The Actual Maximum throughput

 

From running tcpdump on Dublin, we know that the maximum payload on TCP packet is  1448 Byte.

How do you know that?

 

Step 1. Run "tcpdump" on router Dublin.

Step 2. Run "nettest/pingserver" on Prague

Step 3. Run "nettest/pingclient" on Madrid, packet size 1500 Byte ( Max data payload for Ethernet frame )

Step 4. From dublin, we can see the IP fragment is 1448

See the result

[root@dublin /root]# tcpdump -S -i eth2 dst host 192.168.2.2
Kernel filter, protocol ALL, TURBO mode (575 frames), datagram packet socket
tcpdump: listening on eth2
21:45:18.721415 < 192.168.1.2.1641 > 192.168.2.2.1026: S 4098556398:4098556398(0) win 32120 <mss 1460,sackOK,timestamp 51137723 0,nop,wscale 0> (DF)
21:45:18.721640 < 192.168.1.2.1641 > 192.168.2.2.1026: . 4098556399:4098556399(0) ack 4090392485 win 32120 <nop,nop,timestamp 51137723 36471011> (DF)
21:45:18.728551 < 192.168.1.2.1641 > 192.168.2.2.1026: P 4098556399:4098556403(4) ack 4090392485 win 32120 <nop,nop,timestamp 51137724 36471011> (DF)
21:45:18.728706 < 192.168.1.2.1641 > 192.168.2.2.1026: P 4098556403:4098557851(1448) ack 4090392485 win 32120 <nop,nop,timestamp 51137724 36471011> (DF)
21:45:18.736906 < 192.168.1.2.1641 > 192.168.2.2.1026: P 4098557851:4098557903(52) ack 4090392485 win 32120 <nop,nop,timestamp 51137725 36471013> (DF)
21:45:18.737311 < 192.168.1.2.1641 > 192.168.2.2.1026: . 4098557903:4098557903(0) ack 4090393933 win 31856 <nop,nop,timestamp 51137725 36471013> (DF)
21:45:18.741197 < 192.168.1.2.1641 > 192.168.2.2.1026: F 4098557903:4098557903(0) ack 4090393985 win 31856 <nop,nop,timestamp 51137725 36471013> (DF)
21:45:18.741426 < 192.168.1.2.1641 > 192.168.2.2.1026: . 4098557904:4098557904(0) ack 4090393986 win 31856 <nop,nop,timestamp 51137725 36471013> (DF)



[lin@prague nettest]$ pingserver 1026
server: at port 1026 on node prague, internet address 0.0.0.0
server: connected to client at internet address 192.168.1.2, port 1636
server: disconnected from client at internet address 192.168.1.2, port 1636



[lin@madrid nettest]$ pingclient 1026 192.168.2.2 1 1500
pingclient: 1 iterations, 1500 bytes each
pingclient: on host madrid
pingclient: at internet address 192.168.1.2, port 1641
pingclient: sending to server at internet address 192.168.2.2, port 1026
pingclient: finished 1 iterations of 1500 bytes each
Elapsed:    0.01, user:    0.00, sys:     0.00



===============================================================================
The ping test script is:

for i in `seq 1 10`
do
echo " "
echo "pingclient 1026 192.168.2.2 100000 1448"
sleep 3; pingclient 1026 192.168.2.2 100000 1448
done
 

Round Trip Time when packet size is optimal ( 1448 Bytes )      PingTest
  1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th AVG Var RTT
ElapseTime ( Sec ) 63.74 63.77 64.36 64.12 63.67 63.61 63.54 63.81 63.50 63.49 63.761 0.2032 637.61us


ping test Detail Data is HERE



The blast test script is:

for i in `seq 1 10`
do
echo " "
echo "blastclient 1026 192.168.2.2 1000000 1448"
sleep 3; blastclient 1026 192.168.2.2 1000000 1448
done

 

Blast test  when request size is optimal ( 1448 Bytes )      PingTest
  1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th AVG Var Throughput(Byte/Sec)
ElapseTime ( Sec ) 123.21 123.19 123.26 123.35 123.22 123.64 123.32 123.33 123.47 123.17 123.316 0.106 12374712


How do we get 12374712 Byte/Sec

 

Although, we just send useful data 1448 Byte on each request, we actually send TCP and IP header and Ethernet Wrapper. We actually send 1526 Bytes for each frame. ( 1500 for layer 2 data, 26 for ethernet packet wrapper, for header detail, see here )

 

So actual throughput = 1526*1000000/123.316  = 12374712 Bytes/Sec

 

It is (12500000 - 12374712)/12500000 = 1.0023% less than the Ideal Theoretical throughput 100Mb/sec

 

 

Note:

We know 1 Mb/sec in Ethernet specification is 1000000 bit/sec,   1 KB = 1024 Byte, 1 Byte = 8 bits

 

Fast Ethernet 100Mb/Sec = 10^8 bits/Sec  = 12500000 Bytes/sec

 


Blast Test detail Data  is HERE



 

 

Now test the optimal throughput by UDP

The udpserver code is here.

The udpclient code is here.

Updated Makefile is here

 

How do you know that UDP optimal payload is  1472, see experiment here

 

 

Script on client Side( Madrid ): 

[lin@madrid nettest]$ . udp 2>&1 | tee clientInfoOptimal

Where udp is a script file whose content is:

for i in `seq 1 15`;
do
sleep 3
echo " "
echo "/home/lin/nettest/udpclient 1024 192.168.2.2 100000 1490"
/home/lin/nettest/udpclient 1024 192.168.2.2 100000 1490
echo " "
done

 

Script on server side ( Prague ):

[lin@prague nettest]$ . udp 2>&1 | tee serverInfoOptimal

Where udp is a script file whose content is: /home/lin/nettest/udpserver 1024

 

Run tcpdump on prague:  tcpdump dst host 192.168.2.2 and udp

 

 

The result is:

 

UDP test when request size optimal( 1472 Byte ) Sending 1000000 udp packets
  1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th 13th 14th 15th AVG Var Throughput( Byte/sec)
ElapseTime ( Sec ) 123.27 123.48 123.35 123.45 123.45 123.34 123.17 123.19 123.37 123.61 123.20 123.35 123.08 123.11 123.32 123.316 0.1168 12374712
Packet Loss 42 0 0 0 0 108 0 0 0 0 49 0 51 0 61      

 

It's pretty interesting that the Average Elapse Time is the same as that in TCP optimal case.

 

So actual throughput = 1526*1000000/123.316  = 12374712 Bytes/Sec

 

It is (12500000 - 12374712)/12500000 = 1.0023% less than the Ideal Theoretical throughput 100Mb/sec


The detailed experiment data on Elapse Time is here
The detailed experiment data on Packet lost is here