桂林电子科技大学最优化方法B卷答案

B卷答案一、（满分5315分）1--5：FTTTT二、(满分20分)Solution：since134,,xxxareoptimalbasicvariables,wehave（最优基变量,我们有）1211140110,3,0,200100015BNBccb(4分)11011111001,101111211BBN(4分)1312TTBNcBNc(4分)150TBcBb(4分)Therefor（因此）(4分)三、(满分20分)Solution：Itsstandardformisas（它的标准形式）12312341235max234..232340,1,2,3iZSxxxstxxxxPxxxxxi(5分)12312341235max234..232340,1,2,3iZSxxxstxxxxxxxxxiandtheinitialsimplextableauis（和初始单纯形表）（5分）X1X2X3X4X5RHSX4-1-2-110-3X5-21-301-4Z234000ThenwehaveX1X2X3X4X5RHSX40-5/2-1/21-1/2-1X11-1/23/20-1/22Z04101-4(5分)andX1X2X3X4X5RHSX201-1/5-2/51/52/5X1107/5-1/5-2/511/5S009/58/51/5-28/5112280,**555TxSZ(5分)四、Solution:2004042,b=6,2024-6Ac(6points)1112232332323224200440426426426024-6-624246xxxfxAxbxxxxxxxxxx2200042024fxA(6points)Let0fx,i.e.1232324042602460xxxxx,wehave2*33x(6points)Since22000420024fxA,so*xisastrictlyglobalminimizer.(2points)五、Usethesteepestdescendmethodtosolvethefollowingproblem.（使用最速下降法来解决以下问题）：221212min,32fxxxxBeginatthepoint111xandthetolerance（公差）0.（20points）Solution:since122322xfxxthenwehave1122dfxand1d,andmakealinearsearch（线性搜索）110minfxd(6points)Todothat,wedenote（为此,我们表示）221114,124221fxdfLet40200andweobtain10.5and(6points)21111430.5122xxd(3points)2200dfx,20dHence,theoptimalsolution（因此,最优解为）23*2xxand*0f(5points)