Solution

The correct answer is 64 kbps

To determine the maximum throughput for the given satellite channel, we need to consider the protocol window size and the round-trip time (RTT) of the satellite communication. However, the problem statement only provides the error-free channel capacity (64 kbps), the data frame size (512 bytes), and the window size (15).

Given:

• Channel capacity = 64 kbps
• Data frame size = 512 bytes
• Window size = 15

First, convert data frame size to bits: \(512 \text{ bytes} \times 8 \text{ bits/byte} = 4096 \text{ bits}\)

Since the window size is 15, we can have at most 15 frames in transit before an acknowledgment is required. Therefore, the maximum amount of data that can be in transit is: \(15 \text{ frames} \times 4096 \text{ bits/frame} = 61440 \text{ bits}\)

If the channel capacity is 64 kbps (which is 64000 bps), the maximum throughput is constrained by the channel capacity because the theoretical maximum throughput cannot exceed the channel capacity.

Therefore, regardless of other factors, the maximum throughput achievable on this channel, even with a window size of 15, is the same as the channel capacity because frames can be sent continuously without waiting for acknowledgments (assuming very short acknowledgment packets do not significantly impact the bandwidth).

Thus, the maximum throughput is: \(64 \text{ kbps}\)

So, the correct answer is: 3) 64 kbps