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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 ___________

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

by

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?
@abhinav93

200%32 = 8

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

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

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.

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