利用MATLAB求解课本第4.1节的例题： 例２（河流污染与净化问题）某河流边有两个化工厂,流经第一个化工厂的河水流量是每天500万立方米,在两个工厂之间有一条流量为200万立方米的支流。第一个化工厂每天排放工业污水2万立方米,第二个化工厂每天排放工业污水1.4万立方米,从第一个化工厂排出的污水流到第二个化工厂之前,有20%可自然净化。根据环保要求,河流中工业污水的含量应不大于0.2%,因此两个化工厂都必须各自处理净化一部分污水,第一个化工厂处理污水的成本是0.1元/立方米,第二个化工厂处理涯水的成本是0.08元/立方米。问在满足环保要求的条件下,各化工厂每天应处理多少污水,才能使两厂总的处理污水费用最少? 解：该问题的线性规划模型归结为：  Min f=1000x1+800x2 s.t. -x1<=-1 -0.8x1-x2<=-1.6 x1<=2 x2<=1.4 x1,x2>=0 求解程序 %线性规划问题  f=[1000 800]; A=[-1 0;-0.8 -1;1 0;0 1]; b=[-1;-1.6;2;1.4]; lb=zeros(2,1); [x,fval,exitflag]=linprog(f,A,b,[],[],lb) 运行结果 Optimal solution found. x = 1.0000 0.8000 fval = 1640 exitflag = 1 例４（运输问题） 设有以下线性规划模型 min Z=3x11+11x12+3x13+10x14+x21+9x22+2x23+8x24+7x31+4x32+10x33+5x34 s.t. x11+x12+x13+x14=7 x21+x22+x23+x24=4 x31+x32+x33+x34=9 x11+x21+x31=3 x12+x22+x32=6 x13+x23+x33=5 x14+x24+x34=6 程序 c=[3,11,3,10,1,9,2,8,7,4,10,5]; Aeq=[1,1,1,1,0,0,0,0,0,0,0,0; 0,0,0,0,1,1,1,1,0,0,0,0; 0,0,0,0,0,0,0,0,1,1,1,1; 1,0,0,0,1,0,0,0,1,0,0,0; 0,1,0,0,0,1,0,0,0,1,0,0; 0,0,1,0,0,0,1,0,0,0,1,0; 0,0,0,1,0,0,0,1,0,0,0,1]; beq=[7;4;9;3;6;5;6]; lb=[0;0;0;0;0;0;0;0;0;0;0;0]; ub=[Inf;Inf;Inf;Inf;Inf;Inf;Inf;Inf;Inf;Inf;Inf;Inf]; [x,fval]=linprog(c,[],[],Aeq,beq,lb,ub) 结果 Optimal solution found. x = 0 0 5 2 3 0 0 1 0 6 0 3 fval = 85 例７（分配问题） 设有以下线性规划模型 min Z=3x11+11x12+3x13+10x14+x21+9x22+2x23+8x24+7x31+4x32+10x33+5x34 s.t. x11+x12+x13+x14<=7 x21+x22+x23+x24<=4 x31+x32+x33+x34<=9 -x11-x21-x31<=-3 -x12-x22-x32<=-6 -x13-x23-x33<=-5 -x14-x24-x34<=-6 x11,x12,x13,x14,x21,x22,x23,x24,x31,x32,x33,x34>=0 程序： f=[3,11,3,10,1,9,2,8,7,4,10,5]; A=[1,1,1,1,0,0,0,0,0,0,0,0; 0,0,0,0,1,1,1,1,0,0,0,0; 0,0,0,0,0,0,0,0,1,1,1,1; -1,0,0,0,-1,0,0,0,-1,0,0,0; 0,-1,0,0,0,-1,0,0,0,-1,0,0; 0,0,-1,0,0,0,-1,0,0,0,-1,0; 0,0,0,-1,0,0,0,-1,0,0,0,-1]; b=[7;4;9;-3;-6;-5;-6]; lb=zeros(12,1); [x,fval,exitflag]=linprog(f,A,b,[],[],lb) 结果 Optimal solution found. x = 2 0 5 0 1 0 0 3 0 6 0 3 fval = 85 exitflag = 1