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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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Here transmission time Tt = L/B = 1000/10^6 = 1 ms

i.e  to transmit 1 frame we need 1 ms.

Now we are given that propagation time Tp=25 ms i.e in Tp time we can send 1*25 = 25 frames only.

Therefore no of bits required for sequence no = ceil( log2(25) ) = 5 bits.

 

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A possible ans:

Every 1 ms we hv a new frame on the line. (BW 10^6 bps and frame size 1000 b).

See it as the first bit of the frame, the first bit touches the other end at 25 ms delay.

How many frames have we pushed on the line during this 25 ms as we were pushing 1 frame each 1 ms?

25 frames on the line at a time.

To identify 25 frames, we need 5 bits.

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L=1000b,B=1000000b/s

Tt = L/B = 1ms

Tp = 25 ms

a  = Tp/Tt

For maximising the efficiency,no of frames that must be transmitted is 1+2a

thus no. of frames is 51

so no of bits required to represent 51 is 5(ans).
Answer:

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