否定阿姆达尔定律的速度会加快吗? [英] Negative speed up in Amdahl's law?
问题描述
阿姆达尔定律指出，整个系统的速度提高了
Amdahl’s law states that a speed up of the entire system is
an_old_time / a_new_time
其中 a_new_time 可以表示为(1-f)+ f/s'，其中 f 是系统的一部分通过一些修改可以增强，而 s'是系统的这一部分得到增强的数量.但是，在解决了 s 的等式后，似乎在很多情况下 s 为负，这在物理上没有意义.
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.
假设 s = 2 (整个系统的速度提高了100％)和 f = 0.1 (10％的系统受到以下影响)一些速度增强 s')，我们通过设置 an_old time = 1 和s'= f/(f +1/s-1).
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 ).
使用 f 和 s 的值，我们发现:
s'= 0.1/(0.1 + 0.5-1)= 0.1/-0.4
表示 s'的值为负.
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.
这怎么可能?它的物理含义是什么?另外，在回答此类问题时，如何避免出现负的 s 值?
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?
推荐答案
阿姆达尔定律(也称为阿姆达尔定律)用于在仅改善过程的一部分时，为整个过程找到最大的预期改进.
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.
>
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.由于代数， s +(1-s)== 1 ，s是<0.0 .. 1.0> ，这里没有机会获得负值.
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.
它通常应用于本文对过程改进保持了总体看法.
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.
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
(执行时间从左到右，从 [T0] ..到 [T0 + ts1 + tp1] .
草绘的顺序此处选择 [SEQ] ， [PAR] 部分只是出于说明目的，原则上可以相反，因为流程流部分的持续时间顺序是可交换的原理)
( 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 )
{程序|流程，是由于在并行计算中使用多个处理器而产生的(可能令听众惊讶)主要受时间的限制，对于处理的未改进部分，通常是程序处理的顺序部分，仍以纯 [SERIAL] 进程调度方式执行(因为由于未并行处理)本身或本质上不可比拟的.)
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 ).
例如，如果一个程序需要使用单个处理器内核的20小时，并且该程序需要花费一个小时才能执行的特定部分将无法并行化(已经以纯 [SERIAL] 进程调度方式)，而剩余的19小时(95％)的执行时间可以并行化(使用true- [PARALLEL] (不是 [CONCURRENT] )进程调度)，那么毫无疑问，最小可达到的执行时间不能小于该时间(首先)关键一小时，无论有多少处理器专用于该程序其余部分的并行处理执行.
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.
因此，即使 [PARALLEL]可以使用无限数量的处理器，可实现的 -流程的一部分. Speedup 原则上最多限制为20倍.
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.
另请参阅:
CRI UNICOS has a useful command amlaw(1) which does simple
number crunching on Amdahl's Law.
------------
在CRI系统上，键入: man amlaw .
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-分数增强)+(分数增强/Speedup_enhanced))
parallel@ctc.com上的文章(管理:bigrigg@ctc.com)
存档: http://www.hensa.ac.uk/parallel/internet/usenet/comp.parallel
Articles to parallel@ctc.com (Administrative: bigrigg@ctc.com)
Archive: http://www.hensa.ac.uk/parallel/internet/usenet/comp.parallel
批评:
尽管Amdahl已制定了以过程为导向的加速比，但许多教育工作者仍在重复该公式，就像它被假定用于多处理过程重新安排一样，而没有考虑到以下主要问题:
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:
- 处理的原子性(即使可以使用更多的处理资源并且流程调度程序可以自由使用，处理的某些部分也无法进一步分割-参见资源有限，进一步不可分割的部分，上图1中的原子处理部分)
- 附加间接费用，主要存在且与任何新流程创建，其调度程序的重新分配，进程间通信，处理结果的重新收集以及远程进程资源的释放相关联和终止(它对
N的比例依赖性尚未得到广泛证实，参见JL Gustafson博士，Jack Dongarra等人)，他声称在N中比线性缩放更好的方法)
- 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
Nis not widely confirmed, ref. Dr. J. L. Gustafson, Jack Dongarra, et el, who claimed approaches with better than linear scaling inN)
这两个因素中的两个因素都必须纳入对开销的限制，对资源敏感的阿姆达尔定律的重新表述中，如果它能很好地将苹果与当代并行计算领域中的苹果进行比较的话.Gene M. Amdahl博士在他的论文中没有提出过任何形式的使用天真幼稚的公式的结果，但是教条主义的结果到目前为止(教法参考文献)尚未提出，并且将苹果与橙子进行比较从未带来任何积极的结果.任何严格领域的科学话语.
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.
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
开销上限和资源感知的重新配制:
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 /
可实现最大有效加速的交互式工具:
由于上述原因，一张图片可能价值一百万个字.尝试此，其中使用开销限制的阿姆达尔定律的完全交互式工具是交叉链接的.
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.
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