8核处理器和4核处理器

The advent of economical consumer grade multi-core processors raises the question for many users: how do you effectively calculate the real speed of a multi-core system? Is a 4-core 3Ghz system really 12Ghz? Read on as we investigate.

经济型消费级多核处理器的出现为许多用户提出了一个问题：如何有效地计算多核系统的实际速度？ 4核3Ghz系统真的是12Ghz吗？ 在我们进行调查时请继续阅读。

Today’s Question & Answer session comes to us courtesy of SuperUser—a subdivision of Stack Exchange, a community-drive grouping of Q&A web sites.

今天的问答环节由SuperUser提供，它是Stack Exchange的一个分支，它是Q＆A网站的社区推动组织。

问题 (The Question)

SuperUser reader NReilingh was curious how to the processor speed for a multi-core system is actually calculated:

超级用户阅读器NReilingh好奇如何实际计算多核系统的处理器速度：

Is it correct to say, for example, that a processor with four cores each running at 3GHz is in fact a processor running at 12GHz?

例如，一个具有四个运行于3GHz的内核的处理器实际上是运行于12GHz的处理器，这是否正确？

I once got into a “Mac vs. PC” argument (which by the way is NOT the focus of this topic… that was back in middle school) with an acquaintance who insisted that Macs were only being advertised as 1Ghz machines because they were dual-processor G4s each running at 500MHz.

我曾经遇到一个“ Mac vs. PC”的论点(顺便说一下，这并不是本主题的重点……那是在中学时代)，他认识一个人，因为Macs是双重的，因此它们只被宣传为1Ghz机器。 -处理器G4各自以500MHz运行。

At the time I knew this to be hogwash for reasons I think are apparent to most people, but I just saw a comment on this website to the effect of “6 cores x 0.2GHz = 1.2Ghz” and that got me thinking again about whether there’s a real answer to this.

当时我知道这很麻烦，原因是我认为这对大多数人来说都是显而易见的，但是我在该网站上仅看到“ 6核x 0.2GHz = 1.2Ghz”的评论，这让我再次考虑有一个真正的答案。

So, this is a more-or-less philosophical/deep technical question about the semantics of clock speed calculation. I see two possibilities:

因此，这是关于时钟速度计算语义的一个或多或少的哲学/深层技术问题。 我看到两种可能性：

1. Each core is in fact doing x calculations per second, thus the total number of calculations is x(cores).实际上每个内核每秒执行x次计算，因此计算总数为x(内核)。
2. Clock speed is rather a count of the number of cycles the processor goes through in the space of a second, so as long as all cores are running at the same speed, the speed of each clock cycle stays the same no matter how many cores exist. In other words, Hz = (core1Hz+core2Hz+…)/cores.时钟速度实际上是处理器在一秒钟内经历的周期数的计数，因此，只要所有内核都以相同的速度运行，则无论存在多少个内核，每个时钟周期的速度都保持不变。 换句话说，Hz =(core1Hz + core2Hz +…)/核心。

So what is the appropriate way to denote the total clock speed and, more importantly, is it even possible to use single-core speed nomenclature on a multi-core system?

那么，表示总时钟速度的合适方法是什么？更重要的是，甚至有可能在多核系统上使用单核速度术语吗？

答案 (The Answer)

SuperUser contributors Mokubai helps clear things up. He writes:

超级用户贡献者Mokubai帮助您解决问题。 他写：

The main reason why a quad-core 3GHz processor is never as fast as a 12GHz single core is to do with how the task running on that processor works, i.e. single-threaded or multi-threaded. Amdahl’s Law is important when considering the types of tasks you are running.

四核3GHz处理器从来没有像12GHz单核那么快的主要原因是与该处理器上运行的任务的工作方式有关，即单线程或多线程。 在考虑您正在运行的任务类型时， 阿姆达尔定律很重要。

If you have a task that is inherently linear and has to be done precisely step-by-step such as (a grossly simple program)

如果您的任务本质上是线性的，并且必须逐步完成，例如(一个非常简单的程序)

10: a = a + 1

10: a = a + 1

20: goto 10

Then the task depends highly on the result of the previous pass and cannot run multiple copies of itself without corrupting the value of 'a' as each copy would be getting the value of 'a' at different times and writing it back differently. This restricts the task to a single thread and thus the task can only ever be running on a single core at any given time, if it were to run on multiple cores then the synchronisation corruption would happen. This limits it to 1/2 of the cpu power of a dual core system, or 1/4 in a quad core system.

然后，该任务在很大程度上取决于前一遍的结果，并且无法运行其自身的多个副本而不会破坏'a'的值，因为每个副本将在不同的时间获取'a'的值并以不同的方式写回。 这将任务限制为单个线程，因此该任务只能在任何给定时间在单个内核上运行，如果要在多个内核上运行，则将发生同步损坏。 这将其限制为双核系统的cpu功率的1/2，或四核系统的1/4。

Now take a task such as:

现在执行以下任务：

10: a = a + 1

10: a = a + 1

20: b = b + 1

30: c = c + 1

30: c = c + 1

40: d = d + 1

50: goto 10

50: goto 10

All of these lines are independent and could be split into 4 separate programs like the first and run at the same time, each one able to make effective use of the full power of one of the cores without any synchronisation problem, this is where Amdahl’s Law comes into it.

所有这些线路都是独立的，可以像第一个程序一样分成4个单独的程序，并同时运行，每个程序都可以有效利用其中一个内核的全部功能而不会出现任何同步问题，这就是阿姆达尔定律进入它。

So if you have a single threaded application doing brute force calculations the single 12GHz processor would win hands down, if you can somehow make the task split into separate parts and multi-threaded then the 4 cores could come close to, but not quite reach, the same performance, as per Amdahl’s Law.

因此，如果您有一个单线程应用程序进行蛮力计算，那么单个12GHz处理器将胜出，如果您能够以某种方式将任务拆分为单独的部分并使用多线程，那么4个内核可能会接近但不能完全实现，表现与阿姆达尔定律相同。

The main thing that a multi CPU system gives you is responsiveness. On a single core machine that is working hard the system can seem sluggish as most of the time could be being used by one task and the other tasks only run in short bursts in between the larger task, resulting in a system that seems sluggish or juddery. On a multi-core system the heavy task gets one core and all the other tasks play on the other cores, doing their jobs quickly and efficiently.

多CPU系统为您提供的主要功能是响应能力。 在努力工作的单核计算机上，系统似乎很呆滞，因为大部分时间可能被一个任务占用，而其他任务仅在较大的任务之间短暂运行，从而导致系统显得呆滞或摇摇欲坠。 在多核系统上，繁重的任务获得一个核心，而其他所有任务则在另一个核心上发挥作用，从而快速高效地完成工作。

The argument of “6 cores x 0.2GHz = 1.2Ghz” is rubbish in every situation except where tasks are perfectly parallel and independant. There are a good number of tasks that are highly parallel, but they still require some form of synchronsation. Handbrake is a video trancoder that is very good at using all the CPUs available but it does require a core process to keep the other threads filled with data and collect the data that they are done with.

“ 6核x 0.2GHz = 1.2Ghz”的论点在每种情况下都是垃圾，除非任务是完全并行和独立的。 有很多任务是高度并行的，但是它们仍然需要某种形式的同步。 Handbrake是一种视频转码器，非常善于使用所有可用的CPU，但是它确实需要一个核心进程来保持其他线程充满数据并收集处理完的数据。

1. Each core is in fact doing x calculations per second, thus the total number of calculations is x(cores).实际上每个内核每秒执行x次计算，因此计算总数为x(内核)。

Each core is capable of doing x calculations per second, assuming the workload is suitable parallel, on a linear program all you have is 1 core.

假设工作负载适合并行运行，则每个内核每秒可以执行x个计算，在线性程序上，您只有1个内核。

1. Clock speed is rather a count of the number of cycles the processor goes through in the space of a second, so as long as all cores are running at the same speed, the speed of each clock cycle stays the same no matter how many cores exist. In other words, Hz = (core1Hz+core2Hz+…)/cores.时钟速度实际上是处理器在一秒钟内经历的周期数的计数，因此，只要所有内核都以相同的速度运行，则无论存在多少个内核，每个时钟周期的速度都保持不变。 换句话说，Hz =(core1Hz + core2Hz +…)/核心。

I think it is a fallacy to think that 4 x 3GHz = 12GHz, granted the maths works, but you’re comparing apples to oranges and the sums just aren’t right, GHz can’t simply be added together for every situation. I would change it to 4 x 3GHz = 4 x 3GHz.

考虑到数学原理，我认为认为4 x 3GHz = 12GHz是一个谬论，但是您正在将苹果与橘子进行比较，并且总和是不正确的，不能简单地将GHz每种情况相加。 我将其更改为4 x 3GHz = 4 x 3GHz。

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翻译自: https://www.howtogeek.com/127950/how-do-you-calculate-processor-speed-on-multi-core-processors/

8核处理器和4核处理器