* Revise the machine selection algorithm. Previously it chose the first

machine with the lowest number of running jobs.  This worked when the
  clients were all roughly equivalent, but schedules poorly when there
  are some that are much more powerful (e.g. 8-core machines vs UP machines)

* We now compute the ratio of running jobs to maximum jobs and schedule on
  the machine with lowest occupation fraction.  This populates the machines
  to equal fractions of their capacity.

Tools/portbuild/scripts/getmachine

@@ -17,20 +17,28 @@ if [ "$1" = "*" ]; then
   exit 1
 fi
 
-min=9999
+
+min=999999
 while [ $# -gt 0 ]; do
   m=$1
-  num=$(cat $m)
-  if [ $num -lt $min ]; then
+
+  # Pull in maxjobs
+  . ${pb}/${arch}/portbuild.conf
+  test -f ${pb}/${arch}/portbuild.${m} && . ${pb}/${arch}/portbuild.${m}
+
+  curjobs=$(cat $m)
+  weight=$((${curjobs}*1000/${maxjobs}))
+
+  if [ $weight -lt $min ]; then
     mach=$m
-    min=$num
-  elif [ $num -eq $min ]; then
+    min=$weight
+  elif [ $weight -eq $min ]; then
     mach="${mach} ${m}"
   fi
   shift
 done
 
-if [ "$min" = 9999 -o -z "${mach}" ]; then
+if [ "$min" = 999999 -o -z "${mach}" ]; then
   echo ""
   exit 1
 fi
@@ -45,11 +53,12 @@ fi
 . ${pb}/${arch}/portbuild.conf
 test -f ${pb}/${arch}/portbuild.${mach} && . ${pb}/${arch}/portbuild.${mach}
 
+curjobs=$(cat $mach)
 # Now that we've found a machine, register our claim in the queue
-if [ "$((${min}+1))" -ge "${maxjobs}" ]; then
+if [ "$((${curjobs}+1))" -ge "${maxjobs}" ]; then
   rm ${mach}
 else
-  echo $(($min+1)) > ${mach}
+  echo $(($curjobs+1)) > ${mach}
 fi
 
 # Report to caller