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+3 votes
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packets are being transmitted using GB5 and here every 4th packet is lost.How many packets need to be transmitted to transmit 10 packets?

my answr 18 given :20
in Computer Networks by Boss (18k points)
edited by | 639 views
+1
Try again. It should be 19.
0
can u tell how to solve this type of questions

3 Answers

+7 votes
Since, window size is 5, all first 5 packets will be sent continuously. After receiving acknowledgements, window will be slided. When timeout occurs, entire window will be retransmitted again.

12345

[45678] - 4th packet in previous transmission is lost. When timeout occurs, window will be like this. So, this window is retransmitted as below.

[45678] - 8th packet loss in previous transmission will be ignored as expected ack is for 4th packet.

[78910] - 7th packet in previous transmission is lost. When timeout occurs, window will be like this. So, this entire window is retransmitted.

[78910] - Again 7th packet in previous transmission is lost.

[78910]

[78910]...

The last window will be in transmission forever. So question is wrong.
by (83 points)
0
@Aghori, can you explain how 19, mine comes as 20.

same as explaind by @asu.
0
Should be $20$.

sequence must be $1,2,3,4 \, \text{lost},5,6,7,8,4,5,6,7\text{lost},8,9,10,7,8,9,10\text{lost},10$

$\text{total}=20$
+2
Hello sourav ,

see veeresh R joladal's answer , he answered it correctly...

It would be like

after successful transmission of P1 the SW=23456 , after successful transmission of P2 the SW=34567 , after successful transmission of P3 the SW=45678 , but here as 4th packet failed , SW won't slide and as it's GB so send 45678 once again (last SW 45678 have two failed transmission 4,8 but as whole SW would be send again ignore 8) now resending 45678 , after p4 ACK SW=56789 now after P5 successful transmission SW=678910 , now no more new packet to send after P6 succession but here P7 would lost this time and so 78910 would be send , now again P7 lost and 78910  would be sent again again P7 lost and this will go on...

so neither 18 correct nor 20....
+1
yes, question is wrong. Those doing test series; please do not try to make up explanations to get the given answer -- will not work in GATE.
0
@ShobhitAsati This solution is correct. I was wrong.
+1 vote
123(45678)[now 4 lost so again transmit the whole window size]

       45678910[ here the 4th transmission 7 will be lost]

             78910[ now again the 4th transmission packet lost i.e 10 so transmit 10]

                     10

so total 20
by Boss (11.1k points)
+1
@asu this procedure is wrong..if first loss is at 4 then next will be at 8...u have counted after transmitting first window...
+3
here window size is greater them ith packet lost ..so this will send packets forever
+1
actually it depends on transmission and propogation time...if it will be very slow then next loss will happen outside window...so its a wrongly framed question...
+3
Actually this question is wrong!! if you follow the standard procedure,it will loop forever
0
^ how?
+4
it will be like 1,2,3,4,5,6,7,8,4,5,6,7,8,4,5,6,7,8... so on
0 votes

Explanation:
1 2 3 (4 5 6 7 8) 9 10 – first loss (Here total Transmission is 8)

4 5 6 (7 8 9 10) – second loss  (Here total Transmission is 7 )

7 8 9 (10 .. ) – Third lost   10 (Here total Transmission is  5 )

(HERE TOTAL TRANSMISSION IS 8+7+5=20)

Hence a total of 20 transmissions will be required.
 

by Active (1.8k points)
edited by

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