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+2 votes

0

@ hacker16

For n=1,

ceil(log(1+1)) - 1 = ceil(1) - 1 =0

But, it should be 1

floor(logn)+1 should be the answer

0

here they are saying that congestion window increases by 1 segment each time it receives an acknowledgment, doesn't this mean that in slow start phase window size increases linearly?

0

No, its not the linear growth but rather exponential.

Consider, initial window size=1

1->1(acknowledge)-> window size=2

2->2(acknowledge)-> window size=4

4->4(acknowledge)-> window size=8

& so on

Consider, initial window size=1

1->1(acknowledge)-> window size=2

2->2(acknowledge)-> window size=4

4->4(acknowledge)-> window size=8

& so on

+1

@charul 1st the sender sends 1 segment and get an ack. So now congestion window increases by 1.

Now there are 2 segments. The sender sends these 2 segments and get ack. for each of them. Each segment increases by 1 segment so 2+2 =4 is the current window size.

Next these 4 segments are sent and 1 ack is received for each segment. So 4+4 =8.

I think what they meant by this---

congestion window increases by 1 segment each time it receives an acknowledgment,

is that the congestion window increases by 1 segment for each segment present in the current window. This is what happens in slow start phase. Tell me if i am wrong.

+1

you are right @MiNiPanda, this is what happens in slow start phase, but here they have explicitly saying that "each time congestion window size increases by one", that why i'm not sure

0

@charul i understand your doubt and i had the same when i was reading this for the 1st time. But after seeing how it actually works i now know what is the real thing happening.

Read the line marked with star. It's exactly the same as the one you gave. It is basically the definition of slow start phase.

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