As it is full duplex communication link, to achieve optimal window size , sender need to send data during RTT i.e during 60ms.
Now, Data bytes send during 60ms is = 512KB*60ms = 512*60 byte.
so, Window size = no of packet send = 512*60/64 = 480
Now if the channel is given as half duplex in that case:
For half duplex communication link, to achieve optimal window size , sender need to send data during one way progagation time i.e during 60/2 =30ms .
Now, Data bytes send during 30ms is = 512KB*30ms = 512*30 bytes.
so, Window size = no of packet send = 512*30/64 = 240
@Bikram or Arjun Sir Can you please verify If the above procedure is correct or not?