Amdahl’s law states that a speed up of the entire system is
an_old_time / a_new_time
where the a_new_time can be represented as ( 1 - f ) + f / s’, where f is the fraction of the system that is enhanced by some modification, and s’ is the amount by which that fraction of the system is enhanced. However, after solving this equation for s’, it seems like there are many cases in which s’ is negative, which makes no physical sense.
Taking the case that s = 2 (a 100% increase in the speed for entire system) and f = 0.1 (a 10% of the system is impacted by some speed enhancement s’), we solve for s’ by setting an_old time = 1 and s’ = f / ( f + 1 / s - 1 ).
Plugging on the values for f and s, we find that : s’ = 0.1 / ( 0.1 + 0.5 - 1 ) = 0.1 / -0.4 which means that the s’ value is negative.
How can this be possible, and what is the physical meaning of this? Also, how can I avoid negative s’ values when answering questions like these?
Amdahl's Law, also known as Amdahl's argument, is used to find the maximum expected improvement to an overall process when only a part of the process is improved.
1 | where S is the maximum theoretical Speedup achievable
S = __________________________; | s is the pure-[SERIAL]-section fraction
( 1 - s ) | ( 1 - s ) a True-[PARALLEL]-section fraction
s + _________ | N is the number of processes doing the [PAR.]-part
N |
Due to the algebra, the s + ( 1 - s ) == 1, s being anything from < 0.0 .. 1.0 >, there is no chance to get negative values here.
The full context of the Amdahl's argument& the contemporary criticism,adding all principal add-on overheads factors&a better handling of an atomicity-of-work
It is often applied in the field of parallel-computing to predict the theoretical maximum speedup achievable by using multiple processors. The law is named after Dr. Gene M. AMDAHL ( IBM Corporation ) and was presented at the AFIPS Spring Joint Computer Conference in 1967.
His paper was extending a prior work, cited by Amdahl himself as "... one of the most thorough analyses of relative computer capabilities currently published ...", published in 1966/Sep by prof. Kenneth E. KNIGHT, Stanford School of Business Administration.
The paper keeps a general view on process improvement.
Fig.1:
a SPEEDUP
BETWEEN
a <PROCESS_B>-[SEQ.B]-[PAR.B:N]
[START] and
[T0] [T0+tsA] a <PROCESS_A>-[SEQ.A]-ONLY
| |
v v
| |
PROCESS:<SEQ.A>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>|
| |
+-----------------------------------------+
| |
[T0] [T0+tsB] [T0+tsB+tpB]
| | |
v v v
|________________|R.0: ____.____.____.____|
| |R.1? ____.____| :
| |R.2? ____| : :
| |R.3? ____| : :
| |R.4? : : :
| |R.5? : : :
| |R.6? : : :
| |R.7? : : :
| | : : :
PROCESS:<SEQ.B>>>>>>>>>>|<PAR.B:4>: : :
| |<PAR.B:2>:>>>>: :
|<PAR.B:1>:>>>>:>>>>>>>>>: ~~ <PAR.B:1> == [SEQ]
: : :
: : [FINISH] using 1 PAR-RESOURCE
: [FINISH] if using 2 PAR-RESOURCEs
[FINISH] if using 4 PAR-RESOURCEs
( Execution time flows from left to right, from [T0] .. to [T0 + ts1 + tp1].The sketched order of [SEQ], [PAR] sections was chosen just for illustrative purpose here, can be opposite, in principle, as the process-flow sections' durations ordering is commutative in principle )
The speedup of a { program | process }, coming from using multiple processors in parallel computing, was derived to be ( maybe to a surprise of audience ) principally limited by the very fraction of time, that was consumed for the non-improved part of the processing, typically the sequential fraction of the program processing, executed still in a pure [SERIAL] process-schedulling manner ( be it due to not being parallelised per-se, or non-parallelisable by nature ).
For example, if a program needs 20 hours using a single processor core, and a particular portion of the program which takes one hour to execute cannot be parallelized ( having been processed in a pure-[SERIAL] process-scheduling manner ) , while the remaining 19 hours (95%) of execution time can be parallelized ( using a true-[PARALLEL] ( not a "just"-[CONCURRENT] ) process-scheduling ), then out of the question the minimum achievable execution time cannot be less than that ( first ) critical one hour, regardless of how many processors are devoted to a parallelized process execution of the rest of this program.
Hence the Speedup achievable is principally limited up to 20x, even if an infinite amount of processors would have been used for the [PARALLEL]-fraction of the process.
See also:
CRI UNICOS has a useful command amlaw(1) which does simple
number crunching on Amdahl's Law.
------------
On a CRI system type: man amlaw.
1 1
S = lim ------------ = ---
P->oo 1-s s
s + ---
P
S = speedup which can be achieved with P processors
s (small sigma) = proportion of a calculation which is serial
1-s = parallelizable portion
Speedup_overall
= 1 / ( ( 1 - Fraction_enhanced ) + ( Fraction_enhanced / Speedup_enhanced ) )
Articles to parallel#ctc.com (Administrative: bigrigg#ctc.com)
Archive: http://www.hensa.ac.uk/parallel/internet/usenet/comp.parallel
Criticism:
While Amdahl has formulated process-oriented speedup comparison, many educators keep repeating the formula, as if it were postulated for the multiprocessing process rearrangement, without taking into account also the following cardinal issues:
atomicity of processing ( some parts of the processing are not further divisible, even if more processing-resources are available and free to the process-scheduler -- ref. the resources-bound, further indivisible, atomic processing-section in Fig. 1 above )
add-on overheads, that are principally present and associated with any new process creation, scheduler re-distribution thereof, inter-process communication, processing results re-collection and remote-process resources' release and termination ( it's proportional dependence on N is not widely confirmed, ref. Dr. J. L. Gustafson, Jack Dongarra, et el, who claimed approaches with better than linear scaling in N )
Both of these group of factors have to be incorporated in the overhead-strict, resources-aware Amdahl's Law re-formulation, if it ought serve well to compare apples to apples in contemporary parallel-computing realms. Any kind of use of an overhead-naive formula results but in a dogmatic result, which was by far not formulated by Dr. Gene M. Amdahl in his paper ( ref. above ) and comparing apples to oranges have never brought anything positive to any scientific discourse in any rigorous domain.
Overhead-strict re-formulation of the Amdahl's Law speedup S:
1
S = __________________________; where s, ( 1 - s ), N were defined above
( 1 - s ) pSO:= [PAR]-Setup-Overhead add-on
s + pSO + _________ + pTO pTO:= [PAR]-Terminate-Overhead add-on
N
Overhead-strict and resources-aware re-formulation:
1 where s, ( 1 - s ), N
S = ______________________________________________ ; pSO, pTO
/ ( 1 - s ) \ were defined above
s + pSO + max| _________ , atomicP | + pTO atomicP:= further indivisible duration of atomic-process-block
\ N /
Interactive Tool for a maximum effective speedup :
Due to reasons described above, one picture might be worth million words here. Try this, where a fully interactive tool for using the overhead-strict Amdahl's Law is cross-linked.
Related
I am accelerating a MPI program using cuBlas function. To evaluate the application's efficiency, I want to know the FLOPS, memory usage and other stuff of GPU after the program has ran, especially FLOPS.
I have read the relevant question:How to calculate Gflops of a kernel. I think the answers give two ways to calculate the FLOPS of a program:
The model count of an operation divided by the cost time of the operation
Using NVIDIA's profiling tools
The first solution doesn't depend on any tools. But I'm not sure the meaning of model count. It's O(f(N))? Like the model count of GEMM is O(N^3)? And if I multiply two matrices of 4 x 5 and 5 x 6 and the cost time is 0.5 s, is the model count 4 x 5 x 6 = 120? So the FLOPS is 120 / 0.5 = 240?
The second solution uses nvprof, which is deprecated now and replaced by Nsight System and Nsight Compute. But those two tools only work for CUDA program, instead of MPI program launching CUDA function. So I am wondering whether there is a tool to profile the program launching CUDA function.
I have been searching for this question for two days but still can't find an acceptable solution.
But I'm not sure the meaning of model count. It's O(f(N))? Like the
model count of GEMM is O(N^3)? And if I multiply two matrices of 4 x 5
and 5 x 6 and the cost time is 0.5 s, is the model count 4 x 5 x 6 =
120? So the FLOPS is 120 / 0.5 = 240?
The standard BLAS GEMM operation is C <- alpha * (A dot B) + beta * C and for A (m by k), B (k by n) and C (m by n), each inner product of a row of A and a column of B multiplied by alpha is 2 * k + 1 flop and there are m * n inner products in A dot B and another 2 * m * n flop for adding beta * C to that dot product. So the total model FLOP count is (2 * k + 3) * (m * n) when alpha and beta are both non-zero.
For your example, assuming alpha = 1 and beta = 0 and the implementation is smart enough to skip the extra operations (and most are) GEMM flop count is (2 * 5) * (4 * 6) = 240, and if the execution time is 0.5 seconds, the model arithmetic throughput is 240 / 0.5 = 480 flop/s.
I would recommend using that approach if you really need to calculate performance of GEMM (or other BLAS/LAPACK operations). This is the way that most of the computer linear algebra literature and benchmarking has worked since the 1970’s and how most reported results you will find are calculated, including the HPC LINPACK benchmark.
The Using the CLI to Analyze MPI Codes states clearly how to use nsys to collect MPI program runtime information.
And the gitlab Roofline Model on NVIDIA GPUs uses ncu to collect real time FLOPS and memory usage of the program. The methodology to compute these metrics is:
Time:
sm__cycles_elapsed.avg / sm__cycles_elapsed.avg.per_second
FLOPs:
DP: sm__sass_thread_inst_executed_op_dadd_pred_on.sum + 2 x
sm__sass_thread_inst_executed_op_dfma_pred_on.sum +
sm__sass_thread_inst_executed_op_dmul_pred_on.sum
SP: sm__sass_thread_inst_executed_op_fadd_pred_on.sum + 2 x
sm__sass_thread_inst_executed_op_ffma_pred_on.sum +
sm__sass_thread_inst_executed_op_fmul_pred_on.sum
HP: sm__sass_thread_inst_executed_op_hadd_pred_on.sum + 2 x
sm__sass_thread_inst_executed_op_hfma_pred_on.sum +
sm__sass_thread_inst_executed_op_hmul_pred_on.sum
Tensor Core: 512 x sm__inst_executed_pipe_tensor.sum
Bytes:
DRAM: dram__bytes.sum
L2: lts__t_bytes.sum
L1: l1tex__t_bytes.sum
I have systems that have a large number of cores as well as a cluster. For a particular task for which no serial implementation is available, I can only benchmark w.r.t. time taken for tasks running on different input sizes. I see that even when data size was increased by a factor of 10 times, the time for completion is less than 10 times while using identical resources. I would like to know how to measure the performance, as this does not appear to fall under typical definitions of strong/weak scaling. This appears to be related to efficiency, but I am not certain. From what I could gather about the three:
Strong scaling (Amdhal's law): speedup = 1 / ( s + p / N ) = T( 1 ) / T( N )
Weak scaling (Gustafson’s law): scaled speedup = s + p × N
Efficiency: speedup / N
As I don't have speedup due to lack of serial implementation and that N a is constant, I can only think of finding ratios of efficiencies using strong scaling. Is such a parameter used in CS?
Apache Spark on workloads on 250-500 GB data. B/M was done with 100% and 10% data sets. Jobs run between 250-3000s depending on the type and size. I can force number of executors to be 1 with 1 executor core, but that would be wrong as theoretically only optimum serial job should be written.
– Quiescent 24 mins ago( URL added )
Thanks for this note. The problem gets ground to answer it :
Q :... "Is such a parameter used in CS ?"
The answer to the questions about the observations on the above depicted problem has nothing to do with DATA-size per-se, the DATA-sizing is important, yet the core understanding is related to the internal functioning of the distributed-computing where overheads matter :
SMALL RDD-DATA
+-------------------E-2-E ( RDD/DAG Spark-wide distribution
|s+------+o | & recollection
|e| | v s| Turn-Around-Time )
|t| DATA | e d |
|u|1x | r a |
|p+------+ h e |
+-------------------+
| |
| |
|123456789.123456789|
Whereas :
LARGER RDD-DATA
+--------:------:------:------:-------------------E-2-E ( RDD/DAG Spark-wide TAT )
|s+------:------:------:------:------+o + |
|e| : : : : | v s v|
|t| DATA : DATA : DATA : DATA : DATA | e d a|
|u|1x :2x :3x :4x :5x | r a r|
|p+------:------:------:------:------+ h e .|
+--------:------:------:------:-------------------+
| |
| | |
|123456789.123456789| |
| |
|123456789.123456789.123456789.123456789.123456789|
( not a multiple of 5x the originally observed E-2-E for "small" DATA ( Spark-wide TAT )
yet a ( Setup & Termination overheads stay about same ~ const. )
a ( a DATA-size variable part need-not yet may grow )
now
show an E-2-E of about ~ 50 TimeUNITs for 5-times more DATA,
that is
for obvious
reasons not 5-times ~ 20 TimeUNITs
as was seen
during the E-2-E TAT from processing in "small"-DATA use-case
as not
all system-wide overheads accumulation
scale with DATA size
For further reading on Amdahl's argument & Gustafson/Barsis promoted scaling, feel free to continue here.
suppose I have algorithm 1 and 2, their sequential execution time is ts1 and ts2. their parallel execution time is tp1 and tp2.
Now when calculating the speed up for both algorithm, which of the following is true?
min(ts1,ts2)/tp1 for algorithm 1
min(ts1,ts2)/tp2 for algorithm 2
or
ts1/tp1 for algorithm 1
ts2/tp2 for algorithm 2
in other words, for numerator, should I use the best sequential time or their own sequential time?
Short Version:
None of the above
Fig.1:
a SPEEDUP
BETWEEN
a BLACK-BOX <PROCESS_2>
[START] and
+-----------------------------------------+ a BLACK-BOX <PROCESS_1>
| |
[T0] [T0+ts1] [T0+ts1+tp1]
| | |
| | |
v v v
|________________|R.0: ____.____.____.____| ~~ <PAR.1:1> == [SEQ]
| |R.1? ____.____| :
| |R.2? ____| : :
| |R.3? ____| : :
| | : : :
|<SEQ.1>>>>>>>>>>| : : :
| |<PAR.1:N>: : :
| : : :
: : :
: : [FINISH] using 1 PAR-RESOURCE
: [FINISH] if using 2 PAR-RESOURCEs
[FINISH] if using 4 PAR-RESOURCEs
( Execution time flows from left to right, from [T0] .. to [T0 + ts1 + tp1]. The sketched order of [SEQ], [PAR] sections was chosen just for illustrative purpose here, can be opposite, as process-flow sections' durations ordering is commutative in principle )
A TL;DR; Version:
a bit formal simplification of the [SEQ]+[PAR] process-flows above may help to both answer and also understand why.
Needless to tell any HPC planners, that Amdahl Law rules ( the better if extended form of Amdahl, the overhead + atomicity aware formulation were used ).
We see, the more resources R.i were used in [PAR]-section of the PROCESS_1, the shorter the tp1 may get. Here is the power of [PAR]-processing.
Given just the pair of tuples ( ts1, tp1 ) and ( ts2, tp2 ), no one can assume any potential Amdahl Law -- resources-driven ( as demonstrated in Fig.1 ) -- speedup, but if one strives to just compare the two postulated implementations, having potentially different internal processing, the possible speedup S can be formulated as:
max( [ ts1 + tp1 ], [ ts2, tp2 ] )
S = ______________________________________
min( [ ts1 + tp1 ], [ ts2, tp2 ] )
There is a fundamental issue with your question. That is why you are stuck. The issue is that Speedup is defined for processors, not algorithms.
In computer architecture, speedup is a process for increasing the performance between two systems processing the same problem. More technically, it is the improvement in speed of execution of a task executed on two similar architectures with different resources.
Definition taken from Wikipedia.
I’ve got a question about trying to decide the big O notation for a Rabin-Miller algorithm I have implemented.
When generating a
512 bits prime, I get a runtime of 22 seconds,
1024 takes 237 seconds,
2048 takes 2942 seconds.
How can I determine the big O notation for these values? It seems to me that the runtime increases by roughly 10 times everytime the bitsize increases by 2. Does that mean that it’s O(10n)?
You have too few points to estimate the function (and O(function) as well) from the experiment.
x | f(x)
-----------
512 | 22
1024 | 237
2048 | 2942
If we test O(n) (O(10n) is in fact O(n)) as f(x) = Ax + B guess with a help of Least Squares method, we'll get a good fit
A = 2.0
B = -1330.5
R = 0.964 (Correlation)
However, many alternative functions have better support
f(x) = Ax**4 + B with correlation R = 0.99990 <- actual best fit
f(x) = Ax**3 + B -/- R = 0.99937 <- expected
f(x) = Ax**2 + B -/- R = 0.99232
you want more points to find out the right function: when having three values only Ax**4 + B (which corresponds to O(x**4)) is the leader so far, but we can't reject the expected complexity which is Ax**3 + B.
Finally, we can guess (we can't conclude with three points only) that the implementation is suboptimal: O(x**4) instead of expected O(x**3)
If our guess O(x**4) is correct one, than we might expect that doubling x: x -> 2 * x we increase the time 16-fold (2**4 == 16)
I'm working on code where a star rating system was implemented, allowing users to rate between 1 & 5 stars. Instead of displaying an item's actual rating it uses this algorithm:
( rating_votes / ( rating_votes+10 ) ) * ( rating_total/rating_votes ) ) + ( 10 / ( rating_votes+10 ) ) * 4
Based on my intuition it seems like the intent of this is to default the rating to "4 stars" and to not drop the rating too quickly when there's under 10 votes.
Does anyone know what the mathematical name of this algorithm is called? Also can it's implementation be simplified and still produce the same output?
I got:
(rating_votes / ( rating_votes +10 )) * ( rating_total / rating_votes ) +
( 10 / ( rating_votes +10 ) ) *4
= (rating_total / (rating_votes + 10)) + (40 / (rating_votes + 10))
= (rating_total + 40) / (rating_votes + 10)
... you seem to have missed an opening bracket but is that what you meant? If so then your intuition is correct — it pretends that 10 people voted '4' before anyone else jumped in.
Other than integer rounding, depending on your language, simplification should produce the same result.
This is known as a "Bayesian average", a variant of additive smoothing. The basic idea is, you front-load a new estimator with a prior estimate of what the "real" average probably is, and then additional votes are added to that existing evidence. It means that it takes a lot of votes to move the average up or down.
And yes, its implementation can be simplified. See https://en.wikipedia.org/wiki/Bayesian_average for the basic formula.
You are right - the final rating can be rewritten as
final = (1-f) * rating + f * 4
Where the factor f decides how much the "actual" rating matters vs the "default" of 4.
Now you just have to convince yourself that f can be written as
f = 10 / (votes + 10)