[scyld-users] Code execution time

[Harshavardhan Ghuge]

Hi,
I have executed a Fortran code on scyld beowulf cluster by two ways. In
the first method I store the input data file in a subdirectory on the
master node. The binary is stored in the same sub directory. Following
is the command I use-
bpsh 6 sh -c "time ./binary"
The program writes data to a output data file which gets stored in the
same sub directory on the master node.
In this method the input data file is 4 GB.  It is used through the
NFS.The output data file is not very big.
Time required for the code execution is- 10 mins 33.073 sec

In the second method I store the input data file in a directory /scratch
on compute node 6. This directory is not on the NFS. I want the output
data file to be written in the /scratch on compute node 6. So I have
specified the path of the input data file and the output data file in
the Fortran code itself. I use the following command- 
bpsh 6 sh -c "time ./binary"
Note the binary is still stored on the master node. The input data file
stored in the /scratch directory is used and the output data file is
created in the /scratch directory. The input data file which is 4 GB is
not used though the NFS as it is stored on the node itself. Same is the
case for the output data file.
The time required for code execution is - 10 min 42.068 sec 

I expected that the code execution time for the second method will be
less than the first method. I thought that in the first method the huge
input data file is used  through NFS and therefore the code execution
time will be more. But the results are other way round. Can somebody
tell me why?

Harshal