lingo混合投资问题语句编写
2024-05-06 15:46
1. lingo混合投资问题语句编写
1 max = 1.1*a5+1.25*b4+1.4*c+1.55*d+1.1*a4+1.25*b3+1.1*a3+1.25*b2+1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d)-(a3+b3+c)-(a4+b4); a1+b1<=200; a2+b2+d<=1.1*a1+(200-a1-b1); a3+b3+c<=1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d); a4+b4<=1.1*a3+1.25*b2+1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d)-(a3+b3+c); a5<=1.1*a4+1.25*b3+1.1*a3+1.25*b2+1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d)-(a3+b3+c)-(a4+b4); b1<=30;b2<=30;b3<=30;b4<=30; c<=80;d<=100; 2 min=(a1+a2+a3+a4+a5)+3*(b1+b2+b3+b4)+4*c+5.5*d; 1.1*a5+1.25*b4+1.4*c+1.55*d+1.1*a4+1.25*b3+1.1*a3+1.25*b2+1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d)-(a3+b3+c)-(a4+b4)>=330; a1+b1<=200; a2+b2+d<=1.1*a1+(200-a1-b1); a3+b3+c<=1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d); a4+b4<=1.1*a3+1.25*b2+1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d)-(a3+b3+c); a5<=1.1*a4+1.25*b3+1.1*a3+1.25*b2+1.1*a2+1.25*b1+1.1*a1+(200-a1-b1)-(a2+b2+d)-(a3+b3+c)-(a4+b4); b1<=30;b2<=30;b3<=30;b4<=30; c<=80;d<=100;
2. 理财投资分配——lingo怎样编程
代码： model: ys1=10-a1-d1;!第1年年初的资金; yt1=ys1+1.06*d1;!第1年年末的资金; ys2=yt1-a2-c2-d2;!第2年年初的资金; yt2=ys2+1.15*a1+1.06*d2;!第2年年末的资金; ys3=yt2-a3-b3-c3-d3;!第3年年初的资金; yt3=ys3+1.15*a2+1.06*d3;!第3年年末的资金; ys4=yt3-a4-b4-c4-d4;!第4年年初的资金; yt4=ys4+1.15*a3+1.06*d4;!第4年年末的资金; ys5=yt4-b5-c5-d5;!第5年年初的资金; yt5=ys5+1.15*a4+1.25*(b3+b4+b5)+1.40*(c2+c3+c4+c5)+1.06*d5;!第5年年末的资金; c2+c3+c4+c5<3; max=yt5; end 运行结果： Global optimal solution found at iteration: 10 Objective value: 16.98125 Variable Value Reduced Cost YS1 0.000000 0.9357311E-01 A1 10.00000 0.000000 D1 0.000000 0.000000 YT1 0.000000 0.000000 YS2 0.000000 0.1220519 A2 0.000000 0.000000 C2 0.000000 0.3095519 D2 0.000000 0.3580189E-01 YT2 11.50000 0.000000 YS3 0.000000 0.8136792E-01 A3 11.50000 0.000000 B3 0.000000 0.1875000 C3 0.000000 0.1875000 D3 0.000000 0.000000 YT3 0.000000 0.000000 YS4 0.000000 0.1061321 A4 0.000000 0.2061321 B4 0.000000 0.1061321 C4 0.000000 0.1061321 D4 0.000000 0.3113208E-01 YT4 13.22500 0.000000 YS5 0.000000 0.2500000 B5 10.22500 0.000000 C5 3.000000 0.000000 D5 0.000000 0.1900000 YT5 16.98125 0.000000 Row Slack or Surplus Dual Price 1 0.000000 1.653125 2 0.000000 1.559552 3 0.000000 1.559552 4 0.000000 1.437500 5 0.000000 1.437500 6 0.000000 1.356132 7 0.000000 1.356132 8 0.000000 1.250000 9 0.000000 1.250000 10 0.000000 1.000000 11 0.000000 0.1500000 12 16.98125 1.000000 这里的a1-a4代表A项目每年的投资，其他符号意义类似。 希望对你有所帮助！
3. 如何用lingo编程
简单的指派问题 model: sets: project/A,B,C,D,E/:I,R,X; endsets max=@sum(project(m):X(m)*R(m)); @sum(project(m):X(m)*I(m))<500; @for(project(m):@bin(X(m))); X(1)+X(4)=1; X(5)<X(2)*99999; data: I=210,180,120,240,160; R=150,135,95,195,135; enddata end
4. 某银行经理计划用一笔资金进行有价证券的投资怎么编程序lingo
可以这样编辑 设投资ABCDE分别x1x2x3x4x5 根据题列出各个方程（目标函数和约束函数） ----程序---- max=0.043*x1+0.027*x2+0.025*x3+0.022*x4+0.045*x5; x2+x3+x4>=4; x1+x2+x3+x4+x5 反毁灭a17_幎 2014-11-07
5. 投资的收益和风险问题，要用lingo软件进行编程
代码：
model:
ys1=10-a1-d1;!第1年年初的资金;
yt1=ys1+1.06*d1;!第1年年末的资金;
ys2=yt1-a2-c2-d2;!第2年年初的资金;
yt2=ys2+1.15*a1+1.06*d2;!第2年年末的资金;
ys3=yt2-a3-b3-c3-d3;!第3年年初的资金;
yt3=ys3+1.15*a2+1.06*d3;!第3年年末的资金;
ys4=yt3-a4-b4-c4-d4;!第4年年初的资金;
yt4=ys4+1.15*a3+1.06*d4;!第4年年末的资金;
ys5=yt4-b5-c5-d5;!第5年年初的资金;
yt5=ys5+1.15*a4+1.25*(b3+b4+b5)+1.40*(c2+c3+c4+c5)+1.06*d5;!第5年年末的资金;
c2+c3+c4+c5<3;
max=yt5;
end
运行结果：
Global optimal solution found at iteration: 10
Objective value: 16.98125
Variable Value Reduced Cost
YS1 0.000000 0.9357311E-01
A1 10.00000 0.000000
D1 0.000000 0.000000
YT1 0.000000 0.000000
YS2 0.000000 0.1220519
A2 0.000000 0.000000
C2 0.000000 0.3095519
D2 0.000000 0.3580189E-01
YT2 11.50000 0.000000
YS3 0.000000 0.8136792E-01
A3 11.50000 0.000000
B3 0.000000 0.1875000
C3 0.000000 0.1875000
D3 0.000000 0.000000
YT3 0.000000 0.000000
YS4 0.000000 0.1061321
A4 0.000000 0.2061321
B4 0.000000 0.1061321
C4 0.000000 0.1061321
D4 0.000000 0.3113208E-01
YT4 13.22500 0.000000
YS5 0.000000 0.2500000
B5 10.22500 0.000000
C5 3.000000 0.000000
D5 0.000000 0.1900000
YT5 16.98125 0.000000
Row Slack or Surplus Dual Price
1 0.000000 1.653125
2 0.000000 1.559552
3 0.000000 1.559552
4 0.000000 1.437500
5 0.000000 1.437500
6 0.000000 1.356132
7 0.000000 1.356132
8 0.000000 1.250000
9 0.000000 1.250000
10 0.000000 1.000000
11 0.000000 0.1500000
12 16.98125 1.000000
这里的a1-a4代表A项目每年的投资，其他符号意义类似。
希望对你有所帮助！
6. 这个问题lingo编程怎么解决？？？追加~
看下对不对 MAX=130*Y11+18*Y12+50*Y23+0.25*X54+90*Y35+1.2*X56+1.15*X57; X11-220*Y11=0; X21-220*Y21=0; Y11-Y21=0; X12-70*Y12=0; X23-180*Y23=0; X14<=80; X24-X14<=15; X34-X24<=15; X44-X34<=15; X54-X44<=15; X35-320*Y35=0; X16>=60; X26>=60; X36>=60; X46>=60; X56>=60; 220*Y11+70*Y12+X14+X16+X17=350; 0.25*X14+1.2*X16+1.15*X17+300-X21-X23-X24-X26-X27=0; 60*Y11+18*Y12+0.25*X24+1.2*X26+1.15*X27+150-X34-320*Y35-X36-X37=0; 130*Y11+18*Y12+50*Y23+0.25*X34+1.2*X36+1.15*X37-X44-X46-X47=0; 130*Y11+18*Y12+50*Y23+0.25*X44+90*Y35+1.2*X46+1.15*X47-X54-X56-X57=0; @gin(Y11); @gin(Y12); @gin(Y23); @gin(Y35);
7. lingo求最优投资组合？
没有可行解， MODEL: SETS: SEC/1..5/:RETURN,WEIGHT; LINK(SEC,SEC):COV; ENDSETS DATA: RETURN=0.0275,0.0510,0.0526,0.0455,0.2467; COV= 0.187 0.236 0.110 -0.020 -7.243 0.236 1.036 0.789 0.624 -43.903 0.110 0.789 1.333 1.085 -46.298 -0.020 0.624 1.085 2.101 -30.715 -7.243 -43.903 -46.298 -30.715 3837.690; ENDDATA @SUM(LINK(I,J):WEIGHT(I)*WEIGHT(J)*COV(I,J))<0.01; @SUM(SEC(I)|I#EQ#1:WEIGHT(I))>0.1; @SUM(SEC(I)|I#LE#2:WEIGHT(I))>0.5; @SUM(SEC(I)|(I#GE#3) #AND# (I#LE#4):WEIGHT(I))<0.1; @SUM(SEC(I)|I#EQ#5:WEIGHT(I))<0.2; MAX=@SUM(SEC(I):WEIGHT(I)*RETURN(I)); END
8. lingo投资问题；最后得出结果无论加权系数p取何值均不变，而且只有决策变量x（a7）有值。
你写在init里面当然没用 写在data段里面才行
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