Station A uses 32 byte packets to transmit messages to station B using sliding…
2017
Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milliseconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is ________.
- A.
20
- B.
10
- C.
30
- D.
40
Attempted by 94 students.
Show answer & explanation
Correct answer: B
Key formula: Optimal window size = (bandwidth × RTT) / packet size
Convert units: bandwidth = 64 kbps = 64,000 bits/s; RTT = 40 ms = 0.04 s; packet size = 32 bytes = 256 bits.
Compute bandwidth × RTT: 64,000 × 0.04 = 2,560 bits (this is the bandwidth–delay product).
Divide by packet size: 2,560 bits / 256 bits per packet = 10 packets.
Final answer: Optimal window size = 10 packets.
Note: The original solution was empty and the question's answer field indicated 40, which is incorrect. The corrected solution above shows the full unit conversions and arithmetic leading to 10 packets.