Here is my understanding about the reason behind the Minimum Ethernet Frame Size (64 Bytes).
The old Ethernet LAN spread across the campus area of the university. The normal distance is around 2500meters. The network segmented by 500meter cables. The speed of this LAN is 10Mbps.
Ethernet is a broadcast medium and at a time only one node should sent its frame in that. If more then one node is trying to send the frames, then collision will happen. To avoid the collision the Ethernet is using the CSMA/CD technique.
What is collision? Ethernet device will check the Ethernet medium for some signal. If it finds any, then it will understand that, some other node is already transmitting in the medium and it will not send any frame/packet.
Now we will go into some more details, refer the image. Now machine A see no signal in the medium and start sending the 100 bytes packet. When it is sending the 48 bytes, it is see some signal on the medium and immediately abort the transmission. Is it Ok?? No!?
Ethernet device sending the frame bits for X uSec and pause the transmission and check for a collision on the medium. When it is doing like this, the 1st bit signal of current packet hit the other end of network and echoed back in the opposite direction. Our device is still busy sending the packet and checking for collision. Now what will happen, if it detect it's own frame's echo as the collision?? It will wrongly stop its transmission. If our device does like this, it will never be able to transmit the complete the packet. So it should ignore this collision and continue its transmission.
There is no way to differentiate the collisions, then how it can ignore the collision, which is happening due to its own frame echo. The trick is, by the time, when the its own frame echo reaching back, the all computer in the network identified a collision and stop their transmission. So that the device can safely ignore this collision. Now the important info is, After how much time, I should start ignoring the collision??
We find this using the following steps:
LAN Length (L) = 500 x 5 = 2500 Meters
Propagation Time in the Cable (S) = 200 Meters/uSec
Delay added by repeater (D) = ~3uSec x 2 (Bi-Direction) x 4 Repeaters = 24uSec
Round Trip Delay (RTD) = (Total Distance/S) + Other Delays = (2L/S) + D = (5000/200) + 24 = 49uSec
So we should start ignoring the collision after the RTD Time, 49uSec.
To everyone notice the collision, we should keep transmitting the bits above the RTD Time. So our minimum frame size must be greater then the no. of bits that can be transmitted with in the RTD.
LAN Capacity (C) = 10Mbps
Bits transmitted with-in RTD = RTD x LAN Capacity = 49 x (10^-6) x 10 x (10^6) = 490 Bits.
Minimum packet > 490 Bits
We will take the nearest big number, with power of 8, is 512 Bits.
The minimum Ethernet Frame Size is 512/8 = 64 Bytes
The Maximum frame size is decided to prevent a particular node from hogging the network for long time. For Ethernet, 1500.
To keep the backward compatibility, when network speed increases the LAN length is reduced. But recent technologies are using different technique to over the distance issue.
