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Frames of 1000 bits are sent over a 106 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 2l, where I is the numbers of bits as mentioned earlier and acknowledgements are always piggy backed. After sending 2I 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

This is already discussed but i din find the answers to some doubts of mine 

doubt 1 - we are calculating the waiting time here..which is the time for which the sender waits to send the next frame

transmit time of 32 frames is 32ms

now RTT = transmit frame time + propogation frame time +  transmit ack time + propogation ack time

my doubt is we have already calculated the tranmit time for frames as 32ms then why is it again included for RTT ?

doubt 2 -  Full duplex means sender and receiver can send and receive at the same time....but only if the packets sent by the sender is received by the receiver , the reciever can send ACK back(just like half duplex)....so whats the use of it being full duplex here?

doubt 3 - https://gateoverflow.in/1340/gate2009-57-isro2016-75 in this when 25 frames are sent...now ack comes for it...when ack is arriving ...at the same time sender can send 25 packets again(full duplex)...so fully transit the link by 51 packets right ? why asssuming only for one way propogation as 25 frames

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