lingo常见问题求解模板

(1)线性规划LP MODEL: MIN=2*x1-2*x2; 2*x1+x2+x3=3; -x1+x2+x4=-1; @GIN(x1);@GIN(x2);@GIN(x3);@GIN(x4); END (2)线性规划LP MODEL: MIN=-7*x1-12*x2; 9*x1+4*x2+x3=360; 4*x1+5*x2+x4=200; 3*x1+10*x2+x5=300; @GIN(x1);@GIN(x2);@GIN(x3);@GIN(x4); END (3)线性规划LP MODEL: MIN=2*x1+x2-x4; x1+x2+x3=5; -x1+x2+x4=6; 6*x1+2*x2<=21; END (4)指派问题AP MODEL: SETS: r/1..5/:; c/1..5/:; link(r,c):time,x; ENDSETS DATA: time= 8 6 10 9 2 9 12 7 11 9 7 4 3 5 8 9 5 8 11 8 4 6 7 5 11; ENDDATA MIN=@SUM(link:x*time); @FOR(link:@BIN(x)); @FOR(r(i):@SUM(c(j):x(i,j))=1); @FOR(c(j):@SUM(r(i):x(i,j))=1); End (5)纯整数规划PILP MODEL: MAX=40*x1+90*x2; 9*x1+7*x2<=56; 7*x1+20*x2<=70; @GIN(x1);@GIN(x2); END (6)混合整数规划ILP MODEL: MAX=40*x1+90*x2; 9*x1+7*x2<=56; 7*x1+20*x2<=70; @GIN(x1); END (7)0—1型整数规划ZILP MODEL: MAX=5*x1-x2+7*x3; x1+4*x2+x3<=5; x1+2*x2-x3<=2; 2*x1+x2<=3; 4*x2+3*x3<=6; @BIN(x1);@BIN(x2);@BIN(x3); END (8)连续函数的最值问题 最大值MAX MODEL: MAX=(x-x^3+y^3-y^5)*@EXP(-(x^2+y^2)); @BND(-2,x,2);@BND(-3,y,3); END 最小值MIN MODEL: MIN=(x-x^3+y^3-y^5)*@EXP(-(x^2+y^2)); @BND(-2,x,2);@BND(-3,y,3); END (9)非线性规划问题NLP MODEL: SETS: a/1..6/:horizontal,vertical,distance; b/1,2/:positionx,positiony; link(a,b):c; ENDSETS DATA: horizontal=1.25 8.75 0.5 5.75 3 7.25; vertical=1.25 0.75 4.75 5 6.5 7.75; distance=3 5 4 7 6 11; positionx=5 2; !分别为P,Q的横坐标; positiony=1 7; !分别为P,Q的纵坐标; ENDDATA MIN=@sum(b(j):@SUM(a(i):c(i,j)*@SQRT((positionx(j)-horizontal(i))^2+(positiony(j)-vertical(i))^2))); @FOR(a(i):@SUM(b(j):c(i,j))=distance(i)); @FOR(b(j):@SUM(a(i):c(i,j))<=20); END (10)动态规划问题 最短路线问题、生产计划问题以及资源分配问题。 (一)最短路线问题: MODEL: SETS: cities/A,B1,B2,B3,C1,C2,C3,D1,D2,E/:L; !属性L(i)示城市A到城市i的最优行驶路线长; POADS(cities,cities)/ !派生集合POADS表示的是网络中的道路; A,B1 A,B2 A,B3 B1,C1 B1,C2 B1,C3 B2,C1 B2,C2 B2,C3 B3,C1 B3,C2 B3,C3 C1,D1 C1,D2 C2,D1 C2,D2 C3,D1 C3,D2 D1,E D2,E/:P; !属性P(i,j)是城市i到 城市j的直接距离; ENDSETS DATA: P=2 5 3 7 5 6 3 2 4 5 1 5 1 4 6 3 3 3 3 4; L=0,,,,,,,,,; !因为L(A)=0; ENDDATA @FOR(cities(i)|i#GT#@index(A):L(i)=@MIN(POADS(j,i):L(j)+P(j,i));); !这行中'@index'可以直接写成1; End (二)最小运输费用问题 6个出发地8个目的地的最小运输费用问题 MODEL: SETS: warehouses/wh1..wh6/: capacity; vendors/v1..v8/: demand; links(warehouses,vendors): cost, volume; ENDSETS !目标函数; MIN=@sum(links: cost*volume); !需求约束; @FOR(vendors(J): @SUM(warehouses(I): volume(I,J))=demand(J)); !产量约束; @FOR(warehouses(I): @SUM(vendors(J): volume(I,J))<=capacity(I)); !这里是数据; DATA: capacity=60 55 51 43 41 52; demand=35 37 22 32 41 32 43 38; cost=6 2 6 7 4 2 9 5 4 9 5 3 8 5 8 2 5 2 1 9 7 4 3 3 7 6 7 3 9 2 7 1 2 3 9 5 7 2 6 5 5 5 2 2 8 1 4 3; ENDDATA END