Ram real time latency - performance

I read somewhere that to find the real latency of the ram you can use the following rule :
1/((RAMspeed/2)/1000) x CL = True Latency in nanoseconds
ie for DDR1 with 400Mhz clock speed, is it logical to me to divide by 2 to get the FSB speed or the real bus speed which is 200Mhz in this case. So the rule above seems to be correct for the DDR1.
From the other side, DDR2 also doubles the freq of the bus in relation to the previous DDR1 generation (ie 4 bits per clock cycle) according the article "What Every Programmer Should Know About Memory".
So, in the case of a DDR2 with a 800Mhz clock speed, to find the "True Latency" the above rule should be accordingly changed to
1/((RAMspeed/4)/1000) x CL = True Latency in nanoseconds
Is that correct? Because in all the cases I read that the correct way is to take RAMspeed/2 no matter if it's DDR, DDR2, DDR3 or DDR4.
Which is the correct way to get the true latency?

The CAS latency is in memory-bus clock cycles. This is always one half the transfers-per-second number. e.g. DDR3-1600 has a memory clock of 800MHz, doing 1600M transfers per second (during a burst transfer).
DDR2, DDR3, and DDR4 still use a double-pumped 64-bit memory bus (transferring data on the rising and falling edges of the clock signal), not quad-pumped. This is why they're still called Double Data-Rate (DDR) SDRAM.
The FSB speed has nothing to do with it.
On old CPUs without integrated memory controllers, i.e. systems that actually have an FSB, its frequency is often configurable (in the BIOS) separately from the memory speed. See Front Side Bus and RAM speed; on even older systems, the FSB and memory clocks were synchronous.
Normally systems were designed with a fast enough FSB to keep up with the memory controller. Running the FSB at the same clock speed as the memory can reduce latency by avoiding buffering between clock domains.
So yes, the CAS latency in seconds is cycle_count / frequency, or more like your formula
1000ns/us * CL / RAMspeed * 2 transfers/clock, where RAMspeed is in mega-transfers per second.
Higher CL numbers at a higher memory frequency often work out to a similar absolute latency (in seconds). In other words, modern RAM has higher CAS latency timing numbers because more clock cycles happen in the same amount of time.
Bandwidth has vastly improved, while latency has stayed nearly constant, according to these graphs from Crucial which explain CL vs. frequency.
Of course this is not "the memory latency", or the "true" memory latency.
It's the CAS latency of the DRAM itself, and is the most important factor in latency between the memory controller and the DRAM, but is only a part of the latency between a CPU core and memory. There is non-negligible latency inside the CPU between the core and uncore (L3 and memory controller). Uncore is Intel terminology; IDK what AMD calls the parts of the memory hierarchy in their various microarchitectures.
Especially many-core Xeon CPUs have significant latency to L3 / memory controller, because of the large ring bus(es) connecting all the cores. A many-core Xeon has worse L3 and memory latency than a similar dual or quad-core with the same memory and CPU clock frequencies.
This extra latency actually limits single-thread / single-core bandwidth on a big Xeon to worse than a laptop CPU, because a single core can't keep enough requests in flight to fill the memory pipeline with that much latency. Why is Skylake so much better than Broadwell-E for single-threaded memory throughput?.

Ok I found the answer.
Everytime the manufacturers increased the memory clock speed they did it in a constant rate which always was the double (2x) of the FSB clock speed. ie
MEM CLK FSB
-------------------
DDR200 100 MHz
DDR266 133 MHz
DDR333 166 MHz
DDR400 200 MHz
DDR2-400 200 MHz
DDR2-533 266 MHz
DDR2-667 333 MHz
DDR2-800 400 MHz
DDR2-1066 533 MHz
DDR3-800 400 MHz
DDR3-1066 533 MHz
DDR3-1333 666 MHz
DDR3-1600 800 MHz
So, the memory module has always the double speed of the FSB.

Related

Cache miss latency in clock cycles

To measure the impact of cache-misses in a program, I want to latency caused by cache-misses to the cycles used for actual computation.
I use perf stat to measure the cycles, L1-loads, L1-misses, LLC-loads and LLC-misses in my program. Here is a example output:
467 769,70 msec task-clock # 1,000 CPUs utilized
1 234 063 672 432 cycles # 2,638 GHz (62,50%)
572 761 379 098 instructions # 0,46 insn per cycle (75,00%)
129 143 035 219 branches # 276,083 M/sec (75,00%)
6 457 141 079 branch-misses # 5,00% of all branches (75,00%)
195 360 583 052 L1-dcache-loads # 417,643 M/sec (75,00%)
33 224 066 301 L1-dcache-load-misses # 17,01% of all L1-dcache hits (75,00%)
20 620 655 322 LLC-loads # 44,083 M/sec (50,00%)
6 030 530 728 LLC-load-misses # 29,25% of all LL-cache hits (50,00%)
Then my question is:
How to convert the number of cache-misses into a number of "lost" clock cycles?
Or alternatively, what is the proportion of time spent for fetching data?
I think the factor should be known by the constructor. My processor is Intel Core i7-10810U, and I couldn't find this information in the specifications nor in this list of benchmarked CPUs.
This related problem describes how to measure the number of cycles lost in a cache-miss, but is there a way to obtain this as hardware information? Ideally, the output would be something like:
L1-hit: 3 cycles
L2-hit: 10 cycles
LLC-hit: 30 cycles
RAM: 300 cycles
Out-of-order exec and memory-level parallelism exist to hide some of that latency by overlapping useful work with time data is in flight. If you simply multiplied L3 miss count by say 300 cycles each, that could exceed the total number of cycles your whole program took. The perf event cycle_activity.stalls_l3_miss (which exists on my Skylake CPU) should count cycles when no uops execute and there's an outstanding L3 cache miss. i.e. cycles when execution is fully stalled. But there will also be cycles with some work, but less than without a cache miss, and that's harder to evaluate.
TL:DR: memory access is heavily pipelined; the whole core doesn't stop on one cache miss, that's the whole point. A pointer-chasing benchmark (to measure latency) is merely a worst case, where the only work is dereferencing a load result. See Modern Microprocessors
A 90-Minute Guide! which has a section about memory and the "memory wall". See also https://agner.org/optimize/ and https://www.realworldtech.com/haswell-cpu/ to learn more about the details of out-of-order exec CPUs and how they can continue doing independent work while one instruction is waiting for data from a cache miss, up to the limit of their out-of-order window size. (https://blog.stuffedcow.net/2013/05/measuring-rob-capacity/)
Re: numbers from vendors:
L3 and RAM latencies aren't a fixed number of core clock cycles: first, core frequency is variable (and independent of uncore and memory clocks), and second because of contention from other cores, and number of hops over the interconnect. (Related: Is cycle count itself reliable on program timing? discusses some effects of core frequency independent of L3 and memory)
That said, Intel's optimization manual does include a table with exact latencies for L1 and L2, and typical for L3, and DRAM on Skylake-server. (2.2.1.3 Skylake Server Microarchitecture Cache Recommendations)
https://software.intel.com/content/www/us/en/develop/articles/intel-sdm.html#optimization - they say SKX L3 latency is typically 50-70 cycles. DRAM speed depends some on the timing of your DIMMs.
Other people have tested specific CPUs, like https://www.7-cpu.com/cpu/Skylake.html.

Why is the CPU slower for calculations then the GPU when only Memory should matter?

A modern CPU has a ethash hashrate from under 1MH/s (source: https://ethereum.stackexchange.com/questions/2325/is-cpu-mining-even-worth-the-ether ) while GPUs mine with over 20MH/s easily. With overclocked memory they reach rates up to 30MH/s.
GPUs have GDDR Memory with Clockrates of about 1000MHz while DDR4 runs with higher clock speeds. Bandwith of DDR4 seems also to be higher (sources: http://www.corsair.com/en-eu/blog/2014/september/ddr3_vs_ddr4_synthetic and https://en.wikipedia.org/wiki/GDDR5_SDRAM )
It is said for Dagger-Hashimoto/ethash bandwith of memory is the thing that matters (also experienced from overclocking GPUs) which I find reasonable since the CPU/GPU only has to do 2x sha3 (1x Keccak256 + 1x Keccak512) operations (source: https://github.com/ethereum/wiki/wiki/Ethash#main-loop ).
A modern Skylake processor can compute over 100M of Keccak512 operations per second (see here: https://www.cryptopp.com/benchmarks.html ) so then core count difference between GPUs and CPUs should not be the problem.
But why don't we get about ~50Mhash/s from 2xKeccak operations and memory loading on a CPU?
See http://www.nvidia.com/object/what-is-gpu-computing.html for an overview of the differences between CPU and GPU programming.
In short, a CPU has a very small number of cores, each of which can do different things, and each of which can handle very complex logic.
A GPU has thousands of cores, that operate pretty much in lockstep, but can only handle simple logic.
Therefore the overall processing throughput of a GPU can be massively higher. But it isn't easy to move logic from the CPU to the GPU.
If you want to dive in deeper and actually write code for both, one good starting place is https://devblogs.nvidia.com/gpu-computing-julia-programming-language/.
"A modern Skylake processor can compute over 100M of Keccak512 operations per second" is incorrect, it is 140 MiB/s. That is MiBs per second and a hash operation is more than 1 byte, you need to divide the 140 MiB/s by the number of bytes being hashed.
I found an article addressing my problem (the influence of Memory on the algorithm).
It's not only the computation problem (mentioned here: https://stackoverflow.com/a/48687460/2298744 ) it's also the Memorybandwidth which would bottelneck the CPU.
As described in the article every round fetches 8kb of data for calculation. This results in the following formular:
(Memory Bandwidth) / ( DAG memory fetched per hash) = Max Theoreticical Hashrate
(Memory Bandwidth) / ( 8kb / hash) = Max Theoreticical Hashrate
For a grafics card like the RX470 mentioned this results in:
(211 Gigabytes / sec) / (8 kilobytes / hash) = ~26Mhashes/sec
While for CPUs with DDR4 this will result in:
(12.8GB / sec) / (8 kilobytes / hash) = ~1.6Mhashes/sec
or (debending on clock speeds of RAM)
(25.6GB / sec) / (8 kilobytes / hash) = ~3.2Mhashes/sec
To sum up, a CPU or also GPU with DDR4 ram could not get more than 3.2MHash/s since it can't get the data fast enough needed for processing.
Source:
https://www.vijaypradeep.com/blog/2017-04-28-ethereums-memory-hardness-explained/

Why can't my ultraportable laptop CPU maintain peak performance in HPC

I have developed a high performance Cholesky factorization routine, which should have peak performance at around 10.5 GFLOPs on a single CPU (without hyperthreading). But there is some phenomenon which I don't understand when I test its performance. In my experiment, I measured the performance with increasing matrix dimension N, from 250 up to 10000.
In my algorithm I have applied caching (with tuned blocking factor), and data are always accessed with unit stride during computation, so cache performance is optimal; TLB and paging problem are eliminated;
I have 8GB available RAM, and the maximum memory footprint during experiment is under 800MB, so no swapping comes across;
During experiment, no resource demanding process like web browser is running at the same time. Only some really cheap background process is running to record CPU frequency as well as CPU temperature data every 2s.
I would expect the performance (in GFLOPs) should maintain at around 10.5 for whatever N I am testing. But a significant performance drop is observed in the middle of the experiment as shown in the first figure.
CPU frequency and CPU temperature are seen in the 2nd and 3rd figure. The experiment finishes in 400s. Temperature was at 51 degree when experiment started, and quickly rose up to 72 degree when CPU got busy. After that it grew slowly to the highest at 78 degree. CPU frequency is basically stable, and it did not drop when temperature got high.
So, my question is:
since CPU frequency did not drop, why performance suffers?
how exactly does temperature affect CPU performance? Does the increment from 72 degree to 78 degree really make things worse?
CPU info
System: Ubuntu 14.04 LTS
Laptop model: Lenovo-YOGA-3-Pro-1370
Processor: Intel Core M-5Y71 CPU # 1.20 GHz * 2
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
Byte Order: Little Endian
CPU(s): 4
On-line CPU(s) list: 0,1
Off-line CPU(s) list: 2,3
Thread(s) per core: 1
Core(s) per socket: 2
Socket(s): 1
NUMA node(s): 1
Vendor ID: GenuineIntel
CPU family: 6
Model: 61
Stepping: 4
CPU MHz: 1474.484
BogoMIPS: 2799.91
Virtualisation: VT-x
L1d cache: 32K
L1i cache: 32K
L2 cache: 256K
L3 cache: 4096K
NUMA node0 CPU(s): 0,1
CPU 0, 1
driver: intel_pstate
CPUs which run at the same hardware frequency: 0, 1
CPUs which need to have their frequency coordinated by software: 0, 1
maximum transition latency: 0.97 ms.
hardware limits: 500 MHz - 2.90 GHz
available cpufreq governors: performance, powersave
current policy: frequency should be within 500 MHz and 2.90 GHz.
The governor "performance" may decide which speed to use
within this range.
current CPU frequency is 1.40 GHz.
boost state support:
Supported: yes
Active: yes
update 1 (control experiment)
In my original experiment, CPU is kept busy working from N = 250 to N = 10000. Many people (primarily those whose saw this post before re-editing) suspected that the overheating of CPU is the major reason for performance hit. Then I went back and installed lm-sensors linux package to track such information, and indeed, CPU temperature rose up.
But to complete the picture, I did another control experiment. This time, I give CPU a cooling time between each N. This is achieved by asking the program to pause for a number of seconds at the start of iteration of the loop through N.
for N between 250 and 2500, the cooling time is 5s;
for N between 2750 and 5000, the cooling time is 20s;
for N between 5250 and 7500, the cooling time is 40s;
finally for N between 7750 and 10000, the cooling time is 60s.
Note that the cooling time is much larger than the time spent for computation. For N = 10000, only 30s are needed for Cholesky factorization at peak performance, but I ask for a 60s cooling time.
This is certainly a very uninteresting setting in high performance computing: we want our machine to work all the time at peak performance, until a very large task is completed. So this kind of halt makes no sense. But it helps to better know the effect of temperature on performance.
This time, we see that peak performance is achieved for all N, just as theory supports! The periodic feature of CPU frequency and temperature is the result of cooling and boost. Temperature still has an increasing trend, simply because as N increases, the work load is getting bigger. This also justifies more cooling time for a sufficient cooling down, as I have done.
The achievement of peak performance seems to rule out all effects other than temperature. But this is really annoying. Basically it says that computer will get tired in HPC, so we can't get expected performance gain. Then what is the point of developing HPC algorithm?
OK, here are the new set of plots:
I don't know why I could not upload the 6th figure. SO simply does not allow me to submit the edit when adding the 6th figure. So I am sorry I can't attach the figure for CPU frequency.
update 2 (how I measure CPU frequency and temperature)
Thanks to Zboson for adding the x86 tag. The following bash commands are what I used for measurement:
while true
do
cat /sys/devices/system/cpu/cpu0/cpufreq/scaling_cur_freq >> cpu0_freq.txt ## parameter "freq0"
cat sys/devices/system/cpu/cpu1/cpufreq/scaling_cur_freq >> cpu1_freq.txt ## parameter "freq1"
sensors | grep "Core 0" >> cpu0_temp.txt ## parameter "temp0"
sensors | grep "Core 1" >> cpu1_temp.txt ## parameter "temp1"
sleep 2
done
Since I did not pin the computation to 1 core, the operating system will alternately use two different cores. It makes more sense to take
freq[i] <- max (freq0[i], freq1[i])
temp[i] <- max (temp0[i], temp1[i])
as the overall measurement.
TL:DR: Your conclusion is correct. Your CPU's sustained performance is nowhere near its peak. This is normal: the peak perf is only available as a short term "bonus" for bursty interactive workloads, above its rated sustained performance, given the light-weight heat-sink, fans, and power-delivery.
You can develop / test on this machine, but benchmarking will be hard. You'll want to run on a cluster, server, or desktop, or at least a gaming / workstation laptop.
From the CPU info you posted, you have a dual-core-with-hyperthreading Intel Core M with a rated sustainable frequency of 1.20 GHz, Broadwell generation. Its max turbo is 2.9GHz, and it's TDP-up sustainable frequency is 1.4GHz (at 6W).
For short bursts, it can run much faster and make much more heat than it requires its cooling system to handle. This is what Intel's "turbo" feature is all about. It lets low-power ultraportable laptops like yours have snappy UI performance in stuff like web browsers, because the CPU load from interactive is almost always bursty.
Desktop/server CPUs (Xeon and i5/i7, but not i3) do still have turbo, but the sustained frequency is much closer to the max turbo. e.g. a Haswell i7-4790k has a sustained "rated" frequency of 4.0GHz. At that frequency and below, it won't use (and convert to heat) more than its rated TDP of 88W. Thus, it needs a cooling system that can handle 88W. When power/current/temperature allow, it can clock up to 4.4GHz and use more than 88W of power. (The sliding window for calculating the power history to keep the sustained power with 88W is sometimes configurable in the BIOS, e.g. 20sec or 5sec. Depending on what code is running, 4.4GHz might not increase the electrical current demand to anywhere near peak. e.g. code with lots of branch mispredicts that's still limited by CPU frequency, but that doesn't come anywhere near saturating the 256b AVX FP units like Prime95 would.)
Your laptop's max turbo is a factor of 2.4x higher than rated frequency. That high-end Haswell desktop CPU can only upclock by 1.1x. The max sustained frequency is already pretty close to the max peak limits, because it's rated to need a good cooling system that can keep up with that kind of heat production. And a solid power supply that can supply that much current.
The purpose of Core M is to have a CPU that can limit itself to ultra low power levels (rated TDP of 4.5 W at 1.2GHz, 6W at 1.4GHz). So the laptop manufacturer can safely design a cooling and power delivery system that's small and light, and only handles that much power. The "Scenario Design Power" is only 3.5W, and that's supposed to represent the thermal requirements for real-world code, not max-power stuff like Prime95.
Even a "normal" ULV laptop CPU is rated for 15W sustained, and high power gaming/workstation laptop CPUs at 45W. And of course laptop vendors put those CPUs into machines with beefier heat-sinks and fans. See a table on wikipedia, and compare desktop / server CPUs (also on the same page).
The achievement of peak performance seems to rule out all effects
other than temperature. But this is really annoying. Basically it says
that computer will get tired in HPC, so we can't get expected
performance gain. Then what is the point of developing HPC algorithm?
The point is to run them on hardware that's not so badly thermally limited! An ultra-low-power CPU like a Core M makes a decent dev platform, but not a good HPC compute platform.
Even a laptop with an xxxxM CPU, rather than a xxxxU CPU, will do ok. (e.g. a "gaming" or "workstation" laptop that's designed to run CPU-intensive stuff for sustained periods). Or in Skylake-family, "xxxxH" or "HK" are the 45W mobile CPUs, at least quad-core.
Further reading:
Modern Microprocessors
A 90-Minute Guide!
[Power Delivery in a Modern Processor] - general background, including the "power wall" that Pentium 4 ran into.
(https://www.realworldtech.com/power-delivery/) - really deep technical dive into CPU / motherboard design and the challenges of delivering stable low-voltage to very bursty demands, and reacting quickly to the CPU requesting more / less voltage as it changes frequency.

Cycles/cost for L1 Cache hit vs. Register on x86?

I remember assuming that an L1 cache hit is 1 cycle (i.e. identical to register access time) in my architecture class, but is that actually true on modern x86 processors?
How many cycles does an L1 cache hit take? How does it compare to register access?
Here's a great article on the subject:
http://arstechnica.com/gadgets/reviews/2002/07/caching.ars/1
To answer your question - yes, a cache hit has approximately the same cost as a register access. And of course a cache miss is quite costly ;)
PS:
The specifics will vary, but this link has some good ballpark figures:
Approximate cost to access various caches and main memory?
Core i7 Xeon 5500 Series Data Source Latency (approximate)
L1 CACHE hit, ~4 cycles
L2 CACHE hit, ~10 cycles
L3 CACHE hit, line unshared ~40 cycles
L3 CACHE hit, shared line in another core ~65 cycles
L3 CACHE hit, modified in another core ~75 cycles remote
L3 CACHE ~100-300 cycles
Local DRAM ~30 ns (~120 cycles)
Remote DRAM ~100 ns
PPS:
These figures represent much older, slower CPUs, but the ratios basically hold:
http://arstechnica.com/gadgets/reviews/2002/07/caching.ars/2
Level Access Time Typical Size Technology Managed By
----- ----------- ------------ --------- -----------
Registers 1-3 ns ?1 KB Custom CMOS Compiler
Level 1 Cache (on-chip) 2-8 ns 8 KB-128 KB SRAM Hardware
Level 2 Cache (off-chip) 5-12 ns 0.5 MB - 8 MB SRAM Hardware
Main Memory 10-60 ns 64 MB - 1 GB DRAM Operating System
Hard Disk 3M - 10M ns 20 - 100 GB Magnetic Operating System/User
Throughput and latency are different things. You can't just add up cycle costs. For throughput, see Load/stores per cycle for recent CPU architecture generations - 2 loads per clock throughput for most modern microarchitectures. And see How can cache be that fast? for microarchitectural details of load/store execution units, including showing load / store buffers which limit how much memory-level parallelism they can track. The rest of this answer will focus only on latency, which is relevant for workloads that involve pointer-chasing (like linked lists and trees), and how much latency out-of-order exec needs to hide. (L3 Cache misses are usually too long to fully hide.)
Single-cycle cache latency used to be a thing on simple in-order pipelines at lower clock speeds (so each cycle was more nanoseconds), especially with simpler caches (smaller, not as associative, and with a smaller TLB for caches that weren't purely virtually addressed.) e.g. the classic 5-stage RISC pipeline like MIPS I assumes 1 cycle for memory access on a cache hit, with address calculation in EX and memory access in a single MEM pipeline stage, before WB.
Modern high-performance CPUs divide the pipeline up into more stages, allowing each cycle to be shorter. This lets simple instructions like add / or / and run really fast, still 1 cycle latency but at high clock speed.
For more details about cycle-counting and out-of-order execution, see Agner Fog's microarch pdf, and other links in the x86 tag wiki.
Intel Haswell's L1 load-use latency is 4 cycles for pointer-chasing, which is typical of modern x86 CPUs. i.e. how fast mov eax, [eax] can run in a loop, with a pointer that points to itself. (Or for a linked list that hits in cache, easy to microbench with a closed loop). See also Is there a penalty when base+offset is in a different page than the base? That 4-cycle latency special case only applies if the pointer comes directly from another load, otherwise it's 5 cycles.
Load-use latency is 1 cycle higher for SSE/AVX vectors in Intel CPUs.
Store-reload latency is 5 cycles, and is unrelated to cache hit or miss (it's store-forwarding, reading from the store buffer for store data that hasn't yet committed to L1d cache).
As harold commented, register access is 0 cycles. So, for example:
inc eax has 1 cycle latency (just the ALU operation)
add dword [mem], 1 has 6 cycle latency until a load from dword [mem] will be ready. (ALU + store-forwarding). e.g. keeping a loop counter in memory limits a loop to one iteration per 6 cycles.
mov rax, [rsi] has 4 cycle latency from rsi being ready to rax being ready on an L1 hit (L1 load-use latency.)
http://www.7-cpu.com/cpu/Haswell.html has a table of latency per cache (which I'll copy here), and some other experimental numbers, including L2-TLB hit latency (on an L1DTLB miss).
Intel i7-4770 (Haswell), 3.4 GHz (Turbo Boost off), 22 nm. RAM: 32 GB (PC3-12800 cl11 cr2).
L1 Data cache = 32 KB, 64 B/line, 8-WAY.
L1 Instruction cache = 32 KB, 64 B/line, 8-WAY.
L2 cache = 256 KB, 64 B/line, 8-WAY
L3 cache = 8 MB, 64 B/line
L1 Data Cache Latency = 4 cycles for simple access via pointer (mov rax, [rax])
L1 Data Cache Latency = 5 cycles for access with complex address calculation (mov rax, [rsi + rax*8]).
L2 Cache Latency = 12 cycles
L3 Cache Latency = 36 cycles
RAM Latency = 36 cycles + 57 ns
The top-level benchmark page is http://www.7-cpu.com/utils.html, but still doesn't really explain what the different test-sizes mean, but the code is available. The test results include Skylake, which is nearly the same as Haswell in this test.
#paulsm4's answer has a table for a multi-socket Nehalem Xeon, including some remote (other-socket) memory / L3 numbers.
If I remember correctly it's about 1-2 clock cycles but this is an estimate and newer caches may be faster. This is out of a Computer Architecture book I have and this is information for AMD so Intel may be slightly different but I would bound it between 5 and 15 clock cycles which seems like a good estimate to me.
EDIT: Whoops L2 is 10 cycles with TAG access, L1 takes 1 to two cycles, my mistake :\
Actually the cost of the L1 cache hit is almost the same as a cost of register access. It was surprising for me, but this is true, at least for my processor (Athlon 64). Some time ago I written a simple test application to benchmark efficiency of access to the shared data in a multiprocessor system. The application body is a simple memory variable incrementing during the predefined period of time. To make a comapison, I benchmarked non-shared variable at first. And during that activity I captured the result, but then during application disassembling I found that compiler was deceived my expectations and apply unwanted optimisation to my code. It just put variable in the CPU register and increment it iterativetly in the register without memory access. But real surprise was achived after I force compliler to use in-memory variable instead of register variable. On updated application I achived almost the same benchmarking results. Performance degradation was really negligeble (~1-2%) and looks like related to some side effect.
As result:
1) I think you can consider L1 cache as an unmanaged processor registers pool.
2) There is no any sence to apply brutal assambly optimization by forcing compiler store frequently accesing data in processor registers. If they are really frequently accessed, they will live in the L1 cache, and due to this will have same access cost as the processor register.

Cache bandwidth per tick for modern CPUs

What is a speed of cache accessing for modern CPUs? How many bytes can be read or written from memory every processor clock tick by Intel P4, Core2, Corei7, AMD?
Please, answer with both theoretical (width of ld/sd unit with its throughput in uOPs/tick) and practical numbers (even memcpy speed tests, or STREAM benchmark), if any.
PS it is question, related to maximal rate of load/store instructions in assembler. There can be theoretical rate of loading (all Instructions Per Tick are widest loads), but processor can give only part of such, a practical limit of loading.
For nehalem: rolfed.com/nehalem/nehalemPaper.pdf
Each core in the architecture has a 128-bit write port and a
128-bit read port to the L1 cache.
128 bit = 16 bytes / clock read
AND
128 bit = 16 bytes / clock write
(can I combine read and write in single cycle?)
The L2 and L3 caches each have a 256-bit port for reading or writing,
but the L3 cache must share its port with three other cores on the chip.
Can L2 and L3 read and write ports be used in single clock?
Each integrated memory controller has a theoretical bandwidth
peak of 32 Gbps.
Latency (clock ticks), some measured by CPU-Z's latencytool or by lmbench's lat_mem_rd - both uses long linked list walk to correctly measure modern out-of-order cores like Intel Core i7
L1 L2 L3, cycles; mem link
Core 2 3 15 -- 66 ns http://www.anandtech.com/show/2542/5
Core i7-xxx 4 11 39 40c+67ns http://www.anandtech.com/show/2542/5
Itanium 1 5-6 12-17 130-1000 (cycles)
Itanium2 2 6-10 20 35c+160ns http://www.7-cpu.com/cpu/Itanium2.html
AMD K8 12 40-70c +64ns http://www.anandtech.com/show/2139/3
Intel P4 2 19 43 200-210 (cycles) http://www.arsc.edu/files/arsc/phys693_lectures/Performance_I_Arch.pdf
AthlonXP 3k 3 20 180 (cycles) --//--
AthlonFX-51 3 13 125 (cycles) --//--
POWER4 4 12-20 ?? hundreds cycles --//--
Haswell 4 11-12 36 36c+57ns http://www.realworldtech.com/haswell-cpu/5/
And good source on latency data is 7cpu web-site, e.g. for Haswell: http://www.7-cpu.com/cpu/Haswell.html
More about lat_mem_rd program is in its man page or here on SO.
Widest read/writes are 128 bit (16 byte) SSE load/store. L1/L2/L3 caches have different bandwidths and latencies and these are of course CPU-specific. Typical L1 latency is 2 - 4 clocks on modern CPUs but you can usually issue 1 or 2 load instructions per clock.
I suspect there's a more specific question lurking here somewhere - what is it that you are actually trying to achieve ? Do you just want to write the fastest possible memcpy ?

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