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here TOTAL 2000 segments need to be sent, and after x RTT, it will send 2001 segments but for total we shall take count of addition of all the previous also na?

initially window size(WS) = 1, so we can transfer 1 packet.

after first RTT  WS =2, so we can transfer 2 packet.

after second RTT WS = 4 , so in this we can transfer 4 packet.

after third RTT WS = 8, we can transfer 8 packet.

after fourth RTT WS = 16, transfer 16 packet

after 5th RTT WS = 32, transfer 32 packet.

after 6th RTT WS = 64, transfer 64 packet.

after 7th RTT WS = 128,  transfer 128 packet.

after 8th RTT WS = 256, transfer 256 packet.

after 9th RTT WS will become= 512, so in 10th RTT we transfer 512 packet.

after the 10th RTT total packet send=1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512

= 1023

but we have to send 2000 packet.

no of packet still remaining=2000-1023=977

so after 10th RTT WS has become = 1024, so remaining 977 packet will be send in 11th RTT.

so RTT require = 11

### 1 comment

@1997_satya Very well explained.
their explanation seems confusing, but your explanation cleared it.
after each RTT window size refers to the number of packets sent after that RTT means in the next RTT,
so for total we need to check sum of all previous also.
this is the only concept right?

1 vote