9 comments to Midterm Reminder and Seating Arrangement

1. Ooi Wei Tsang says:

   March 9, 2014 at 10:47 AM

   Questions from email:

   Q: For rdt 2.2 and after, there is no longer NAK so what if the very first packet 0 received is corrupted? It cannot send ACK any number since it has no last correctly received packet right? So, what does the receiver do?
   
   A: The “last correctly received packet” is initialized to 1 so that the protocol works correctly.

   The details can be found in the FSM in the textbook.

   Q: Does the receiver receive the checksum with 1’s compliment already done? Or does the receiver receive the original checksum then computes the 1’s complement itself?

   A: Already done. All the receiver needs is to add up all the numbers and see if it gives all 1s.
   
   Q: If a question in a test or in the exam mentions checksum without mentioning anything about 1’s complement, do we assume that it is the original checksum before computing 1’s compliment? Or can we assume that the checksum may be the result of 1’s compliment?

   A: Checksum is always 1-complemented.
   

   Q: For propagation-delay, has the receiver received the first bit by the time of the delay? Or is the first bit just arriving, but not arrived?

   A: Depending on when you start counting.

   If you count from the time the first bit just started to leave the sender, then the delay is counted until the first bit just arrives at the receiver.

   If you count from the time the first bit has completely leave the sender, then the delay is counted until the first is received by the receiver.

   Q: And finally, what is bandwidth-delay product? I’m not sure if I understand it correctly but does it mean the number of bits on the link by the time the first bit arrives at the receiver? If yes, is there a limit for the number of bits that can be on the link?

   A: The Bandwidth delay product is the product of the link rate and propagation delay. It means the maximum number of bits that can fit onto the link at any one time. Yes, there is, unfortunately, a limit to how many bits can fit on the link at one time.

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2. Oh Shunhao says:

   March 9, 2014 at 8:32 PM

   Some last-minute questions:
   Hopefully they can be answered before the midterm :)
   (All of them happen to be about TCP)

   1) For TCP, If the sequence number of the timer is 500, but it has already sent up to packets 600, 700, until 1000, when TCP retransmits (upon timeout or fast retransmit), does it send only packet 500, and wait until it receives an acknowledgement, or does it resend 500, 600, … 1000 together?

   2) The textbook (ch3.5.3 pg239) says “TCP never computes a SampleRTT for a segment that has been retransmitted; it only measures SampleRTT for segments that have been transmitted once.”
   What if the current RTO is set far too low, such that every packet times out? doesn’t that mean that TCP will never recalculate its RTO due to no SampleRTTs being taken at all, meaning the RTO will never be readjusted?

   3) When TCP times out and retransmits, its timeout value temporarily doubles. When it receives a correct acknowledgement, it reverts to its EstimatedRTT value for its timeout. But does it revert to EstimatedRTT if it receives a duplicate acknowledgement?

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   • Chan Jun Wei says:

     March 9, 2014 at 9:41 PM

     I asked your question no1 in Tutorial before and the answer is:
     TCP is more optimism, which means it will only retransmit 500. :)

     Good luck in Mid-Term

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     • Ooi Wei Tsang says:

       March 10, 2014 at 12:15 PM

       1) Jun Wei is right. TCP only retransmit one segment.
       2) Remember that TCP doubles RTO every time it times out (for the same packet). So eventually it should not time out anymore.
       3) Duplicate acknowledgement for the same packet that is timing out? If so, yes.

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       • Chan Jun Wei says:

         March 10, 2014 at 12:35 PM

         For question 3, but when the pkt has received acknowledgement, it should have stopped timer already right?

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         • Ooi Wei Tsang says:

           March 10, 2014 at 1:16 PM

           Yes. TCP then restarts a timer for the earliest unacknowledged packet with the EstimatedRTT value.

         • Oh Shunhao says:

           March 10, 2014 at 1:21 PM

           Actually, I was thinking about a situation like this:
           http://i61.tinypic.com/1556x4p.jpg
           If the RTO reverts back to its original EstimatedRTT every time an acknowledgement comes in, then the RTO won’t be able to double, causing the problem in Question 2, where every segment times out, and so would be considered “segments that have been retransmitted”, so the EstimatedRTT will never be altered.

3. Oh Shunhao says:

   March 10, 2014 at 9:17 AM

   Oh yeah, I have another question, haha.

   (Assume start with sequence number 0 and the receiver buffer is 1000 bytes).
   What if the Sender sends 100 byte packets from 0 to 900, but all packets are lost except for the last packet (900)?
   That means only the last 100 bytes of the receiver buffer will be filled up, so the receiver sends a receiver window size = 0 back to the sender with ACK#=0.
   This forces the sender ensure (last byte sent – last byte ACK’d) <= 0, so the sender can't send any bytes to the receiver, and the receiver can never clear its buffer until the sender sends byte 0.
   So what happens then?

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   • Ooi Wei Tsang says:

     March 10, 2014 at 12:18 PM

     (Sorry I misunderstood the question — ignore my previous reply)

     The sender can still retransmit, although it can no longer send new data after the receiver tells the sender that rwin is 0.

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