LINGO_程序范例

线性规划LINGO程序集 向浩 2011-7-20程序一：Model:Max= 4.8*x11 + 4.8*x21 + 5.6*x12 + 5.6*x22 - 10*x1 - 8*x2 - 6*x3;x11+x12 x + 500;x21+x22 0; 0.4*x12 - 0.6*x22 0;x=x1+x2+x3; (x1 - 500) * x2=0; (x2 - 500) * x3=0; bnd(0,x1, 500);bnd(0,x2, 500);bnd(0,x3,500);End程序二：Model:SETS:Points/1.4/: b, c, y, z;! 端点数为4，即分段数为3;ENDSETSDATA:b=0 500 1000 1500;c=0 5000 9000 12000;y=,0;! 增加的虚拟变量y(4)=0;ENDDATAMax= 4.8*x11 + 4.8*x21 + 5.6*x12 + 5.6*x22 - sum(Points: c*z);x11+x12 x + 500;x21+x22 0; 0.4*x12 - 0.6*x22 0;sum(Points: b*z)=x;for(Points(i)|i#eq#1: z(i) = y(i);for(Points(i)|i#ne#1: z(i) = y(i-1)+y(i);sum(Points: y)=1;sum(Points: z)=1;for(Points: bin(y);End程序三：MODEL:TITLE 瓶颈设备的多级生产计划;! 从文本文件exam0502.LDT中读取数据;SETS:! PART = 项目集合, Setup = 生产准备费，Hold = 单件库存成本， A = 对瓶颈资源的消耗系数;PART/ FILE( F:lindolindo书ch05exam0502.LDT)/ : Setup, Hold, A;! TIME = 计划期集合，Capacity = 瓶颈设备的能力;TIME / FILE( F:lindolindo书ch05exam0502.LDT)/ : Capacity;! USES = 项目结构关系，Req = 项目之间的消耗系数;USES( PART, PART) : Req;! PXT = 项目与时间的派生集合，Demand = 外部需求, X = 产量（批量）, Y = 0/1变量，INV = 库存;PXT( PART, TIME): Demand, X, Y, Inv;ENDSETS! 目标函数;OBJ Min = sum(PXT(i,t): setup(i)*Y(i,t) + hold(i)*Inv(i,t) );! 物流平衡方程;FOR( PXT(i, t) | t #NE# 1 : Bal Inv(i,t-1)+X(i,t)-Inv(i,t) = Demand(i, t) + SUM( USES(i,j): Req(i,j)*X(j,t) );FOR( PXT(i, t) | t #eq# 1 : Ba0 X(i,t)-Inv(i,t) = Demand(i, t) + SUM( USES(i,j): Req(i,j)*X(j,t) );! 能力约束;FOR( TIME(t): Cap SUM( PART(i): A(i)*X(i,t) ) Capacity(t) ); ! 其他约束;M = 25000;FOR( PXT(i,t): X(i,t) NUM(I) ); !满足需求约束;FOR(CUTS(J): SUM(NEEDS(I): LENGTH(I)*R(I,J) ) CAPACITY -MIN(NEEDS(I):LENGTH(I) ); !合理切割模式约束;SUM(CUTS(I): X(I) ) 26; SUM(CUTS(I): X(I) ) X(I+1) ); !人为增加约束;FOR(CUTS(J): GIN(X(J) ) ;FOR(PATTERNS(I,J): GIN(R(I,J) );end程序五：model:Title 面试问题;SETS:! Person = 被面试者集合，Stage = 面试阶段的集合;Person/FILE(exam0505.txt)/;Stage/FILE(exam0505.txt)/;! T = 已知的面试所需要的时间，X = 面试开始时间;PXS(Person,Stage): T, X;! Y(i,k) = 1: k排在i前，0：否则;PXP(Person,Person)|&1 #LT# &2: Y;ENDSETSDATA:T=FILE(exam0505.txt);ENDDATAobj min =MAXT;! MAXT是面试的最后结束时间;MAXT = max(PXS(i,j)|j#EQ#size(stage): x(i,j)+t(i,j);! 只有参加完前一个阶段的面试后才能进入下一个阶段;for(PXS(i,j)|j #LT# size(stage): ORDER x(i,j) + t(i,j) x(i,j+1) );! 同一时间只能面试1名同学;for(Stage(j): for(PXP(i,k): SORT1 x(i,j) + t(i,j) - x(k,j) MAXT*Y(i,k) );for(PXP(i,k): SORT2 x(k,j) + t(k,j) - x(i,j) = TARGET;END程序九：MODEL:Title 含有国库券的投资组合模型;SETS: STOCKS/ A, B, C, D/: Mean,X; STST(Stocks,stocks): COV;ENDSETSDATA: TARGET = 1.1; ! 1.15;! Mean是收益均值,COV是协方差矩阵; mean=1.089083 1.213667 1.234583 1.05; COV=0.01080754 0.01240721 0.01307513 0 0.01240721 0.05839170 0.05542639 0 0.01307513 0.05542639 0.09422681 0 0 0 0 0;ENDDATAOBJ MIN = sum(STST(i,j): COV(i,j)*x(i)*x(j);ONE SUM(STOCKS: X) = 1;TWO SUM(stocks: mean*x) = TARGET;END程序十：MODEL:Title 考虑交易费的投资组合模型;SETS: STOCKS/ A, B, C/: C,Mean,X,Y,Z; STST(Stocks,stocks): COV;ENDSETSDATA: TARGET = 1.15;! 股票的初始份额; c=0.5 0.35 0.15; ! Mean是收益均值,COV是协方差矩阵; mean=1.089083 1.213667 1.234583; COV=0.01080754 0.01240721 0.01307513 0.01240721 0.05839170 0.05542639 0.01307513 0.05542639 0.09422681;ENDDATAOBJ MIN = sum(STST(i,j): COV(i,j)*x(i)*x(j);ONE SUM(STOCKS: X+0.01*Y+0.01*Z) = 1;TWO SUM(stocks: mean*x) = TARGET;FOR(stocks: ADD x = c - y + z);END程序十一：MODEL:Title 利用股票指数简化投资组合模型;SETS: STOCKS/A, B, C/: u, b, s2, x;ENDSETSDATA:! mean0,s20,u,b,s2是线性回归的结果数据;mean0=1.191458;s20 = 0.02873661;s2 = 0.005748320,0.01564263,0.03025165;u = 0.5639761, -0.2635059,-0.5809590;b = 0.4407264, 1.239802, 1.523798;ENDDATAOBJ MIN = s20*y*y + sum(stocks: s2*x*x); !OBJ MIN = s20*sqr(y) + sum(stocks: s2*sqr(x); sum(stocks: b*x)=y;sum(stocks: x)=1;sum(stocks: (u+b*mean0)*x)1.15;END程序十二：MODEL:TITLE 新产品的市场预测;SETS: PROD/ A B C D/: P; LINK(PROD, PROD): T;ENDSETSDATA: ! 转移概率矩阵; T = .75 .1 .05 .1 .4 .2 .1 .3 .1 .2 .4 .3 .2 .2 .3 .3;ENDDATAFOR(PROD(I)| I #LT# SIZE(PROD): !去掉了一个冗余约束; P(I)=SUM(LINK(J,I): P(J)* T(J,I) );SUM(PROD: P) = 1;FOR(PROD(I): WARN( 输入矩阵的每行之和必须是1, ABS( 1 - SUM(LINK(I,J): T(I,J) #GT# .000001); );END程序十三：MODEL:TITLE 产品属性的效用函数;SETS: PRICE /H, M, L/ : P; SAFETY/2, 1, 0/ : Q; M(safety, PRICE) : C0; MM(M, M)|C0(&1,&2) #LT# C0(&3,&4): ERROR;ENDSETSDATA: C0 = 7 8 9 3 4 6 1 2 5;ENDDATAFOR( MM( i, j, k, l): ERROR( i, j, k, l) = 1 + ( P( j) + Q( i) - ( P( l) + Q( k) );obj MIN = SUM(mm: ERROR);END程序十四：MODEL:TITLE 最小二乘法计算产品属性的效用函数;SETS: PRICE /H, M, L/ : P; SAFETY/2, 1, 0/ : Q; M(safety, PRICE) : C0, ERROR, sort;ENDSETSDATA: C0 = 7 8 9 3 4 6 1 2 5;ENDDATAFOR(M(i,j):sort(i,j)=p(j)+q(i); ERROR(i,j)= sort(i,j) -C0(i,j) );MIN = SUM(M: sqr(ERROR) );!obj MIN = SUM(M: ERROR2 );FOR(M(i,j): FREE(ERROR) );!FOR(price: gin(P) );!FOR(safety: gin(Q) );END程序十五：MODEL:TITLE 机票销售计划;SETS: route /AH,AB,AC,HB,HC/:a,b,p,q,x,y;ENDSETSDATA: a p b q =331905690242444319312261671994414069801618617103; c1 c2 c3 = 120 100 110;ENDDATAobj Max = SUM(route: p*x+q*y );SUM(route(i)|i#ne#4#and#i#ne#5:x(i)+y(i) c1;SUM(route(i)|i#eq#2#or#i#eq#4:x(i)+y(i) c2;SUM(route(i)|i#eq#3#or#i#eq#5:x(i)+y(i) c3;FOR(route: bnd(0,x,a);bnd(0,y,b) );END程序十六：! 3 Warehouse, 4 Customer Transportation Problem;sets: Warehouse /1.3/: a; Customer /1.4/: b; Routes( Warehouse, Customer) : c, x;endsets! Here are the parameters;data: a = 30, 25, 21; b = 15, 17, 22, 12; c = 6, 2, 6, 7, 4, 9, 5, 3, 8, 8, 1, 5;enddata! The objective;OBJ min = sum( Routes: c * x);! The supply constraints;for( Warehouse(i): SUP sum( Customer(j): x(i,j) = a(i);! The demand constraints;for( Customer(j): DEM sum( Warehouse(i): x(i,j) = b(j);程序十七：! Assignment Problem Model;sets: Flight/1.6/; Assign(Flight, Flight): c, x;endsets! Here is income matrix;data: c = 20 15 16 5 4 7 17 15 33 12 8 6 9 12 18 16 30 13 12 8 11 27 19 14 -99 7 10 21 10 32 -99 -99 -99 6 11 13;enddata! Maximize valve of assignments;max = sum(Assign: c*x);for(Flight(i): ! Each i must be assigned to some j; sum(Flight(j): x(i,j) = 1;! Each I must receive an assignment; sum(Flight(j): x(j,i) = 1;);程序十八：! 2 plants, 3 warehouses and 4 customers Transshipment Problem;sets: Plant /A, B/: produce; Warhouse /x, y, z/; Customer /1.4/: require; LinkI ( Plant, Warhouse): cI, xI; LinkII ( Warhouse, Customer): cII, xII;endsets! Here are the parameters;data: produce = 9, 8; require = 3, 5, 4, 5; cI = 1, 2, 100, 3, 1, 2; cII = 5, 7, 100, 100, 9, 6, 7, 100, 100, 8, 7, 4;enddata! The objective;OBJ min = sum( LinkI: cI * xI) + sum( LinkII: cII * xII);! The supply constraints;for( Plant(i): SUP sum( Warhouse(j): xI(i,j) = produce(i);! The warhouse constraints;for( Warhouse(j): MID sum( Plant(i): xI(i,j)=sum( Customer(k): xII(j,k);! The demand constraints;for( Customer(k): DEM sum( Warhouse(j): xII(j,k) = require(k);! 2 plants, 3 warehouses and 4 customers Transshipment Problem;sets: Plant /A,B/ : produce; Warhouse /x,y,z/; Customer /1.4/ : require; Link ( Plant, Warhouse, Customer) : poss, cost, x;endsets! Here are the parameters;data: produce = 9, 8; require = 3, 5, 4, 5; poss = 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1; cost = 6, 8, 0, 0, 11, 8, 9, 0, 0, 0, 0, 0, 8,10, 0, 0, 10, 7, 8, 0, 0,10, 9, 6;enddata! The objective;OBJ min = sum( Link: poss * cost * x);! The supply constraints;for( Plant(i): SUP sum( Warhouse(j): sum( Customer(k): poss(i,j,k) * x(i,j,k)=1;!For city i, except the base (city 1);for(cities(i) | i #gt# 1 :! It must be entered; sum(cities(j)| j #ne# i: x(j,i)=1;! level(j)=levle(i)+1, if we link j and i; for(cities(j)| j #gt# 1 #and# j #ne# i : level(j) = level(i) + x(i,j) - (n-2)*(1-x(i,j) + (n-3)*x(j,i); );! The level of city is at least 1 but no more n-1, and is 1 if it links to base (city 1); bnd(1,level(i),999999); level(i)=1;!For city i, except the base (city 1);for(cities(i) | i #gt# 1 :! It must be entered; sum(cities(j)| j #ne# i: x(j,i)=1;! level(j)=levle(i)+1, if we link j and i; for(cities(j)| j #gt# 1 #and# j #ne# i : level(j) = l