Sliding window protocol

Anmol_Binani 2017-12-01 11:14:55 UTC #1

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The answer given is 5 which we get when we solve it using the formula 2*a.
My doubt is when do we use the formula (1 + 2*a) to find the optimal window size ?

bravitheja143 2017-12-02 12:53:50 UTC #2

propagation : 10^6/(210^8)sec = 5 mSec
round trip time : 10millisec
number of bytes we can transmit in 10millisec : 1010^(-3) * 128 * (10^3)*(1/8) = 160Bytes
number packets is 160B/32B = 5packets.
It is optimal window
But i dont know about that formula how to use .

Anmol_Binani 2017-12-02 13:11:16 UTC #3

go to this link and see, for max utilisation the window size should be (1 + 2a)
but here we are only considering 2a