If each call to compute $\_$ next $\_$ val requires different amounts of work $($ For example, if k $=1$and the first call $($ i $=0)$requires $2$milliseconds$,$ the second call $($ i $=1)$would require $4$milliseconds$,$ the third call $($ i $=2)6$milliseconds$,$ and so on $),$you would adjust the formulas for my $\_$ first $\_$ i and my $\_$ last $\_$ i accordingly

these are the formulas:

$*$my$\_$first$\_$i $=($n $/$ p$)*$ core$\_$id

$*$my$\_$last$\_$i $=*$my$\_$first$\_$i $+[($n $/$ p$)-1]$