made easy

Question

Assume scenario where the 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

CW = congestion window

Comments

5600 mili sec ??

Answer 1

threshold=40kB/2=20KB

2 | 4 | 8 | 16 | 20 |  22 | 24   | 26 | 28 |  30 | 32 | 34 |36 | 38 | 40         

Because Its TOT at 40 KB so its slow start therfore start exponentially from starting MSS ie.,given 2 KB till threshold (20KB)

After threshold go linearly ie increment by MSS (given 2KB) till Max window size (given 40 KB)

NO OF RTT=14

RTT = 2 * Tp = 400ms

therfore 14*400= 5600 ms after it can send fully

Comments

in made easy solution  no. of RTT =15

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hope this link helps - https://gateoverflow.in/146634/congestion-control 

Answer 2

threshold=40kB/2=20KB

MSS=2KB

2KB 4KB 8KB 16KB 18KB 20KB 22KB 24KB 26KB 28KB 30KB 32KB 34KB 36KB 38KB 40KB

NO OF RTT=15

RTT =400ms

SO 15*400=6000ms after it can send fully

Comments

no. of .RTT = 16  why u take 15?

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after 16 why 18 ???why not 20 diretly...

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because max segment size=2KB given

now actually packets used send in segment form.

so first it send 1 segment in the network which is 2KB..

next 2 segments --->4KB

4segment----->8KB

8segment------>16KB<-----THRESHOLD

then after it will not send 16 segment....now because slow phase is completed

9segment------->9*2KB=18KB

10 segment----->20KB.....ITS increasing 1 by 1 segment.........................................20segment(max)CW can hold=40KB.

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see from 2kb to 4kb ----1RTT

4KB TO 8KB ----2ND RTT

LIKE THAT IF U COUNT 38KB TO 40 KB ---15 NUMBER RTT

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kk.....

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same question why have you taken window size to 18kb after 16 kb rather than 20 kb???