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

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

Suppose that the sliding window protocol is used with the sender window size of $2^{I}$, where $I$ is the numbers of bits as mentioned earlier and acknowledgements are always piggy backed. After sending $2^I$ frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time)

  1. $16ms$
  2. $18ms$
  3. $20ms$
  4. $22ms$
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7 Answers

2 votes
2 votes
Since I = 5 from previous question

Window size of sender = 2^5 = 32

Scenario 1 :

For 1st frame out of these 32 frames :

Tx = 1 ms , Tp = 25 ms and so Tx(ack) = 1ms (duplex so receiver is also sending some data and putting ack of received frame in it .i.e. piggybacking) , Tp(ack) = 25 ms

We are getting ack for first frame in time = 1 + 25 + 1 + 25 = 52 ms

After this time this first frame can be discarded from window and window can be slided to transmit frame number 33.

Scenario 2 :

1 frame takes 1ms to transmit

since window size is 32 we can transmit 32 frames on cable without waiting for ack of first frame.

which takes 32*1ms = 32ms

After sending for 32ms sender will sit idle till ack of 1st frame is not received.

 

Conclusion :  52ms - 32ms = 20ms is the time for which sender is waiting before transmiting next frame (frame 33).
1 votes
1 votes

This question would've been a lot easier, if they asked the maximum waiting time.
It'd simply be, $T_{p} + T_{t} + T_{p} = 51ms$

But they asked the minimum waiting time.
It'll be the total time for 1 packet - time for which the sender is busy.
= ($T_{t} + T_{p} + T_{t} + T_{p}) - 32 * T_{t}$
= 52ms - 32ms
= 20ms.

Option C

PS: The green coloured line depicts the total time for 1 packet.

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Detailed solution of Q57 and 58 combined is in the pic below.
Q 57 & 58 combined answer

Answer:

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