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In a particular scenario I found that a code has taken 20 CPU Years and 4 real Months time. My goal is to approximate the amount of processing power utilized considering the fact that all the processors were on 100% usage all the time. So, my approach is as follows,
20 CPU Years = 20 * 365 * 24 CPU Hours = 175,200 CPU Hours.
Now, 1 CPU Year means 1 GFLOP machine working for 1 real Hour. Which means, in this case, the work done is, 1 GFLOP machine working for 175,200 real Hours. But in reality it took 4 * 30 * 24 = 2,880 real hours. So, approximately 175,200/2,880 =(approx.) 61 GLFOP machine.
My question is am I doing the approximation correctly or misunderstanding some particular term as per the calculations given above ? Or I am mixing GFLOPS and GFLOP together ?
Definitions
My question is am I doing the approximation correctly or misunderstanding some particular term as per the calculations given above ?
"100% usage" may mean the CPU spent 20% of its time doing nothing waiting for data to be transferred to/from RAM (and/or branch mispredictions or other stalls), 10% of its time running faster than normal because other CPUs where actually doing nothing, and 15% of its time running slower than normal for power/temperature management reasons; and (depending on where you got that "100% usage" statistic) "100% usage" may be significantly more confusing (e.g. http://www.brendangregg.com/blog/2017-08-08/linux-load-averages.html ).
Depending on context; GFLOPS is either "theoretical maximum under perfect conditions that will never occur in practice" (worthless marketing hype); or a direct measurement of a specific case that ignores most of the work a CPU did (everything involving integers, all control flow, all data transfer, all memory management, ...)
In a particular scenario I found that a code has taken 20 CPU Years and 4 real Months time. My goal is to approximate the amount of processing power utilized.
From this; you might (or might not) be able to say "most of the work that CPUs did was discarded due to lockless algorithm retries and/or transactions that couldn't be committed; and (partly because the bottleneck was RAM bandwidth and partly because of the way SMT works on this system) it would have been 4 times as fast if half as many CPUs were used."
TL;DR: Approximating processor power is just an inconvenient way to obfuscate the (more useful) information that you started with (e.g. that a specific piece of code running on a specific piece of hardware that was working on a specific piece of data happened to take 4 months of real time).
Your Calculation:
Yes; you're mixing GFLOP and GFLOPS (e.g. GFLOPS = GFLOP per second; and a "1 GFLOP machine" is a computer that can do a billion floating point operations in an infinite amount of time, which is every computer), and the web page you linked to is making the same mistake (e.g. saying "a 1 GFLOP reference machine" when it should be saying "a 1 GFLOPS reference machine").
Note that there's no need to care about GFLOPS or GFLOP for the calculation you're doing: If something was supposed to take 20 "reference CPU years" and actually took 4 months (or 4/12 years); then you'd say that your hardware is equivalent to "20 / (4/12) = 60 reference CPUs". Of course this is horribly silly and it'd make more sense to say that your hardware happened to achieve 60 GFLOPS without bothering with the misleading "reference CPU" nonsense.
When I try to optimize my code, for a very long time I've just been using a rule of thumb that addition and subtraction are worth 1, multiplication and division are worth 3, squaring is worth 3 (I rarely use the more general pow function so I have no rule of thumb for it), and square roots are worth 10. (And I assume squaring a number is just a multiplication, so worth 3.)
Here's an example from a 2D orbital simulation. To calculate and apply acceleration from gravity, first I get distance from the ship to the center of earth, then calculate the acceleration.
D = sqrt( sqr(Ship.x - Earth.x) + sqr(Ship.y - Earth.y) ); // this is worth 19
A = G*Earth.mass/sqr(D); // this is worth 9, total is 28
However, notice that in calculating D, you take a square root, but when using it in the next calculation, you square it. Therefore you can just do this:
A = G*Earth.mass/( sqr(Ship.x - Earth.x) + sqr(Ship.y - Earth.y) ); // this is worth 15
So if my rule of thumb is true, I almost cut in half the cycle time.
However, I cannot even remember where I heard that rule before. I'd like to ask what is the actual cycle times for those basic arithmetic operations?
Assumptions:
everything is a 64-bit floating number in x64 architecture.
everything is already loaded into registers, so no worrying about hits and misses from caches or memory.
no interrupts to the CPU
no if/branching logic such as look ahead prediction
Edit: I suppose what I'm really trying to do is look inside the ALU and only count the cycle time of its logic for the 6 operations. If there is still variance within that, please explain what and why.
Note: I did not see any tags for machine code, so I chose the next closest thing, assembly. To be clear, I am talking about actual machine code operations in x64 architecture. Thus it doesn't matter whether those lines of code I wrote are in C#, C, Javascript, whatever. I'm sure each high-level language will have its own varying times so I don't wanna get into an argument over that. I think it's a shame that there's no machine code tag because when talking about performance and/or operation, you really need to get down into it.
At a minimum, one must understand that an operation has at least two interesting timings: the latency and the throughput.
Latency
The latency is how long any particular operation takes, from its inputs to its output. If you had a long series of operations where the output of one operation is fed into the input of the next, the latency would determine the total time. For example, an integer multiplication on most recent x86 hardware has a latency of 3 cycles: it takes 3 cycles to complete a single multiplication operation. Integer addition has a latency of 1 cycle: the result is available the cycle after the addition executes. Latencies are generally positive integers.
Throughput
The throughput is the number of independent operations that can be performed per unit time. Since CPUs are pipelined and superscalar, this is often more than the inverse of the latency. For example, on most recent x86 chips, 4 integer addition operations can execute per cycle, even though the latency is 1 cycle. Similarly, 1 integer multiplication can execute, on average per cycle, even though any particular multiplication takes 3 cycles to complete (meaning that you must have multiple independent multiplications in progress at once to achieve this).
Inverse Throughput
When discussing instruction performance, it is common to give throughput numbers as "inverse throughput", which is simply 1 / throughput. This makes it easy to directly compare with latency figures without doing a division in your head. For example, the inverse throughput of addition is 0.25 cycles, versus a latency of 1 cycle, so you can immediately see that you if you have sufficient independent additions, they use only something like 0.25 cycles each.
Below I'll use inverse throughput.
Variable Timings
Most simple instructions have fixed timings, at least in their reg-reg form. Some more complex mathematical operations, however, may have input-dependent timings. For example, addition, subtraction and multiplication usually have fixed timings in their integer and floating point forms, but on many platforms division has variable timings in integer, floating point or both. Agner's numbers often show a range to indicate this, but you shouldn't assume the operand space has been tested extensively, especially for floating point.
The Skylake numbers below, for example, show a small range, but it isn't clear if that's due to operand dependency (which would likely be larger) or something else.
Passing denormal inputs, or results that themselves are denormal may incur significant additional cost depending on the denormal mode. The numbers you'll see in the guides generally assume no denormals, but you might be able to find a discussion of denormal costs per operation elsewhere.
More Details
The above is necessary but often not sufficient information to fully qualify performance, since you have other factors to consider such as execution port contention, front-end bottlenecks, and so on. It's enough to start though and you are only asking for "rule of thumb" numbers if I understand it correctly.
Agner Fog
My recommended source for measured latency and inverse throughput numbers are Agner's Fogs guides. You want the files under 4. Instruction tables: Lists of instruction latencies, throughputs and micro-operation breakdowns for Intel, AMD and VIA CPUs, which lists fairly exhaustive timings on a huge variety of AMD and Intel CPUs. You can also get the numbers for some CPUs directly from Intel's guides, but I find them less complete and more difficult to use than Agner's.
Below I'll pull out the numbers for a couple of modern CPUs, for the basic operations you are interested in.
Intel Skylake
Lat Inv Tpt
add/sub (addsd, subsd) 4 0.5
multiply (mulsd) 4 0.5
divide (divsd) 13-14 4
sqrt (sqrtpd) 15-16 4-6
So a "rule of thumb" for latency would be add/sub/mul all cost 1, and division and sqrt are about 3 and 4, respectively. For throughput, the rule would be 1, 8, 8-12 respectively. Note also that the latency is much larger than the inverse throughput, especially for add, sub and mul: you'd need 8 parallel chains of operations if you wanted to hit the max throughput.
AMD Ryzen
Lat Inv Tpt
add/sub (addsd, subsd) 3 0.5
multiply (mulsd) 4 0.5
divide (divsd) 8-13 4-5
sqrt (sqrtpd) 14-15 4-8
The Ryzen numbers are broadly similar to recent Intel. Addition and subtraction are slightly lower latency, multiplication is the same. Latency-wise, the rule of thumb could still generally be summarized as 1/3/4 for add,sub,mul/div/sqrt, with some loss of precision.
Here, the latency range for divide is fairly large, so I expect it is data dependent.
Hi I have a question regarding inherent parallelism.
Let's say we have a sequential program which takes 20 seconds to complete execution. Suppose the execution time consists of 2 seconds of setup time at the beginning and 2 seconds of finalization time at the end of the execution, and the remaining work can be parallelized. How do we calculate the inherent parallelism of this program?
How do you define "inherent parallelism"? I've not heard the term. We can talk about "possible speedup".
OP said "remaining work can be parallelized"... to what degree?
Can it run with infinite parallelism? If this were possible (it isn't practical), then the total runtime would be 4 seconds with a speedup of 20/4 --> 5.
If the remaining work can be run on N processors perfectly in parallel,
then the total runtime would be 4+16/N. The ratio of that to 20 seconds is 20/(4+16/N) which can have pretty much any degree of speedup from 1 (no speedup) to 5 (he the limit case) depending on the value of N.
Suppose I have a program that has an instruction to add two numbers and that operation takes 10 nanoseconds(constant, as enforced by the gate manufactures).
Now I have 3 different processors A, B and C(where A< B < C in terms of clock cycles). A's one clock cycle has 15 nanosec, B has 10 nanosec and C has 7 nanosec.
Firstly am I correct on my following assumptions-
1. Add operation takes 1 complete cycle of processor A(slow processor) and wastes rest of 5 ns of the cycle.
2. Add operation takes 1 complete cycle of processor B wasting no time.
3. Add operation takes 2 complete cycles(20 ns) of processor C(fast processor) wasting rest of the 20-14=7 ns.
If the above assumptions are correct then isn't this a contradiction to the regular assumption that processors with high clock cycles are faster. Here processor C which is the fastest actually takes 2 cycles and wastes 7ns whereas, the slower processor A takes just 1 cycle.
Processor C is fastest, no matter what. It takes 7 ns per cycle and therefore performs more cycles than A and B. It's not C's fault that the circuit is not fast enough. If you would implement the addition circuit in a way that it gives result in 1 ns, all processors will give the answer in 1 clock cycle (i.e. C will give you the answer in 7ns, B in 10ns and A in 15ns).
Firstly am I correct on my following assumptions-
1. Add operation takes 1 complete cycle of processor A(slow processor) and wastes rest of 5 ns of the cycle.
2. Add operation takes 1 complete cycle of processor B wasting no time.
3. Add operation takes 2 complete cycles(20 ns) of processor C(fast processor) wasting rest of the 20-7=13 ns.
No. It is because you are using incomplete data to express the time for an operation. Measure the time taken to finish an operation on a particular processor in clock cycles instead of nanoseconds as you are doing here. When you say ADD op takes 10 ns and you do not mention the processor on which you measured the time for the ADD op, the time measurement in ns is meaningless.
So when you say that ADD op takes 2 clock cycles on all three processors, then you have standardized the measurement. A standardized measurement can then be translated as:
Time taken by A for addition = 2 clock cycles * 15 ns per cycle = 30 ns
Time taken by B for addition = 2 clock cycles * 10 ns per cycle = 20 ns
Time taken by C for addition = 2 clock cycles * 07 ns per cycle = 14 ns
In case you haven't noticed, when you say:
A's one clock cycle has 15 nanosec, B has 10 nanosec and C has 7 nanosec.
which of the three processors is fastest?
Answer: C is fastest. It's one cycle is finished in 7ns. It implies that it finishes 109/7 (~= 1.4 * 108) cycles in one second, compared to B which finishes 109/10 (= 108) cycles in one second, compared to A which finishes only 109/15 (~= 0.6 * 108) cycles in one second.
What does a ADD instruction mean, does it purely mean only and only ADD(with operands available at the registers) or does it mean getting
the operands, decoding the instruction and then actually adding the
numbers.
Getting the operands is done by MOV op. If you are trying to compare how fast ADD op is being done, it should be compared by time to perform ADD op only. If you, on the other hand want to find out how fast addition of two numbers is being done, then it will involve more operations than simple ADD. However, if it's helpful, the list of all Original 8086/8088 instructions is available on Wikipedia too.
Based on the above context to what add actually means, how many cycles does add take, one or more than one.
It will depend on the processor because each processor may have the adder differently implemented. There are many ways to generate addition of two numbers. Quoting Wikipedia again - A full adder can be implemented in many different ways such as with a custom transistor-level circuit or composed of other gates.
Also, there may be pipelining in the instructions which can result in parallelizing of the addition of the numbers resulting in huge time savings.
Why is clock cycle a standard since it can vary with processor to processor. Shouldn't nanosec be the standard. Atleast its fixed.
Clock cycle along with the processor speed can be the standard if you want to tell the time taken by a processor to execute an instruction. Pick any two from:
Time to execute an instruction,
Processor Speed, and
Clock cycles needed for an instruction.
The third can be derived from it.
When you say the clock cycles taken by ADD is x and you know the processor speed is y MHz, you can calculate that the time to ADD is x / y. Also, you can mention the time to perform ADD as z ns and you know the processor speed is same y MHz as earlier, you can calculate the cycles needed to execute ADD as y * z.
I'm no expert BUT I'd say ...
the regular assumption that processors with high clock cycles are faster FOR THE VAST MAJORITY OF OPERATIONS
For example, a more intelligent processor might perform an "overhead task" that takes X ns. The "overhead task" might make it faster for repetitive operations but might actually cause it to take longer for a one-off operation such as adding 2 numbers.
Now, if the same processor performed that same operation 1 million times, it should be massively faster than the slower less intelligent processor.
Hope my thinking helps. Your feedback on my thoughts welcome.
Why would a faster processor take more cycles to do the same operation than a slower one?
Even more important: modern processors use Instruction pipelining, thus executing multiple operations in one clock cycle.
Also, I don't understand what you mean by 'wasting 5ns', the frequency determines the clock speed, thus the time it takes to execute 1 clock. Of course, cpu's can have to wait on I/O for example, but that holds for all cpu's.
Another important aspect of modern cpu's are the L1, L2 and L3 caches and the architecture of those caches in multicore systems. For example: if a register access takes 1 time unit, a L1 cache access will take around 2 while a normal memory access will take between 50 and 100 (and a harddisk access would take thousands..).
This is actually almost correct, except that on processor B taking 2 cycles means 14ns, so with 10ns being enough the next cycle starts 4ns after the result was already "stable" (though it is likely that you need some extra time if you chop it up, to latch the partial result). It's not that much of a contradiction, setting your frequency "too high" can require trade-offs like that. An other thing you might do it use more a different circuit or domino logic to get the actual latency of addition down to one cycle again. More likely, you wouldn't set addition at 2 cycles to begin with. It doesn't work out so well in this case, at least not for addition. You could do it, and yes, basically you will have to "round up" the time a circuit takes to an integer number of cycles. You can also see this in bitwise operations, which take less time than addition but nevertheless take a whole cycle. On machine C you could probably still fit bitwise operations in a single cycle, for some workloads it might even be worth splitting addition like that.
FWIW, Netburst (Pentium 4) had staggered adders, which computed the lower half in one "half-cycle" and the upper half in the next (and the flags in the third half cycle, in some sense giving the whole addition a latency of 1.5). It's not completely out of this world, though Netburst was over all, fairly mad - it had to do a lot of weird things to get the frequency up that high. But those half-cycles aren't very half (it wasn't, AFAIK, logic that advanced on every flank, it just used a clock multiplier), you could also see them as the real cycles that are just very fast, with most of the rest of the logic (except that crazy ALU) running at half speed.
Your broad point that 'a CPU will occasionally waste clock cycles' is valid. But overall in the real world, part of what makes a good CPU a good CPU is how it alleviates this problem.
Modern CPUs consist of a number of different components, none of whose operations will end up taking a constant time in practice. For example, an ADD instruction might 'burst' at 1 instruction per clock cycle if the data is immediately available to it... which in turn means something like 'if the CPU subcomponents required to fetch that data were immediately available prior to the instruction'. So depending on if e.g. another subcomponent had to wait for a cache fetch, the ADD may in practice take 2 or 3 cycles, say. A good CPU will attempt to re-order the incoming stream of instructions to maximise the availability of subcomponents at the right time.
So you could well have the situation where a particular series of instructions is 'suboptimal' on one processor compared to another. And the overall performance of a processor is certainly not just about raw clock speed: it is as much about the clever logic that goes around taking a stream of incoming instructions and working out which parts of which instructions to fire off to which subcomponents of the chip when.
But... I would posit that any modern chip contains such logic. Both a 2GHz and a 3GHz processor will regularly "waste" clock cycles because (to put it simply) a "fast" instruction executed on one subcomponent of the CPU has to wait for the result of the output from another "slower" subcomponent. But overall, you will still expect the 3GHz processor to "execute real code faster".
First, if the 10ns time to perform the addition does not include the pipeline overhead (clock skew and latch delay), then Processor B cannot complete an addition (with these overheads) in one 10ns clock cycle, but Processor A can and Processor C can still probably do it in two cycles.
Second, if the addition itself is pipelined (or other functional units are available), then a subsequent non-dependent operation can begin executing in the next cycle. (If the addition was width-pipelined/staggered (as mentioned in harold's answer) then even dependent additions, logical operations and left shifts could be started after only one cycle. However, if the exercise is constraining addition timing, it presumably also prohibits other optimizations to simplify the exercise.) If dependent operations are not especially common, then the faster clock of Processor C will result in higher performance. (E.g., if a dependence stall occurred every fourth cycle, then, ignoring other effects, Processor C can complete four instructions every five 7ns cycles (35 ns; the first three instruction overlap in execution) compared to 40ns for Processor B (assuming the add timing included pipelining overhead).) (Note: Your assumption 3 is incorrect, two cycles for Processor C would be 14ns.)
Third, the extra time in a clock cycle can be used to support more complex operations (e.g., preshifting one operand by a small immediate value and even adding three numbers — a carry-save adder has relatively little delay), to steal work from other pipeline stages (potentially reducing the number of pipeline stages, which generally reduces branch misprediction penalties), or to reduce area or power by using simpler logic. In addition, the extra time might be used to support a larger (or more associative) cache with fixed latency in cycles, reducing miss rates. Such factors can compensate for the "waste" of 5ns in Processor A.
Even for scalar (single issue per cycle) pipelines clock speed is not the single determinant of performance. Design choices become even more complex when power, manufacturing cost (related to yield, adjusted according to sellable bins, and area), time-to-market (and its variability/predictability), workload diversity, and more advanced architectural and microarchitectural techniques are considered.
The incorrect assumption that clock frequency determines performance even has a name: the Megahertz myth.
simple problem from Wilkinson and Allen's Parallel Programming: Techniques and Applications Using Networked Workstations and Parallel Computers. Working through the exercises at the end of the first chapter and want to make sure that I'm on the right track. The full question is:
1-11 A multiprocessor consists of 10 processors, each capable of a peak execution rate of 200 MFLOPs (millions of floating point operations per second). What is the performance of the system as measured in MFLOPs when 10% of the code is sequential and 90% is parallelizable?
I assume the question wants me to find the number of operations per second of a serial processor which would take the same amount of time to run the program as the multiprocessor.
I think I'm right in thinking that 10% of the program is run at 200 MFLOPs, and 90% is run at 2,000 MFLOPs, and that I can average these speeds to find the performance of the multiprocessor in MFLOPs:
1/10 * 200 + 9/10 * 2000 = 1820 MFLOPs
So when running a program which is 10% serial and 90% parallelizable the performance of the multiprocessor is 1820 MFLOPs.
Is my approach correct?
ps: I understand that this isn't exactly how this would work in reality because it's far more complex, but I would like to know if I'm grasping the concepts.
Your calculation would be fine if 90% of the time, all 10 processors were fully utilized, and 10% of the time, just 1 processor was in use. However, I don't think that is a reasonable interpretation of the problem. I think it is more reasonable to assume that if a single processor were used, 10% of its computations would be on the sequential part, and 90% of its computations would be on the parallelizable part.
One possibility is that the sequential part and parallelizable parts can be run in parallel. Then one processor could run the sequential part, and the other 9 processors could do the parallelizable part. All processors would be fully used, and the result would be 2000 MFLOPS.
Another possibility is that the sequential part needs to be run first, and then the parallelizable part. If a single processor needed 1 hour to do the first part, and 9 hours to do the second, then it would take 10 processors 1 + 0.9 = 1.9 hours total, for an average of about (1*200 + 0.9*2000)/1.9 ~ 1053 MFLOPS.