Little tip about sliding windows

little tip about sliding windows

@ (computer network)

If a conclusion is made: Window size = Send window size + receive window size.

I don't know how many people think this is doing.

For n-bit numbered data frames, it has been discussed that the Send window + receive window ≤2n send window + receive window \leq 2^n can distinguish between the old and new rounds.

Http://blog.csdn.net/u011240016/article/details/52653923?locationNum=2&fps=1

This is a little bit of a very painful thought that night, because there is no suitable argument, there is not enough abstraction to draw such a conclusion, so I can only guess, and then I want to understand that.

Then understand the new and old round of the calculation formula, it is easy to understand the above conclusions, but this is also very easy to overlook the point, that the sending window and receive window apart, love who, what is the relationship. The actual size of the two is the same, and the discrete state of n bits is used as its own window data number.

With these, you can be happy to judge the following several conclusions.

I. For sliding windows with a window size of n, up to n frames have been sent but not confirmed.
II. Assume that the frame sequence number has 3 bits, using continuous ARQ protocol, the maximum value of the sending window is 4.
III. In the GBN protocol, if the sending window size is 16, a minimum of 4-bit serial number is required to guarantee that the protocol is not faulted. It's all wrong, you can believe it.

Analysis: Window size, well, not send window. The receiving window is at least 1, the sending window maximum n-1, the most, the sky can only send n-1 frames continuously. I, properly and wrongly. Concept digging pits.
3 bits, discrete state there are 8, continuous ARQ, can be, let the receiving window is the smallest, 1. The Send window will monopolize 7 states with a size of 7. II, properly and wrongly.
GBN, fallback n frames, n is the size of the sending window. Why fallback to N frames. Receive slow chant. Can only receive one at a time (receive window is 1), once the error will GBN back N, re-come. 2n≥16+1→n=5 (at least) 2^n \geq 16+1 \rightarrow n = 5 (at\ least). III. properly and wrongly.

So I thought:

There are some seemingly correct mistakes.

Some of the right crap.

All of them need to be filtered.