数值最优化算法与理论第11章习题.pdf

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1、Æ8aH.0G2007-12-11ihminf(x)s.t.gI≥0,I=1,···,m1,hE(x)=0,E=m1,···,m.(k)nVA#x;minFµk(x)x∈R$(k)Xx¯;x
_x¯;

2、k−1)s(xk)−s(xk−1)≤ukuk−1[s(x)−s(x)](u−1−u−1)≤0kk−1kk−1TD{uk}`
>s(xk)≥s(xk−1).b
(2)(k)VFuxSF(x(k+1))≤F(x(k))uk+1uk+1=f(x(k))+u−1s(x)k+1k≤f(x(k))+u−1s(x)kk=F(x(k))uk(k)>{Fuk(x)}"
b
(3)(k−1(k)VFuk(x)≥Fuk(x)fx(k−1)+u−1s(x)≥f(x(k))+u−1s(x)kk−1kkf(x(k))≤fx(k−1)+u−1(s(x)−s(x))≤fx(k−1)kk−1kk>

3、{f(x)}"
b
3ihTDf;C=gi,i∈I;=_5Rx1,x2∈D3α∈[0,1]Wf(αx1+(1−α)x2)≤αf(x1)+(1−α)f(x2)gi(αx1+(1−α)x2)≥αgi(x1)+(1−α)gi(x2),i∈I27G(x)=logx,_G(x)=logx;=3`_log[gi(αx1+(1−α)x2)]≥log[αgi(x1)+(1−α)gi(x2)]≥αloggi(x)+(1−α)loggi(x2)XTDXF(αx+(1−α)x)=f(αx+(1−α)x)−u−1logg(αx+(1−α)x)u1212i12i∈EX≤αf(x

4、)+(1−α)f(x)−u−1[αlogg(x)+(1−α)logg(x)]12i1i2i∈EXX=α[f(x)−u−1logg(x)]+(1−α)[f(x)−u−1logg(x)]1i12i2i∈Ei∈E=αFu(x1)+(1−α)Fu(x2)>Fu℄D01;C=
b
4(1)gZE?!Y122minf(x)=x1+x22g1(x)=−3x1−x2+10≥0s.t.g2(x)=x1−1≥0g3(x)=4−x1≥0x['D9=1Fu(x)=f(x)+us(x)21222=f(x)+u[(3x1+x2−10)+(x1−1)+(x1−4)]2_!x1+3u(3x1+x2−1

5、0)+u(x1−1)+u(x1−4)∇Fu(x)=2x1+u(3x1+x2−10)3,∇Fu(x)=0$5u15u+5Txu=(,10−)2u−52u−555T∗u→∞9xu→(2,2)=x(2)gZE?!Y122minf(x)=x1−x1x2+x2−2x12g1(x)=−3x1−x2+2≥0s.t.g2(x)=1−x1≥0x['D9=1Fu(x)=f(x)+us(x)2122=f(x)+u[(3x1+x2−2)+(x1−1)]2_!x1−x2−2+3u(3x1+x2−2)+u(x1−1)∇Fu(x)=−x1+2x2+u(3x1+x2−2),∇Fu(x)=0$2uu+2−u

6、2+3u+2−u2+3u+2Txu=(−,)3u−13u−1u2+33u−1u2+33u−1T∗u→∞9xu→(1,−1)=x5g^d=XF(x)=f(x)−µ−1logg(x)µii∈I122−1=x1+x2−2x1−µ(log(x1+x2−2)+logx1+log(1−x1))24n$F

7、∗1∗2∗Lµk(x)≤Lµk(x)=f(x)+λhE(x)+µkkhE(x)k=f(x)2(k)(k)(k)T(k)1(k)2∗Lµk(x)=f(x)+λhE(x)+µkkhE(x)k≤f(x)2(k)22(f(x∗)−f(x(k))−λ(k)ThE(x(k)))0≤khE(x)k≤µkkTD{λ};W%+{µk}`3µk→+∞,6:*,k→∞,(k)xk∈K→x,µ¯k→∞,S(¯x)=0Tx¯;E?11.8$