Computer Networking Basics
A TCP connection has a fixed receive window of 65536 bytes and an RTT of 10 ms. The link bandwidth is ample and there is no loss. Which is closest to the upper limit on this connection's throughput?
- AAbout 52 Mbps
- BAbout 6.6 Mbps
- CAbout 524 Mbps
Show answer and explanation
Answer: A. About 52 Mbps
Under window control, at most one window of data can be sent per round trip (RTT), the time until ACKs return. The throughput limit is therefore window ÷ RTT = 65536 bytes × 8 ÷ 0.01 s ≈ 52.4 Mbps (about 52 Mbps), and it goes no higher even on a 1 Gbps link. To fill a fast, distant link, the window must be at least the product of bandwidth and RTT (the bandwidth-delay product): for 1 Gbps and an RTT of 10 ms, 10^9 × 0.01 = 10^7 bits = 1.25 MB. The window field in the TCP header is 16 bits, for a maximum of 65535 bytes, but the window scale option can enlarge it.
- A. Correct. 65536 × 8 ÷ 0.01 ≈ 52.4 Mbps.
- B. This is 65536 ÷ 0.01 ≈ 6.6 MB/s in bytes per second, not multiplied by 8 to get bits per second.
- C. This is the value for an RTT of 1 ms.