Station A uses 32 byte packets to transmit messages to Station B using a…
2006
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 milliseconds 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
Attempted by 93 students.
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Correct answer: B
Solution: Find the window that fills the path (bandwidth–delay product).
Compute the bandwidth–delay product (BDP):
BDP = bandwidth × RTT = 128 kbps × 0.08 s = 128,000 bits/s × 0.08 s = 10,240 bits.
Convert packet size to bits:
32 bytes = 32 × 8 = 256 bits.
Calculate the required window in packets:
Window = BDP / packet size = 10,240 bits / 256 bits = 40 packets.
Therefore, the optimal window size to fully utilize the 128 kbps link with an 80 ms round-trip time and 32-byte packets is 40 packets.
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