Minimum frame length for hub regeneration and maximum distance between hosts

minimum frame length for hub regeneration and maximum distance between hosts

@ (computer network)

First look at a question. Related to a small knowledge point think for a long time, check a lot of books, information, finally can only say a word, find effective information really difficult. Lasted three hours not to seek, went out to eat a bit of things, come back to find almost have their own understanding.

And then came back and took a look at the front. Think of the minimum frame length, and notice the details of the contention period: it is a round-trip spread of Shiyan and. This in the subconscious I feel very natural concept, a bit opened the mind of doubts. Since the contention period is at two points directly between the cable, then now only a little pause in the middle, in essence, but slightly extended a little bit of propagation delay, then add not to.

Such a thought, the following question is very simple.

Remember one point:

Minimum frame length = contention period ∗ data transfer rate minimum frame length = contention period * Data rate

Note: Hub is a 100base-t hub

(2016.36) If the hub regeneration bit stream process, will produce 1.535us delay, signal propagation speed of 200m/s, regardless of the Ethernet frame preamble, the H3 and H4 theoretically can be separated from the furthest distance is: B
a.200m
B. 205m
C. 359m
D. 512m

Analysis: This problem if you do not understand, you can also have a great chance to choose the right. Since 100BASE-T Ethernet is the transmission of 100mb/s baseband signals (digital signals) on a twisted pair, the minimum frame length remains the same with the CSMA/CD protocol. However, the maximum cable length of a network segment is reduced to 100m.

A network segment is 100m, then because the hub regeneration, you can run 100m. So the H3 and H4 can probably be 200m.

If it is 200m, then the hub regeneration time, and signal propagation speed is used, is redundant conditions, so, guess 205 best.

And then the answer is B, it's done right.

Consider this article to review the concept of the shortest frame length.

Http://blog.csdn.net/u011240016/article/details/52719183?locationNum=1&fps=1

But, to think carefully, the problem is very simple. The same is the shortest frame length of 64B, without a preamble, the preamble is 8 B.

Now between H3 and H4, the hub-connected network segment belongs to a bus. Then the propagation delay between the two +hub is 1.535us.

The propagation speed 200m/us has been told.

So we can solve the problem, set the maximum distance of two points is x m.
2⋅ (x200m/μs+1.535μs) ⋅100mb/s=64b 2\cdot (\frac{x}{200 m/\mu s}+1.535\mu s) \cdot 100mb/s = 64B

The x = 205m is obtained

Explain the meaning of the column: the frame from H3 to H4, first to the hub, after a period of propagation delay, hub regeneration with a little time, and then after a period of propagation delay to reach H4. When a collision occurs at the H4, a 48bit congestion signal is sent, and the signal is regenerated by the hub, so the delay to H3 is also regenerated. Therefore, the first time to go back is the propagation delay + regeneration delay.

The point at which a collision is chosen is due to a better understanding of how the hub functions when it returns. No collision occurs, according to the CSMA/CD protocol, always in the listening channel, but listen to what the specific signal, and can not be determined through the hub regeneration. The critical point, the two states of the turning point, similar to the function of the continuous, left and right limit equal, then can be considered as the hub regeneration delay calculation two times.

In addition, the topic said that the regenerative bitstream time 1.535us, is not the regeneration of 1bit when so much. Regeneration 64B will 64*1.535*8.

It takes too long to regenerate, and 1.535us only regenerates a bit too slowly.