<br><font size=2 face="sans-serif">I used cyclets_to_units() get </font>
<br>
<br><font size=2 face="sans-serif">Bandwidth peak (#0 to #972): 1852.72
MByte/sec</font>
<br><font size=2 face="sans-serif">Bandwidth average: 1852.67 MByte/sec</font>
<br>
<br><font size=2 face="sans-serif">Which is far away from the 1X (2.5Gbit/s)
throughtput.</font>
<br>
<br><font size=2 face="sans-serif">This per secs cycles value is incorrect.
Either change the measurment to gettimeofday() to measure MB/s or</font>
<br><font size=2 face="sans-serif">measure the how many cycles per sec
by doing something like that:</font>
<br>
<br><font size=2 face="sans-serif">#define MEASURE_TIME 16</font>
<br>
<br><font size=2 face="sans-serif">get_cycles(tbegin);</font>
<br><font size=2 face="sans-serif">sleep(MEASURE_TIME);</font>
<br><font size=2 face="sans-serif">get_cycles(tend)</font>
<br><font size=2 face="sans-serif">cycles_to_units = (tend-tbegin)/MEASURE_TIME</font>
<br>
<br><font size=2 face="sans-serif">to caculate the right MB/s throughput.
</font>
<br><font size=2 face="sans-serif"><br>
Thanks<br>
Shirley Ma<br>
IBM Linux Technology Center<br>
15300 SW Koll Parkway<br>
Beaverton, OR 97006-6063<br>
Phone(Fax): (503) 578-7638<br>
<br>
</font>