Over a link with an RTT of 50 ms and a bandwidth of 100 Mbps, a request that returns a 10 KB response is repeated 100 times, each sent only after the previous response has been fully received. Which is closest to the total time? Take the time per request as the RTT plus the response transfer time (data size ÷ bandwidth), 1 KB = 1000 bytes, and ignore server processing time.

Answer: C. About 5.1 s

Each request takes one round trip, 50 ms, from sending the request until the response starts to arrive, plus the time to receive 10 KB = 80000 bits at 100 Mbps: 80000 ÷ (100 × 10^6) = 0.8 ms. The total is (50 + 0.8) × 100 = 5080 ms ≈ 5.1 s, almost all of it waiting for round trips. Even with ten times the bandwidth, 1 Gbps, it only drops to (50 + 0.08) × 100 ≈ 5.0 s. Light in optical fiber travels about 200 km per millisecond, so distance sets a lower bound on RTT. To speed up this kind of processing, batch requests, send several requests in parallel without waiting for responses, or place servers close to users.

• A. This is only the 0.8 ms transfer time × 100, without the RTT waits.
• B. This uses 25 ms, the one-way time, instead of the RTT; a full round trip is needed.
• C. Correct. (50 + 0.8) × 100 ≈ 5.1 s, mostly RTT.