Station A uses 32 byte packets to transmit messages to Station B using a…
2017
Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip time delay between A and B is 40 ms and the bottleneck bandwidth on the path A and B is 64 kbps. What is the optimal window size that A should use?
- A.
5
- B.
10
- C.
40
- D.
80
Attempted by 126 students.
Show answer & explanation
Correct answer: B
To find the optimal window size, calculate the bandwidth-delay product and divide by the packet transmission time. First, convert units: Bandwidth = 64 kbps = 64,000 bits/sec. RTT = 40 ms = 0.04 sec. Packet size = 32 bytes = 256 bits. The bandwidth-delay product is 64,000 * 0.04 = 2560 bits. The optimal window size is 2560 / 256 = 10 packets.
A video solution is available for this question — log in and enroll to watch it.