Iperf: Transfer of data -


i have question in order understand how iperf working, using following command. dont understand "how can 6945 datagrams send?" because if 9.66 mbytes transfered, 9.66m/1458 = 6625 data grams should tranfereded according understanding.

if 10.125mbytes (2.7mbps * 30 sec) have been transfered 6944 data grams have been send (excluding udp , other header)

please clerify if 1 knows ..

(also have used wireshark on both client , server , checked , there number of packets greater number of packets shown iperf) 

umar@umar-vpceb11fm:~$ iperf -t 30 -c 192.168.3.181 -u -b 2.7m -l 1458 ------------------------------------------------------------ client connecting 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 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%)

iperf uses base 2 m , k, meaning k = 1024 , m = 1024*1024.

when math way, 9.66 mb / 1458 b/d = 6947 datagrams within precision error (you have max resolution of 0.01 mb means rounding error of 0.005 mb ~= 3.6 datagrams).


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