Speed Scaling to Manage Energy and Temperature Nikhil Bansal (IBM Research) Tracy Kimbrel (IBM) and...
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Transcript of Speed Scaling to Manage Energy and Temperature Nikhil Bansal (IBM Research) Tracy Kimbrel (IBM) and...
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