运筹学实验报告

.运筹学实验报告专 业： 班 级： 姓 名： 学 号: 指导教师： 数学与应用数学专业2015-12-18实 验 目 录一、实验目的3二、实验要求3三、实验内容31、 线性规划32、 整数规划63、 非线性规划134、 动态规划145、 排队论19四、需用仪器设备26五、MATLAB优化工具箱使用方法简介26六、LINGO优化软件简介26七、实验总结27一、实验目的 1、 会利用适当的方法建立相关实际问题的数学模型；2、 会用数学规划思想及方法解决实际问题；3、 会用排队论思想及方法解决实际问题；4、 会用决策论思想及方法解决实际问题；5、 掌握MATLAB、LINGO等数学软件的应用；2、 实验要求1、 七人一组每人至少完成一项实验内容；2、 每组上交一份实验报告；3、 每人进行12分钟实验演示；4、 实验成绩比例： 出 勤：40% 课堂提问：20% 实验报告：30% 实验演示：10%。3、 实验内容1、 线性规划例运筹学74页14题 Min z=-2x1-x2 s.t. 2x1+5x260 x1+x218 3x1+x244 X210 X1,x20用matlab运行后得到以下结果:the program is with the linear programmingPlease input the constraints number of the linear programming m=6m = 6Please input the variant number of the linear programming n=2n = 2Please input cost array of the objective function c(n)_T=-2,-1c = -2 -1Please input the coefficient matrix of the constraints A(m,n)=2,5;1,1;3,1;0,1;-1,0;0,-1A = 2 5 1 1 3 1 0 1 -1 0 0 -1Please input the resource array of the program b(m)_T=60,18,44,10,0,0b = 60 18 44 10 0 0Optimization terminated.The optimization solution of the programming is:x = 13.0000 5.0000The optimization value of the programming is:opt_value = -31.0000LINDO程序在命令窗口键入以下内容:max -2x-ysubject to2x+5y=60x+y=183x+y=44y=10end按solve键在reports window出现:Global optimal solution found. Objective value: 0.000000 Total solver iterations: 0 Variable Value Reduced Cost X 0.000000 2.000000 Y 0.000000 1.000000 Row Slack or Surplus Dual Price 1 0.000000 1.000000 2 60.00000 0.000000 3 18.00000 0.000000 4 44.00000 0.000000 5 10.00000 0.0000002、 整数规划课本第二章79页1题 Max z=100x1+180x2+70x3 s.t. 40x1+50x2+60x310000 3 x1+6x2+ 2x3600 x1130 X280 x3200 x1 x2 x30程序运行及结果：biprogramtheprogramiswiththebinarylinearprogrammingPleaseinputtheconstraintsnumberoftheprogrammingm=5m=5Pleaseinputthevariantnumberoftheprogrammingn=5n=5Pleaseinputcostarrayoftheobjectivefunctionc(n)_T=100,180,70c=10018070PleaseinputthecoefficientmatrixoftheconstraintsA(m,n)=40,50,60;3,6,2;1,0,0;0,1,0;0,0,1A=405060362100010001Pleaseinputtheresourcearrayoftheprogramb(m)_T=10000;600;130;80;200b=1000060013080200Optimizationterminated.Theoptimizationsolutionoftheprogrammingis:x=000Theoptimizationvalueoftheprogrammingis:opt_value= 0程序名：intprogram b程序说明：%theprogrammiswiththeintegerlinearprogrammingusebranchandboundmethod!%这个程序是用分支定界法解决整数规划问题%pleaseinputtheparametersinthemainfunctioninthecommandwinows%请在命令窗口输入这个主要定义函数的参数functionx,f=ILp(c,A,b,vlb,vub,x0,neqcstr,pre)%minf=c*x,s.t.A*x=b,vlb=x=vub%f的最小值等于c的转置乘以x，A乘以x小于等于b,x大于等于vlb小于等于vub%thevectorsofxisrequiredasintegersaswhole%x是整个的整数需要%x0istheinitialization,isalsook%x0是初始值,也可以是。%neqcstristhenumberofequationalconstraints,when0canbedelete%neqcstr是平均约束条件的数目，当0能删除时%preistheconciserate%pre是简明率%xistheintegeroptimizationandfistheoptimalvalue%x是整数规划，f是最优值%ifnargin8,pre=0;%narginisthefactuallyinputvariantsnumber（这个参数是实际输入的变量个数）ifnargin7,neqcstr=0;ifnargin6,x0=;ifnargin5,vub=;ifnarginpre);mtemp=length(temp2);ifisempty(temp2)x_f_b=xtemp;ftemp;vlb;vub;whilej=mmi=1;whilei=mtemp%ifx_f_b(nvars+1,j)pre);ifisempty(templ2)xall=xall,xtemp;fall=fall,ftemp;fvub=min(fvub,fall);elseifftemp=fvubx_f_b=x_f_b,xtemp;ftemp;vlbl;vubl;endendend%ifx_f_b(nvars+1,j)pre);ifisempty(tempr2)xall=xall,xtemp;fall=fall,ftemp;fvub=min(fvub,fall);elseifftempmmbreakend%theendbecausethebreak（因为中断而结束）temp0=round(xint(:,j);temp1=floor(xint(:,j);temp2=find(abs(xint(:,j)-temp0)pre);mtemp=length(temp2);end%theendofwhile（结束当前）else%correspondthesecondif（符合第一个如果）x=xtemp;f=ftemp;end%theendofsecondif（第二个如果的结束）%5ifisempty(fall)fmin=min(fall);nmin=find(fall=fmin);x=xall(:,nmin);f=fmin;endelse%correspondthefirstif（符合第一个如果）x=nan*ones(1,nvars);end LINDO程序例99页第6题第二问在命令窗口键入以下内容:max -11x1-4x2st-x1+2x2=45x1+2x2=162x1-x2 In fminunc at 265Optimization terminated: relative infinity-norm of gradient less than options.TolFun.x = 1.0e-006 * 0.2541 -0.2029fval = 1.3173e-0134、 动态规划程序名： dynamic; dynfun1_1, dynfun1_2, dynfun1_3;例180页第一题程序说明：dynamic程序：% the programm is with the dynamic programming use the recurisive method for the last to first% this is the main function of the methodfunctionp_opt,fval,u=dynprog(x,DecisFun,ObjFun,TransFun)% the function is to solve the dynamic example in the textbook% x is the situation variant and its column number represent the stage situation% subfunction DecisFun(k,x) is to solve the decision variant of k stage variant x % subfunction ObjFun(k,x,u) is to stage index function% subfunction TransFun(k,x,u) is the stage transformation function,u is the corresponding decision variant% p_opt has four output,the first is the number of the stage,the second is the optimal road of decision% the third is the optimal stategies of the decision ,the forth is the index function group.% fval is a column vector,is to represent the optimal value correspend to the initial stage is x % k=length(x(1,:);f_opt=nan*ones(size(x);d_opt=f_opt;t_vubm=inf*ones(size(x);x_isnan=isnan(x);t_vub=inf;% to caculate the teminate valuestmp1=find(x_isnan(:,k);tmp2=length(tmp1);for i=1:tmp2 u=feval(DecisFun,k,x(i,k); tmp3=length(u);for j=1:tmp3 tmp=feval(ObjFun,k,x(tmp1(i),k),u(j);if tmp=t_vub f_opt(i,k)=tmp; d_opt(i,k)=u(j); t_vub=tmp;endendend% recurisivefor ii=k-1:-1:1 tmp10=find(x_isnan(:,ii); tmp20=length(tmp10);for i=1:tmp20 u=feval(DecisFun,ii,x(i,ii); tmp30=length(u);for j=1:tmp30 tmp00=feval(ObjFun,ii,x(tmp10(i),ii),u(j); tmp40=feval(TransFun,ii,x(tmp10(i),ii),u(j); tmp50=x(:,ii+1)-tmp40; tmp60=find(tmp50=0);if isempty(tmp60),if nargin5 tmp00=tmp00+f_opt(tmp60(1),ii+1);if tmp00x=nan*ones(3,5);x(1,1)=1;x(1:3,2)=(2:4);x(1:2,3)=5,6;x(1:3,4)=(7:9);x(1,5)=10; p,f=dynamic(x,dynfun2_1,dynfun2_2,dynfun2_3)得到以下结果:p = 1 1 3 2 2 3 5 1 3 5 7 2 4 7 10 3 5 10 10 0f = 8该结果表明 最短路线为1 3 5 7 10最短路程为85排队论程序：queue M/M/1/k系统 课本270页例8.2.5程序及结果如下：queuethe program is with queueing theoryPlease input the system pattern:M/M/1/inf=1,M/M/1/k=2,M/M/c/inf=3,M/M/c/m/m=4! Pattern=2Pt = 2（该数字表示选择M/M/1/k系统）Please input the average arrival number in unit time lapta=10lapta = 10Please input the average service number in unit time mu=30mu = 30Please input the parameter k=2k = 2输出结果如下:The service intensity(Untruth) of the system is:ru = 0.3333The average queue length is:L = 0.3846（平均队长）The average waiting length is:Lq = 0.0769（平均等待队长）The lost possibility is:p_k = 0.0769The average lost customer number in unit time is:lapta_L = 0.7692The truth average arrival customer number in unit time is:lapta_e = 9.2308The average delay time is:W = 0.0417（平均逗留时间）The average waiting time is:Wq = 0.0083（平均等待时间）The average service intensity(efficent and truth) of the system is:ru_e =0.3077若按照甲方案，则如下结果：the program is with queueing theoryPlease input the system pattern:M/M/1/inf=1,M/M/1/k=2,M/M/c/inf=3,M/M/c/m/m=4! Pattern=2Pt = 2Please input the average arrival number in unit time lapta=10lapta = 10Please input the average service number in unit time mu=30mu = 30Please input the parameter k=3k = 3输出结果如下:The service intensity(Untruth) of the system is:ru = 0.25The average queue length is:L = 0.3176The average waiting length is:Lq = 0.0706The lost possibility is:p_k = 0.0118The average lost customer number in unit time is:lapta_L = 0.1176The truth average arrival customer number in unit time is:lapta_e = 9.8824The average delay time is:W = 0.0321The average waiting time is:Wq = 0.0071The average service intensity(efficent and truth) of the system is:ru_e =0.2471the program is with queueing theoryPlease input the system pattern:M/M/1/inf=1,M/M/1/k=2,M/M/c/inf=3,M/M/c/m/m=4! Pattern=2Pt = 2Please input the average arrival number in unit time lapta=10lapta = 10Please input the average service number in unit time mu=40mu = 40Please input the parameter k=2k = 2输出结果如下:The service intensity(Untruth) of the system is:ru = 0.25The average queue length is:L = 0.2857The average waiting length is:Lq = 0.0476The lost possibility is:p_k = 0.0476The average lost customer number in unit time is:lapta_L = 0.4762The truth average arrival customer number in unit time is:lapta_e = 9.5238The average delay time is:W = 0.03The average waiting time is:Wq = 0.005The average service intensity(efficent and truth) of the system is:ru_e =0.2381the program is with queueing theoryPlease input the system pattern:M/M/1/inf=1,M/M/1/k=2,M/M/c/inf=3,M/M/c/m/m=4! Pattern=2Pt = 2Please input the average arrival number in unit time lapta=30lapta = 30Please input the average service number in unit time mu=30mu = 30Please input the parameter k=2k = 2输出结果如下:The service intensity(Untruth) of the system is:ru = 1The average queue length is:L = 1The average waiting length is:Lq = 0.3333The lost possibility is:p_k = 0.3333The average lost customer number in unit time is:lapta_L = 10The truth average arrival customer number in unit time is:lapta_e = 20The average delay time is:W = 0.05The average waiting time is:Wq = 0.0167The average service intensity(efficent and truth) of the system is:ru_e =0.6667四、需用仪器设备 PC i5、 windows XP、MATLABR2007a、LINGO115、 MATLAB优化工具箱使用方法简介MATLAB优化工具箱具有强大的科学计算能力,在工程设计领域 得到了广泛的应用.简要介绍了MATLAB优化工具箱,通过对MATtAB优化工具箱中fmincon函数的语法进行分析,提出了结构优化设计的通用求解 方法.首先,合理设置优化目标函数和约束条件.然后,使用MATLAB优化工具箱进行编程计算.结果显示,与其他方法相比,使用MATLAB优