lingo编程练习题.doc

一、编写lingo程序求解下列方程（组）1、 syms x; x=solve(x=(cos(x)+sin(x)/4)x =0.315182442838735901956487060939832、 syms x; x=solve(x=4-2x)x =4 - lambertw(0, 16*log(2)/log(2) x=eval(x)x = 1.3861669800714943、 求方程在中的根的近似值Matlab: syms x; y=x3-2*x2-4*x-7; x=eval(solve(y)x = 3.6320 + 0.0000i -0.8160 - 1.1232i -0.8160 + 1.1232iOR: syms x; a=1,-2,-4,-7; x=roots(a)x = 3.6320 + 0.0000i -0.8160 + 1.1232i -0.8160 - 1.1232iLingo:x3-2*x2-4*x-7=0;bnd(3,x,4);Feasible solution found. Infeasibilities: 0.5329071E-14 Extended solver steps: 5 Total solver iterations: 36 Variable Value X 3.631981Matlab:function y=f(m,n)syms x xk;a=m;b=n;ff=x3-2*x2-4*x-7;while b-a0.00001 xk=(a+b)/2; fx=subs(ff,x,xk); fa=subs(ff,x,a); if fx=0 y=xk; break; elseif fa*fx x4=solve(x42-3*x4-4=0,x4)x4 = 4 -1lingox12-3*x1-4=0;x22-3*x2-4=0;free(x2);Variable Value X1 4.000000 X2 -1.0000005、 syms x1 x2 x3 x4; x1,x2,x3,x4=solve(2*x1+x2-x3+x4,3*x1-2*x2+x3-x4,x1+4*x2-3*x3+5*x4,x1,x2,x3,x4)x1 =z/7x2 =(5*z)/7x3 =z x4 =06、 a=5,2,1;-1,4,2;2,-3,10; b=-12;20;3; x=inv(a)*bx = -4.000000000000000 3.000000000000000 2.000000000000000二、编写lingo程序求解下列最优化问题1、4*x1-x2+2*x3-x4=-2;x1+x2-x3+2*x42;free(x4);Feasible solution found. Infeasibilities: 0.000000 Total solver iterations: 2 Variable Value X1 0.000000 X2 1.000000 X3 0.000000 X4 1.0000002、3、max=3*x1-x2;3*x1-x210;2*x1+x21;x1-x2+6*x3+4*x48;5*x1+3*x3+x45;bin(x1);bin(x2);bin(x3);bin(x4);Global optimal solution found. Objective value: 0.000000 Objective bound: 0.000000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost X 0.000000 3.000000 X2 0.000000 7.000000 X3 1.000000 -1.000000 X4 1.000000 1.000000 X1 1.000000 0.0000006、求图中点到各点的最短路（不可逆行）74v2v4v3v1v5v6v725324633421v8model:SETS: points/v1,v2,v3,v4,v5,v6,v7,v8/: L; roads(points,points)/v1,v2 v1,v3 v1,v4v2,v3 v2,v5 v3,v6 v4,v3v6,v5 v6,v8 v7,v6 v7,v4 v8,v5 v8,v7/: d;ENDSETSDATA:d=2 5 3 2 4 6 4 3 4 2 7 3 1; L=0,;ENDDATA FOR(points(i)|i#GT#index(v1): L(i) = MIN( roads(j, i): L(j) + d(j, i););end Linearization components added: Constraints: 24 Variables: 14 Integers: 10 Feasible solution found. Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value L( V1) 0.000000 L( V2) 2.000000 L( V3) 4.000000 L( V4) 3.000000 L( V5) 6.000000 L( V6) 10.00000 L( V7) 15.00000 L( V8) 14.00000 D( V1, V2) 2.000000 D( V1, V3) 5.000000 D( V1, V4) 3.000000 D( V2, V3) 2.000000 D( V2, V5) 4.000000 D( V3, V6) 6.000000 D( V4, V3) 4.000000 D( V6, V5) 3.000000 D( V6, V8) 4.000000 D( V7, V6) 2.000000 D( V7, V4) 7.000000 D( V8, V5) 3.000000 D( V8, V7) 1.000000三、先建立问题的数学模型，再编写lingo程序求解1、某厂每日8小时的产量不低于1800件为了进行质量控制，计划聘请两种不同水平的检验员一级检验员的标准为：速度25件/小时，正确率98%，计时工资4元/小时；二级检验员的标准为：速度15小时/件，正确率95%，计时工资3元/小时检验员每错检一次，工厂要损失2元为使总检验费用最省，该工厂应聘一级、二级检验员各几名？min=8*x1*(4+0.02*25*2)+8*x2*(3+0.05*15*2);8*x1*25+8*x2*15=1800;gin(x1);gin(x2); Global optimal solution found. Objective value: 360.0000 Objective bound: 360.0000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0 Variable Value Reduced Cost X1 9.000000 40.00000 X2 0.000000 36.000002、某饲料场饲养动物出售，设每头动物每天至少需700克蛋白质、30克矿物质、100毫克维生素现有5种饲料可供选用，各种饲料每公斤营养成分含量及单价如表所示：饲料蛋白质（g）矿物质（g）维生素（mg）价格（元/kg）1234532161810.50.220.50.51.00.220.80.20.70.40.30.8要求确定既满足动物生长的营养需要，又使费用最省的选用饲料的方案model:sets:f/1.5/:x,p;e/1.3/;s(f,e):c;endsetsdata:p=0.2 0.7 0.4 0.3 0.8;c=3 1 0.52 0.5 11 0.2 0.26 2 218 0.5 0.8;enddatamin=sum(f(i):p(i)*x(i);sum(f(i):c(i,1)*x(i)700;sum(f(i):c(i,2)*x(i)30;sum(f(i):c(i,3)*x(i)100;Global optimal solution found. Objective value: 32.43590 Infeasibilities: 0.000000 Total solver iterations: 2 Variable Value Reduced Cost X( 1) 0.000000 0.5961538E-01 X( 2) 0.000000 0.5935897 X( 3) 0.000000 0.3525641 X( 4) 39.74359 0.000000 X( 5) 25.64103 0.000000 P( 1) 0.2000000 0.000000 P( 2) 0.7000000 0.000000 P( 3) 0.4000000 0.000000 P( 4) 0.3000000 0.000000 P( 5) 0.8000000 0.0000003、某医院护士值班班次、每班工作时间及各班所需护士数如表所示每班护士值班开始时向病房报到，并连续工作8小时试决定该医院最少需要多少名护士，以满足轮班需要班次工作时间所需护士数1234566：0010：0010：0014：0014：0018：0018：0022：0022：002：002：006：00607060502030min=x1+x2+x3+x4+x5+x6;x1+x6=60;x2+x1=70;x2+x3=60;x3+x4=50;x5+x4=20;x5+x6=30;gin(x1);gin(x2);gin(x3);gin(x4);gin(x5);gin(x6);Global optimal solution found. Objective value: 150.0000 Objective bound: 150.0000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 4 Variable Value Reduced Cost X1 60.00000 1.000000 X2 10.00000 1.000000 X3 50.00000 1.000000 X4 0.000000 1.000000 X5 30.00000 1.000000 X6 0.000000 1.0000004、一艘货轮分前、中、后三个舱位，它们的容积与最大允许载重量如表1所示现有三种货物待运，已知有关数据列于表2为了航运安全，前、中、后舱的实际载重量上大体保持各舱最大允许载重量的比例关系具体要求：前、后舱分别与中舱之间载重量比例上偏差不超过15%，前后舱之间不超过10%问该货轮应装载A，B，C各多少件运费收入才最大？表1前 舱中 舱后 舱最大允许载重量（t）容 积（m3）200040003000540015001500表2商 品数 量（件）每件体积（m3/件）每件重量（t/件）运 价（元/件）ABC6001000800105786510007006005、某市有三个面粉厂，它们供给三个面食加工厂所需的面粉各面粉厂的产量、各面食加工厂加工面粉的能力、各面食加工厂和各面粉厂之间的单位运价如下表所示假定在第1，2和3面食加工厂制作单位面粉食品的利润分别为12元、16元和11元，试确定使总收益最大的面粉分配计划食品厂面粉厂123面粉厂产量IIIIII348101111284203020食品厂需量152520max=12*(x11+x21+x31)+16*(x12+x22+x32)+11*(x13+x23+x33)-(x11*3+10*x12+2*x13)-(4*x21+11*x22+8*x23)-(8*x31+11*x32+4*x33);3*x11+4*x21+8*x3115;10*x12+11*x22+11*x3225;2*x13+8*x23+4*x3320;3*x11+10*x12+2*x1320;4*x21+11*x22+8*x2330;8*x31+11*x32+4*x3320;Global optimal solution found. Objective value: 151.3636 Infeasibilities: 0.000000 Total solver iterations: 5 Variable Value Reduced Cost X11 0.000000 4.500000 X21 6.250000 0.000000 X31 0.000000 12.00000 X12 0.000000 23.54545 X22 0.4545455 0.000000 X32 1.818182 0.000000 X13 10.00000 0.000000 X23 0.000000 13.00000 X33 0.000000 1.0000006、1，2，3三个城市每年需分别供应电力320，250和350单位，由I，II两个电站提供，它们的最大可供电量分别为400个单位和450个单位，单位费用如下表所示由于需要量大于可供量，决定城市1的供应量可减少0单位30单位，城市2的供应量不变，城市3的供应量不能少于270单位，试求总费用最低的分配方案（将可供电量用完）城 市电 站123III152118252216min=15*x11+18*x12+22*x13+21*x21+25*x22+16*x23;x11+x12+x13=400;x21+x22+x23=450;x11+x21290;x12+x22=250;x13+x23270;Global optimal solution found. Objective value: 14010.00 Infeasibilities: 0.000000 Total solver iterations: 3 Variable Value Reduced Cost X11 150.0000 0.000000 X12 250.0000 0.000000 X13 0.000000 12.00000 X21 140.0000 0.000000 X22 0.000000 1.000000 X23 270.0000 0.0000007、有三种资源被用于生产三种产品，资源量、产品单件可变费用、单件售价、资源单耗量及组织三种产品生产的固定费用见下表要求制定一个生产计划，使总收益最大产品单耗量资源IIIIII资源量A248500B234300C123100单件可变费用456固定费用100150200单件售价81012max=if(x1#eq#0,if(x2#eq#0,4*x1+5*x2+6*x3-200,if(x3#eq#0,4*x1+5*x2+6*x3-150,4*x1+5*x2+6*x3-350),if(x2#eq#0,if(x3#eq#0,4*x1+5*x2+6*x3-100,4*x1+5*x2+6*x3-300),if(x3#eq#0,4*x1+5*x2+6*x3-250,4*x1+5*x2+6*x3-450);2*x1+4*x2+8*x3500;2*x1+3*x2+4*x3300;x1+2*x2+3*x3=1;x6=if(x1#eq#1#AND#x4#eq#1,0,1);x2+x8=2;x1+x2+x3+x4+x5+x6+x7+x8=5;bin(x1);bin(x2);bin(x3);bin(x4);bin(x5);bin(x6);bin(x7);bin(x8); Linearization components added: Constraints: 36 Variables: 26 Integers: 20 Global optimal solution found. Objective value: 1.864000 Objective bound: 1.864000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 9 Variable Value Reduced Cost X1 0.000000 -0.3840000 X2 1.000000 -0.3800000 X3 1.000000 -0.3760000 X4 1.000000 -0.3720000 X5 1.000000 -0.3700000 X6 1.000000 -0.3660000 X7 0.000000 -0.3600000 X8 0.000000 -0.356000010、有5项设计任务可供选择各项设计任务的预期完成时间分别为3，8，5，4，10周，设计报酬分别为7，17，11，9，21万元设计任务只能一项一项地进行，总的期限是20周选择任务时必须满足下面要求：（1）至少完成3项设计任务；（2）若选择任务1，必须同时选择任务2；（3）任务3和任务4不能同时选择应当选择那些设计任务，才能使总的设计报酬最大？max=7*x1+17*x2+11*x3+9*x4+21+x5;3*x1+8*x2+5*x3+4*x4+10*x5=3;bin(x1);bin(x2);bin(x3);bin(x4);bin(x5); Linearization components added: Constraints: 48 Variables: 30 In