Optimization and Simulation

Transcription

1 Optimization and Simulation Variance reduction Michel Bierlaire Transport and Mobility Laboratory School of Architecture, Civil and Environmental Engineering Ecole Polytechnique Fédérale de Lausanne M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 1/ 23

2 Outline Other techniques M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 2/ 23

3 Example Use simulation to compute I = 1 0 e x dx We know the solution: e 1= Simulation: consider draws two by two Let r 1,...,r R be independent draws from U(0, 1). Let s 1,...,s R be independent draws from U(0, 1). I 1 R R e r i + e s i i=1 2 Use R = (that is, a total of draws) Mean over R draws from (e r i + e s i )/2: 1.720,variance: M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 3/ 23

4 Example Now, use half the number of draws Idea: if X U(0, 1), then (1 X ) U(0, 1) Let r 1,...,r R be independent draws from U(0, 1). I 1 R R e r i + e 1 r i i=1 2 Use R = Mean over R draws of (e r i + e 1 r i )/2: ,variance: Compared to: mean of (e r i + e s i )/2: 1.720,variance: M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 4/ 23

5 Example Independent Antithetic e M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 5/ 23

6 Antithetic draws Let X 1 and X 2 i.i.d r.v. with mean θ Then ( ) X1 + X 2 Var = (Var(X 1)+Var(X 2 ) + 2 Cov(X 1, X 2 )) If X 1 and X 2 are independent, then Cov(X 1, X 2 ) = 0. If X 1 and X 2 are negatively correlated, then Cov(X 1, X 2 ) < 0, and the variance is reduced. M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 6/ 23

7 Back to the example Independent draws X 1 = e U, X 2 = e U Var(X 1 )=Var(X 2 ) = E[e 2U ] E[e U ] 2 = 1 0 e 2x dx (e 1) 2 = e (e 1) 2 = Cov(X 1, X 2 )=0 ( ) X1 + X 2 Var = 1 ( )) = M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 7/ 23

8 Back to the example Antithetic draws X 1 = e U, X 2 = e 1 U Var(X 1 )=Var(X 2 )= Cov(X 1, X 2 ) = E[e U e 1 U ] E[e U ]E[e 1 U ] = e (e 1)(e 1) = ( ) X1 + X 2 Var = 1 ( )) = M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 8/ 23

9 Antithetic draws: generalization Suppose that X 1 = h(u 1,...,U m ) where U 1,...U m are i.i.d. U(0, 1). Define X 2 = h(1 U 1,...,1 U m ) X 2 has the same distribution as X 1 If h is monotonic in each of its coordinates, then X 1 and X 2 are negatively correlated. If h is not monotonic, there is no guarantee that the variance will be reduced. M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 9/ 23

10 Another example Antithetic draws: X 1 = I = The covariance is positive: 1 0 ( x 1 2) 2 dx ( U 1 2 (, X 2 = (1 U) 2) 1 ) 2 2 Cov(X 1, X 2 )= > 0. The variance will therefore be (slightly) increased! M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 10 / 23

11 Another example 2000 Independent M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 11 / 23

12 Another example 2000 Antithetic M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 12 / 23

13 Another example Independent Antithetic M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 13 / 23

14 Outline Other techniques M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 14 / 23

15 We use simulation to estimate θ =E[X ], where X is an output of the simulation Let Y be another output of the simulation, such that we know E[Y ]=µ We consider the quantity: By construction, E[Z] =E[X ] Its variance is Z = X + c(y µ). Var(Z) =Var(X + cy )=Var(X )+c 2 Var(Y )+2c Cov(X, Y ) Find c such that Var(Z) isminimum M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 15 / 23

16 First derivative: Zero if Second derivative: We use Its variance 2c Var(Y ) + 2 Cov(X, Y ) c = Cov(X, Y ) Var(Y ) 2Var(Y ) > 0 Z = X Cov(X, Y ) (Y µ). Var(Y ) Var(Z )=Var(X ) Cov(X, Y )2 Var(X ) Var(Y ) M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 16 / 23

17 In practice... Cov(X, Y )andvar(y ) are usually not known. We can use their sample estimates: Ĉov(X, Y )= 1 n 1 R (X r X )(Y r Ȳ ) r=1 and Var(Y )= 1 n 1 R (Y r Ȳ ) 2. r=1 M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 17 / 23

18 In practice... Alternatively, use linear regression X = ay + b + ε where ε N(0,σ 2 ). The least square estimators of a and b are â = R r=1 (X r X )(Y r Ȳ ) R r=1 (Y r Ȳ ) 2 ˆb = X âȳ. Therefore c = â M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 18 / 23

19 Moreover, ˆb +âµ = X âȳ +âµ = X â(ȳ µ) = X + c (Ȳ µ) = θ Therefore, the control variate estimate θ of θ is obtained by the estimated linear model, evaluated at µ. M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 19 / 23

20 Back to the example Use simulation to compute I = X = e U 1 Y = U, E[Y ]=1/2, Var(Y )=1/12 Cov(X, Y )=(3 e)/ Therefore, the best c is 0 e x dx c = Cov(X, Y ) Var(Y ) = 6(3 e) 1.69 Test with R = Result of the regression: â = , ˆb = Estimate: ˆb + â/2 = , Variance: (compared to 0.24) M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 20 / 23

21 Back to the example No control Control M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 21 / 23

22 Other techniques Outline Other techniques M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 22 / 23

23 Other techniques Variance reductions techniques Other techniques Conditioning Stratified sampling Importance sampling Draw recycling In general Correlation helps! M. Bierlaire (TRANSP-OR ENAC EPFL) Optimization and Simulation 23 / 23