Question on congestion window and slow start of TCP

Question

Given solution:

I couldn't understand from reaching 1KB to 16KB why it took 50ms not 40ms. Please check

Answer 1

1||2|| 4|| 8 ||16|| 17|| 18|| 19|| 20|| 21|| 22|| 23|| 24|| 25|| 26|| 27|| 28|| 29|| 30|| 31|| 32

Just Count the no. of vertical lines thats the ans ie., 20*10=200 msec irrespective what is all given in solution just stick to ur concept :)

Comments

The answer should be 210 m sec because when the window size is 32 MSS has sent and after that when you will get the ack then only you will be sure that your data has successfully reached,

In your answer you are not taking the ack of 32 MSS data.

So answer should be 210 m sec.

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NO Answer is exact 200 since when 31 KB recived sucessfully with Ack we get know congesion window size 32kb so no need to send and get ack for that.

Answer 2

Th=32/2=16

Slow startPhase=1 |10ms| 2 |10ms|, 4 |10ms| ,8|10ms| ,16 |10ms|

So Time taken to slow start phase=50ms

now turn of Congetion avoidance Phase=17 |10ms| 18|10ms| ..............................32|10ms|

Time=160ms

So total time=210ms

Comments

no need for extra 10ms after reaching 32...so 200 is ans

Answer 3

200 MS IS ANSWER

Answer 4

200 MS IS ANSWER APPROPIATE ONE