Consider an error-free 64kbps satellite channel used to send 512 byte data…

Answer summary: the correct maximum throughput cannot be determined from the information given unless the round‑trip time (RTT) is specified. Below is the general derivation and numeric checks that explain when each listed throughput would apply.

• Step 1 — Transmission time per frame: 512 bytes = 512 × 8 = 4096 bits. At 64 kbps (64,000 bps) the transmission time is 4096 / 64,000 = 0.064 s (64 ms).

• Step 2 — Window capacity (bits in flight): window_size × frame_bits = 15 × 4096 = 61,440 bits. This is the amount of data the sender can have outstanding before waiting for ACKs.

• Step 3 — General throughput formula for sliding window: throughput = min(channel_rate, (window_size × frame_size_in_bits) / RTT).

• Numeric checks:

  • If RTT = 0.54 s (typical GEO satellite RTT ≈ 540 ms), then (61,440 bits) / 0.54 s ≈ 113,777 bps, so the channel limits the rate to 64 kbps.

  • To obtain 48 kbps: required RTT = (61,440 bits) / 48,000 bps = 1.28 s. Thus 48 kbps is correct only if RTT = 1.28 s.

  • 70 kbps is impossible because it exceeds the physical channel capacity of 64 kbps.

• Conclusion: The provided solution was missing. The correct throughput depends on RTT: with typical GEO RTT (~0.54 s) the maximum is 64 kbps; 48 kbps occurs only if RTT = 1.28 s. The question should state the RTT (or propagation delay) to make the answer unambiguous.