In csma/cd the minimum frame size that should be transmitted in twice the propagation time for the standard Ethernet cable (10 base 2 or 10 base 5) . But how to maintain the same frame size for the fast Ethernet (100 mbps) and gigabit Ethernet (1Gb/sec) ?

2 Answers 2


Due to collision detection requirements, the collision domain for Fast Ethernet was significantly reduced: only two class-ii repeaters (96 bit-times delay or less) are allowed between any two nodes (or a single class-i repeater). Also, all Fast Ethernet variants are of the link-segment type (using duplex signaling with twisted pair or fiber), allowing for faster collision detection.

For Gigabit Ethernet, reducing the collision domain according to the single-repeater rule is not enough, so frames need to be extended to the full slot time (4096 bits) as minimum. Alternatively, multiple frames can be sent back-to-back without releasing the carrier.

While half-duplex operation was defined for Gigabit Ethernet (with a single repeater), repeaters and hubs failed to emerge - switches had become so cheap that the restrictions with half-duplex operation didn't make sense any more.

To understand the underlying mechanisms it might help to look at an earlier question and read the IEEE 802.3 specifications, especially Clauses 2, 4, 13, 29, and 42.


I have tried to find my own question answers @Zac67 please correct me.

We know transmission time, T.T = 2 * propagation time ,T.P

Which is frame size / bandwidth =( 2 *length of cable) / velocity.

Given bandwidth BW= 10 mbps.

Let length of cable and velocity =d and v respectively.

Now 100 mbps= 10BW and 1 gbps= 100BW.

For first Ethernet,

                     frame size / bandwidth =( 2 *length of cable) / velocity
                 =>L1 / 10BW = 2*(d/v) 

                => L1=2*(d/v) *10BW

               => L1 = 10k bits where k = 2*(d/v) BW

For gigabit Ethernet,

                     frame size / bandwidth =( 2 *length of cable) / velocity
                 =>L2 / 100BW = 2*(d/v) 

                => L2=2*(d/v) *100BW

               => L2 = 100k bits where k = 2*(d/v) BW

Now we have to keep the frame size constant, where as BW is increasing, so we need to minimize the length of the cable because velocity is constant.

If L is the length of the cable in standard Ethernet then to maintain the same frame size for fast Ethernet the length of cable can be reduced by factor of 10 and then for a gigabit Ethernet the length of the cable is decreased by 100.

  • 10BASE-T has a maximum cable length of 100 m. You can't decrease that by a factor of 100 for 1000BASE-T to 1 m. Instead, you need to look at the total propagation latency.
    – Zac67
    Commented Aug 25, 2021 at 21:05
  • @Zac67 could you edit my answer I don't understand..
    – S. M.
    Commented Aug 25, 2021 at 21:18
  • Sorry, I can't do your homework.
    – Zac67
    Commented Aug 25, 2021 at 21:31
  • In fact, 10BASE-T, 100BASE-TX and 1000BASE-T all have the same reach of 100 m.
    – Zac67
    Commented Aug 25, 2021 at 21:33

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