运筹学实例分析及lingo求解

1、运筹学实例分析及lingo求解一、线性规划某公司有6个仓库，库存货物总数分别为60、55、51、43、41、52，现有8个客户各要一批货，数量分别为35，37，22，32，41，32，43，38。各供货仓库到8个客户处的单位货物运输价见表V1V2V3V4V5V6V7V8W162674258W249538582W352197433W476739271W523957265W655228143试确定各仓库到各客户处的货物调运数量，使总的运输费用最小。解：设表示从第个仓库到第个客户的货物运量。表示从第个仓库到第个客户的单位货物运价，表示第个仓库的最大供货量，表示第个客户的订货量。目标函数是使总运输费用

2、最少，约束条件有三个：1、各仓库运出的货物总量不超过其库存数2、各客户收到的货物总量等于其订货数量3、非负约束数学模型为：编程如下：model:Sets:Wh/w1.w6/:ai;Vd/v1.v8/:dj;links(wh,vd):c,x;endsetsData: ai=60,55,51,43,41,52;dj=35,37,22,32,41,32,43,38;c=6,2,6,7,4,2,5,9 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;EnddataMin=sum(links(

3、i,j):c(i,j)*x(i,j);for(wh(i):sum(vd(j):x(i,j)=ai(i);for(vd(j):sum(wh(i):x(i,j)=dj(j);end Global optimal solution found. Objective value: 664.0000 Total solver iterations: 0 Variable Value Reduced Cost AI( W1) 60.00000 0. AI( W2) 55.00000 0. AI( W3) 51.00000 0. AI( W4) 43.00000 0. AI( W5) 41.00000 0.

4、 AI( W6) 52.00000 0. DJ( V1) 35.00000 0. DJ( V2) 37.00000 0. DJ( V3) 22.00000 0. DJ( V4) 32.00000 0. DJ( V5) 41.00000 0. DJ( V6) 32.00000 0. DJ( V7) 43.00000 0. DJ( V8) 38.00000 0. C( W1, V1) 6. 0. C( W1, V2) 2. 0. C( W1, V3) 6. 0. C( W1, V4) 7. 0. C( W1, V5) 4. 0. C( W1, V6) 2. 0. C( W1, V7) 5. 0.

5、C( W1, V8) 9. 0. C( W2, V1) 4. 0. C( W2, V2) 9. 0. C( W2, V3) 5. 0. C( W2, V4) 3. 0. C( W2, V5) 8. 0. C( W2, V6) 5. 0. C( W2, V7) 8. 0. C( W2, V8) 2. 0. C( W3, V1) 5. 0. C( W3, V2) 2. 0. C( W3, V3) 1. 0. C( W3, V4) 9. 0. C( W3, V5) 7. 0. C( W3, V6) 4. 0. C( W3, V7) 3. 0. C( W3, V8) 3. 0. C( W4, V1)

6、7. 0. C( W4, V2) 6. 0. C( W4, V3) 7. 0. C( W4, V4) 3. 0. C( W4, V5) 9. 0. C( W4, V6) 2. 0. C( W4, V7) 7. 0. C( W4, V8) 1. 0. C( W5, V1) 2. 0. C( W5, V2) 3. 0. C( W5, V3) 9. 0. C( W5, V4) 5. 0. C( W5, V5) 7. 0. C( W5, V6) 2. 0. C( W5, V7) 6. 0. C( W5, V8) 5. 0. C( W6, V1) 5. 0. C( W6, V2) 5. 0. C( W6

7、, V3) 2. 0. C( W6, V4) 2. 0. C( W6, V5) 8. 0. C( W6, V6) 1. 0. C( W6, V7) 4. 0. C( W6, V8) 3. 0. X( W1, V1) 0. 5. X( W1, V2) 19.00000 0. X( W1, V3) 0. 5. X( W1, V4) 0. 7. X( W1, V5) 41.00000 0. X( W1, V6) 0. 2. X( W1, V7) 0. 2. X( W1, V8) 0. 10.00000 X( W2, V1) 1. 0. X( W2, V2) 0. 4. X( W2, V3) 0. 1

8、. X( W2, V4) 32.00000 0. X( W2, V5) 0. 1. X( W2, V6) 0. 2. X( W2, V7) 0. 2. X( W2, V8) 0. 0. X( W3, V1) 0. 4. X( W3, V2) 11.00000 0. X( W3, V3) 0. 0. X( W3, V4) 0. 9. X( W3, V5) 0. 3. X( W3, V6) 0. 4. X( W3, V7) 40.00000 0. X( W3, V8) 0. 4. X( W4, V1) 0. 4. X( W4, V2) 0. 2. X( W4, V3) 0. 4. X( W4, V

9、4) 0. 1. X( W4, V5) 0. 3. X( W4, V6) 5. 0. X( W4, V7) 0. 2. X( W4, V8) 38.00000 0. X( W5, V1) 34.00000 0. X( W5, V2) 7. 0. X( W5, V3) 0. 7. X( W5, V4) 0. 4. X( W5, V5) 0. 2. X( W5, V6) 0. 1. X( W5, V7) 0. 2. X( W5, V8) 0. 5. X( W6, V1) 0. 3. X( W6, V2) 0. 2. X( W6, V3) 22.00000 0. X( W6, V4) 0. 1. X

10、( W6, V5) 0. 3. X( W6, V6) 27.00000 0. X( W6, V7) 3. 0. X( W6, V8) 0. 3. Row Slack or Surplus Dual Price 1 664.0000 -1. 2 0. 3. 3 22.00000 0. 4 0. 3. 5 0. 1. 6 0. 2. 7 0. 2. 8 0. -4. 9 0. -5. 10 0. -4. 11 0. -3. 12 0. -7. 13 0. -3. 14 0. -6. 15 0. -2.由以上结果可以清楚的看到由各仓库到各客户处的货物调运数量，由此得出的符合条件的最佳运货方案，而使运

11、费最低，最低为664。二、运输规划重庆有三家电子厂分别是新普,隆宇和恒华,生产的笔记本电脑将要运向北京,天津,广东,上海四个城市销售,其产量和销售量见下表：（单位：万台）表：1-1北京天津广东上海产量新普626730隆宇495325恒华881521销量15172212-问：哪种销售方案将会取得最少的运输费用,费用为多少？针对该运输问题,为了方便计算,可以设新普（A1）,隆宇（A2）和恒华（A3）分别销往北京（B1）、天津（B2）、广东（B3）和上海（B4）四个城市销售量为.建立以下模型：表：1-2B1B2B3B4产量A1626730A2495325A3881521销量15172212-目标（T

12、he objective）最少费用：约束条件：供应限制（The supply constrains） 指标约束（The damand constrains） LINGO模型：model:sets:origin/1.3/:a;sale/1.4/:b;routes(origin,sale):c,x;endsetsdata:a=30,25,21;b=15,17,22,12;c=6,2,6,7,4,9,5,3,8,8,1,5;enddataOBJmin=sum(routes:c*x);for(origin(i):SUPsum(sale(j):x(i,j)= 0); for(station(k): su

13、m(txs(i,k):t(i)*x(i,k)=100;x1*r21+x2*r22+x3*r23=100;x1*r31+x2*r32+x3*r33=100;x1*r41+x2*r42+x3*r43=70;x1+x2+x3=105;x1+x2+x3=5.9;2.9*r12+2.1*r22+1.5*r32+1.8*r42=5.9;2.9*r13+2.1*r23+1.5*r33+1.8*r43=5.9;2.9*r11+2.1*r21+1.5*r31+1.8*r41=7.4;2.9*r12+2.1*r22+1.5*r32+1.8*r42=7.4;2.9*r13+2.1*r23+1.5*r33+1.8*r43=x2;x2=x3;gin(x1);gin(x2);gin(x3);gin(r11);gin(r21);gin(r31);gin(r41);gin(r12);gin(r22);gin(r32);gin(r42);gin(r13);gin(r23);gin(r33);gin(r43);End运行结果为：Feasible solution found. Extended solver steps: 12 Total solver iterations: 346 Model Title: 钢管下料；Variable Value Row Slack or Sur