g A computing facility has to perform n processing jobs {J1 , J2 , … , Jn}. Each Ji demands three types of computers: at first, a supercomputer for the time ri , then a regular desktop for the time qi and after this, a specialized computer for the time ti . The facility has only one supercomputer that can perform only one job after another but it has more than n desktops and more than n specialized computers. Design an efficient algorithm allowing finding the schedule to do all jobs with the minimal time and estimate the complexity of this algorithm.

I’ve tried and doesn’t let me either

Idont under stand dat > > > > >

The answer is c. 36======analyzing the for-loop, we see that the control variable, k, starts at 0 and increments by after every loop iteration. the condition  tells us that the loop will stop when k is above 7; in this case, when k is 8.0 1 2 3 4 5 6 7 are the values for k that will be used. each number is for each iteration; therefore, the for-loop will have 8 iterations in total.the code inside the loop itself goes like this: 1st iteration (k=0)starting with sum=0, val=1    means that add  to  .since sum is 0 and val is 1, 0 + 1 = 1. the new value of sum is 1.    means to increment  by 1.since val is 1, the new value of val of 2.end of iteration with  k=0; k is increased by one to become  k=12nd iteration (k=1)starting with sum=1, val=2    ======let us establish a pattern. this is moreso logic than programming: we can  rewrite so that we have the elements from the previous iteration. the 1 came from 0 + 1 before. so we can write it as      .and if we did the third iteration, by the end, sum would look like      notice how the number of plus signs is equal to the iteration you are on (which is just k+1), and what is being added is increasing by one each time.  for example, the third iteration (k=2)  ends up with  with 3 plus signs.the second iteration (k=1) ends up with  with two plus signs.the first iteration (k=0) ends up with  with one plus sign.this is a pattern. the nth iteration will have n plus signs.since we established before that there are 8 iterations of this loop in total, the value of sum at k=7 (the 8th and last iteration) will have sum with 8 plus signs.    so the value of sum at the end of the loop is 36.