Lecture 25 CSE 331 Oct 29, 2010

HW 6 due today All questions in one pile I will not take any HW after 1:15pm

Blog posts/Group Scribe leader Please sign up if you have not If I have a pick a blogger/leader I will only pick THREE/lecture Will lose out on 5%-10% of your grade There are many lectures with <3 students

Solutions to HW 6 END of the lecture “Model” answers

HW 7 Posted on the blog/webpage

Two definitions for schedules f=s For every i in 1..n do Schedule job i from s(i)=f to f(i)=f+t i f=f+t i Idle time Inversion Max “gap” between two consecutively scheduled tasks i i j j Idle time =1Idle time =0 (i,j) is an inversion if i is scheduled before j but d i > d j i i j j i i j j What is the idle time and max # inversion for greedy schedule?

Proving greedy is optimal Any two schedules with 0 idle time and 0 inversions have the same max lateness Greedy schedule has 0 idle time and 0 inversions There is an optimal schedule with 0 idle time and 0 inversions

HW 6 due today All questions in one pile I will not take any HW after 1:15pm

Optimal schedule with 0 idle time = = = = ≥ ≥ ≥ ≥ Lateness “Only” need to convert a 0 idle optimal ordering to one with 0 inversions (and 0 idle time)

Today’s agenda “Exchange” argument to convert an optimal solution into a 0 inversion one