There
is an obvious reason why the frame payload size was chosen to be 1500
bytes. A frame size of 1500 bytes, offers, maximum efficiency or
throughput.
As you know, ethernet frame has 8 byte
preamble, 6 byte source and 6 byte destination mac address, mac type of 2
bytes, and 4 bytes CRC. Assuming the MTU payload to be 1500 the total
number of bytes comes to 1500 + 8 + 6 + 6 + 2 + 4 = 1526 bytes. Now
between each frame there is a inter frame gap of 12 bytes which
constitues 9.6micro seconds gap between each frame. This is essential so
that frames dont mix up. So the total size of each frame going out of a
host is 1538 bytes.
So at 10 Mbps rate, the frame rate is 10 Mbps / 1538 bytes = 812.74 frames / second.
Now
we can find the throughput or efficiency of link, to transmit 1500
bytes of payload. by multiplying the frame rate with the number of bytes
of the payload.
So efficiency = 812.74 * 1500 * 8 = 9752925.xxxxx bps which is 97.5 percent efficient ( comparing with 10 MBps)
I
guess I have gone too much with mathematics of Ethernet, but the
interesting thing to notice is that, as the number of bytes in the
payload increases, the frame rate is decreasing. See that for an MTU of
1500 bytes on payload, the frame rate has reduced to 812 frames per
second. If you increase it above 1500, frame rate would become less
than 812.
Also there is a minimum limit for the MTU which
is actually 46 bytes. If you calculate the size of the frame for a 46
byte payload it would come to 12+8+6+6+2+46+4 = 84 bytes. Now
calculating the frame rate we get it as =
10mbps/ (84 * 8 bytes) =
14880 frames per second. We could have gone to a frame size even lesser
than this, which could increase the frame rate even more, but I guess
during those times, when IEEE made the standards, the routers didnt have
that much frame forwarding capability.
So I think due to
above reasons, and considering maximum efficiency, IEEE would have fixed
the min and max size of payload as 46 bytes and 1500 bytes.
There is an obvious reason why the frame payload size was chosen to be 1500 bytes. A frame size of 1500 bytes, offers, maximum efficiency or throughput.
As you know, ethernet frame has 8 byte preamble, 6 byte source and 6 byte destination mac address, mac type of 2 bytes, and 4 bytes CRC. Assuming the MTU payload to be 1500 the total number of bytes comes to 1500 + 8 + 6 + 6 + 2 + 4 = 1526 bytes. Now between each frame there is a inter frame gap of 12 bytes which constitues 9.6micro seconds gap between each frame. This is essential so that frames dont mix up. So the total size of each frame going out of a host is 1538 bytes.
So at 10 Mbps rate, the frame rate is 10 Mbps / 1538 bytes = 812.74 frames / second.
Now we can find the throughput or efficiency of link, to transmit 1500 bytes of payload. by multiplying the frame rate with the number of bytes of the payload.
So efficiency = 812.74 * 1500 * 8 = 9752925.xxxxx bps which is 97.5 percent efficient ( comparing with 10 MBps)
I guess I have gone too much with mathematics of Ethernet, but the interesting thing to notice is that, as the number of bytes in the payload increases, the frame rate is decreasing. See that for an MTU of 1500 bytes on payload, the frame rate has reduced to 812 frames per second. If you increase it above 1500, frame rate would become less than 812.
Also there is a minimum limit for the MTU which is actually 46 bytes. If you calculate the size of the frame for a 46 byte payload it would come to 12+8+6+6+2+46+4 = 84 bytes. Now calculating the frame rate we get it as =
10mbps/ (84 * 8 bytes) = 14880 frames per second. We could have gone to a frame size even lesser than this, which could increase the frame rate even more, but I guess during those times, when IEEE made the standards, the routers didnt have that much frame forwarding capability.
So I think due to above reasons, and considering maximum efficiency, IEEE would have fixed the min and max size of payload as 46 bytes and 1500 bytes.