CompTIA Network+ (N10-009)Network TroubleshootingMedium
A technician transfers a 250 MB file across a link rated at 100 Mbps, and the transfer completes in exactly 25 seconds. What is the actual throughput of the transfer, and what does this result suggest?
- A100 Mbps; the link is performing at its theoretical maximum with no overhead
- B20 Mbps; the link is severely degraded and likely has a duplex mismatch
- C250 Mbps; the throughput exceeds link capacity due to compression
- D80 Mbps; this is a reasonable throughput reflecting normal protocol overhead
Show answer & explanationAnswer & explanation
Correct answer: D. 80 Mbps; this is a reasonable throughput reflecting normal protocol overhead
250 MB equals 2,000 megabits (250 x 8). Dividing by 25 seconds gives a throughput of 80 Mbps. On a 100 Mbps link, this is a normal result, as actual throughput is always somewhat lower than rated bandwidth due to protocol overhead, headers, and acknowledgments.
Why the other options are wrong
- A. 100 Mbps would mean zero overhead, which is unrealistic for real transfers.
- B. 20 Mbps is an incorrect calculation and doesn't match the given numbers.
- C. 250 Mbps is mathematically incorrect and exceeds the link's rated bandwidth, which is not physically possible.
Bandwidth vs. Throughput
Bandwidth is the theoretical maximum data rate of a link, while throughput is the actual measured rate achieved, which is typically lower due to overhead.
- Formula: Throughput = (file size in bits) / (time in seconds)
- 1 byte = 8 bits, so convert MB to Mb before dividing
- Throughput close to but below rated bandwidth is normal and expected
Memory trick: Multiply by 8 to go from bytes to bits, then divide by time to get real speed.