In a Stop-and-wait system, the bandwidth of the line is 1 Mbps, and 1 bit…
2025
In a Stop-and-wait system, the bandwidth of the line is 1 Mbps, and 1 bit takes 30 milliseconds to make a round trip. The bandwidth-delay product is
- A.
15,000 bits
- B.
20,000 bits
- C.
30,000 bits
- D.
60,000 bits
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Correct answer: C
Answer: 30,000 bits
Calculation:
Bandwidth = 1 Mbps = 1,000,000 bits/s.
Round-trip time = 30 milliseconds = 0.03 seconds.
Bandwidth-delay product = bandwidth × round-trip time = 1,000,000 × 0.03 = 30,000 bits.
Note: The bandwidth-delay product represents how many bits can be 'in flight' on the link. For a stop-and-wait protocol (window size = 1 bit or 1 frame), this shows that the channel can hold many bits at once, so stop-and-wait is inefficient unless frames are much larger or the RTT is much smaller.
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