A( 1) 480.0000 0.000000 A( 2) 520.0000 0.000000 B( 1) 400.0000 0.000000 B( 2) 600.0000 0.000000 X( 1, 1) 240.0000 0.000000 X( 1, 2) 240.0000 0.000000 X( 2, 1) 160.0000 0.000000 X( 2, 2) 360.0000 0.000000 C( 1, 1) 20.00000 0.000000 C( 1, 2) 18.00000 0.000000 C( 2, 1) 25.00000 0.000000 C( 2, 2) 20.00000 0.000000
Row Slack or Surplus Dual Price 1 0.000000 -18.00000
2 0.000000 -18.00000 3 0.000000 -2.000000 4 0.000000 0.000000 5 0.000000 0.000000 6 0.000000 -5.000000 7 0.000000 0.000000 8 0.000000 0.000000 9 0.000000 0.000000 10 0.000000 -2.000000 11 0.000000 0.000000 12 0.000000 0.000000 13 40.00000 0.000000 14 40.00000 0.000000 15 0.000000 -2.800000 16 0.000000 0.000000 17 200.0000 0.000000 18 200.0000 0.000000 19 0.000000 -18.00000 20 0.000000 0.000000 21 0.000000 0.000000
14
22 0.000000 -1.200000 23 20320.00 -1.000000
附件二：生产调运问题lingo程序及结果 Lingo程序：
model: sets:
ck/1..2/:;!定义两个仓库m; cj/1..2/:;!定义两个车间k; xm/1..4/:;!定义四个项目n; gc/1..3/:;!定义三种钢材的种类j; gj/1..2/:;!定义两种钢架构件类型i;
cj_gj(cj,gj):c,d,x;!c为车间生产成本矩阵,d为各车间生产能力限制矩阵,x为k车间生产i类构件数量; gc_ck(gc,ck):a2;!a2仓库的钢材拥有量矩阵; ck_cj(ck,cj):f1;!f1仓库运往车间单位重量运费矩阵; cj_xm(cj,xm):f2;!f2车间运往项目单位运费矩阵; gj_gc(gj,gc):b;!b消耗量矩阵;
xm_gj(xm,gj):bs,bx,z,p;!项目对构件的需求上(bs)、下限(bx),p项目对单位构件的售价矩阵; ck_cj_gc(ck,cj,gc):x1;!从m仓库运往k车间的j型钢。材量; cj_xm_gj(cj,xm,gj):x2;!由k车间生产运往项目n的i类构件数量;
endsets data: c=6000 8000 6500 7300; d=100 200 200 300; a2=15000 5000 3000 6000 5000 7000; f1=40 80 100 20; f2=100 150 80 70
15
50 60 80 90; p=9000 9500 9200 9600 9100 9500 9000 9350; b=15 10 20 20 10 15; bs=80 120 70 90 120 220 60 90; bx=50 100 30 60 100 200 50 80; enddata
max=@sum(xm(n):@sum(gj(i):@sum(cj(k):p(n,i)*x2(k,n,i))))-@sum(cj(k):@sum(gj(i):@sum(xm(n):c(k,i)*x2(k,n,i))))-@sum(cj(k):@sum(ck(m):@sum(gc(j):f1(m,k)*x1(m,k,j))))-@sum(cj(k):@sum(xm(n):@sum(gj(i):f2(k,n)*x2(k,n,i))));!最大收益=总售价-成本-运费，其中第一个是总售价，第二个是成本，第三、四个是运费; @for(gj(i):@for(cj(k):@gin(x(k,i))));!令X为整数;
@for(gc(j):@for(cj(k):@sum(gj(i):x(k,i)*b(i,j))=@sum(ck(m):x1(m,k,j))));!k车间生产2种构件所需j型钢材=由2个仓库运往k车间的j型钢材量;
@for(gc(j):@for(ck(m):@sum(cj(k):x1(m,k,j))<=a2(j,m)));!由m仓库运往k个车间的j型钢材量之和<=m仓库j型钢材拥有量;
@for(cj(k):@for(gj(i):x(k,i)<=d(k,i)));!k车间生产i类构件数量<=k车间生产i构件类型能力; @for(gj (i):@for(cj(k):@sum(xm(n):x2(k,n,i))<=x(k,i)));!由k车间运往项目n的i型构件量的和<=k车间生产的i型构件量;
@for(gj(i):@for(xm(n):@sum(cj(k):x2(k,n,i))=z(n,i)));!需求量; @for(gj(i):@for(xm(n):@bnd(bx(n,i),z(n,i),bs(n,i))));!需求上下限; end
Lingo结果：
Global optimal solution found.
Objective value: 472600.0 Extended solver steps: 0
16
Total solver iterations: 27
Variable Value Reduced Cost C( 1, 1) 6000.000 0.000000 C( 1, 2) 8000.000 0.000000 C( 2, 1) 6500.000 0.000000 C( 2, 2) 7300.000 0.000000 D( 1, 1) 100.0000 0.000000 D( 1, 2) 200.0000 0.000000 D( 2, 1) 200.0000 0.000000 D( 2, 2) 300.0000 0.000000 X( 1, 1) 100.0000 -620.0000 X( 1, 2) 200.0000 860.0000 X( 2, 1) 170.0000 480.0000 X( 2, 2) 240.0000 760.0000 A2( 1, 1) 15000.00 0.000000 A2( 1, 2) 5000.000 0.000000 A2( 2, 1) 3000.000 0.000000 A2( 2, 2) 6000.000 0.000000 A2( 3, 1) 5000.000 0.000000 A2( 3, 2) 7000.000 0.000000 F1( 1, 1) 40.00000 0.000000 F1( 1, 2) 80.00000 0.000000 F1( 2, 1) 100.0000 0.000000 F1( 2, 2) 20.00000 0.000000 F2( 1, 1) 100.0000 0.000000 F2( 1, 2) 150.0000 0.000000 F2( 1, 3) 80.00000 0.000000 F2( 1, 4) 70.00000 0.000000 F2( 2, 1) 50.00000 0.000000 F2( 2, 2) 60.00000 0.000000 F2( 2, 3) 80.00000 0.000000 F2( 2, 4) 90.00000 0.000000 B( 1, 1) 15.00000 0.000000
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B( 1, 2) 10.00000 0.000000 B( 1, 3) 20.00000 0.000000 B( 2, 1) 20.00000 0.000000 B( 2, 2) 10.00000 0.000000 B( 2, 3) 15.00000 0.000000 BS( 1, 1) 80.00000 0.000000 BS( 1, 2) 120.0000 0.000000 BS( 2, 1) 70.00000 0.000000 BS( 2, 2) 90.00000 0.000000 BS( 3, 1) 120.0000 0.000000 BS( 3, 2) 220.0000 0.000000 BS( 4, 1) 60.00000 0.000000 BS( 4, 2) 90.00000 0.000000 BX( 1, 1) 50.00000 0.000000 BX( 1, 2) 100.0000 0.000000 BX( 2, 1) 30.00000 0.000000 BX( 2, 2) 60.00000 0.000000 BX( 3, 1) 100.0000 0.000000 BX( 3, 2) 200.0000 0.000000 BX( 4, 1) 50.00000 0.000000 BX( 4, 2) 80.00000 0.000000 Z( 1, 1) 50.00000 70.00000 Z( 1, 2) 100.0000 90.00000 Z( 2, 1) 70.00000 -120.0000 Z( 2, 2) 60.00000 0.000000 Z( 3, 1) 100.0000 0.000000 Z( 3, 2) 200.0000 120.0000 Z( 4, 1) 50.00000 90.00000 Z( 4, 2) 80.00000 260.0000 P( 1, 1) 9000.000 0.000000 P( 1, 2) 9500.000 0.000000 P( 2, 1) 9200.000 0.000000 P( 2, 2) 9600.000 0.000000 P( 3, 1) 9100.000 0.000000 P( 3, 2) 9500.000 0.000000
18