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Try:
#!/bin/ksh
ERE=${1:-"^G_TP"}
file=${2:-"f16_may_25_16_23_04_fsw1952.txt"}

awk -F, -v ERE="$ERE" '
NR == 1 {
for(i = 1; i <= NF; i++)
if($i ~ ERE)
of[++nof] = i
}
NR < 3 {for(i...

Anything that will always sort higher than any value that can appear in the time field you're processing. If you were processing dates of the form YYYY/MM/DD HH:MM:SS you might choose something like...

Would this come close to what you want (may need some polishing):
awk '
NR == 1 {getline HD1 < F1
HD2 = $0
next
}

$1 >= T[1] {do...

Hmmm. Since you're not familiar with perl, a solution won't be easy for you. One solution for the sorting problem is to insert a marker(and remove it later) at line1/file1, so header of file1 will...

After finally checking out your original try, I'm realizing that creating an epoch timestamp is absolutely not necessary here. The date given in file1/file2 is sortable without transformation.

Ok....

Here another possibillity, a bit more stream oriented ...

perl -MTime::Local -pe '
BEGIN{$fc=1}
s#(([0-9]{4})/([0-9]{2})/([0-9]{2})...