R help - Dec 2005 - Snow & rvpm

In the example at
http://www.stat.uiowa.edu/~luke/R/cluster/cluster.html clusterCall
divides the parameter R by the number of nodes -- what you've done is
calculate 999 bootstraps on each node, and compared the execution time
to 999 bootstraps on one node.

You probably want 'clusterCall' to be smart enough to partition the
bootstraps between nodes, and then collate the results. It doesn't do
that.

Martin

vittorio <vdemart1 at tin.it> writes:
> At office, using the internal LAN at my disposal,  I'm having a go at
parallel
> computing - to begin with - with pvm, rpvm & snow.
> The two boxes are as follows
>
> Remote machine uffbsd:
> CPU: Intel(R) Pentium(R) 4 CPU 2.00GHz (1994.13-MHz 686-class CPU)
>   Origin = "GenuineIntel"  Id = 0xf24  Stepping = 4
>  real memory  = 260046848 (248 MB)
>
> This machine NbBSD:
> CPU: Mobile Intel(R) Pentium(R) 4 - M CPU 2.00GHz (1993.54-MHz 686-class
CPU)
> real memory  = 536674304 (511 MB)
>
> And starting library snow under R I have the following situation
>
> clusterCall(cl, function() Sys.info()[c("sysname", 
> "release","nodename","machine")])
> [[1]]
>       sysname       release      nodename       machine
>     "FreeBSD" "5.4-RELEASE" "uffbsd.myd"     
"i386"
>
> [[2]]
>          sysname          release         nodename          machine
>        "FreeBSD"    "6.0-RELEASE" "NbBSD.myd"
"i386"
>
> NOW,
> using the example of boot in the end of page 
> http://www.stat.uiowa.edu/~luke/R/cluster/cluster.html
>
> I find this amazing result:
>
>> system.time(cl.nuke.boot<- 
>
clusterCall(cl,boot,nuke.data,nuke.fun,R=999,m=1,fit.pred=new.fit,x.pred=new.data))
> [1]  0.0078125  0.0078125 27.9609375  0.0000000  0.0000000
>>  system.time(nuke.boot<- 
> boot(nuke.data,nuke.fun,R=999,m=1,fit.pred=new.fit,x.pred=new.data))
> [1] 26.976562  0.109375 28.484375  0.000000  0.000000
>> 
> There's not that much gain in time between my cluster and the local 
> computation, isn't it.
> Now my question is:
> What could have been gone wrong and what should I verify?
>
> Ciao
> Vittorio
>
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> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html

At office, using the internal LAN at my disposal,  I'm having a go at
parallel
computing - to begin with - with pvm, rpvm & snow.
The two boxes are as follows

Remote machine uffbsd:
CPU: Intel(R) Pentium(R) 4 CPU 2.00GHz (1994.13-MHz 686-class CPU)
  Origin = "GenuineIntel"  Id = 0xf24  Stepping = 4
 real memory  = 260046848 (248 MB)

This machine NbBSD:
CPU: Mobile Intel(R) Pentium(R) 4 - M CPU 2.00GHz (1993.54-MHz 686-class CPU)
real memory  = 536674304 (511 MB)

And starting library snow under R I have the following situation

clusterCall(cl, function() Sys.info()[c("sysname", 
"release","nodename","machine")])
[[1]]
      sysname       release      nodename       machine
    "FreeBSD" "5.4-RELEASE" "uffbsd.myd"       
"i386"

[[2]]
         sysname          release         nodename          machine
       "FreeBSD"    "6.0-RELEASE" "NbBSD.myd"     
"i386"

NOW,
using the example of boot in the end of page 
http://www.stat.uiowa.edu/~luke/R/cluster/cluster.html

I find this amazing result:
> system.time(cl.nuke.boot<- 
clusterCall(cl,boot,nuke.data,nuke.fun,R=999,m=1,fit.pred=new.fit,x.pred=new.data))
[1]  0.0078125  0.0078125 27.9609375  0.0000000 
0.0000000
>  system.time(nuke.boot<- 
boot(nuke.data,nuke.fun,R=999,m=1,fit.pred=new.fit,x.pred=new.data))
[1] 26.976562  0.109375 28.484375  0.000000  0.000000
> 
There's not that much gain in time between my cluster and the local 
computation, isn't it.
Now my question is:
What could have been gone wrong and what should I verify?

Ciao
Vittorio