocation strategy from best fit to address order best fit, which could help squeeze more blocks into first carriers and second one decreases maximum multiblock carrier size, which makes carriers smaller (this should make freeing them easier).
Related
I am writing an algorithm which all blocks are reading a same address. Such as we have a list=[1, 2, 3, 4], and all blocks are reading it and store it to their own shared memory...My test shows the more blocks reading it, the slower it will be...I guess no broadcast happen here? Any idea I can make it faster? Thank you!!!
I learnt from previous post that this can be broadcast in one wrap, seems can not happen in different wrap....(Actually in my case, the threads in one wrap are not reading a same location...)
Once list element is accessed by first warp of a SM unit, the second warp in same SM unit gets it from cache and broadcasts to all simt lanes. But another SM unit's warp may not have it in L1 cache so it fetches from L2 to L1 first.
It is similar in __constant__ memory but it requires same address to be accessed by all threads. Its latency is closer to register access. __constant__ memory is like instruction cache, you get more performance when all threads do same thing.
For example, if you have a Gaussian-filter that iterates over same coefficient-list of filter on all threads, it is better to use constant memory. Using shared memory does not have much advantage as the filter array is not scanned randomly. Shared memory is better when the filter array content is different per block or if it needs random access.
You can also combine constant memory and shared memory. Get half of list from constant memory, then the other half from shared memory. This should let 1024 threads hide latency of one memory type hidden behind the other.
If list is small enough, you can use registers directly (has to be compile-time known indices). But it increases register pressure and may decrease occupancy so be careful about this.
Some old cuda architectures (in case of fma operation) required one operand fetched from constant memory and the other operand from a register to achieve better performance in compute-bottlenecked algorithms.
In a test with 12000 floats as filter to be applied on all threads inputs, shared memory version with 128 threads-per-block completed work in 330 milliseconds while constant-memory version completed in 260 milliseconds and the L1 access performance was the real bottleneck in both versions so the real constant-memory performance is even better, as long as it is similar-index for all threads.
I am trying to use some of the uncore hardware counters, such as: skx_unc_imc0-5::UNC_M_WPQ_INSERTS. It's supposed to count the number of allocations into the Write Pending Queue. The machine has 2 Intel Xeon Gold 5218 CPUs with cascade lake architecture, with 2 memory controllers per CPU. linux version is 5.4.0-3-amd64. I have the following simple loop and I am reading this counter for it. Array elements are 64 byte in size, equal to cache line.
for(int i=0; i < 1000000; i++){
array[i].value=2;
}
For this loop, when I map memory to DRAM NUMA node, the counter gives around 150,000 as a result, which maybe makes sense: There are 6 channels in total for 2 memory controllers in front of this NUMA node, which use DRAM DIMMs in interleaving mode. Then for each channel there is one separate WPQ I believe, so skx_unc_imc0 gets 1/6 from the entire stores. There are skx_unc_imc0-5 counters that I got with papi_native_avail, supposedly each for different channels.
The unexpected result is when instead of mapping to DRAM NUMA node, I map the program to Non-Volatile Memory, which is presented as a separate NUMA node to the same socket. There are 6 NVM DIMMs per-socket, that create one Interleaved Region. So when writing to NVM, there should be similarly 6 different channels used and in front of each, there is same one WPQ, that should get again 1/6 write inserts.
But UNC_M_WPQ_INSERTS returns only around up 1000 as a result on NV memory. I don't understand why; I expected it to give similarly around 150,000 writes in WPQ.
Am I interpreting/understanding something wrong? Or is there two different WPQs per channel depending wether write goes to DRAM or NVM? Or what else could be the explanation?
It turns out that UNC_M_WPQ_INSERTS counts the number of allocations into the Write Pending Queue, only for writes to DRAM.
Intel has added corresponding hardware counter for Persistent Memory: UNC_M_PMM_WPQ_INSERTS which counts write requests allocated in the PMM Write Pending Queue for Intel® Optane™ DC persistent memory.
However there is no such native event showing up in papi_native_avail which means it can't be monitored with PAPI yet. In linux version 5.4, some of the PMM counters can be directly found in perf list uncore such as unc_m_pmm_bandwidth.write - Intel Optane DC persistent memory bandwidth write (MB/sec), derived from unc_m_pmm_wpq_inserts, unit: uncore_imc. This implies that even though UNC_M_PMM_WPQ_INSERTS is not directly listed in perf list as an event, it should exist on the machine.
As described here the EventCode for this counter is: 0xE7, therefore it can be used with perf as a raw hardware event descriptor as following: perf stat -e uncore_imc/event=0xe7/. However, it seems that it does not support event modifiers to specify user-space counting with perf. Then after pinning the thread in the same socket as the NVM NUMA node, for the program that basically only does the loop described in the question, the result of perf kind of makes sense:
Performance counter stats for 'system wide': 1,035,380 uncore_imc/event=0xe7/
So far this seems to be the the best guess.
From time to time I face with the concepts like zero garbage or efficient use of memory etc. As an example in the section Features of well-known package httprouter you can see the following:
Zero Garbage: The matching and dispatching process generates zero bytes of garbage. In fact, the only heap allocations that are made, is by building the slice of the key-value pairs for path parameters. If the request path contains no parameters, not a single heap allocation is necessary.
Also this package shows very good benchmark results compared to standard library's http.ServeMux:
BenchmarkHttpServeMux 5000 706222 ns/op 96 B/op 6 allocs/op
BenchmarkHttpRouter 100000 15010 ns/op 0 B/op 0 allocs/op
As far as I understand the second one has (from the table) no heap memory allocation and zero average number of allocations made per repetition.
The question: I want to learn a basic understanding of memory management. When garbage collector allocates/deallocates memory. What does the benchmark numbers means (the last two columns of the table) and how people know when heap is allocating?
I'm absolutely new in memory management, so it's really difficult to understand what's going on "under the hood". The articles I've read:
https://golang.org/ref/mem
https://golang.org/doc/effective_go.html
http://gribblelab.org/CBootcamp/7_Memory_Stack_vs_Heap.html
http://en.wikipedia.org/wiki/Garbage_collection_(computer_science)
The garbage collector doesn't allocate memory :-), it just deallocates. Go's garbage collector is evolving, for the details have a look at the design document https://docs.google.com/document/d/16Y4IsnNRCN43Mx0NZc5YXZLovrHvvLhK_h0KN8woTO4/preview?sle=true and follow the discussion on the golang mailing lists.
The last two columns in the benchmark output are dead simple: How many bytes have been allocated in total and how many allocations have happened during one iteration of the benchmark code. (This allocation is done by your code, not by the garbage collector). As any allocation is a potential creation of garbage reducing these numbers might be a design goal.
When are things allocated on the heap? Whenever the Go compiler decides to! The compiler tries to allocate on the stack, but sometimes it must use the heap, especially if a value escapes from the local stack-bases scopes. This escape analysis is currently undergoing rework, so it is not easy to tell which value will be heap- or stack-allocated, especially as this is changing from compiler version to version.
I wouldn't be too obsessed with avoiding allocations until your benchmarking show too much GC overhead.
I am implementing a spiking neural network using the CUDA library and am really unsure of how to proceed with regard to the following things:
Allocating memory (cudaMalloc) to many different arrays. Up until now, simply using cudaMalloc 'by hand' has sufficed, as I have not had to make more than 10 or so arrays. However, I now need to make pointers to, and allocate memory for thousands of arrays.
How to decide how much memory to allocate to each of those arrays. The arrays have a height of 3 (1 row for the postsynaptic neuron ids, 1 row for the number of the synapse on the postsynaptic neuron, and 1 row for the efficacy of that synapse), but they have an undetermined length which changes over time with the number of outgoing synapses.
I have heard that dynamic memory allocation in CUDA is very slow and so toyed with the idea of allocating the maximum memory required for each array, however the number of outgoing synapses per neuron varies from 100-10,000 and so I thought this was infeasible, since I have on the order of 1000 neurons.
If anyone could advise me on how to allocate memory to many arrays on the GPU, and/or how to code a fast dynamic memory allocation for the above tasks I would have more than greatly appreciative.
Thanks in advance!
If you really want to do this, you can call cudaMalloc as many times as you want; however, it's probably not a good idea. Instead, try to figure out how to lay out the memory so that neighboring threads in a block will access neighboring elements of RAM whenever possible.
The reason this is likely to be problematic is that threads execute in groups of 32 at a time (a warp). NVidia's memory controller is quite smart, so if neighboring threads ask for neighboring bytes of RAM, it coalesces those loads into a single request that can be efficiently executed. In contrast, if each thread in a warp is accessing a random memory location, the entire warp must wait till 32 memory requests are completed. Furthermore, reads and writes to the card's memory happen a whole cache line at a time, so if the threads don't use all the RAM that was read before it gets evicted from the cache, memory bandwidth is wasted. If you don't optimize for coherent memory access within thread blocks, expect a 10x to 100x slowdown.
(side note: The above discussion is still applicable with post-G80 cards; the first generation of CUDA hardware (G80) was even pickier. It also required aligned memory requests if the programmer wanted the coalescing behavior.)
Admittedly I don't get it. Say you have a memory with a memory word of length of 1 byte. Why can't you access a 4 byte long variable in a single memory access on an unaligned address(i.e. not divisible by 4), as it's the case with aligned addresses?
The memory subsystem on a modern processor is restricted to accessing memory at the granularity and alignment of its word size; this is the case for a number of reasons.
Speed
Modern processors have multiple levels of cache memory that data must be pulled through; supporting single-byte reads would make the memory subsystem throughput tightly bound to the execution unit throughput (aka cpu-bound); this is all reminiscent of how PIO mode was surpassed by DMA for many of the same reasons in hard drives.
The CPU always reads at its word size (4 bytes on a 32-bit processor), so when you do an unaligned address access — on a processor that supports it — the processor is going to read multiple words. The CPU will read each word of memory that your requested address straddles. This causes an amplification of up to 2X the number of memory transactions required to access the requested data.
Because of this, it can very easily be slower to read two bytes than four. For example, say you have a struct in memory that looks like this:
struct mystruct {
char c; // one byte
int i; // four bytes
short s; // two bytes
}
On a 32-bit processor it would most likely be aligned like shown here:
The processor can read each of these members in one transaction.
Say you had a packed version of the struct, maybe from the network where it was packed for transmission efficiency; it might look something like this:
Reading the first byte is going to be the same.
When you ask the processor to give you 16 bits from 0x0005 it will have to read a word from 0x0004 and shift left 1 byte to place it in a 16-bit register; some extra work, but most can handle that in one cycle.
When you ask for 32 bits from 0x0001 you'll get a 2X amplification. The processor will read from 0x0000 into the result register and shift left 1 byte, then read again from 0x0004 into a temporary register, shift right 3 bytes, then OR it with the result register.
Range
For any given address space, if the architecture can assume that the 2 LSBs are always 0 (e.g., 32-bit machines) then it can access 4 times more memory (the 2 saved bits can represent 4 distinct states), or the same amount of memory with 2 bits for something like flags. Taking the 2 LSBs off of an address would give you a 4-byte alignment; also referred to as a stride of 4 bytes. Each time an address is incremented it is effectively incrementing bit 2, not bit 0, i.e., the last 2 bits will always continue to be 00.
This can even affect the physical design of the system. If the address bus needs 2 fewer bits, there can be 2 fewer pins on the CPU, and 2 fewer traces on the circuit board.
Atomicity
The CPU can operate on an aligned word of memory atomically, meaning that no other instruction can interrupt that operation. This is critical to the correct operation of many lock-free data structures and other concurrency paradigms.
Conclusion
The memory system of a processor is quite a bit more complex and involved than described here; a discussion on how an x86 processor actually addresses memory can help (many processors work similarly).
There are many more benefits to adhering to memory alignment that you can read at this IBM article.
A computer's primary use is to transform data. Modern memory architectures and technologies have been optimized over decades to facilitate getting more data, in, out, and between more and faster execution units–in a highly reliable way.
Bonus: Caches
Another alignment-for-performance that I alluded to previously is alignment on cache lines which are (for example, on some CPUs) 64B.
For more info on how much performance can be gained by leveraging caches, take a look at Gallery of Processor Cache Effects; from this question on cache-line sizes
Understanding of cache lines can be important for certain types of program optimizations. For example, the alignment of data may determine whether an operation touches one or two cache lines. As we saw in the example above, this can easily mean that in the misaligned case, the operation will be twice slower.
It's a limitation of many underlying processors. It can usually be worked around by doing 4 inefficient single byte fetches rather than one efficient word fetch, but many language specifiers decided it would be easier just to outlaw them and force everything to be aligned.
There is much more information in this link that the OP discovered.
you can with some processors (the nehalem can do this), but previously all memory access was aligned on a 64-bit (or 32-bit) line, because the bus is 64 bits wide, you had to fetch 64 bit at a time, and it was significantly easier to fetch these in aligned 'chunks' of 64 bits.
So, if you wanted to get a single byte, you fetched the 64-bit chunk and then masked off the bits you didn't want. Easy and fast if your byte was at the right end, but if it was in the middle of that 64-bit chunk, you'd have to mask off the unwanted bits and then shift the data over to the right place. Worse, if you wanted a 2 byte variable, but that was split across 2 chunks, then that required double the required memory accesses.
So, as everyone thinks memory is cheap, they just made the compiler align the data on the processor's chunk sizes so your code runs faster and more efficiently at the cost of wasted memory.
Fundamentally, the reason is because the memory bus has some specific length that is much, much smaller than the memory size.
So, the CPU reads out of the on-chip L1 cache, which is often 32KB these days. But the memory bus that connects the L1 cache to the CPU will have the vastly smaller width of the cache line size. This will be on the order of 128 bits.
So:
262,144 bits - size of memory
128 bits - size of bus
Misaligned accesses will occasionally overlap two cache lines, and this will require an entirely new cache read in order to obtain the data. It might even miss all the way out to the DRAM.
Furthermore, some part of the CPU will have to stand on its head to put together a single object out of these two different cache lines which each have a piece of the data. On one line, it will be in the very high order bits, in the other, the very low order bits.
There will be dedicated hardware fully integrated into the pipeline that handles moving aligned objects onto the necessary bits of the CPU data bus, but such hardware may be lacking for misaligned objects, because it probably makes more sense to use those transistors for speeding up correctly optimized programs.
In any case, the second memory read that is sometimes necessary would slow down the pipeline no matter how much special-purpose hardware was (hypothetically and foolishly) dedicated to patching up misaligned memory operations.
#joshperry has given an excellent answer to this question. In addition to his answer, I have some numbers that show graphically the effects which were described, especially the 2X amplification. Here's a link to a Google spreadsheet showing what the effect of different word alignments look like.
In addition here's a link to a Github gist with the code for the test.
The test code is adapted from the article written by Jonathan Rentzsch which #joshperry referenced. The tests were run on a Macbook Pro with a quad-core 2.8 GHz Intel Core i7 64-bit processor and 16GB of RAM.
If you have a 32bit data bus, the address bus address lines connected to the memory will start from A2, so only 32bit aligned addresses can be accessed in a single bus cycle.
So if a word spans an address alignment boundary - i.e. A0 for 16/32 bit data or A1 for 32 bit data are not zero, two bus cycles are required to obtain the data.
Some architectures/instruction sets do not support unaligned access and will generate an exception on such attempts, so compiler generated unaligned access code requires not just additional bus cycles, but additional instructions, making it even less efficient.
If a system with byte-addressable memory has a 32-bit-wide memory bus, that means there are effectively four byte-wide memory systems which are all wired to read or write the same address. An aligned 32-bit read will require information stored in the same address in all four memory systems, so all systems can supply data simultaneously. An unaligned 32-bit read would require some memory systems to return data from one address, and some to return data from the next higher address. Although there are some memory systems that are optimized to be able to fulfill such requests (in addition to their address, they effectively have a "plus one" signal which causes them to use an address one higher than specified) such a feature adds considerable cost and complexity to a memory system; most commodity memory systems simply cannot return portions of different 32-bit words at the same time.
On PowerPC you can load an integer from an odd address with no problems.
Sparc and I86 and (I think) Itatnium raise hardware exceptions when you try this.
One 32 bit load vs four 8 bit loads isnt going to make a lot of difference on most modern processors. Whether the data is already in cache or not will have a far greater effect.