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Ethernet: max number of packets/sec?

Posted on 2003-03-06
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Last Modified: 2013-12-07
Hi,

I want an algorithm/formula to evaluate max number of packets what can pass ethernet network per second when I give packet size.

I need this for:
- 100Mbps shared
- 100Mbps switched (store-and-forward technology)
- 1Gbps shared/switched
- 10Mbps shared/switched

As an example I want to get this number for tcp/ip packets with 20 bytes of payload data.
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Question by:monas
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6 Comments
 

Expert Comment

by:sw1tch_n1nja
ID: 8081771
your really gonna have a hard time measureing shared bandwidth, that really depends on the size of the network and number of broadcasts
other wise the theoretical maximum fps rates for switched Ethernet are calculated by adding 20 bytes to each frame size to account for the 0.096 microsecond interframe gap equivalent to 12 bytes) and the preamble (eight bytes). Thus, the theoretical maximum Gigabit Ethernet frame rate in fps for a frame of X bytes is defined by the following formula: (1,000,000,000bits/s)/((eight bits/
byte)* (X+20)) for gigabit. and (1,000,000bits/s)/((eight bits/byte)* (X+20)) for 100 meg.
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Author Comment

by:monas
ID: 8082868
Well, shared is more important to me :-)

Let's assume there are no broadcasts. [I have seen very few broadcasts in the packet dump I have grabbed.]

And let's assume that for a few seconds only two nodes are talking to each other (other 40 are silent, if that number is  taken  into account when NIC chooses packet output time).

What times between packets are in this case?
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Expert Comment

by:djluff
ID: 8102589
If you're talking TCP payload (ie. TCP data) you also need to add (at least) 40 bytes of IP and TCP headers. So 20 bytes of TCP data makes (at least) a 68 byte ethernet frame (= 424 bits). Plus the time for the inter-frame gap.

Divide the bandwidth in bits/second by the number of bits in the packet to get the max fps.

On a shared hub this number is the maximum total of all traffic.

In a switch each independant source/destination should be able to achieve this throughput, so the total can be much higher.

The (slight) disadvantage with the switch is that it introduces a delay for each frame, because the switch has to buffer at least up to the end of the destination MAC address before it can start forwarding.
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Author Comment

by:monas
ID: 8102987
djluff,

I know this. Yet I don't know how long this inter-frame gap last. Could you confirm that in shared case this gap is 0.096 microseconds log as sw1tch_n1nja wrote?

That value seems too small for me, because during 9.6e-8 s light travels ~3e8 m/s * 9.6e-8 s = 28.8 m. This is much less than allowed distance of 205m between any nodes.

Or do I forgot completely how ethernet avoids collisions?
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Accepted Solution

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djluff earned 800 total points
ID: 8104147
The inter-frame gap is always 96 bit times (for 10M, 100M, 1000M). So it's 9.6uS at 10M, 960nS at 100M, and 96nS at 1000M.

The signal propagation speed through copper is only around 0,6c (180,000,000m/s). So each bit covers around 1.8m of cable at 100M. 96 bit times * 1.8m = 172m.

But 'normal' collisions are detected during packet transmission, not in the inter-frame gap, as long as the cable length is short enough that the first bit of the frame has time to get to the end of the cable and back before the transmitter finishes sending the last bit.
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Author Comment

by:monas
ID: 8117677
Thanks a lot for explanations.

sw1tch_n1nja, to reward you I opened question: http://www.experts-exchange.com/Networking/Q_20547415.html
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