Station A uses 32 byte packets to transmit messages to Station B using a…
2018
Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 ms and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?
- A.
20
- B.
40
- C.
160
- D.
320
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Correct answer: B
To find the optimal window size, calculate the Bandwidth-Delay Product (BDP) and divide by the packet size. First, convert bandwidth to bits per second: 128 kbps = 128,000 bps. Convert RTT to seconds: 80 ms = 0.08 s. Calculate BDP: 128,000 bps * 0.08 s = 10,240 bits. Convert packet size to bits: 32 bytes * 8 = 256 bits. Finally, divide BDP by packet size: 10,240 / 256 = 40 packets. The optimal window size is 40.
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