Hacker News new | past | comments | ask | show | jobs | submit login

A 1500 byte (octet) MTU frame is 1538 bytes “on the wire”.

7 byte preamble

1 byte SFD

6 byte dst MAC

6 byte src MAC

2 byte ethertype or length

46-1500 bytes of payload (ignoring “Jumbo” frames and 802.1q tags)

4 byte CRC

12 byte IFG (which is silence, but still counts for time on the wire)

Add it up and you have 1538 bytes “on the wire”.

TCP overhead for IPv4 is 20 bytes for IP(v4) (no options) and 20 bytes for TCP (again, no options).

So 1460 bytes of data for 1538 bytes on the wire. 1460/1538 = 0.949284

So for 100M Ethernet, 94.9284Mbps is “perfect”.




Consider applying for YC's Spring batch! Applications are open till Feb 11.

Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: