A different re-execution speed can help

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1 A different re-execution speed can help Anne Benoit, Aurélien Cavelan, alentin Le Fèvre, Yves Robert, Hongyang Sun LIP, ENS de Lyon, France PASA orkshop, in conjunction with ICPP 16 August 16, 2016 A different re-execution speed can help PASA 16 1 / 25

2 Motivation: Resilience Large-scale platforms: increasingly subject to errors Major challenge for Exascale: frequent striking of silent errors How to deal with these errors? erification + Checkpoint/Restart erification mechanism: general-purpose (replication, triplication) or application-specific erified checkpoints: a verification is performed just before each checkpoint C C C Time Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 2 / 25

3 Silent vs Fail-stop errors C: time to checkpoint; λ: error rate (platform MTBF µ = 1/λ); : time to verify; R: time to recover Optimal checkpointing period for fail-stop errors (Young/Daly): = 2C/λ ( = 0) Fail-stop error C? R C C Time Silent errors: = ( + C)/λ (C + C; missing factor 2) Silent error Detection C R C C Time Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 3 / 25

4 Motivation: Energy consumption Power requirement of current petascale platforms = small town Need to reduce energy consumption of future platforms Popular technique: dynamic voltage and frequency scaling (DFS) Lower speed energy savings: when computing at speed σ, power proportional to σ 3 and execution time proportional to 1/σ (dynamic) energy proportional to σ 2 Also account for static energy: trade-offs to be found Realistic approach: minimize energy while guaranteeing a performance bound Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 4 / 25

5 Motivation: Energy consumption Power requirement of current petascale platforms = small town Need to reduce energy consumption of future platforms Popular technique: dynamic voltage and frequency scaling (DFS) Lower speed energy savings: when computing at speed σ, power proportional to σ 3 and execution time proportional to 1/σ (dynamic) energy proportional to σ 2 Also account for static energy: trade-offs to be found Realistic approach: minimize energy while guaranteeing a performance bound At which speed should we execute the workload? Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 4 / 25

6 Outline of the talk Model and optimization problem Optimal pattern size and speeds Simulations Extensions: both fail-stop and silent errors Conclusion A different re-execution speed can help PASA 16 5 / 25

7 Framework Divisible-load applications Subject to silent data corruption Checkpoint/restart strategy: periodic patterns that repeat over time erified checkpoints Is it better to use two different speeds rather than only one? hat are the optimal checkpointing period and optimal execution speeds? A different re-execution speed can help PASA 16 6 / 25

8 Model Set of speeds S = {s 1,..., s K }: σ 1 S speed for first execution, σ 2 S speed for re-executions Silent error Detection C R σ2 C σ2 C Time ith a silent error Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 7 / 25

9 Model Set of speeds S = {s 1,..., s K }: σ 1 S speed for first execution, σ 2 S speed for re-executions Silent errors: exponential distribution of rate λ Silent error Detection C R σ2 C σ2 C Time ith a silent error Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 7 / 25

10 Model Set of speeds S = {s 1,..., s K }: σ 1 S speed for first execution, σ 2 S speed for re-executions Silent errors: exponential distribution of rate λ erif.: units of work; checkpointing: time C; recovery: time R Silent error Detection C R σ2 C σ2 C Time ith a silent error Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 7 / 25

11 Model Set of speeds S = {s 1,..., s K }: σ 1 S speed for first execution, σ 2 S speed for re-executions Silent errors: exponential distribution of rate λ erif.: units of work; checkpointing: time C; recovery: time R P idle and P io constant; and P cpu (σ) = κσ 3 Silent error Detection C R σ2 C σ2 C Time ith a silent error Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 7 / 25

12 Model Set of speeds S = {s 1,..., s K }: σ 1 S speed for first execution, σ 2 S speed for re-executions Silent errors: exponential distribution of rate λ erif.: units of work; checkpointing: time C; recovery: time R P idle and P io constant; and P cpu (σ) = κσ 3 Energy for units of work at speed σ: σ (P idle + κσ 3 ) Energy of a verification at speed σ: σ (P idle + κσ 3 ) Energy of a checkpoint: C(P idle + P io ) Energy of a recovery: R(P idle + P io ) Silent error Detection C R σ2 C σ2 C Time ith a silent error Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 7 / 25

13 Problem Optimization problem BiCrit: Minimize E(, σ 1, σ 2 ) s.t. T (, σ 1, σ 2 ) ρ, E(, σ 1, σ 2 ) is the expected energy consumed to execute units of work at speed σ 1, with eventual re-executions at speed σ 2 T (, σ 1, σ 2 ) is the expected execution time to execute units of work at speed σ 1, with eventual re-executions at speed σ 2 ρ is a performance bound, or admissible degradation factor Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 8 / 25

14 Computing expected execution time Proposition 1 For the BiCrit problem with a single speed, ( ) T (, σ, σ) = C + e λ + ) σ + (e λ σ 1 R σ Proposition 2 For the BiCrit problem, T (, σ 1, σ 2 ) = C + + ) + (1 e λ σ 1 e λ σ 2 (R + + σ 1 σ 2 ) Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA 16 9 / 25

15 Proof of Proposition 1 Proof. The recursive equation to compute T (, σ, σ) writes: T (, σ, σ) = + σ + p( /σ) (R + T (, σ, σ)) + (1 p( /σ))c, where p( /σ) = 1 e λ σ. The reasoning is as follows: e always execute units of work followed by the verification, in time + σ ; ith probability p( /σ), a silent error occurred and is detected, in which case we recover and start anew; Otherwise, with probability 1 p( /σ), we simply checkpoint after a successful execution. Solving this equation leads to the expected execution time. Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

16 Proof of Proposition 2 Proof. The recursive equation to compute T (, σ 1, σ 2 ) writes: T (, σ 1, σ 2 ) = + σ 1 + p( /σ 1 ) (R + T (, σ 2, σ 2 )) + (1 p( /σ 1 ))C, where p( /σ 1 ) = 1 e λ σ 1. The reasoning is as follows: e always execute units of work followed by the verification, in time + σ 1 ; ith probability p( /σ 1 ), a silent error occurred and is detected, in which case we recover and start anew at speed σ 2 ; Otherwise, with probability 1 p( /σ 1 ), we simply checkpoint after a successful execution. Solving this equation leads to the expected execution time. Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

17 Computing expected energy consumption Proposition 3 For the BiCrit problem, E(, σ 1, σ 2 ) = ( ) ) C + (1 e λ σ 1 e λ σ 2 R (P io + P idle ) + + σ 1 (κσ P idle ) + + σ 2 (1 e λ σ 1 )e λ σ 2 (κσ2 3 + P idle ) Power spent during checkpoint or recovery: P io + P idle ; power spent during computation and verification at speed σ: P cpu (σ) + P idle = κσ 3 + P idle. From Proposition 2, we get the expression of E(, σ 1, σ 2 ). Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

18 Finding optimal pattern length (1) To get closed-form expression for optimal value of, use of first-order approximations, using Taylor expansion e λ = 1 + λ + O(λ 2 2 ): T (, σ 1, σ 2 ) = 1 + λ + λr + λ + C + /σ 1 + O(λ 2 ) (1) σ 1 σ 1 σ 2 σ 1 σ 1 σ 2 E(, σ 1, σ 2 ) = κσ3 1 + P idle σ 1 + λ σ 1 σ 2 (κσ P idle ) + λr σ 1 (P io + P idle ) + λ σ 1 σ 2 (κσ P idle ) + C(P io + P idle ) + (κσ P idle)/σ 1 + O(λ 2 ) (2) Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

19 Finding optimal pattern length (2) Theorem 1 Given σ 1, σ 2 and ρ, consider ( the ) equation a 2 + b + c = 0, where a = λ σ 1 σ 2, b = 1 σ 1 + λ R + σ 1 σ 2 ρ and c = C + σ 1. If there is no positive solution to the equation, i.e., b > 2 ac, then BiCrit has no solution. Otherwise, let 1 and 2 be the two solutions of the equation with 1 2 (at least 2 is positive and possibly 1 = 2 ). Then, the optimal pattern size is opt = min(max( 1, e ), 2 ), (3) where e = C(P io + P idle ) + σ 1 (κ 3 + P idle) λ σ 1 σ 2 (κσ2 3 + P. (4) idle) Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

20 Finding optimal pattern length (3) Proof. Neglecting lower-order terms, Equation (2) is minimized when = e given by Equation (4). Two cases: ρ is too small no solution 2 > 0: e < 1 1 e 2 e > 2 Using that the energy overhead is a convex function, we get the result ( opt is in the interval [ 1, 2 ]) Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

21 Finding optimal speed pair Speed pair (s i, s j ), with 1 i, j K: ρ i,j is the minimum performance bound for which the BiCrit problem with σ 1 = s i and σ 2 = s j admits a solution For each speed pair, compute 1, 2 the roots of a 2 + b + c; discard pairs with ρ < ρ i,j For each remaining speed pair (σ 1, σ 2 ), compute opt and associated energy overhead Select speed pair (, σ 2 ) that minimizes energy overhead Time O(K 2 ), where K is the number of available speeds, usually a small constant Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

22 Simulation setup Platform parameters, based on real platforms Platform λ C = R Hera 3.38e-6 300s 15.4 Atlas 7.78e-6 439s 9.1 Coastal 2.01e s 4.5 Coastal SSD 2.01e s Power parameters, determined by the processor used Processor Normalized speeds P(σ) (m) Intel Xscale 0.15, 0.4, 0.6, 0.8, σ Transmeta Crusoe 0.45, 0.6, 0.8, 0.9, σ Default values: P io equivalent to power used when running at lowest speed; ρ = 3 Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

23 Simulation results, using Hera/XScale configuration A different re-execution speed does help! And all speed pairs can be optimal solutions (depending on ρ)! E( σ 1 Best σ 2 opt,σ 1,σ 2 ) opt opt ρ = 8 E( σ 1 Best σ 2 opt,σ 1,σ 2 ) opt opt ρ = 3 E( σ 1 Best σ 2 opt,σ 1,σ 2 ) opt opt ρ = E( σ 1 Best σ 2 opt,σ 1,σ 2 ) opt opt ρ = 1.4 Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

24 Simulations - Impact of the parameters (1) σ 1 σ 2σ opt(σ 1,σ 2) opt(σ,σ) E(opt,σ 1,σ 2)/opt E(opt,σ,σ)/opt Speed Optimal Energy overhead C C C Figure: The optimal solution (speed pair, pattern size, and energy overhead) as a function of the checkpointing time C in Atlas/Crusoe configuration σ 1 σ 2σ opt(σ 1,σ 2) opt(σ,σ) E(opt,σ 1,σ 2)/opt E(opt,σ,σ)/opt Speed Optimal Energy overhead Figure: The optimal solution (speed pair, pattern size, and energy overhead) as a function of the verification time in Atlas/Crusoe configuration. Dotted line: one single speed; up to 35% improvement with two speeds Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

25 Simulations - Impact of the parameters (2) Speed σ 1 σ 2σ Optimal opt(σ 1,σ 2) opt(σ,σ) Energy overhead E(opt,σ 1,σ 2)/opt E(opt,σ,σ)/opt λ λ λ Figure: The optimal solution (speed pair, pattern size, and energy overhead) as a function of the error rate λ in Atlas/Crusoe configuration. Speed σ 1 σ 2σ Optimal opt(σ 1,σ 2) opt(σ,σ) Energy overhead E(opt,σ 1,σ 2)/opt E(opt,σ,σ)/opt ρ ρ ρ Figure: The optimal solution (speed pair, pattern size, and energy overhead) as a function of the performance bound ρ in Atlas/Crusoe configuration. Two speeds: checkpoint less frequently and provide energy savings Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

26 Simulations - Impact of the parameters (3) Speed σ 1 σ 2σ Optimal Energy overhead E(opt,σ 1,σ 2)/opt E(opt,σ,σ)/opt P idle 3600 opt(σ 1,σ 2) opt(σ,σ) P idle P idle Figure: The optimal solution (speed pair, pattern size, and energy overhead) as a function of the idle power P idle in Atlas/Crusoe configuration. Speed σ 1 σ 2σ Optimal Energy overhead E(opt,σ 1,σ 2)/opt E(opt,σ,σ)/opt opt(σ 1,σ 2) opt(σ,σ) P io P io P io Figure: The optimal solution (speed pair, pattern size, and energy overhead) as a function of the I/O power P io in Atlas/Crusoe configuration. Increase of and E with P idle and Pio; P io has no impact on speeds Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

27 Extensions: ith fail-stop errors f : proportion of fail-stop errors s: proportion of silent errors Proposition 4 ith fail-stop and silent errors, ( T (, σ 1, σ 2 ) (f + s) = + f σ 1 σ ( E(, σ 1, σ 2 ) (f + s)(κσ 3 = P idle ) σ 1 σ 2 ) λ + O(λ 2 ). (5) f (κσ3 1 + P ) idle) 2 2 λ + O(λ 2 ) (6) Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

28 Limit of the first-order approximation For BiCrit, the first-order approximation leads to a solution iff ( ( s )) 1/2 σ ( 2 < < s ) f σ 1 f Use second-order approximation? Open problem in the general case! Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

29 Interesting case Theorem 2 hen considering only fail-stop errors with rate λ, the optimal pattern size to minimize the time overhead is T (,σ,2σ) opt = 3 12C λ 2 σ Young/Daly s formula: opt = 2C/λσ = O(λ 1/2 ) Here: opt = O(λ 2/3 ) Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25

30 Conclusion A different re-execution speed indeed helps saving energy while satisfying a performance constraint Silent errors: extension of Young/Daly formula general closed-form solution to get optimal speed pair and optimal checkpointing period (first-order) Extensive simulations: up to 35% energy savings, any speed pair can be optimal BiCrit still open for general case with both silent and fail-stop errors Interesting case with fail-stop errors and double re-execution speed: O(λ 2/3 ) vs O(λ 1/2 ) New methods needed to capture the general case Anne.Benoit@ens-lyon.fr A different re-execution speed can help PASA / 25