lingolingo解目標規(guī)劃

1、Lingo解目標規(guī)劃方法：按目標的優(yōu)先級次序，依次用lingo求解。每次求得的值再作為下一次的約束條件。例1：求解下列目標規(guī)劃min z = p d + + p(d - + d + ) + p d -11222332x1+ x2 11s.t.x1- x2 + d - - d += 011x1+ 2x2 + d - - d += 1022-3- d += 568x1+10x2 + d3x1, x2, d +- 0iLingo代碼min=d31; 2*x1+x2=11;x1-x2+d11-d12=0; x1+2*x2+d21-d22=10; 8*x1+10*x2+d31-d32=56; d12=0

2、;d21+d22=0;例2：min z = p d - + p d + + p(2d -+ d - )1122334x1+ x2 + d - - d += 4011s.t.x1+ x2 + d - - d += 5022x1+ d - - d += 24= 30 033x3 + d - - d +44x1, x2, d +-iLingo代碼min=2*d31+d41; x1+x2+d11-d12=40; x1+x2+d21-d22=50; x1+d31-d32=24; x2+d41-d42=30; d11=0;d22=0;例3：某棉紡車間用甲乙兩種棉花混紡生產(chǎn)A、B兩種棉紗，其相關(guān)數(shù)據(jù)如下表：

3、若利潤指標為1755元，A種棉花要生產(chǎn)650km，問A,B兩種棉紗各應該生產(chǎn)多少？產(chǎn)品資源單耗（kg/kmA）B棉花庫存量甲綿(kg)0.50.3300乙綿(kg)0.10.3180利潤（元/km）2.53.2目標規(guī)劃模型：設(shè)分別生產(chǎn)A、B兩種棉紗x1, x2km，則：300180d += 17550122x1, x2, d 0imin z = p d - + p d -11220.5x1+ 0.3x2 0.1x1+ 0.3x2 2.5x1+ 3.2x2 + d - -1x1+ d - - d += 65Lingo代碼：min=d21; 0.5*x1+0.3*x2=300;0.1*x1+0.3

4、*x2=180;2.5*x1+3.2*x2+d11-d12=1755; x1+d21-d22=650;d11=0;例4：已知有三個產(chǎn)地給四個銷地供應某種產(chǎn)品，產(chǎn)銷地之間的供需量和單位運價如下表：銷地產(chǎn)地B1B2B3B4產(chǎn)量A15267300A23546200A34523400銷量200100450250900/1000有關(guān)部門在研究調(diào)運方案時依次考慮以下七項目標， 并規(guī)定其相應的優(yōu)先等級：P1-B4是重點保證單位，必須全部滿足其需求; P2-A3向B1提供的產(chǎn)量不少于100；P3-每個銷地供應量不小于其需要的80%；P4-所訂調(diào)運方案的總運費不超過最小運費的10%； P5-因路段問題，盡量避免

5、安排將A2產(chǎn)品往B4； P6-給B1,B2的供應率要相同；P7-力求總運費最省。試求滿意的調(diào)運方案？不考慮目標，運輸問題的數(shù)學模型為：設(shè)xij 表示從第i個產(chǎn)地向第j個銷地的運量。則：min z = cij xiji=1j =1344 xijj =1= ai , i = 1,2,33 xij bj , j = 1,2,3,4i=1xij 0Lingo代碼：sets:cd/1.3/:a;xd/1.4/:b; links(cd,xd):c,x; endsetsdata:a=300 200 400;b=200 100 450 250;c=5 2 6 73 5 4 64 5 2 3;enddata m

6、in=sum(links(i,j):c(i,j)*x(i,j);for(cd(i):sum(xd(j):x(i,j)=b(j);考慮目標規(guī)劃：供應約束：x11 + x12 + x13 + x14 300 x21 + x22 + x23 + x24 200 x31 + x32 + x33 + x34 400 需求約束：x11+ x21+ x31 200x12 + x22 + x32 100x13 + x23 + x33 450B4的需求量必須全部滿足：x14 + x24 + x34 + d - - d + = 25044A3向B1提供的產(chǎn)品不少于100x31+ d - - d + = 10055

7、每個銷地的供應量不小 于其需求量的80%x11 + x21 + x31 + d - - d + = 200 * 0.866x12 + x22 + x32 + d - - d += 100 * 0.8= 450 * 0.8= 250 * 0.877x13 + x23 + x33 + d - - d +88x14 + x24 + x34 + d - - d +99調(diào)運方案的總運費不超 過最小費用的10%34cx+d - - d += 2950 *(1+10%)ijij1010i=1j =1因路段原因，盡量避免將A2的產(chǎn)品往B4x24 + d - - d += 01111給B1和B2的供應率要相同x

8、11+ x21+ x31- (200 / 450)(x13 + x23 + x33) + d - - d += 01212力求運費最省：34cx+ d -+ d += 2950ijij1313i=1j =1目標函數(shù)：min z = Pd - + P d - + P (d - + d -+ d - + d - ) + P d +142535678410+ P d + P (d -+ d + ) + P d +51161212713Lingo代碼sets:cd/1.3/:a;xd/1.4/:b; links(cd,xd):c,x;px/1.13/:d1,d2; endsetsdata:a=300

9、200 400;b=200 100 450 250;c=5 2 6 73 5 4 64 5 2 3;enddatamin=d2(13); for(cd(i):sum(xd(j):x(i,j)=a(i);for(xd(j):sum(cd(i):x(i,j)=b(j);x(1,4)+x(2,4)+x(3,4)+d1(4)-d2(4)=250; x(3,1)+d1(5)-d2(5)=100;for(xd(j):sum(cd(i):x(i,j)+d1(j+5)-d2(j+5)=b(j)*0.8);sum(links(i,j):c(i,j)*x(i,j)+d1(10)-d2(10)=2950*1.1;

10、x(2,4)+d1(11)-d2(11)=0;(x(1,1)+x(2,1)+x(3,1)-(200/450)*(x(1,3)+x(2,3)+x(3,3)+d1(12)-d2(12)=0;sum(links(i,j):c(i,j)*x(i,j)+d1(13)-d2(13)=2950;y=sum(links(i,j):c(i,j)*x(i,j); d1(4)=0;d1(5)=0; d1(6)+d1(7)+d1(8)+d1(9)=0; d2(10)=115;d2(11)=0;d1(12)+d2(12)=30; d1(1)+d1(2)+d1(3)+d2(1)+d2(2)+d2(3)=0;vvvvvvv

11、vvvvvvvvvvvvvvvvvvvvvvvv注:關(guān)于算法的復雜性問題：P112? 度量方法：當問題的規(guī)模為n時，利用該算法求解此問題需要做的加減乘除四則運算的次數(shù)。第五章整數(shù)規(guī)劃1. 整數(shù)規(guī)劃的數(shù)學模型及解的特點2. 分支定界法、割平面法3.0-1整數(shù)規(guī)劃4.指派問題1.整數(shù)規(guī)劃問題的提出整數(shù)規(guī)劃數(shù)學模型的一般形式一部分或全部決策變量取整數(shù)值的規(guī)劃問題整數(shù)規(guī)劃整數(shù)規(guī)劃中不考慮整數(shù)條件是對應的規(guī)劃問題該整數(shù)規(guī)劃的松弛問題松弛問題為線性規(guī)劃的整數(shù)規(guī)劃問題整數(shù)線性規(guī)劃v 整數(shù)線性規(guī)劃一般形式：nmax(min)z = cj xjLL(a)j =1nax (=, )b(b)ijjiLLj =1xj 0LL(c)x, x,L, x中部分或全部取整數(shù)LL(d )12n整數(shù)線性規(guī)劃的幾種類型v 純整數(shù)線性規(guī)劃v 混合整數(shù)線性規(guī)劃v 0-1型整數(shù)線性規(guī)劃例：max z = 20x1+10x25x1+ 4x2 242x1+ 5x2 13s.t.x1, x2 0 x1, x2取整數(shù)不考慮整數(shù)約束時求解max=20*