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I have a couple of confusing problems on my course, so I decided to ask here. The question is:

What is the minimum size of Ethernet frame for a network speed of 100MBps* and maximal segment of 100 meters?

*Could be that this is actually 100Mbps (100BaseT), but doesn't matter a lot.

I thought 100 meters is the distance, so 200 meters is all around. I wanted to divide that by speed of signal, so I can get the time for one bit to come back, so I got 1 microsecond, and for 1 microsecond this network can send 100 Bytes (or 100bits if 100Mbps), which is what I thought was an answer.

My professor answered me that it's 500 meters in total, based on 5-4-3-2-1 rule, but I still don't get it. How do I measure it and is it 5 segments of 100 meters resulting in 500? Should I get 64B (Ethernet standard) as answer?

Also, if there is site more fitting for these kind of questions, feel free to redirect me.

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The minimum Ethernet frame size is 64 bytes or 512 bits, by definition.

To derive that size you'd need to calculate the serialization delay (frame size / link rate) plus the propagation delay (length / (speed of light * velocity factor)) plus possible repeater (SOP) and collision detection (SOJ) delays, from one end of the network to the other end and back. You also need to account for the preamble and SFD byte (eight additional bytes).

SOP and SOJ vary with the Ethernet variant, so for a concrete calculation you'd need to calculate a whole table (you can find all values in IEEE 802.3 Clause 9).

The 5-4-3 rule generally only applies to a half-duplex, single collision domain, 10 Mbit/s Ethernet network. It's a rule of thumb that works in almost all cases. Half-duplex Fast Ethernet with 100 Mbit/s only allows for a single repeater class I or two class II repeaters. Gigabit Ethernet and faster is switched only = the minimum frame size exists for compatibility only.

Note that half-duplex operation and repeaters are long obsolete and more or less historical trivia today, even for 10/100 links. Unfortunately, there's a lot of nonsense propagated and taught in various books and courses. You can read up on the actual definition, proven and reviewed many times in the original IEEE 802.3 specification, available at http://ieeexplore.ieee.org/browse/standards/get-program/page/series?id=68 (free after registration).

IEEE 802.3 also features a more detailed calculation for minimal frame size and maximum network size in Clause 13.4 (Clause 13.3 is the 5-4-3 rule).

See also:

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  • Well that question was meant to be for a csma/cd 5-4-3 network. So what would you say is an answer I need and how should I calculate it? Was my calculation wrong? What about that 500 meters thing?
    – bezdoman
    Jun 14 at 10:16
  • 500 meters is the maximum length for a 10BASE5 segment. The maximum size for 100 Mbit/s CSMA/CD is 3x 100BASE-FX = 3*412 m (HDX with class II repeaters).
    – Zac67 ♦
    Jun 14 at 10:27
  • 5-4-3 is 10 Mbit/s only, so the maximum size is 5x 10BASE-FL = 5*2000 m.
    – Zac67 ♦
    Jun 14 at 10:36
  • On twisted pair, it would be 10BASE-T: 5*100 m or 100BASE-TX: 3*100 m (class II repeaters).
    – Zac67 ♦
    Jun 14 at 11:03

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