The advent of economical c***umer 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.

今天的問答環節是由SuperUser提供的，SuperUser是Stack Exchange的一個分支，它是一個由Q&a網站組成的社群驅動分組。

問題

超級使用者讀者nreligh很好奇如何計算多核系統的處理器速度：

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?

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.

At the time I knew this to be hogwash for reas*** 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.

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

1. Each core is in fact doing x calculati*** per second, thus the total number of calculati*** is x(cores).
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.

那麼，什麼是表示總時鐘速度的合適方法呢？更重要的是，在多核系統上使用單核速度術語是否可能？

答案

超級使用者貢獻者Mokubai幫助解決問題。他寫道：

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 c***idering the types of tasks you are running.

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

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.

Now take a task such as:

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.

So if you have a single threaded application doing brute force calculati*** 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.

The main thing that a multi CPU system gives you is resp***iveness. 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.

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 synchr***ation. 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.

1. Each core is in fact doing x calculati*** per second, thus the total number of calculati*** is x(cores).

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

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.

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.

有什麼要補充的解釋嗎？在評論中發出聲音。想從其他精通技術的Stack Exchange使用者那裡瞭解更多答案嗎？在這裡檢視完整的討論主題。