Database updates are replicated synchronously between two data centers 1,000 km apart. For each update, the system waits for one round trip: the transfer to the other site and the notification that the write is complete. Taking the speed of light in optical fiber as about 200,000 km/s and the fiber route length as equal to the straight-line distance, which is closest to the increase in the response time of one update caused by propagation alone?

Answer: C. 10 ms

In one round trip the signal travels 2 × 1,000 = 2,000 km, so propagation takes 2,000 ÷ 200,000 = 0.01 s = 10 ms. Synchronous replication completes an update only after confirming the write at the other data center, so no data is lost on failure, but this delay is added to every update. Actual routes are longer and equipment adds delay, so between distant sites asynchronous replication, which tolerates some data loss, is generally used.

• A. This is the one-way time (1,000 ÷ 200,000). The system waits for the completion notification to return, so the round trip counts.
• B. This is the round trip calculated with the speed of light in a vacuum (about 300,000 km/s). Light in fiber is slower.
• C. Correct. 2,000 km ÷ 200,000 km/s = 10 ms.