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Station A needs to send a message consisting of 10 packets to station B using a sliding window of size 4. All packets are ready and can be transferred immediately.Selective repeat and GBN are used at 2 different times and every 5th packet get lost for both protocols.(ACK's from B never get lost).Let x and y be the number of transmissions that A has to make in selective repeat and GBN respectively to ensure safe delivery to B. Then x+y= ?
asked in Computer Networks by Active (1.5k points)
edited by | 237 views
0
Answer given is 31 but i think that the ans has to be 21. Please confirm?
0
12 for selective repeat and 19 in case of GBN
31 is correct
0

for GBN, 

1234

          5678 (5 lost)

                   5678 (6 lost)

                               6789 (8 lost)

                                            8910

Total = 19

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thanks
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@Utkarsh , I think you have done some mistake while taking window size of 4.

Total no. of transmissions for GBN will be 26 and sequence will be like this :-

$1,2,3,4, (5*,6,7,8) ,5, (6*,7,8,9) ,6, (7*,8,9,10) ,7, (8*,9,10), 8,9, (10*), 10$

3 Answers

+1 vote
Best answer

In GBN,
First time 4,3,2,1 will be transmitted and received successfully.
Second time 8,7,6,5* will be transmitted and 5 will lost. so this along with the rest packets will be transmitted again.
Third Time 8,7,6*,5 will be transmitted and 6 will lost this time as it is the next fifth. So this along with the rest packets will be transmitted again.
Fourth time 9,8*,7,6 will be transmitted and 8 will lost as it is next fifth. so this along with the rest packets will be transmitted again.
Lastly 10,9,8 will be transmitted and reached successfully.
Therefore total 19 Transmissions in GBN.

In SR,

First time 4,3,2,1 will be transmitted and received successfully.
Second time 8,7,6,5* will be transmitted and 5 will lost and rest will reach. Third time 10,9* 5 will be transmitted and 9 will lost.
Lastly 9 will be transmitted and reaches successfully.
Therefore total 12 Transmissions in SR.

Then x+y = 31.

answered by Active (1.5k points)
edited by
0

https://gateoverflow.in/1822/gate2006-46#139733

Please check this question.

According to your method,

First 1 2 3 will be sent and acknowledged successfully.

Next 4 5* 6 will be sent but 5 will be lost so 5 along with rest of packets will be retransmitted.

So 5 6 7 will be sent and acknowledged successfully.

So next 8* 9 will be sent and 8 will b lost so retransmit 8 along with rest of the packets.

So 8 9 will be sent.

Hence total number of transmissions=13 but correct answer is 16.

0
According yo my method:

1 2 3 4 5 6 7 8 9 (5 lost so retransmit 5 6 7)

1 2 3 4 5 6 7 5 6 7 8 9 (7 lost so retransmit 7 8 9)

1 2 3 4 5 6 7 5 6 7 8 9 7 8 9 (9 lost so retransmit 9)

1 2 3 4 5 6 7 5 6 7 8 9 7 8 9 9 (Successful transmission)

Hence total number of transmissions=16.

So which is the correct method mine or yours?

Please clarify.
0
@Rutvik

Can you please share the packet transmission diagram. That will be really helpful in  Analysing how you are getting 26 for GBN.
+1

I am getting this. Can you please share your transition diagram for this question as well as for https://gateoverflow.in/1822/gate2006-46#139733 this one.It would be very helpful.

Thanks.

0
@sumit  In GBN , when the packet in the window gets lost , then it has to go back n packets and retransmit the whole packets in the window again.

Here Window Size is 4 (GB4) and is told that every 5th packet gets lost. So during every loss of 5th packet whatever packets are in the window should be retransmitted. By doing that iam getting total transmissions as 26. Correct me if am wrong. :)
+1
yeah 26 is correct for GBN and 12 for SR.
0 votes
31 is the correct answer
answered by (11 points)
0
either explain the whole answer or post as a comment
0
For SR i am getting 12 but in GBN why i am getting 26 transmissions ..(total both of Sr and gb4 i am getting 38 ) ..can you please explain @Mk Utkarsh..how GBN are 19 ??
0
I added the new answer, please check that
0
I am also getting 26 transmissions in GBN and Total result as 38 !!
0
Please verify, might be you are making any small mistake.
+1
I am also getting 26 transmissions for GBN:

1 2 3 4 5 6 7 8 9 10 (5 lost so retransmit entire window of consisting of 5 6 7 8)

1 2 3 4 5 6 7 8 5 6 7 8 9 10 (6 lost since 5th packet from 5 is 6 so retransmit entire window of 6 consisting of 6 7 8 9)

1 2 3 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 (7 lost since 5th packet form 6 is 7 so retransmit 7 8 9 10)

1 2 3 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 7 8 9 10 (8 lost since 5th packet from 7 is 8 so retransmit 8 9 10)

1 2 3 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 7 8 9 10 8 9 10 ( 10 lost since 5th packet from 8 is 10 so retransmit 10)

1 2 3 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 7 8 9 10 8 9 10 10

Hence total number of transmissions=26
0
Wrong, you are recounting the frame which has been lost.

See explanation above the 5th packet will be 5,6,8 respectively.
0
I am confused. Let me check again.
0 votes

For GBN

1  2  3  4  *5 6 7 8  5  *6 7 8 9 *7 8 9 10  7  *8 9 10  8  9  *10  10

* indicates every 5th packet.

Total 26.

answered by (29 points)

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