Speed Scaling to Manage Energy

and Temperature

Nikhil Bansal (IBM Research)

Tracy Kimbrel (IBM) and Kirk Pruhs (Univ. of Pittsburgh)

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Reasons for Power Management1) 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

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Energy Problem [ Yao, Demers, Shenker 96] Jobs arrive over time:

Job i: arrives at ri, work wi to do by deadline di

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.

0 1 2Area = Work of a job

Work = 2

Work = 2

0 1 2Cost = 2p + 2p

speed 2

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Energy Problem [ Yao, Demers, Shenker 96] Jobs arrive over time:

Job i: arrives at ri, work wi to do by deadline di

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.

0 1 2Area = Work of a job

Work = 2

Work = 2

Cost = 1p + 3p 0 1 2

speed 1

speed 3

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Energy Problem [ Yao, Demers, Shenker 96] Jobs arrive over time:

Job i: arrives at ri, work wi to do by deadline di

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?]

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Energy Problem (Lower Bound)No 2o(p) competitive algorithm possible

0 1 2Work = 2

Online Cost = 2p Offline Cost = 1p + 1p

Ratio ¼ (2)p/2 = O(2p)

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Energy Problem (Lower Bound)No 2o(p) competitive algorithm possible

0 1 2Work = 2

0 1 20 1 2

Online Cost = 1p + 3p Offline Cost = 2p + 2p

Ratio ¼ (3/2)p

Work = 2

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Previous Work [Yao Demers Shenker

96] Optimum Offline Algorithm

Average: Work on each job independently at rate wi/(di-ri)

Competitive ratio 2 [pp,(2p)p] [complicated spectral analysis]

Open: Is there an O(cp) competitive algorithm ?

Example: Wi =1 , ri=0 , di = i

Opt = 1 + 1 + … + 1 = nAverage ¼ k log (n/k)p

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1/21/3

1/n

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Previous Work

YDS propose another algorithm Opt Available (OA)

Work at minimum feasible speed

Speed(t) = maxx (Unfinished Work w/ deadline <= t+x) / x

Open: Competitive ratio of OA ( YDS show that >= pp)?

Example: Wi =1 , ri=0 , di = i

t=0

OA optimal on this example

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1/21/3

1/n

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Our results for Energy Problem1) Analyze OA, show (tight) competitive ratio of pp

[elementary potential function based proof]

2) Give a 8 ep 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.

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Our results for Energy Problem1) Analyze OA, show (tight) competitive ratio of pp

[elementary potential function based proof]

2) Give a 8 ep 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.

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Opt Available (OA)

OA: Work at minimum feasible speed

Suggests: Be a bit more aggressivePerhaps work twice the minimum feasible speed?

OptimumOA

t=0 t=n

speed = 1/n

t=1

Too fast towards the end. Not cp competitive

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Main Algorithm

t t+xt- (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 ¢ maxx w(t,x) / (ex) = maxx w(t,x)/x

Intuition: If W work totally contained in [a,b], then Opt rate ¸ W/(a-b) on average during [a,b]

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Main Algorithm

t t+xt- (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 ¢ maxx w(t,x) / (ex) = maxx w(t,x)/x

e competitive for max-speed2 (p/(p-1))p ep competitive for energy.Bad for small p. Choose best of OA and this: min (pp , 2(p/(p-1))p ep ) · 8 ep

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Main Algorithm

t t+xt- (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 ¢ maxx w(t,x) / (ex) = maxx w(t,x)/x

Need to show:1) Feasibility : All jobs finish by their deadlines2) Bound the energy

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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 maxx w(t,x)/x behave

Non-trivial inequalities due to Hardy and Littlewood(1920’s)Competitive ratio of 2 (p/(p-1))p ep

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Temperature Problem

Fourier’s Law of Heat Conduction: dT(t)/dt = a P(t) – b (T(t) – Ta) (heating term) (cooling term)

T = Temperaturet = timeP = supplied powerTa = ambient temperature (assume stays constant) Rescale so that Ta = 0

Basic Equation: dT/dt = a P - b T

Problem: Finish all jobs while minimizing Tmax

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Temperature

Exact offline algorithm: Convex Program

Wi,j : work done on job j in interval i.Ti, Ti+1: Temp. at beginning and end of Interval i.

Constraints:Each job j receives wj work in total Temperature always below Tmax

ti+1tiInterval i

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MaxW SubproblemStart: time t0, temp T0

End: time t1, temp T1

What is maximum work you can do?

Constraint: Temperature remains ≤ Tmax?

Tmax

t0=0 t1time

T0

T1

?

?

?

Temp Calculus of Variations

Work = st speed(t) dt

= stPower(t)1/3 dt

= st (dT/dt + bT/a)1/3dt

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MaxW w/ Boundary Constraint T ≤ Tmax

Tmax

t0 t1time

T0

T1

Euler Curve

Euler Curve

T = Tmax

α β T = c exp( -bt) + d exp( -3bt/2)

Show convexity and solve using Ellipsoid AlgorithmCan compute subgradient of MaxW(ti,ti+1,Ti,Ti+1)

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Conclusions

Subsequent Work:

O(1) competitive algorithm for Tmax in the online setting.

Future Directions: Tight competitive ratio for Energy problem. Other bicriteria algorithms that trade off energy vs.

quality of schedule.

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Thank You.

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Feasibility

Suppose a deadline first missed at time d.

Online always working till time d.

EDF => all deadlines <= d

maxx 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+xt- (e-1)x