1. Speed Scaling to Manage Energy and Temperature Nikhil Bansal (IBM Research) Tracy Kimbrel (IBM) and Kirk Pruhs (Univ. of Pittsburgh)

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3. 3 Reasons for Power Management 1) Optimize Energy: Power(t) ¼ c ¢ speed(t) 3 Energy =  t power(t) Power (Pentium 4): 50 W @ 2.60 GHz, 1.8V 7 W @ 1.40 GHz, 1.0V 2) Control Temperature

4. 4 Energy Problem [ Yao, Demers, Shenker 96] Jobs arrive over time: Job i: arrives at r i, work w i to do by deadline d i At time t: Speed s(t) requires power s(t) p, p>1 Goal: Minimize total energy =  t s(t) p, subject to: Finish each job by its deadline. 012 Area = Work of a job Work = 2 012 Cost = 2 p + 2 p speed 2

5. 5 Energy Problem [ Yao, Demers, Shenker 96] Jobs arrive over time: Job i: arrives at r i, work w i to do by deadline d i At time t: Speed s(t) requires power s(t) p, p>1 Goal: Minimize total energy =  t s(t) p, subject to: Finish each job by its deadline. 012 Area = Work of a job Work = 2 Cost = 1 p + 3 p 012 speed 1 speed 3

6. 6 Energy Problem [ Yao, Demers, Shenker 96] Jobs arrive over time: Job i: arrives at r i, work w i to do by deadline d i At time t: Speed s(t) requires power s(t) p, p>1 Goal: Minimize total energy =  t s(t) p, subject to: Finish each job by its deadline. Note: 1) Speed allowed to be arbitrarily fast. 2) Which job: Earliest Deadline First (EDF) [Main issue: How fast to work?]

7. 7 Energy Problem (Lower Bound) No 2 o(p) competitive algorithm possible 012 Work = 2 Online Cost = 2 p Offline Cost = 1 p + 1 p Ratio ¼ (2) p /2 = O(2 p )

8. 8 Energy Problem (Lower Bound) No 2 o(p) competitive algorithm possible 012 Work = 2 012012 Online Cost = 1 p + 3 p Offline Cost = 2 p + 2 p Ratio ¼ (3/2) p Work = 2

9. 9 Previous Work [Yao Demers Shenker 96] Optimum Offline Algorithm Average: Work on each job independently at rate w i /(d i -r i ) Competitive ratio 2 [p p,(2p) p ] [complicated spectral analysis] Open: Is there an O(c p ) competitive algorithm ? Example: W i =1, r i =0, d i = i Opt = 1 + 1 + … + 1 = n Average ¼  k log (n/k) p 1 1/2 1/3 1/n

10. 10 Previous Work YDS propose another algorithm Opt Available (OA) Work at minimum feasible speed Speed(t) = max x (Unfinished Work w/ deadline <= t+x) / x Open: Competitive ratio of OA ( YDS show that >= p p )? Example: W i =1, r i =0, d i = i t=0 OA optimal on this example 1 1/2 1/3 1/n

11. 11 Our results for Energy Problem 1) Analyze OA, show (tight) competitive ratio of p p [elementary potential function based proof] 2) Give a 8 e p competitive online algorithm. 3) Exponent e above is tight. [If p>>1, c.r. determined by max speed, any online algorithm for max speed has c.r. >= e] 4) Tight e competitive algorithm for max speed.

12. 12 Our results for Energy Problem 1) Analyze OA, show (tight) competitive ratio of p p [elementary potential function based proof] 2) Give a 8 e p competitive online algorithm. 3) Exponent e above is tight. [If p>>1, c.r. determined by max speed, any online algorithm for max speed has c.r. >= e] 4) Tight e competitive algorithm for max speed.

13. 13 Opt Available (OA) OA: Work at minimum feasible speed Suggests: Be a bit more aggressive Perhaps work twice the minimum feasible speed? Optimum OA t=0 t=n speed = 1/n t=1 Too fast towards the end. Not c p competitive

14. 14 Main Algorithm tt+x t- (e-1)x w(t,x) : Work arrived by time t, and Totally contained in the interval (t – (e-1)x, t+x) Algorithm: Speed(t) = e ¢ max x w(t,x) / (ex) = max x w(t,x)/x Intuition: If W work totally contained in [a,b], then Opt rate ¸ W/(a-b) on average during [a,b]

15. 15 Main Algorithm tt+x t- (e-1)x w(t,x) : Work arrived by time t, and Totally contained in the interval (t – (e-1)x, t+x) Algorithm: Speed(t) = e ¢ max x w(t,x) / (ex) = max x w(t,x)/x e competitive for max-speed 2 (p/(p-1)) p e p competitive for energy. Bad for small p. Choose best of OA and this: min (p p, 2(p/(p-1)) p e p ) · 8 e p

16. 16 Main Algorithm tt+x t- (e-1)x w(t,x) : Work arrived by time t, and Totally contained in the interval (t – (e-1)x, t+x) Algorithm: Speed(t) = e ¢ max x w(t,x) / (ex) = max x w(t,x)/x Need to show: 1)Feasibility : All jobs finish by their deadlines 2)Bound the energy

17. 17 Bounding the energy Intuition: If W work totally contained in [a,b]. During [a,b], Opt rate >= W/(a-b) on average Problems: 1) Locally very different speeds for online and Opt. 2) How does max x w(t,x)/x behave Non-trivial inequalities due to Hardy and Littlewood(1920’s) Competitive ratio of 2 (p/(p-1)) p e p

18. 18 Temperature Problem Fourier’s Law of Heat Conduction: dT(t)/dt = a P(t) – b (T(t) – T a ) (heating term) (cooling term) T = Temperature t = time P = supplied power T a = ambient temperature (assume stays constant) Rescale so that T a = 0 Basic Equation: dT/dt = a P - b T Problem: Finish all jobs while minimizing T max

19. 19 Temperature Exact offline algorithm: Convex Program W i,j : work done on job j in interval i. T i, T i+1 : Temp. at beginning and end of Interval i. Constraints: Each job j receives w j work in total Temperature always below T max t i+1 titi Interval i

20. 20 MaxW Subproblem Start: time t 0, temp T 0 End: time t 1, temp T 1 What is maximum work you can do? Constraint: Temperature remains ≤ T max ? T max t 0 =0t1t1 time T0T0 T1T1 ? ? ? Temp Calculus of Variations Work = s t speed(t) dt = s t Power(t) 1/3 dt = s t (dT/dt + bT/a) 1/3 dt

21. 21 MaxW w/ Boundary Constraint T ≤ T max T max t0t0 t1t1 time T0T0 T1T1 Euler Curve T = T max αβ T = c exp( -bt) + d exp( -3bt/2) Show convexity and solve using Ellipsoid Algorithm Can compute subgradient of MaxW(t i,t i+1,T i,T i+1 )

22. 22 Conclusions Subsequent Work: O(1) competitive algorithm for T max in the online setting. Future Directions: Tight competitive ratio for Energy problem. Other bicriteria algorithms that trade off energy vs. quality of schedule.

23. 23 Thank You.

24. 24 Feasibility Suppose a deadline first missed at time d. Online always working till time d. EDF => all deadlines <= d max x w(t,x)) /x ¸ w(t,d-t)/(d-t) Proof: Show that total work that arrives during [0,d] with deadline · d. t t+x t- (e-1)x