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Frames of $\text{1000 bits}$ are sent over a $10^6$ $\text{bps}$ duplex link between two hosts. The propagation time is $\text{25 ms}$. Frames are to be transmitted into this link to maximally pack them in transit (within the link).

What is the minimum number of bits $(I)$ that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.

  1. $I=2$
  2. $I=3$
  3. $I=4$
  4. $I=5$
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I think this is the most clear solution that I found on geeksforgeeks…

Transmission delay for 1 frame = 1000/(10^6) = 1 ms
Propagation time = 25 ms
The sender can atmost transfer 25 frames before the first frame reaches the destination.
The number of bits needed for representing 25 different frames = 5
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Transmission time (Tt)=1000/10$^6$ seconds =1 ms
Maximum number of frames that can be transmit to maximally pack them is=(Tt+2Tp)/Tx = (25+1)/1=26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is log$_2$26 =  5.
Answer:

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