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Alright, so I have a question as listed below:

Suppose you are designing a sliding window protocol for a 10 Mbps point-to-point link to the moon, which has a one-way latency of 2.0 seconds. Assuming that each frame carries 5KB of data, what is the minimum number of bits you need for the sequence number if RWS = SWS?

What I have worked out so far is:

SWS = RTT x Bandwidth / Framesize

10,000,000 bps * 4 seconds = 40,000,000 bits is the through put

5000 Bytes = 40,000 bits per frame is the size of each frame

40,000,000 bits/ 40,000 bits = 1000 bits

SWS = 1000 bits

According to the instructor there are two more steps past this to get the answer, and they are: "First, compute MaxSeqNum based on SWS.

Then, compute the number of bits you need for the sequence number. For example, if MaxSeqNum is 1000, you will need at least 10 bits for all possible sequence number, 2^10 =1024 > 1000."

I can't figure out how to calc MaxSeqNum based of SWS

And i'm not sure how in the second part you get 10 bits because 2^10... that part doesn't make any sense either. Why is it only 10 bits when the SWS equals 1000 bits?? Where does 1024 come into play?

What i'm looking for is the formula needed to calculate this, or a similar example solved that I can derive a formula or method of solving this from.

closed as off-topic by Ron Maupin, Teun Vink Oct 20 '15 at 9:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "NE is a site for to ask and provide answers about professionally managed networks in a business environment. Your question falls outside the areas our community decided are on topic. Please visit the help center for more details. If you disagree with this closure, please ask on Network Engineering Meta." – Ron Maupin, Teun Vink
If this question can be reworded to fit the rules in the help center, please edit the question.

  • Unfortunately, homework, and any educational questions in general, for certification or degree programs, are specifically off-topic for this forum. Please see the Help Center for which types of questions are allowed, and which types are not allowed. – Ron Maupin Oct 19 '15 at 17:04
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You were doing fine until you got here:

40,000,000 bits/ 40,000 bits = 1000 bits 
SWS = 1000 bits

The unit of measure of SWS is not bits, but frames. So the question becomes, how big a binary number do you need to represent 1000 (frames)? The answer, as you've stated, is 10 bits. That's because it takes at least 10 bits to represent 1000 (2^10).

  • +1, but, of course, he originally had written that it was a homework question, and he edited after my comment. – Ron Maupin Oct 20 '15 at 1:55
  • Let's discuss this on Meta... – Ron Trunk Oct 20 '15 at 1:57
  • Start on chat, first? – Ron Maupin Oct 20 '15 at 1:59

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