Q63.Marks: +2.0UGC NET Paper 2: Computer Science and Application 26th June 2025 Shift 1
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
1.15,000 bits
2.20,000 bits
3.30,000 bits✓ Correct
4.60,000 bits
Solution
The correct answer isOption 3) 30,000 bits
Key Ideas
Bandwidth–delay product (BDP) is the amount of data that can be “in flight” on the link:
BDP = Bandwidth × Round-Trip Time (RTT)
Bandwidth is given as 1 Mbps = 1,000,000 bits/second.
“1 bit takes 30 ms to make a round trip” means the RTT = 30 ms = 0.03 s.
Step-by-step Calculation
Convert units: 30 ms → 0.03 s.
Apply the formula:
BDP = 1,000,000 (bits/s) × 0.03 (s) = 30,000 bits
Why this matters
BDP tells you the pipeline capacity: to keep a 1 Mbps link fully busy, up to 30,000 bits can be on the wire before the first acknowledgement returns.
In stop-and-wait, if the frame size is much smaller than the BDP, the link stays idle while waiting for ACKs, reducing utilization.
Why other options are incorrect
15,000 bits / 20,000 bits / 60,000 bits: These come from using the wrong time (half-RTT, some other value) or wrong bandwidth unit. The given RTT is 30 ms round trip, and the bandwidth is 1,000,000 bits/s, so only 30,000 bits fits the formula exactly.