Devise formulas for the functions that calculate my first i and my last i in the global sum example. Remember that each core should be assigned roughly the same number of elements of computations in the loop

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

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

We$$ve implicitly assumed that each call to Compute next value requires roughly the same amount of work as the other calls. How would you change your answer to the preceding question if call i $=$ k requires k $+1$ times as much work as the call with i $=0?$ So if the first call $($i $=0)$ requires $2$ milliseconds, the second call $($i $=1)$ requires $4,$ the third $($i $=2)$ requires $6,$ and so on$.$