Test by Bikram | Computer Networks | Test 1 | Question: 26

Question

Consider a sliding window protocol that uses Selective Repeat, the sender sliding window size is $32$. The sequence number of the frame to be sent after sending $200$ frames ___________

Answer 1

When a host initiates a TCP session, its initial sequence number is effectively random; it may be any value between 0 and 4,294,967,295 ( 2^32 - 1 ), inclusive .

The ISN (Initial Sequence Number) of a packet is not zero and assigned random value for the purpose SECURITY, preventing SPOOFING or HACKING or CONFLICTING with other data bytes transmitted over the data communication(TCP)!

BUT By default, the We take  these starting sequence numbers as zero, regardless of what these values really are. The purpose of this is to make sequence and acknowledgement number analysis simpler. These values will be referred to as relative sequence numbers to the connection.

For sender side

It has a send window and the receiver side has a corresponding recieve window

Now as we are using Selective Repeat , this send window size= recieve window size Which is 32

Now minimum number of sequence number(frame numbers) required to communicate without any ambiguity is 64

So my sequence number space range from 0-63
200%32 = 8 (make set of 32,  0 to 31 )
192 will get 31
193 =0
194=1
195 = 2

196=3

197=4

198=5

199=6

200 = 7
 The sequence number of the frame to be sent after sending 200 frames is  8

Reference :

http://www.lovemytool.com/blog/2010/08/practical-tcp-series-sequence-and-acknowledgement-numbers-by-chris-greer.html

Comments

sir, sequence number assign to packet must be mod64. for frame number 228 sequence number must be 35 not 3.

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No this answer is correct,  it is mod 32 ( 0 to 31 ) as for Selec tive Repeat this

sender window size= recieve window size Which is 32 .

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yes sir sender window is equal to reciever window but sequence number must be >= (Ws + Wr) to avoid duplecate data packet problem at reciever

.

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this question have to solve conceptually , without using formula. Please see that answer carefully, it is correct.

you can also check here manoj 

https://www.facebook.com/photo.php?fbid=1714487108878020&set=gm.1349075968457986&type=3&theater

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question says 200 frames means frames are starting from 0 i,e

0th frame > 0 sequence number

1st frame >1 sequence number

....

199th frame which is the 200th frame will get 6 th sequnce number so the next to the 200 must get 7 ans should be 7

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0 to 7 should be 8 !!

read my answer carefully, it is correct.

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Sir the frames are starting from 0 or 1 ??

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please read my given answer for this question clearly and try to understand , i mentioed clearly it is 0 and also why it is 0 .

I discussed this answer with few students after that write this answer. The description i gave is correct.

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"BUT By default, the We take  these starting sequence numbers as zero,

sir any reference for this...i cant find any reference 

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Read each and every comment in given  reference thread .

I discussed there in this thread  @alpha  and @Manoj_Kumar   

Reference 

https://www.facebook.com/photo.php?fbid=1714487108878020&set=gm.1349075968457986&type=3&theater

and @firki lama   This two source:

http://www.lovemytool.com/blog/2010/08/practical-tcp-series-sequence-and-acknowledgement-numbers-by-chris-greer.html

http://packetlife.net/blog/2010/jun/7/understanding-tcp-sequence-acknowledgment-numbers/

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Sir how is 192 having 31 it should have 0

0-0 ,31-0

32-0 ,63-31

64-0 and so on, so 192 should have 0

According to this answer should be 9 Please correct where i am going wrong sir?

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@abhinav93

200%32 = 8

200, 199, 198, 197,196,195,194,193

 so next acknowledgement start from 192 , 192 get ack no 31.

Answer 2

The sequence number of the frame to be sent after sending 200 frames

=> 201th frame is to be sent.

For 201th frame, frame number = 200 because the first frame number = 0.

 

Now, 200 mod 32 = 8.