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 .