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Power Management Algorithms An effort to minimize Processor Temperature and Energy Consumption
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Motivation Microprocessor power consumption is increasing exponentially
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Motivation Battery capacity is increasing linearly Expected battery life increase in the next 5 years: 30 to 40% Chip manufacturers are close to “thermal wall” Increase in speed increase in heat generation Expensive and noisy cooling systems required Intel: Tejas and Jayhawk www.cs.pitt.edu/~kirk/cool.avi www.cs.pitt.edu/~kirk/cool.avi Laptops may damage male fertility due to increased temperature (Reuters: December 9, 2004)
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Motivation Information Technology (IT) consumes about 8% of energy in US Exponential growth 50% of energy consumption Analysis from Intel: 25,000-square-foot server farm with approximately 8,000 servers consumes 2 megawatts -- 25% of the cost of such a facility
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Processor Technologies for Power Management Speed Scaling Processor can operate on multiple speeds o Intel’s SpeedStep — 2 speeds o AMD’s PowerNow — 9 speeds o Intel’s Foxton technology — 64 speeds Power Down Processor can operate on multiple power levels o Can operate on any power level L 0, L 1, …, L n. o L n is normal state. L 0, …, L n-1 are idle states o It costs to bring back processor to L n
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Relationship Between Speed and Energy P = c V 2 s o Minimum voltage V required to run processor at speed s. V is roughly linear to s o Therefore, P = c s 3 o Generalize to P = s p, for some constant p ≥ 1 Energy = ∫ Time P dt Speed goes up(down) Energy consumption goes up (down)
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Relationship Between Speed and Temperature Key Assumption: fixed ambient temperature T a First order approximation of temperature dT/dt = a P – b (T – T a ) = a P – b T T = Temprature t = time P = supplied power a,b some constants For simplicity rescale so that T a = 0
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Problem Formulation Input: A collection of tasks, where task I has: o Release time r i when it arrives in the system o Deadline d i when it must finish by o Work requirement w i (number of cycles) The processor must perform w i units of work between time r i and time d i o Preemption is allowed Objectives o Minimize energy consumption o Minimize maximum temperature For each time, the scheduler must specify both o Job Selection: which job to run may assume Earliest Deadline First policy o Speed Setting: at what speed the processor should run at
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Summary of Results
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Offline YDS Algorithm (1995) Repeat o Find the time interval I with maximum intensity Intensity of time interval I = Σ w i / |I| Where the sum is over tasks i with [r i,d i ] in I o During I speed = to the intensity of I Earliest Deadline First policy o Remove I and the jobs completed in I
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YDS Example Release timedeadline time
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YDS Example First Interval Intensity Second Interval Intensity = green work + blue work Length of solid green line
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YDS Example Final YDS schedule o Height = processor speed YDS theorem: The YDS schedule is optimal for energy, or equivalently for temperature when b = 0. And YDS is optimal for maximum power, or equivalently when b = ∞. o Bansal, Pruhs: Consequence of KKT optimality Bansal, Pruhs: The YDS is at worst 20-competitive with respect to temperature for all cooling parameters b
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Why is YDS optimal? Convex program o They are called KKT optimality conditions The problem has solution if these conditions hold:
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Why is YDS optimal? YDS as convex problem o Break time into intervals t 0,…t m at release times and deadlines o J(i): tasks feasibly executed in I i = [t i,t i+1 ] o W i,j for j in J(i): work done on j during [t i,t i+1 ] KKT optimality conditions hold It took 10 years to prove YDS’s optimality!!!
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Online AVR Algorithm (1995) Each job i has av. rate requirement or density avr i =w i /(d i – r i ) while(t < max d j ) o s(t) = Σavr j (t) o Apply Earliest deadline First policy Yao, Demers, Schenker: 4 ≤ AVR ratio ≤ 8 with respect to energy Bansal, Pruhs: AVR is not O(1)-competitive with respect to temperature AVR(t)
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Online OA Algorithm (1995) After each arrival o Recompute an optimal schedule (YDS alg.) consisting of Newly arrived job j Remaining portions of other jobs Bansal, Pruhs: OA is not O(1)-competitive with respect to temperature
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BKP Algorithm (2004) Algorithm description Speed k(t) at time t = e * maximum over all t 2 > t of Σw i /(t 2 - t 1 ) o Sum is over jobs i with t 1 = et – (e-1)t 2 < r i < t and d i < t 2 Bansal, Pruhs: BKP is O(1)-competitive with respect to temperature tt2t2 riri didi didi t 1 = et – (e-1)t 2 current time Can be computed by an online algorithm
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BKP example Suppose e = 2.7 t = 4 0 1 2 3 4 5 6 4 5 3
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BKP example Suppose e = 2.7 t = 4 For t’ = 5 t 1 = et – (e – 1)t’ = 2.7*4 – (2.7 – 1)*5 = 10.8 – 8.5 = 2.3 0 1 2 3 4 5 6 4 5 3
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BKP example Suppose e = 2.7 t = 4 For t’ = 5 t 1 = et – (e – 1)t’ = 2.7*4 – (2.7 – 1)*5 = 10.8 – 8.5 = 2.3 w(t,t 1,t’) = w(4,2,5) = 4 0 1 2 3 4 5 6 4 5 3
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BKP example Suppose e = 2.7 t = 4 For t’ = 5 t 1 = et – (e – 1)t’ = 2.7*4 – (2.7 – 1)*5 = 10.8 – 8.5 = 2.3 w(t,t 1,t’) = w(4,2,5) = 4 w(t,t 1,t’) /e(t’-t) = w(4,2,5)/2.7(5-4) = 4/2.7 = 1.5 0 1 2 3 4 5 6 4 5 3
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BKP example Suppose e = 2.7 t = 4 For t’ = 6 t 1 = et – (e – 1)t’ = 2.7*4 – (2.7 – 1)*6 = 10.8 – 10.2 = 0 w(t,t 1,t’) = w(4,0,6) = 4 + 5 + 3 = 12 w(t,t 1,t’) /e(t’-t) = w(4,0,6)/2.7*(6-4) = 12/5.4 = 2.22 0 1 2 3 4 5 6 4 5 3
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BKP example Suppose e = 2.7 t = 4 So t 2 = 6 s(4) = e*2.22 = 2.7 * 2.22 = 6 Bansal, Pruhs: BKP is O(1)-competitive with respect to temperature 0 1 2 3 4 5 6 4 5 3
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Future Work Combination of Speed Scaling and Power Down What about multicore processors? What about systems with rejuvinative sources (i.e. solar cells)?
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