in Computer Networks
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2 votes
2 votes

What's the answer ?

in Computer Networks
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2 Comments

is answer c?
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Yes, can you explain?
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1 Answer

2 votes
2 votes
Best answer

Answer C),I am giving the counter example for other options

a)Let we have a sliding window protocol $(GB_{N})$ Window of size $4$.Sender will send $1,2,3,4$.Let packet number $2$ gets lost , so reciever will send Ack no $2$(that it is expecting) $3$ times  even though sender will send packet $1,2,3,4$ .After that sender will send packet $2,3,4,5$


b)same as a) i,e even though sender will recieve ack no $3,3,3$(duplicate ack) , it(sender) will keep on sending packet of all its window i.e$1,2,3,4$.


c) correct 

Any new (previosly unsent) packet with sequence number greater than $W$ is sent by sender if and only if a new (previosly unseen ) packet arrives.

It means sender cannot send packet i+w, until it has recieved ack for packet i where w=window size

edited by

4 Comments

Actually I was able to figure out that a and b options are not correct. I had problem with C option.

In C option, sender can send packet 2 at the beginning without needing any packet from the receiver. How is it correct?
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Take  the case when you start recieving ack ,at minimum sender will recieve ack of packet number $2$(expecting) now sender on recieving will slide its window which have currently packet number $2,3,4,5$  it  can only be able to send packet upto packet number $5$  i.w $w+1$=5 , not more than that
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i have updated C) , check it now
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