北京工业大学 最优化基础实验：使用matlab求解规划问题

1、北京工业大学最优化基础实验：使用matlab求解规划问题 实验一非线性最优化方法实验min f (x)二 x2 + x2 4x 4x + 81. 1 2 1 2s.t: x + 2x 4 f=(x)x(1)A2+x(2F2-4*x(1)-4*x(2)+8; A=1,2b=4A =12 b =4 x,fval=fmincon(f,10,10,A,b) x =1.6000 1.2000 fval =0.8000min f (x) = x 2 + x 2 + x 2 一 (2x + 4x + 6x )1 2 3 1 2 3解： apple=(x)x(1)A2+x (2) A2+x(3)A2-(2*x

2、(1)+4*x(2)+6*x(3)apple =(x)x(1)A2+x(2)A2+x(3)A2-(2*x(1)+4*x(2)+6*x(3) x,fval=fminunc(apple,1,1,4)Warning: Gradient must be provided for trust-region algorithm; using line-search algorithm instead. In fminunc at 341Local minimum found.Optimization completed because the size of the gradient is less th

3、an the default value of the function tolerance.x =1.0000 2.0000 3.0000fval =-14min f (x)二 xs.t: (1 x )3 x 0 12x,x 012解 f3=(x)x(1)f3 =(x)x(1)lb=zeros(2,1)lb =00 x0=1,1x0 = x,optimal=fmincon(objfun,x0,lb,nonlcon)Active inequalities (to within options.TolCon = 1e-006): lower upper ineqlinineqnonlin12x

4、=00 optimal =0min f (x)二(x 3)2 + (x 3)212s.t: 4 x x 012x , x 012 f4=(x)(x(1)-3)A2+(x(2)-3F2f4 =(x)(x(1)-3)A2+(x(2)-3)A2 A=1,1A = b=4lb=zeros(2,1)lb =00 x,fval=fmincon(f4,10,10,A,b,lb,)Warning: Trust-region-reflective algorithm does not solve this type of problem, using active-set algorithm. You coul

5、d also trythe interior-point or sqp algorithms: set the Algorithm option to interior-point or sqp and rerun. For more help, seeChoosing the Algorithm in the documentation. In fmincon at 472Local minimum found that satisfies the constraints.Optimization completed because the objective function is non

6、-decreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.Active inequalities (to within options.TolCon = 1e-006): lower upper ineqlinineqnonlin1x =2.0000 2.0000fval =2.0000min f (x

7、)二(x 2)2 + (x 2)25. 1 2s.t: x + x 6 二 012本题要求输出寻优过程的迭代次数。解： f5=(x)(x(1)-2)A2+(x(2)-2F2f5 =(x)(x(1)-2)A2+(x(2)-2)A2Aeq=1,1Aeq =11beq=6beq =6options=optimset(Display,iter) options =Display: iter MaxFunEvals: MaxIter: TolFun: TolX: FunValCheck: OutputFcn: PlotFcns: ActiveConstrTol: Algorithm: AlwaysHon

8、orConstraints: BranchStrategy: DerivativeCheck: Diagnostics: DiffMaxChange: DiffMinChange: FinDiffType: GoalsExactAchieve: GradConstr: GradObj: HessFcn: Hessian: HessMult: HessPattern: HessUpdate: InitialHessType: InitialHessMatrix: InitBarrierParam: InitTrustRegionRadius: Jacobian: JacobMult: Jacob

9、Pattern: LargeScale: LevenbergMarquardt: LineSearchType: MaxNodes: MaxPCGIter: MaxProjCGIter: MaxRLPIter: MaxSQPIter: MaxTime: MeritFunction: MinAbsMax: NodeDisplayInterval: NodeSearchStrategy: NonlEqnAlgorithm: NoStopIfFlatInfeas: ObjectiveLimit: PhaseOneTotalScaling: Preconditioner: PrecondBandWid

10、th: RelLineSrchBnd: RelLineSrchBndDuration: ScaleProblem: Simplex: SubproblemAlgorithm: TolCon: TolConSQP: TolGradCon: TolPCG: TolProjCG: TolProjCGAbs: TolRLPFun: TolXInteger: TypicalX: UseParallel: x,fval=fmincon(f5,1,3,Aeq,beq,options)Warning: Trust-region-reflective algorithm does not solve this

11、type of problem, using active-set algorithm. You could also trythe interior-point or sqp algorithms: set the Algorithm option to interior-point or sqp and rerun. For more help, seeChoosing the Algorithm in the documentation. In fmincon at 472Max Line search Directional First-orderIter F-countf(x)con

12、straint steplength derivativeoptimality Procedure03 2 2Infeasible start point1648.882e-0161-2.535292.0246901-2.833.78312201-0.3144Local minimum found that satisfies the constraints.Optimization completed because the objective function is non-decreasing in feasible directions, to within the default v

13、alue of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.x =3.0000 3.0000fval =实验二线性规划a )min f (x) = x + 2 x12s.t: x + x 212x + x 112x 012解： f=1,2f =12 A=-1,-11,-10,1A =-1 -11 -101 b=-2;-1;3b =-2-13lb =00 x,fval=linprog(f,A,b,lb) Optimizat

14、ion terminated.x =0.50001.5000fval =3.5000min f (x) = -x + 2x - 3x123s.t: x + x + x = 6123x + x + 2 x = 41232 x + 3x = 1023x 0123解： f=-1,2,-3f =-12-3 A=0,0,1A = b=2aeq=1,1,1,-1,1,20,2,3aeq =111-1 1 2023beq=6;4;10beq =6410 lb=zeros(3,1)lb =000 x,fval=linprog(f,A,b,aeq,beq,lb)Optimization terminated.x =2.00002.00002.0000fval =min f (x) = -2x - 5x12s.t: x + 2 x 812x 41x 012解： f=-2,-5f =-2-5 A=1,21,00,1A =121001 b=8;4;3b =843lb=zeros(2,1)lb =00 x,fval=linprog(f,A,b,lb)Optimization terminated.2.00003.0000fval =-19