They seem quite fond of these statistics. They are also defined in: Structure and expression analysis of rice paleo duplications, where the equation for CIP is described in a slightly different (and clearer) way - "by" is the same as "/":

∑nb ID/HSP/AL x 100

In "Bioperl terms", I would say that:

ID  = $hsp->num_identical
HSP = $hsp->length('total')
AL  = sum of all $hsp->length('total')

So you would first calculate AL from all HSPs. Then you would calculate ID and HSP for each $hsp, divide by AL, multiply by 100 and sum them all together. Basically, it just gives you a percentage identity across the whole sequence by adding together the identities for the HSPs.

Thanks first,"add comment" have a word number limitation，so I write my try by this way.

Here is my perl script:

use warnings;
use strict;
use Bio::SearchIO;

my $infile  = shift;

my $parser = Bio::SearchIO->new(-file   => $infile,
                                -format => 'blast');
while( my $result = $parser->next_result ) {
    print "Query: ",$result->query_name,"\tlength: ",$result->query_length,
          "\tnumber of hits: ",$result->num_hits,"\n";
    while(my $hit=$result->next_hit){
        print "\tHit: ",$hit->name,"\tlength: ",$hit->hit_length,
                  "\te-value: ",$hit->significance,"\n";
        my $al=0;
        my @id=();
        while(my $hsp=$hit->next_hsp){
            print "\t\tHsp length: ",$hsp->hsp_length,"\tID length: ",
                          $hsp->num_identical,"\t idenity: ",
                          $hsp->percent_identity,"\n";
            $al=$al+($hsp->hsp_length);
            my $ai=($hsp->num_identical)/($hsp->hsp_length);
            push @id,$ai;
        }
        my $cip=0;
        foreach (@id){
            $cip=$cip+($_/$al*100);
            }
        my $calp=$al/($result->query_length);
        print "\tcip: ",$cip,"\tcalp: ",$calp,"\n";
    }
}

I parsed the blastn result (cutoff: 1e-40,all genes from rice chromosome 11 vs all genes from Brachypodium distachyon chromosome 4 )

single hsp sample output:

Query: LOC_Os11g01010.1 length: 357 number of hits: 1
    Hit: Bradi4g45390.1 length: 366 e-value: 9e-76
        Hsp length: 337 ID length: 288   idenity: 85.459940652819
    cip: 0.253590328346644  calp: 0.943977591036415

and multiple hsps sample output:

Query: LOC_Os11g01380.1 length: 5127    number of hits: 4
    Hit: Bradi4g26880.1 length: 4821    e-value: 0.0
        Hsp length: 2829    ID length: 2576  idenity: 91.0569105691057
        Hsp length: 1437    ID length: 1298  idenity: 90.3270702853166
        Hsp length: 432 ID length: 387   idenity: 89.5833333333333
        cip: 0.0576771635137837         calp: 0.916325336454067

All the CIP value are too low, and nearly no hit is greater than 60% (cutoff value as the paper said). But truth is the two chromosome were orthologous chromosomes and have a large high conserved segment. If my script is right , the result infers that this explanation of the formula may be unreasonable.