传输数据时的标准差
我正在解决 Gallager 的数据网络中的一些示例，但我不太理解这个特定问题。问题如下：
假设链路上的预期帧长度为 1000 位，标准差为 500 位 求在 9600bps 链路上传输一百万帧所需时间的预期时间和标准差。 随之而来的问题是：帧传输的速率通常定义为预期传输时间的倒数。利用上一题的结果，讨论这个定义是否合理。
我的尝试如下：
发送一帧的时间为：1000/9600 = 0.104秒 因此，发送一百万帧的时间 = 104, 000 秒
我不明白第二部分，其中要求找到标准偏差和后续问题。我也不明白当我们说 500 位的标准偏差是什么意思时，这是否意味着错误损失，这里是 50%？
这不是一个家庭作业问题。几天后我有期中考试，我正在解决这些问题以提高我对这个主题的掌握。
任何帮助/提示将不胜感激，
谢谢， 钱德
I am solving some examples from Data Networks by Gallager and i didnt quite understand this particular question. The question goes as follows:
Suppose the expected frame length on a link is 1000 bits and the standard deviation is 500 bits
Find the expected time and standard deviation of the time required to transfer a million frames on a 9600bps link.
The follow up question is: The rate at which frames can be transmitted is generally defined as the reciprocal of the expected transmission time. Using the result u find in the previous problem, discuss whether this definition is reasonable.
My attempt is as follows:
Time to send one frame is : 1000/9600 = 0.104 seconds
Hence, time to send million frames is = 104, 000 seconds
I did not understand the second part, where it asks to find standard deviation and the follow up question. I also didnt understand what does it mean when we say standard deviation of 500bits, does that mean error loss, which over here is 50%?
This is not a homework problem. I have a midterm in a few days, and im solving these to improve my grip on the subject.
Any help/hints will be appreciated
Thanks,
Chander
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假设分布是正态分布，那么您就有一个正态分布变量的总和。在这种情况下，期望和方差都很容易计算，只需将它们相加即可。
只需将两者除以 9600 即可获得时间。
一个细节是，他们使用正态分布是愚蠢的，因为帧具有负位的概率非零。因此，您有一些非零的机会在负时间发送所有（负）位......因为他们没有指定分布，所以我不知道他们还有什么意思。
Assuming the distributions are normal then you have a sum of normally distributed variables. In this case both the expectation and the variance are easy to compute, you can just add them.
to get the times you just divide both by 9600.
One detail is that it is stupid for them to use a normal distribution since there is some non-zero probability a frame has negative bits. And consequently there is some non-zero chance you will send all of your (negative) bits in negative time... Since they don't specify a distribution I don't see what else they could have meant though.