Rounding Technique We can simplify scheduling problem structure by rounding job sizes, release dates

Rounding Scheme 1 Make all processing times and release dates integral powers of 1+ε –R x = (1+ε) x –I x = [R x, R x+1 ) –Length of I x = ε R x R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7 1 1+ε (1+ε) 2

Effect of rounding on optimal average completion time How much does this rounding increase optimal average completion time? Give argument: R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7 1 1+ε (1+ε) 2

Enforce r j > εp j Make each job j have a release time of at least εp j Argue this causes at most (1+ε) effect on optimal average completion time

PTAS for 1|r j |ΣC j Run SPT until at most 3/ε 7 jobs are left –Each job released at max{r j,p j /ε 2 } Enumerate last jobs optimally R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7

Key idea 1: Time stretching A job is small if p j < εI x where I x is interval that job j is started No small job crosses into interval I x+1 Stretch interval I x by (1+ε) to fit last crossing job in R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7

Observation If all jobs are small in their intervals, with previous time-stretching idea, we achieve optimality as we approximate SRPT which is optimal for the preemptive version of this problem. R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7

Key idea 2: Moving large jobs We can move large jobs that appear before time t = ε 7 OPT back in the schedule by some amount to where they are small Key arguments –Space to fit these jobs (time stretching) –Delay caused by stretching is not too large R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7

Rounding Scheme 2 Round release dates and processing times to integral powers of εP/n 2 where P is the largest job size How much does this rounding affect average flow time? Key observation: all processing times in interval [0, n 2 /ε] and all events occur in interval [0,2n 3 /ε] Leads to a polynomial number of decision times to be represented in a dynamic programming solution