Computer Networking Basics
On 1 Gbps Ethernet, TCP data is sent continuously with IP packets filling the 1500-byte MTU. Taking the IPv4 header as 20 bytes, the TCP header as 20 bytes, the Ethernet header and FCS together as 18 bytes, and the preamble and inter-frame gap together as 20 bytes, which is closest to the upper limit on application-data throughput?
- AAbout 949 Mbps
- BAbout 973 Mbps
- CAbout 1000 Mbps
Show answer and explanation
Answer: A. About 949 Mbps
Each frame carries 1500 − 20 − 20 = 1460 bytes of application data. On the wire, however, each frame takes the time of 1500 + 18 + 20 = 1538 bytes. The efficiency is therefore 1460 ÷ 1538 ≈ 0.949, and the throughput limit is 1000 × 0.949 ≈ 949 Mbps. Using the TCP timestamp option adds 12 bytes to the header and lowers efficiency slightly further. The smaller the packets, the larger the share of headers and other overhead, and the lower the efficiency.
- A. Correct. 1460 ÷ 1538 ≈ 0.949, about 949 Mbps.
- B. This is 1460 ÷ 1500, ignoring the Ethernet header, FCS, preamble and inter-frame gap.
- C. This is the raw link speed, with no overhead considered.