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. What is the maximum throughput for window size of 1

To calculate the maximum throughput for a window size of 1, we need to consider the transmission time of the data frame and the propagation delay of the acknowledgement.

1. First, calculate the transmission time of the data frame. The data frame is 512 bytes, which is 4096 bits. The channel has a rate of 64 kbps. So, the transmission time is 4096 bits / 64,000 bits per second = 0.064 seconds.

2. The acknowledgement is very short, so we can ignore its transmission time. However, we need to consider the propagation delay. The problem does not give us this value, so we cannot calculate it. Let's call it Tp.

3. The throughput is the amount of data sent per unit of time. In this case, with a window size of 1, we can send one frame per round-trip time. The round-trip time is the transmission time of the data frame plus twice the propagation delay (once for the acknowledgement to reach the sender, and once for the next data frame to reach the receiver). So, the round-trip time is 0.064 seconds + 2*Tp.

4. The throughput is then 512 bytes / (0.064 seconds + 2*Tp). To get the maximum throughput, we want to minimize the round-trip time. Since we cannot change the transmission time or the propagation delay, the maximum throughput is achieved when Tp is as small as possible.

Without the value of Tp, we cannot give a numerical value for the maximum throughput. However, we can say that it is less than or equal to 512 bytes / 0.064 seconds = 8000 bytes per second, and it decreases as Tp increases.