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I have a question in order to understand how iperf is working, I am using the following command. What i dont understand is "How can 6945 datagrams are send?" because if 9.66 MBytes are transfered, then 9.66M/1458 = 6625 data grams should be tranfereded according to my understanding.

If 10.125MBytes (2.7Mbps * 30 sec) would have been transfered then 6944 data grams would have been send (excluding udp and other header)

Please clerify if some one knows ..

(Also I have used wireshark on both client and server and checked and there the number of packets is greater then the number of packets shown by iperf) 

umar@umar-VPCEB11FM:~$ iperf -t 30 -c 192.168.3.181 -u -b 2.7m -l 1458
------------------------------------------------------------
Client connecting to 192.168.3.181, UDP port 5001
Sending 1458 byte datagrams
UDP buffer size:  208 KByte (default)
------------------------------------------------------------
[  3] local 192.168.3.175 port 47241 connected with 192.168.3.181 port 5001
[ ID] Interval       Transfer     Bandwidth
[  3]  0.0-30.0 sec  9.66 MBytes  2.70 Mbits/sec
[  3] Sent 6946 datagrams
[  3] Server Report:
[  3]  0.0-92318.4 sec  9.66 MBytes   878 bits/sec   0.760 ms    0/ 6945 (0%)
Umar
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1 Answers1

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iperf uses base 2 for M and K, meaning that K = 1024 and M = 1024*1024.

When you do that math that way, you get 9.66 MB / 1458 B/d = 6947 datagrams which is within precision error (you have a max resolution of 0.01 MB which means a rounding error of 0.005 MB ~= 3.6 datagrams).

justhecuke
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