宇宙射线:它们影响程序的概率是多少? [英] Cosmic Rays: what is the probability they will affect a program?

问题描述

我再一次在进行设计审查时，遇到了声称特定场景的概率小于宇宙射线的风险"影响程序的说法，我突然想到我没有最微弱的知道这个概率是多少.

Once again I was in a design review, and encountered the claim that the probability of a particular scenario was "less than the risk of cosmic rays" affecting the program, and it occurred to me that I didn't have the faintest idea what that probability is.

由于 2^-128 是 340282366920938463463374607431768211456 中的 1，我认为我们有理由在这里冒险，即使这些计算结果相差几十亿……我们我相信，宇宙射线把我们搞砸的风险要大得多."

"Since 2^-128 is 1 out of 340282366920938463463374607431768211456, I think we're justified in taking our chances here, even if these computations are off by a factor of a few billion... We're way more at risk for cosmic rays to screw us up, I believe."

这个程序员正确吗?宇宙射线撞击计算机并影响程序执行的概率是多少?

Is this programmer correct? What is the probability of a cosmic ray hitting a computer and affecting the execution of the program?

推荐答案

来自 维基百科:

IBM 在 1990 年代的研究表明，计算机通常每个月每 256 兆字节 RAM 会遇到大约一个宇宙射线引起的错误.^[15]

Studies by IBM in the 1990s suggest that computers typically experience about one cosmic-ray-induced error per 256 megabytes of RAM per month.^[15]

这意味着概率为 3.7 ×每月每字节 10^-9，或 1.4 ×10^-15 每字节每秒.如果您的程序运行 1 分钟并占用 20 MB RAM，则失败概率为

This means a probability of 3.7 × 10^-9 per byte per month, or 1.4 × 10^-15 per byte per second. If your program runs for 1 minute and occupies 20 MB of RAM, then the failure probability would be

                 60 × 20 × 1024²
1 - (1 - 1.4e-15)                = 1.8e-6 a.k.a. "5 nines"

<小时>

错误检查有助于减少失败的后果.另外，由于乔所评论的芯片尺寸更加紧凑，故障率可能与20年前不同.

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Error checking can help to reduce the aftermath of failure. Also, because of more compact size of chips as commented by Joe, the failure rate could be different from what it was 20 years ago.

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