Lecture 22

1. Lecture 22 CSE 331 Oct 26, 2009

2. Blog posts Please sign up if you have not 32 of you have signed up: 4 lectures with no volunteers If I have a pick a blogger I will only pick ONE/lecture Will lose out on 5% of your grade Nov 6 is now open

3. How to do better in this class Look up the blog post Helpful for job interviews!

4. Two definitions for schedules Idle time Max “gap” between two consecutively scheduled tasks Idle time =1 Idle time =0 f=1 Inversion (i,j) is an inversion if i is scheduled before j but di > dj For every i in 1..n do i Schedule job i from si=f to fi=f+ti j f=f+ti j i 0 idle time and 0 inversions for greedy schedule

5. We proved last lecture Any two schedules with 0 idle time and 0 inversions have the same max lateness

6. 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

7. Will prove today 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

8. 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)

9. Today’s agenda “Exchange” argument to convert an optimal solution into a 0 inversion one Shortest Path problem