Complementary Random Numbers Dagger Sampling. Lecture 4, autumn 2015 Mikael Amelin

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1 Complementary Random Numbers Dagger Sampling Lecture 4, autumn 2015 Mikael Amelin 1

2 Introduction All observations in simple sampling are independent of each other. Sometimes it is possible to increase the probability of a good selection of samples if the samples are not independent. Random number generator U Inverse transform method Y Model g(y) X Sampling M X 2

3 Mean of Two Estimates Consider two estimates M X1 and M X2 of the same expectation value X, i.e., E[M X1 ] = E[M X2 ] = X. Study the mean of these estimates, i.e., M X1 + M X2 M X =

4 Mean of Two Estimates Expectation value: E[M X ] = E M X1 + M X2 1 = -- E M = 2 2 X1 + E M X2 1 = -- = X, 2 X + X i.e., the combined estimate M X is also an estimate of X. Variance: Var[M X ] = Var M X1 + M X2 = = --Var M + 4 X1 + --Var M 4 X2 -- 2Cov M 4 X1 M X2. 4

5 Mean of Two Estimates If M X1 and M X2 are independent estimates obtained with simple sampling and the number of samples is the same in both simulations, then we get Var[M X1 ] = Var[M X2 ] and Cov[M X1, M X2 ] = 0. Hence, Var[M X ] = Var M X1 + M X2 = 2 1 Var M = -- X1 Var M = 4 X1 + Var M X The variance is equivalent to one run of simple sampling using 2n samples; according to theorem 4.2 we should get half the variance when the number of samples is doubled. 5

6 Mean of Two Estimates If M X1 and M X2 are negatively correlated then the covariance term will make Var[M X ] smaller than the corresponding variance for simple sampling using the same number of samples. A negative correlation between estimates can be created using complementary random numbers. 6

7 Complementary Random Numbers The complementary random number of a U(0, 1)-distributed random variable is U* = 1 U. - Correlation coefficient: U, U* = 1. If a random input value is calculated using the inverse transform method, i.e., Y = F 1 Y U, then the complementary input is given by Y* = F 1 Y 1 U. - For symmetrical distributions we get that if Y= Y + then Y* = Y. - Correlation coefficient: Y, Y* U, U* 7

8 Complementary Random Numbers If an output is calculated by analysing a scenario, i.e., X = g(y), then the complementary output is given by X* =g(y*). - Correlation coefficient: X, X* Y, Y*. If an estimate is based on a set of scenarios, i.e., m X = i x i /n, then the complementary estimate is given by m X* = i x i */n. - Correlation coefficient: mx, mx* X, X*. In practice there is no need to differentiate original and complementary values: n m X + m X* 1 m X = = x 2 2n i + x i *. i = 1 8

9 Multiple Inputs Generally, the input vector Y has more than one element. The complementary random numbers of a vector can be created by combining the original and complementary random numbers of each element in the vector. - It can be noted that each scenario y i will have more than one complementary scenario. It is possible to apply complementary random numbers only to a selection of the elements in Y. 9

10 Example: Complementary Scenarios Problem The input vector Y has the three elements A, B, C. The outcomes a i, b i, c i have been randomised for each element, and the corresponding complementary random numbers, a i *, b i *, c i *, have been calculated. Enumerate the original and complementary scenarios if complementary random numbers are applied to the following elements: a) A, B, C. b) A, B. 10

11 Example: Complementary Scenarios Solution a) Randomise three values, a, b, c and calculate their complementary random numbers. The results are combined as follows: [a, b, c], [a, b, c*], [a, b*, c], [a, b*, c*], [a*, b, c], [a*, b, c*], [a*, b*, c], [a*, b*, c*]. b) Randomise six values, a, b, c 1, c 2, c 3, c 4 and calculate the random numbers a* and b*. The results are combined as follows: [a, b, c 1 ], [a, b*, c 2 ], [a*, b, c 3 ], [a*, b*, c 4 ]. 11

12 Effectiveness The variance reduction effect is depending on the correlation between x i = g(y i ) and x i * = g(y i *). This correlation depends on the correlation between Y and X. All inputs may not have a significant correlation to the outputs no use in applying complementary random numbers to these inputs. Sometimes a stronger correlation can be obtained by modifying the input probability distribution. 12

13 Auxiliary Inputs It might be possible to find a subset of the inputs, Y i, i =1,, J such that there is a strong correlation between Y E = h(y 1,,Y J ) (where h is an arbitrary function) and X. When it can be worthwhile to introduce Y E as an auxiliary input variable to the system, and apply complementary random numbers to Y E instead of Y i, i =1,, J. 13

14 Example: Auxiliary Input Problem Gismo Inc. is selling their products to K retailers, and the demand of each retailer, D k, is N( k, k )-distributed. How should complementary random numbers be applied to a Monte Carlo simulation of Gismo Inc.? 14

15 Example: Auxiliary Input Solution The costs of Gismo Inc. has a stronger correlation to the total demand compared to the demand of an individual retailer. Introduce the auxiliary input variable D tot = D D K. The probability distribution of D tot is calculated using the addition theorems for normally distributed random variables. 15

16 Example: Auxiliary Input Solution To generate the original scenario, randomise D tot as well as temporary values of the demand of the individual retailers, D 1,, D K. The temporary demand of individual retailers will generally not match the total demand. Therefore, scale the demands of the individual retailers to match D tot, i.e., let D k = D tot D k. K j = 1 D j 16

17 Example: Auxiliary Input Solution To generate the complementary scenario, calculate the complementary random value of D tot and randomise new temporary values, D 1,, D K. Then scale the individual retailer demands to match D tot *. 17

18 Simulation procedure Complementary random numbers Random number generator U Inverse transform method Y X Model g(y) U* Y* X* Sampling M X Step 1. Generate a random number u k, i and calculate y k, i = F 1 Yk u k i. If applicable to this scenario parameter, also calculate y k, i *= F 1 Yk 1 u ki. Step 2. Repeat step 1 for all scenario parameters. Create all complementary scenarios and calculate x i = g(y i ) for each scenario. 18

19 Simulation procedure Complementary random numbers Step 3. Repeat step 1 and 2 until enough scenarios have been analysed for this batch. Step 4. Update the sums n x i and x2 i i = 1 n i = 1 or store all samples x i (depending on the objective of the simulation). Step 5. Test stopping rule. If not fulfilled, repeat step 1 4 for the next batch. Step 6. Calculate estimates and present results. 19

20 Example: Complementary Random Numbers in Akabuga District Simulation Apply complementary to the total generation capacity and the total load. - Joint probability distribution for the generation capacity of the diesel generator sets: State, k G tot G 1 G Probability, f Gtot (k)

21 Example: Complementary Random Numbers in Akabuga District Simulation Simulation method Enumeration Simple sampling Compl. r.n. (G, D), (G*, D*) Compl. r.n. (G, D), (G, D*), (G*, D), (G*, D*) Average simulation time [ms] M TOC M LOLO Min. Mean Max. Var. Eff. Min. Mean Max. Var. Eff. <

22 Dagger Sampling We have seen from complementary random numbers that a negative correlation between the outputs from different scenarios can have a variance reducing effect. When the input is a two-state probability distribution, we can create a negative correlation by using dagger sampling instead of complementary random numbers. 22

23 Dagger Sampling Consider a two-state input variable Y with the frequency function f Y x = 1 p p 0 x = a, x = b, all other x, where p < 0.. In dagger sampling, the inverse transform method is replaced by a dagger transform function. 23

24 Dagger Transform Theorem 5.1: If U is a U(0, 1)-distributed random number then Y is a two-state random variable with two possible states, a and b, such that P(b) = p, if Y is calculated according to F Yj x = b if j 1 p x jp, a otherwise, for j = 1,, S, where S is the largest integer such that S 1/p. The value S in theorem 5.1 is referred to as the dagger cycle length. One pseudorandom number is used to generate S values of Y. 24

25 Dagger Transform The dagger transform can be illustrated by dividing an interval between 0 and 1 in S sections, where each section has the width p. There will also be a remainder interval if S p <1. U p p p Remainder y k = b if U is found in the k:th section, and for i = 1,, S, i k, we get y i = a. 25

26 Example: Dagger Transform Problem The random variable Y is either equal to 2 (70% probability) or 5 (30% probability). a) Randomise three values of Y using dagger sampling and the random value U =0.34 from a U(0, 1)-distribution. b) Randomise three values of Y using dagger sampling and the random value U =0.94 from a U(0, 1)-distribution. 26

27 Example: Dagger Transform Solution Here p =0.3 S = 3. Moreover, we have a =2 and b =5. a) b) U y 1 y 2 y 1 = 2 y 2 = 5 y 3 = 2 y 1 = 2 y 2 = 2 y 3 = 2 y 3 27

28 Correlation of Dagger Samples The outputs X 1,, X S from the scenarios generated in a dagger cycle will be negatively correlated if the inputs Y 1,, Y S are negatively correlated (but the correlation might be weaker). Are the input values Y i = F Yi U, i = 1,, S negatively correlated, i.e., what is the value of Cov[Y i, Y j ] = E[Y i Y j ] E[Y i ]E[Y j ]? 28

29 Correlation of Dagger Samples There are two possible values for the product Y i Y j : ab or aa. There are two sections there either Y i or Y j will be transformed to b; hence, the probability for this event is 2p. Thus, we get E[Y i Y j ] = 2pab + (1 2p)aa. y 1 y 2 y 3 29

30 Correlation of Dagger Samples Moreover, we have E[Y i ] = E[Y j ] = (1 p)a + pb. The covariance can now be calculated as Cov[Y i, Y j ] = E[Y i Y j ] E[Y i ]E[Y j ] = = 2pab + (1 2p)aa ((1 p)a + pb) 2 = = = = p 2 (a + b) 2 < 0. The input values are negatively correlated and we can therefore expect a negative correlation in the outputs too. 30

31 Multiple Inputs Assume that a computer simulation has several two-state input variables. Which dagger cycle length should be chosen? Alternative 1: Independent dagger cycles. Inputs Scenarios 31

32 Multiple Inputs Alternative 2: Reset all dagger cycles at the end of the longest cycle. Inputs Scenarios 32

33 Multiple Inputs Alternative 3: Reset all dagger cycles at the end of the shortest cycle. Scenarios Inputs 33

34 Simulation procedure Complementary random numbers Random number generator U Dagger transform Y Y S Model g(y) X X S Sampling M X Step 1. Generate a random number u k, i and calculate y i = F Yi u k i for one dagger cycle. Step 2. Repeat step 1 for all scenario parameters. Combine the dagger cycles and calculate x i = g(y i ). Step 3. Repeat step 1 and 2 until enough scenarios have been analysed for this batch. 34

35 Simulation procedure Complementary random numbers Step 4. Update the sums n x i and x2 i i = 1 n i = 1 or store all samples x i (depending on the objective of the simulation). Step 5. Test stopping rule. If not fulfilled, repeat step 1 4 for the next batch. Step 6. Calculate estimates and present results. 35

36 Example: Dagger Sampling of Akabuga District Apply dagger sampling to available generation in the diesel generator sets and to the available transmission capacity. Use independent dagger cycles. 36

37 Example: Dagger Sampling of Akabuga District Simulation method Average simulation time [ms] M TOC M LOLO Min. Mean Max. Var. Eff. Min. Mean Max. Var. Eff. Enumeration < Simple sampling Compl. r.n Dagger sampling

38 Variance Reduction To achieve a variance reduction, we need to have some information of the simulated system. For complementary random numbers, we need to identify the inputs where a negative correlation between scenarios will result in negative correlations for one or more outputs. For dagger sampling, we need to identify two-state inputs where a negative correlation between the scenarios will result in negative correlations for one or more outputs. 38

COMPLEMENTARY RANDOM NUMBERS & DAGGER SAMPLING

INTRODUCTION EG2080 Monte Carlo Methods in Engineering Theme 4 COMPLEMENTARY RANDOM NUMBERS & DAGGER SAMPLING All observations in simple sampling are independent of each other. Sometimes it is possible