1. 1 Server Scheduling in the L p norm Nikhil Bansal (CMU) Kirk Pruhs (Univ. of Pittsburgh)

2. 2 Scheduling Provide service such that users are satisfied

3. 3 Motivation Single Machine Arbitrary arrival or release time ( r j ) Arbitrary processing requirement or size ( p j ) t=0 (r 1 ) r2r2 r3r3

4. 4 Motivation t=0 (r 1 ) r2r2 r3r3 c2c2 c3c3 c1c1 Job 1 preempted Single Machine Arbitrary arrival or release time ( r j ) Arbitrary processing requirement or size ( p j )

5. 5 Flow Time t=0 (r 1 ) r2r2 r3r3 c2c2 c3c3 c1c1 Completion time: c j Flow time: f j = c j -r j (time user waits) Flow Time of Job 1

6. 6 Flow Time t=0 (r 1 ) r2r2 r3r3 c2c2 c3c3 c1c1 Completion time: c j Flow time: f j = c j -r j (time user waits) Flow Time of Job 3

7. 7 Stretch Stretch (i) = Flow time (i) / Size (i) [Bender Chakrabarti Muthukrishnan’98] Jobs willing to tolerate flow time proportional to size Also known as normalized flow time Each job contributes equally

8. 8 Known Results Minimize total flow time (  i f i ) [L 1 norm] Optimal Online algorithm: Work on job with smallest remaining proc. time (SRPT) SRPT is 2 – competitive for total stretch [ Gehrke, Muthukrishnan, Rajaraman, Shaheen’99 ] Concern: Big jobs could be stuck! Starvation! Another end: Minimize maximum flow time [L 1 norm] First Come First Served (FCFS) is optimal But bad average performance [Smalls stuck behind big]

9. 9 Balancing average and maximum L p norms: Penalize outliers, good average performance Online algorithms for minimizing 1) Flow time (i.e. (  j f j p ) 1/p ) 2) Stretch (i.e. (  j s j p ) 1/p ) L p norms, previously studied: Load balancing [ Awerbuch Azar Grove+ ’95, Alon Azar Woeginger Yadid ’97, Avidor Azar Sgall ’01 ] Completion time scheduling [ Epstein,Sgall’99 ]

10. 10 Lower Bound L 3 jobs of size 1 By time L 3, If do not finish size L job, bad! Size L Online: Cost of big (L 3 ) 2 = L 6 Opt: L 3 Smalls delayed by L= O(L 5 ) No n o(1) competitive randomized alg for L p norms of flow time and stretch, for 1<p< 1 SRPT, FCFS optimal algorithms for p=1, 1 Suppose p=2

11. 11 Lower Bound No n o(1) competitive randomized alg for L p norms of flow time and stretch, for 1<p< 1 SRPT, FCFS optimal algorithms for p=1, 1 L 3 jobs of size 1 By time L 3, If do not finish size L job, bad! L 5 jobs of size 1/L 2 Optional Stream of jobs Size L

12. 12 Results Thm: SRPT, SJF (Shortest Job First) are 1+  speed O(1/  ) competitive Resource Augmentation: [Kalyanasundaram, Pruhs 95] Online has more resources s-speed, c-competitive max I Online(I,s) / Opt(I,1) · c No starvation unless close to peak capacity

13. 13 Interpreting Resource Augmentation Load Performance Optimal

14. 14 Interpreting Resource Augmentation Load Performance Optimal Online

15. 15 Proof Sketch L p norm of flow time as ‘weighted’ Flow Time problem Age(j,t) = 0 if t c j t-r j if r j < t < c j Observation: f j 2 /2 =  t age(j,t)  j f j 2 /2=  t  j: alive at time t age(j,t) Proof idea: Show, at all times total age of alive jobs < O(1/  ) ¢ total age under Opt

16. 16 Proof Idea If job J of size x delayed for t units under SJF Then either, 1) Had lot of work of size < x before arrival of J 2) Lot of work arriving continuously since J arrived. In either case, Opt can be shown to have sufficiently many “old’’ unfinished jobs,

17. 17 Non-Clairvoyant Scheduling Non-Clairvoyant Model [ Motwani Phillips Torng ’94 ] Scheduler does not know size of a job Learns size only when job finishes. Cannot do things like Shortest Job First, SRPT What can we do? FCFS, Round Robin (time-sharing), …

18. 18 Measuring Performance Resource Augmentation: s-speed, c-competitive Online non-clairvoyant (s,I) Opt offline clairvoyant (1,I) · cMax I

19. 19 The Algorithm (MLF) Multi-Level Feedback (MLF): Used in Windows NT, Unix Levels L 0,L 1,L 2,… job enters L 0 first In L i, receive 2 i amount of work, then promoted to L i+1 Work on level i, iff no job in level 0.. i-1 L0L0 L1L1 L2L2

20. 20 Previous Results (Non-Clairvoyant) L 1 norm of Flow Time: MLF: 1+  speed, O(1/  ) competitive [ Kalyanasundaram, Pruhs ’95 ] L 1 norm of Stretch: MLF: 1+  speed, O(1/  4 log 2 B) competitive [ B., Dhamdhere, Konemann, Sinha’ 03 ] Any algorithm is 1+  speed  (log B/  ) competitive B = ratio of maximum to minimum job size

21. 21 Our Results MLF is 1+  speed, O(1/  4 ) competitive for all norms of flow time MLF is 1+  speed, O(1/  4 log 2 B) competitive for all norms of stretch

22. 22 Analysis Generic technique: reduce MLF to SJF (Shortest Job First) Reduces a non-clairvoyant problem into a clairvoyant one.

23. 23 Final Result Round Robin: At any time, share processor equally Considered to be fair (each job treated equally) But not good according to the L p norm criteria Round Robin is not 1+  speed n o(1) competitive (for sufficiently small  )

24. 24 Open Problems 1) Offline case totally open NP-Hard? Non-trivial approximation algorithms? 2) Multiple machines 3) Other notions to deal with tradeoff between average vs. max performance

25. 25 Thank You!

26. 26 An example Goal: Minimize  f i 2 (i.e. p=2) a job of size 1 arrives for n time units If n small ( < L 2 ), work on the small jobs If n big ( > L 2 ), at time L 2, shift to finish the big job ……… Good algorithm: A combination of SRPT + FCFS Size L

27. 27 MLF 1 3 7 15

28. 28 MLF 1 3 7 15

29. 29 MLF 1 3 7 15

30. 30 MLF 1 3 7 15 1 3 7

31. 31 MLF 1 3 7 15 1 3 7

32. 32 MLF 1 3 7 15 1 3 7

33. 33 MLF 1 3 7 15 1 3 7

34. 34 MLF 1 3 7 15 1 3 7

35. 35 Connection between MLF and SJF 1 3 7 15 1 248 Instance J : 2 i -1 Instance J’ : 1,2,4,…,2 i-1 J J’

36. 36 MLF and SJF 1 3 7 15 1 248 Instance J : 2 i -1 Instance J’ : 1,2,4,…,2 i-1

37. 37 MLF and SJF 1 3 7 15 1 248 Instance J : 2 i -1 Instance J’ : 1,2,4,…,2 i-1

38. 38 MLF and SJF 1 3 7 15 1 248 1 3 7 1 24 Instance J : 2 i -1 Instance J’ : 1,2,4,…,2 i-1

39. 39 MLF and SJF 1 3 7 15 1 248 1 3 7 1 24 MLF works on a job in level i  SJF works on 2 i size copy of same job MLF works on smallest level SJF works on smallest job

40. 40 MLF and SJF 1 3 7 15 1 248 1 3 7 1 24 MLF works on a job in level i  SJF works on 2 i size copy of same job MLF works on smallest level SJF works on smallest job

41. 41 Analysis Idea 2 Main ideas: 1) MLF(J) can be viewed as SJF (J’) 2) We know that SJF(J’) ¼ Opt(J’), so MLF(J) ¼ Opt(J’) Opt(J) previous clairvoyant result Fairly general technique, usually allows us to reduce a non-clairvoyant problem into a clairvoyant one.