实验十三 线性规划与程序模块

实验十三线性规划与程序模块1、求下列线性规划问题：方法一:用LinerProgramming[c,A,b]做In[5]:=c={0.2,0.7,0.4,0.3,0.5}A={{0.3,2,1,0.6,1.8},{0.1,0.05,0.02,0.2,0.05},{0.05,0.1,0.02,0.2,0.08}}b={70,3,10}LinearProgramming[c,A,b]minf=c.%Out[5]={0.2,0.7,0.4,0.3,0.5}Out[6]={{0.3,2,1,0.6,1.8},{0.1,0.05,0.02,0.2,0.05},{0.05,0.1,0.02,0.2,0.08}}Out[7]={70,3,10}Out[8]={0.,0.,0.,39.7436,25.641}Out[9]=24.7436方法二:用ConstrainedMin来做In[10]:=ConstrainedMin[0.2x1+0.7x2+0.4x3+0.3x4+0.5x5,{0.3x1+2x2+x3+0.6x4+1.8x5≥70,0.1x1+0.05x2+0.02x3+0.2x4+0.05x5≥3,0.05x1+0.1x2+0.02x3+0.2x4+0.08x5≥10},{x1,x2,x3,x4,x5}]ConstrainedMin::deprec:ConstrainedMinisdeprecatedandwillnotbesupportedinfutureversionsofMathematica.UseNMinimizeorMinimizeinstead.More..Out[10]={24.7436,{x1→0.,x2→0.,x2→0.,x4→39.7436,x5→25.641}}方法三:用NMinimize来做In[11]:=NMinimize[{0.2x1+0.7x2+0.4x3+0.3x4+0.5x5,0.3x1+2x2+x3+0.6x4+1.8x5≥70,0.1x1+0.05x2+0.02x3+0.2x4+0.05x5≥3,0.05x1+0.1x2+0.02x3+0.2x4+0.08x5≥10,x1≥0,x2≥0,x3≥0,x4≥0,x5≥0},{x1,x2,x3,x4,x5}]Out[11]={24.7436,{x1→0.,x2→0.,x2→0.,x4→39.7436,x5→25.641}}2、某工厂生产四种不同型号的产品，而每件产品的生产要经过三个车间加工，根据该厂现有设备和劳动力等生产条件，可以确定各车间每日的生产能国（折合成有效工时数表示）。各车间每日可利用的有效工时数、每个产品在各车间加工时所花费的工时数以及每件产品可获得的利润见下表，问每种产品每季度各应该生产多少，才能使这个工厂每季度生产总值最大？车间每件产品所需的加工工时有效工时（h/d）1#2#3#4#IIIIII0.80.81.11.20.60.80.70.80.40.50.70.7160120100利润（元/件）68910解：设各生产件数为:x1,x2,x3,x4,即为:Maxf=6x1+8x2+9x3+10x4条件:0.8x1+0.8x2+1.1x3+1.2x4≤160*90(*一季度为90天*)0.6x1+0.8x2+0.7x3+0.8x4≤120*900.4x1+0.5x2+0.7x3+0.7x4≤100*90方法一：用NMaximizeIn[12]:=NMaximize[{6x1+8x2+9x3+10x4,0.8x1+0.8x2+1.1x3+1.2x4≤160*90,0.6x1+0.8x2+0.7x3+0.8x4≤120*90,0.4x1+0.5x2+0.7x3+0.7x4≤100*90,x1≥0,x2≥0,x3≥0,x4≥0},{x1,x2,x3,x4}]Out[12]={126000.,{x1→0.,x2→4500.,x3→0.,x4→9000.}}方法二:用LinearProgrammingIn[13]:=c={6,8,9,10}A={{0.8,0.8,1.1,1.2},{0.6,0.8,0.7,0.8},{0.4,0.5,0.7,0.7}}b={160,120,100}*90(*一季度为90天*)x0=LinearProgramming[-c,-A,-b]c.x0Out[13]={6,8,9,10}Out[14]={{0.8,0.8,1.1,1.2},{0.6,0.8,0.7,0.8},{0.4,0.5,0.7,0.7}}Out[15]={14400,10800,9000}Out[16]={0.,4500.,0.,9000.}Out[17]=126000.方法三:用ConstrainedMaxIn[18]:=ConstrainedMax[6x1+8x2+9x3+10x4,{0.8x1+0.8x2+1.1x3+1.2x4≤160*90,0.6x1+0.8x2+0.7x3+0.8x4≤120*90,0.4x1+0.5x2+0.7x3+0.7x4≤100*90},{x1,x2,x3,x4}]ConstrainedMax::deprec:ConstrainedMaxisdeprecatedandwillnotbesupportedinfutureversionsofMathematica.UseNMaximizeorMaximizeinstead.More…Out[18]={126000.,{x1→]0.,x2→4500.,x3→0.,x4→9000.}}3、使用两个求解线性规划问题的函数，解：方法一：用ConstrainedMinIn[19]:=Clear[x,y,z]ConstrainedMin[-3x+y+z,{x-2y+z≤11,-4x+y+2z≥3,-2x+z==1},{x,y,z}]Out[19]={-2,{x→4,y→1,z→9}}方法二:用NMinimizeIn[27]:=NMinimize[{-3x+y+z,x-2y+z≤11,-4x+y+2z≥3,-2x+z==1,x>0,y>0,z>0},{x,y,z}]Out[27]={-2.,{x→4.,y→1.,z→9.}}方法三:用LinearProgrammingIn[28]:=c={-3,1,1}A={{-1,2,-1},{-4,1,2},{-2,0,1}}b={-11,3,1}x0=LinearProgramming[c,A,b]c.x0Out[28]={-3,1,1}Out[29]={{-1,2,-1},{-4,1,2},{-2,0,1}}Out[30]={-11,3,1}Out[31]={4,1,9}Out[32]=-24、教材上自己寻找到的问题或者自己感兴趣的问题。今年(2007)湖北省高考数学题(文科11题,理科13题)设变量x,y满足约束条件,求目标函数2x+y的最小值方法一：In[33]:=NMinimize[{2x+y,x-y+3≥0,x+y≥0,x≥-2,x≤3},{x,y}]Out[33]={-1.5,{x→-1.5,y→1.5}}方法二：In[]:=x=x1-2;ConstrainedMin[2x+y,{x-y+3>=0,x+y>=0,x>=