Based on the following passage, answer the Questions : A 3000 km long trunk…

Key insight: the maximum throughput depends on how many frames can be kept in flight (the bandwidth–delay product).

Step 1 — frame transmission time:

• Frame size = 64 bytes = 512 bits.

• Transmission rate = 1.536 Mbps → frame transmission time = 512 / 1.536e6 = 0.00033333 s = 333.33 µs.

Step 2 — propagation delay and RTT:

• One-way propagation = 3000 km × 6 µs/km = 18000 µs = 0.018 s.

• Round-trip time (RTT) ≈ 2 × 0.018 s = 0.036 s.

Step 3 — bandwidth–delay product (frames in flight required to fill the pipe):

• BDP (bits) = bandwidth × RTT = 1.536e6 × 0.036 = 55,296 bits.

• Frames needed = 55,296 / 512 = 108 frames.

Conclusion:

• If the sliding-window size is at least 108 frames, the sender can fully utilize the link (throughput = 100% of 1.536 Mbps).

• If the window is smaller than 108, normalized throughput ≈ (window size) / 108.

• Among the provided choices, 0.901 is the closest to the computed achievable utilization for a realistic window slightly below 108 (97–98 frames gives ≈0.898–0.907). Therefore 0.901 is the best match given the options. Note: the question does not state the actual window limit; if the window limit were ≥108, the true maximum would be 1.0 (full line rate).