Congestion window


#1

Assume scenario where he size of CW of TCP connection is 40 KB when time out occurs ,MSS= 2KB ,Propagation delay = 200 msec , the time taken by TCP connection to get back 40KB CW is …msec

solution : one way P.D= 200 msec
,RTT=400 msec
new threshold =1/2* (40KB)=20 KB
therefore,
2KB
4KB
8KB
16KB
20KB
22KB
24KB
26KB
28KB
30KB
32KB
34KB
36KB
38KB
40KB
15* 400=6000 msec
right?or missed something


#2

I think it should be 14*400 instead of 15*400. because after 14 RTT your window size will be 40KB.


#3

@ RUTURAJ
threshold is 20KB then SS till 16 then after 16 it should be 18 ,20,22,24,24,26,28,30,32,34,36,38,40 right?


#4

Nopes after 16 it would be 20 then 22 24 so on.


#5

why … 20 …


#6

After 16 you can jump to 20 because if Th had not been applied then next size would have been 32.
Ananlogy: If you are allowed to drive the car at a speed of 80 kmps on an empty road, you can easily drive at a speed less than 80.
I hope you got it.