I would like to determine the theoretical number of FLOPs (Floating Point Operations) that my computer can do. Can someone please help me with this. (I would like to compare my computer to some supercomputers just to get an idea of the difference between them)


4 Answers 4


The theoretical peak FLOP/s is given by: $$ \text{Number of Cores} * \text{Average frequency} * \text{Operations per cycle} $$ The number of cores is easy. Average frequency should, in theory, factor in some amount of Turbo Boost (Intel) or Turbo Core (AMD), but the operating frequency is a good lower bound. The operations per cycle is architecture-dependent and can be hard to find (8 for SandyBridge and IvyBridge, see slide 26). It is the subject of this stack overflow question, which includes numbers for a bunch of modern architectures.

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    $\begingroup$ Ok, I have 2 cores, Operating Frequency: 1.8 GHz, Intel Turbo Boost Technology: 3.00 Ghz, I can't find the operations per cycle, here is the website: ark.intel.com/products/75460/… thanks $\endgroup$ Commented Apr 10, 2014 at 2:15
  • $\begingroup$ Haswell can do 16 DP / cycle. I just added a link in the answer body to an SO answer. $\endgroup$ Commented Apr 10, 2014 at 6:00
  • $\begingroup$ Does this mean that my computer can do: 2 x 3,000,000,000 Hz x 16 = 96 Giga FLOPs? $\endgroup$ Commented Apr 10, 2014 at 20:05
  • $\begingroup$ It means that it could do between 2*1.8 GHz*16 DP = 57.6 GFLOP/s and 96 GFLOP/s, depending on the actual average frequency. If you need to use a single number, 57.6 is the more fair one, IMO. $\endgroup$ Commented Apr 10, 2014 at 22:42
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    $\begingroup$ FLOP rates are generally a poor measure of the 'goodness' of a processor. See scicomp.stackexchange.com/questions/114/… for example. You might want to think about the limiting costs of your task (e.g. compute bound vs memory bound vs disk bound) and focus on the relevant hardware (compute system, memory system, I/O). $\endgroup$ Commented Apr 13, 2014 at 21:21

I understand that you asked for the theoretical value, but as this is nearly always inaccessible by any real code, even LINPACK, you might want to just run (optimized) DGEMM for very large matrices. The reason that I prefer this method is that it exposes some of the shortcomings of certain processors that prevent them from achieving their theoretical peak flop value.

For example, circa 2014, NVIDIA GPUs did integer and floating-point operations on the same pipeline. This means that you could only achieve the theoretical peak flop/s if you do no integer computation whatsoever. As array indexing and any other form of data access requires integer arithmetic somewhere, no code would achieve the theoretical peak flop/s on an NVIDIA GPU. In most cases, one saw ~80% as the upper bound. For CPUs that issue integer and floating-point operations simultaneously, this is a non-issue.

On some GPU-like multicore processors like Intel Knights Corner and Blue Gene/Q, it is harder to achieve the peak flop/s than on traditional CPUs for similar pipeline issues (although both can achieve ~90% of peak in large DGEMM at least).

  • $\begingroup$ Why no code can achieve the theoretical peak flop/s on an NVIDIA GPU? $\endgroup$
    – skytree
    Commented Jul 2, 2019 at 4:53
  • $\begingroup$ See paragraph 2. However, all of this was relevant to older architectures. More recent NVIDIA architectures can execute integer and floating-point concurrently and achieve >90% of peak in large DGEMM. $\endgroup$ Commented Jul 2, 2019 at 4:56

You will need to know the model and vendor of the CPUs in your machine. Once you have that, you can look up on the vendor's website (or maybe on Wikipedia) the clock rate, number of chips/sockets, number of cores per chip, number of floating point operations per cycle, and the vector width of those operations. Then, you simply multiply.

Take, for example, the Intel Xeon E5-2680 "Sandy Bridge" processors in Stampede where I work. The specs are:

  • 2.7GHz
  • 2 chips/node, 8 cores/chip
  • 2 vector instructions/cycle
  • 256-bit wide AVX instructions (4 simultaneous double-precision operands)

Multiplying those gives 345.6 GF/node or 2.2 PF for the un-accelerated part of the system.

We usually think in terms of double-precision (64-bit) operations, because that's the precision required for the vast majority of our users, but you can can redo the calculation in single-precision terms if you like. This usually only changes the last factor, say 8 SP Flops/instruction instead of 4 DP Flops/inst, but it can be wildly different from that. Older GPUs, for example, only did DP at about 1/8th the rate of SP. If you ever quote a number for your system, you should be explicit about which you used if it's not double-precision because people will assume it was, otherwise.

Also, if your chip supports fused multiply-add (FMA) instructions, and it can do them at full rate, then most people count this as 2 floating-point operations though a hardware performance counter might count it as only one instruction.

Finally, you can also do this for any accelerators that might exist in your system (like a GPU or Xeon Phi) and add that performance to the CPU performance to get a theoretical total.

  • $\begingroup$ It's not enough to know the CPU model, one needs to find out the actual operating frequencies $\endgroup$ Commented Apr 10, 2014 at 2:36
  • $\begingroup$ @Aksakal, for a theoretical analysis, it's probably OK to pick the nominal frequency. It's hard to know what frequency your chips will actually run at since that can depend on the workload and the quality of your air conditioning. $\endgroup$
    – Bill Barth
    Commented Apr 10, 2014 at 2:57

What I suggest you is to take a look at the Intel MKL benchmark suit

Here one can find binaries for Windows, Linux and Mac OS. When solving linear system of equations, it estimates FLOPs. The information is given as,

=================== Timing linear equation system solver ===================

Size   LDA    Align. Time(s)    GFlops   Residual     Residual(norm) Check
2000   2000   4      0.038      139.8416 3.396561e-12 2.702137e-02   pass
5000   5008   4      0.457      182.4536 2.230932e-11 2.943595e-02   pass
5000   5008   4      0.477      174.6359 2.230932e-11 2.943595e-02   pass

Hope it helps.


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