Population-Based Incremental Learning for Multiobjective - PowerPoint PPT Presentation

Population-Based Incremental Learning for Multiobjective Optimisation Sujin Bureerat and Krit Sriworamas Department of Mechanical Engineering, Faculty of Engineering, Khon Kaen University, Thailand, 40002 Sujbur@kku.ac.th

Outlines � Introduction � Multiobjective Optimisation � Multiobjective PBIL � Comparative Performance Tests � Results � Conclusions and Discussion

Introduction � Using EAs are advantageous in that they are simple to use � more suitable for global optimisation � They can deal with all kinds of design variables � The search procedure can hardly stall � can search for a Pareto optimal set within one simulation run � Disadvantage: complete lack of consistency & low convergence rate � no convergence guaranteed, � the results obtained are classified as an approximated Pareto front.

Multiobjective Optimisation Math Problem Definition Find x such that Min: f = { f 1 ( x ),…, f m ( x )} Subject to g i ( x ) ≤ 0 h i ( x ) = 0

Multiobjective Optimisation Ex. Illustaration of Bi-objective Optimisation Problem Pareto front f 2 Feasible region f 1

MO-PBIL Probability vectors representing populations in PBIL Note: binary design solution is a row of the populations population 1 population 2 population 3 0 0 1 1, 0 1 1 0, 0 1 0 1 1 1 0 0, 1 1 0 1, 1 0 0 1 0 0 1 1, 1 0 1 0, 0 0 0 1 1 1 0 0, 0 0 0 1, 0 1 0 0 Probability Vectors [0.5, 0.5, 0.5, 0.5] [0.5, 0.5, 0.5, 0.5] [0.25, 0.5, 0, 0.75]

MO-PBIL PBIL for Single objective optimisation Initial P , probability vector 1. Generate binary solutions { b i } from P 2. Find corresponding objective function 3. values { f i } Find b i that gives the best f i 4. If termination criterion is met stop; 5. otherwise, update P using b i and go to 2.

MO-PBIL Updating Equations � Updating a probability vector P = − + new old P P ( 1 LR ) b LR Eq. 1 i i i � Mutation = − + new old Eq. 2 P P ( 1 ms ) rand ( 0 or 1 ). ms i i

MO-PBIL PBIL for multiobjective optimisation Initial P , probability matrix & Pareto archive Pareto 1. Generate binary solutions { b i } from P 2. Find objective function values { f i } of { b i } 3. Replace Pareto with a new non-dominated set 4. obtained from sorting { b i } ∪ old Pareto If the new Pareto’s size is too big, discard some of 5. them using the adaptive grid algorithm If termination criterion is met stop; otherwise, 6. update P using updating Scheme1 or Scheme2 and go to 2.

MO-PBIL Updating SCHEME1 � For the number of row of P � Select some of the members in Pareto at random leading to a set of solution � { c i } � Find b where j th element of b is mean (the j th elements of { c i }) � Update the i th row of P using b � Repeat for all rows of P

MO-PBIL Updating SCHEME2 � For the number of row of P � Generate weighting factors w i randomly such that ∑ w i = 1 � Find a solution b from Pareto, b has the best F value & F = ∑ w i f i � Update the i th row of P using b � Repeat for all rows of P

MO-PBIL Flowchart Initialisati on k = 0 P k , i,j = 0.5, Pareto k = { } Find X k , F k from P k k = k + 1 Find Update Pareto k + 1 P k from Pareto k ∪ using no Pareto k Stop ? X k yes Post- processing

MO-PBIL Ex. Problem 1 Find x (one design variable) Min: f 1 = x 2 f 2 = ( x -2) 2 x ∈ [-1,3] No. of binary strings = 5 No. of solutions = 8 No. of probability vectors = 2 Thus, one prob. vector creates 4 solution

MO-PBIL Ex. Problem 2 Initial P (2x5) = [0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5] { b i } 1 generated from the 1 st row of P = [0 0 1 1 1 0 1 0 0 1 0 1 1 0 0 1 0 0 1 1] * *Note: binary design solution is a column of the population

MO-PBIL Ex. Problem 3 Initial P (2x5) = [0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5] { b i } 2 generated from the 2 nd row of P = [0 1 0 1 1 0 0 1 0 1 1 0 0 1 0 1 0 1 0 1] * *Note: binary design solution is a column of the population

MO-PBIL Ex. Problem 4 Initial P (2x5) = [0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5] The initial population { b i } = { b i } 1 ∪ { b i } 2 = [0 0 1 1 | 0 1 0 1 1 0 1 0 | 1 0 0 1 0 1 0 1 | 0 1 1 0 1 0 0 1 | 0 1 0 1 0 0 1 1 | 0 1 0 1] * Decoded to be x x :0.2903 -0.4839 1.4516 2.7419 -0.7419 2.7419 -0.4839 2.4839 f 1 :0.0843 0.2341 2.1072 7.5182 0.5505 7.5182 0.2341 6.1696 f 2 :2.9230 6.1696 0.3007 0.5505 7.5182 0.5505 6.1696 0.2341 *Note: binary design solution is a column of the population

MO-PBIL Ex. Problem 5 Non-dominated sorting b 1 , b 3 and b 8 are saved to the archive as Pareto = { Pareto 1 Pareto 2 Pareto 3 } Pareto = [ 0 1 1 1 1 1 0 0 0 1 0 1 0 1 1 ]

MO-PBIL Ex. Problem 6 Updating probability SCHEME1 1 st row of P ( P 1 ), use c ={ b 1 b 3 } � mean( c ) = [0.5 1 0 0.5 0.5] P 1 = [0.5 0.75 0.25 0.5 0.5] (use Eq. 1 LR = 0.5) 2 st row of P ( P 2 ), use c ={ b 1 b 3 } � mean( c ) = [0.5 1 0 0.5 0.5] P 2 = [0.5 0.75 0.25 0.75 0.5] (use Eq. 1 LR = 0.5) Updated P = [0.5 0.75 0.25 0.5 0.5 0.5 0.75 0.25 0.75 0.5]

MO-PBIL Ex. Problem 7 New population according to the updated probability P by SCHEME1 P = [0.5 0.75 0.25 0.5 0.5 0.5 0.75 0.25 0.75 0.5] b = [0 1 1 0 | 1 0 1 0 1 1 0 1 | 1 1 1 0 1 0 0 0 | 0 1 0 0 0 1 0 1 | 1 1 0 1 1 0 1 0 | 0 1 1 0] * *Note: binary design solution is a column of the population

MO-PBIL Ex. Problem 8 Updating probability SCHEME2 1 st row of P ( P 1 ), random w ={0.2207 0.7793} � b 3 = [1 1 0 0 1] gives the min F = w 1 f 1 + w 2 f 2 P 1 = [0.5 0.75 0.25 0.5 0.5] (use Eq. 1 LR = 0.5) 2 st row of P ( P 2 ), random w ={0.6158 0.3842} � b 1 = [0 1 0 1 0] gives the min F = w 1 f 1 + w 2 f 2 P 2 = [0.5 0.75 0.25 0.75 0.5] (use Eq. 1 LR = 0.5) Updated P = [0.75 0.75 0.25 0.25 0.75 0.25 0.75 0.25 0.75 0.25]

MO-PBIL Ex. Problem 9 New population according to the updated probability P by SCHEME2 P = [0.75 0.75 0.25 0.25 0.75 0.25 0.75 0.25 0.75 0.25] b = [1 0 1 1 | 0 1 0 0 0 1 1 1 | 1 1 0 1 0 1 0 0 | 0 0 1 0 1 0 0 0 | 1 0 1 1 1 1 0 1 | 0 0 1 0] * *Note: binary design solution is a column of the population

Comparative Performance Tests 1 [8] F 1 : convex Pareto front = ( ) f x x 1 1 where n = 30 and x i ∈ [0,1]. ∑ = n = + ⋅ − g ( ) 1 9 x /( n 1 ) x i i 2 = − h ( f , g ) 1 f / g 1 1 F 2: nonconvex counterpart to F1 = f ( x ) x 1 1 ∑ = where n = 30 and x i ∈ [0,1]. n = + ⋅ − g ( x ) 1 9 x /( n 1 ) i i 2 ( ) = − 2 h ( f , g ) 1 f / g 1 1 F 3 : non-contiguous convex Pareto front = f ( x ) x 1 1 where n = 30 and x i ∈ [0,1]. ∑ = = + ⋅ n − g ( x ) 1 9 x /( n 1 ) i i 2 ( ) = − − π h ( f , g ) 1 f / g f / g sin( 10 f ) 1 1 1 1

Comparative Performance Tests 2 F 4 : Pareto front with multimodality = f ( x ) x 1 1 ( ) ( ) ∑ = = + − + n − π 2 g ( x ) 1 10 n 1 x 10 cos( 4 f ) i 1 i 2 = − where n = 30 and x 1 ∈ [0,1] and x 2, …, xn ∈ [-5,5]. h ( f , g ) 1 f / g 1 1 F 5 : deceptive problem = + where x i represents a binary string, u ( x i ) f ( x ) 1 u ( x ) 1 1 gives the number of ones in the bit vector x i , ∑ = = n g ( x ) v ( u ( x )) + < ⎧ 2 u ( x ); u ( x ) 5 i i 2 = i i ⎨ v ( u ( x )) = = i h ( f , g ) 1 / f ⎩ 1 ; u ( x ) 5 1 1 i and n = 11, x 1 ∈ {0,1}30, and x 2,…, xn ∈ {0,1}. F 6 : non-uniformly distributed Pareto front = − − π 6 f ( x ) 1 exp( 4 x ) sin ( 6 x ) 1 1 1 ( ) ∑ = 0 . 25 = + ⋅ n − g ( x ) 1 9 x /( n 1 ) where n = 30 and x i ∈ [0,1]. i i 2 ( ) = − 2 h ( f , g ) 1 f / g 1 1

Comparative Performance Tests 3 F 7 [14] min ( ) ⎛ ⎞ x 3 ∑ 2 = − ⎜ − − ⎟ f 1 exp x 1 3 1 i ⎝ ⎠ = i 1 ( ) ⎟ ⎛ ⎞ 3 ∑ 2 = − ⎜ − + f 1 exp x 1 3 2 i ⎝ ⎠ = i 1 ∈ [ − 4 , 4 ] x i F 8 [15] min ∑ = + − = x 2 2 L f x ( x 1 ) ; i { 2 , 3 , , 30 } 1 i 1 i ∑ = + − = 2 2 L f x ( x 1 ) ; i { 1 , 3 , , 30 } 2 i 2 i ∈ [ − x 5 , 5 ] i