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Ethernet speed - some basic questions

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engr_joni_ee

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Hi,

I am wondering about 1G Ethernet. I guess 1G Ethernet is 1 Giga bits per second which is 125 Mega bytes per second but this is just transmission rate of Ethernet cable and not the data rate (symbol rate) because of package header, right ?

What is the maximum possible data rate (symbol rate) if the PC is directly connected to Ethernet switch ? because if the PC is connected through Ethernet hub then the transmission rate is shared between all the devices connected to the hub, right ?

I guess this also applies to internet downloading speed. Normally we get downloading speed around 10 M bits per second or little more but it can never be more then the Ethernet speed, right ?
 

Yes, 125 MByte/s is the raw speed. For ipv4 udp, 42 bytes are lost per packet for 3 different levels of headers. For tcp it's similar.
To maximize throughput, you want 8K jumbo frames instead of the standard 1500 byte frames, but the jumbo frames must be accepted by all links in the chain.
There are no hubs for 1G ethernet. Switches have a maximum "backplane" speed that is shared by all ports. A good switch can handle full speed on all ports simultaneously.
Your internet speed can not be higher than the ethernet speed of your PC, but some compression techniques are available to make downloading of web pages faster.
 

Thanks for reply.

If for UDP 42 packets are lost from the raw data then what is the size of each packet in bytes ?

I guess the modern Ethernet Networks have TCP/IP and not UDP, is that right ?

I guess one TCP packet header is around 20-60 bytes, so how many packet headers are lost from the raw data per second ?
 

If for UDP 42 packets are lost from the raw data then what is the size of each packet in bytes ?

I guess the modern Ethernet Networks have TCP/IP and not UDP, is that right ?

I guess one TCP packet header is around 20-60 bytes, so how many packet headers are lost from the raw data per second ?

Normal ethernet packets are 64-1500 bytes. Jumbo frames/packets are up to about 8000 bytes.
42 bytes are lost in each packet for udp, but there are 3 levels:
ethernet header: 14 bytes
ipv4 header: 20 bytes
udp header: 8 bytes

The ethernet and ipv4 headers are the same size for tcp, so in total 54-94 bytes are lost per packet (20-60 bytes in the tcp header).

Both udp and tcp are used in "modern Ethernet Networks".
 

Normal ethernet packets are 64-1500 bytes. Jumbo frames/packets are up to about 8000 bytes.

Is this header or packet (which include header and actual information) ?


42 bytes are lost in each packet for udp, but there are 3 levels:
ethernet header: 14 bytes
ipv4 header: 20 bytes
udp header: 8 bytes

This means that for UDP there are 42 bytes which are allocated to packet header and the remaining (125 M bytes - 42 bytes) is the actual data rate per second ?


The ethernet and ipv4 headers are the same size for tcp, so in total 54-94 bytes are lost per packet (20-60 bytes in the tcp header).

If IPV4 packet header size is upto 60 bytes then the remaining (125 M bytes - 60 bytes) is the actual data rate per second ?
 

The headers always exist in every packet, so for udp is 42 bytes "lost" in each packet of 64-1500 bytes.
 

The headers always exist in every packet, so for udp is 42 bytes "lost" in each packet of 64-1500 bytes.

I guess the packet size (64-1500 bytes) depend on the amount of data to transmit, right ? and 42 byes are reserved for header regardless of packet size for UDP, right ?

This means to increase the throughput (data or message rate) the packet size should be as large as possible, right ?

Let's suppose the packet size is 1K byte i.e., 1024 byes and out of which 42 bytes are reserved for header and the remaining (1024 - 42) bytes contains the actual data which is about 95 %.

This means theoretically 95 % of 125 M bytes per second is the useful data (actual message information) for 1G Ethernet, i.e. 118 M bytes per second, right ?
 

This means theoretically 95 % of 125 M bytes per second is the useful data (actual message information) for 1G Ethernet, i.e. 118 M bytes per second, right ?
Not quite. 1 GBPS (4 pairs a 250 MBPS) is raw bit rate, by 8B/10B encoding, the netto data rate is 100 MByte/s.
 

Not quite. 1 GBPS (4 pairs a 250 MBPS) is raw bit rate, by 8B/10B encoding, the netto data rate is 100 MByte/s.

All types of gigabit ethernet have a payload ("netto data rate") of 125 Mbyte/s. Only some types use 8B/10B encoding and for those types the line rate must be 1.25 Gbit/s.
The common 1000BASE-T does not use 8B/10B coding. The symbol rate for 1000BASE-T is 125 MHz and 8 payload bits are transferred in each 12-bit symbol (3 bits per twisted pair).
 
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All types of gigabit ethernet have a payload ("netto data rate") of 125 Mbyte/s. Only some types use 8B/10B encoding and for those types the line rate must be 1.25 Gbit/s.
The common 1000BASE-T does not use 8B/10B coding. The symbol rate for 1000BASE-T is 125 MHz and 8 payload bits are transferred in each 12-bit symbol (3 bits per twisted pair).

I don't understand symbol rate 125 Mbyte/s for 1Gbps.

I guess symbol information is contained within the packet and each packet for UDP is between 64 and 1500 bytes, where in each packet 42 bytes are reserved for header, right ?

This means to increase the throughput (symbol rate) the packet size should be as large as possible, right ?

Let's suppose the packet size is 1K byte i.e., 1024 byes and out of which 42 bytes are reserved for header and the remaining (1024 - 42) bytes contains the actual data (symbol) which is about 95 %.

This means theoretically 95 % of 125 M bytes per second is the useful data (actual symbol information) for 1G Ethernet, i.e. 118 M bytes per second, right ?
 

Hi,

I have found a useful tutorial regarding network performance and frame sizes.

https://www.cisco.com/c/en/us/about/security-center/network-performance-metrics.html

For 1G bps Ethernet

The minimum Frame size is 84 bytes including frame gaps out of which 46 bytes is the payload (data bytes) which means 46 / 84 x 125 M bytes per second = 68.4524 M bytes per second is the actual data rate, the symbol rate, right ?

The maximum Frame size is 1538 bytes including frame gaps out of which 1500 bytes is the payload (data bytes) which means 1500 / 1538 x 125 M bytes per second = 121.9116 M bytes per second is the actual data rate, the symbol rate, right ?
 

@joniengr, symbol rate isn't the best term here. I think you are talking about useable data rates. I think some of the previous posts may be confusing because they focus on the media-dependent, PHY-level symbol rate vs "symbol == byte".

Usable data rate would be what you describe -- the data rate after overhead is removed. Your example removes the 802.3 ethernet overhead. But not overhead for things like IPv4 or UDP. practical systems will have these in most cases. Most applications will also have some amount of overhead as well. The result is that you can get somewhere in this ~118MB/s rate.

But in practice you can either get more or less. To get to 118+ you basically need control of the network. But if you have control of the network you can use jumbo frames. Then you just use 4k/8k packets to make the packet overhead tiny. But if you don't control the network suddenly you have issues beyond just packet headers.
 
I was working this out for 10G ethernet. To get the actual bandwidth, I use iperf in Linux to get actual data rate between two PCs connected to a 10G switch.
 

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