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Station $A$ needs to send a message consisting of $9$ packets to Station $B$ using a sliding window (window size $3$) and go-back-$n$ error control strategy. All packets are ready and immediately available for transmission. If every $5$th packet that $A$ transmits gets lost (but no acks from $B$ ever get lost), then what is the number of packets that $A$ will transmit for sending the message to $B$? 

  1. $12$
  2. $14$
  3. $16$
  4. $18$
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In GBN we send entire window again if any frame lost 

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SInce it is said that Go Back N Protocol is used and the window size is 3 it means that if sender doesnot get an acknowledgment for the given packet then it would resend the packet based on the timeout and here timeout will be 3 packets as Window size is 3.

Now according to the question every 5th packet is lost.

Suppose we have 9 packets named: 1, 2, 3, 4, 5, 6, 7, 8, 9 

Now initial 5 packets are send successfully

1, 2, 3, 4, 5

But when 5 is send it gets lost.

So not a problem till now,

Now notice that A will keep sending the packets as it is getting acknowledgment of every packet till 1 to 4 and because packet 5 is lost A will not get it’s acknowlegement.

So when packet 6 would be send there will be no problem when packet 7 will be send there will be no problem but as soon as packet 8 would be ready in the queue for sending to B, A will realize that hey all things are good but where’s the acknowledgement of 5 ?

As soon as A will realize that 5th packet acknowlegment is not received it will start sending packets from 5th packet onward. 

Now notice before sending 5th packet again A has already sent packet 6 and packet 7 so in the queue 

packet 6, 7, 5, 6 are send again and packet 7 will get lost again as it is the 5th packet in the queue.

 

As 7 is lost it’s not a problem packet 8 and 9 are sent.but after sending 9 A will realize that hey packet 7 acknowledgement is not yet received and hence need to resend packet 7 again following by packets 8 and 9

Again before sending packet 7, packet 8 and 9 are already in the queue and and now 

8, 9, 7, 8, 9 are send 

 

 

so total packets send are:

1 2 3 4 5 -→ 5th packet lost

6, 7, A realize acknowledgement of 5th packet is not received.

it will again send 5, 6, 7, also 6 and 7 are also in the queue

so queue will be 6, 7, 5, 6 ,7 

frame 7 is lost

8 , 9 are send but A will realize 7 acknowledgement is not received.

so it will send again 7, 8, and 9 also 8 and 9 are already in the queue and since 9 is the 5th packet it will get lost and A will resend packet 9 again.

 

Therefore total packets send = 1, 2, 3, 4, 5, 6, 7, 5, 6 ,7, 8, 9, 7, 8, 9, 9

= 16 packets.

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