Consider an error free 64-Kbps satellite channel used to send 512 byte data frames in one direction with very short acknowledgements coming back the other way?

The transmission starts at t = 0. At t = 4096/64000 sec = 64 msec, the last bit is sent. At t = 334 msec, the last bit arrives at the satellite and the very short ACK is sent. At t = 604 msec(270 + 334), the ACK arrives at the earth. The data rate here is 4096 bits in 604 msec or about 6781 bps (4096/604msec) (window size is 1). With a window size of 7 frames, transmission time is 448 msec(512*8*7/64000) for the full window, at which time the sender has to stop. At 604 msec, the first ACK arrives and the cycle can start again. Here we have 7 × 4096 = 28,672 bits in 604 msec. The data rate is 47,470.2 bps (28,672/604msec) (window size is 7). Continuous transmission can only occur if the transmitter is still sending when the first ACK gets back at t = 604 msec. In other Words, if the window size is greater than 604 msecworth of transmission, it can run at full speed. For a window size of 10 or greater, this condition is met, so for any window size of 10 or greater (e.g., 15 or 127), the data rate is 64 kbps.