運籌學(xué)實例分析及l(fā)ingo求解

1、運籌學(xué)實例分析及l(fā)ingo求解一、線性規(guī)劃某公司有6個倉庫，庫存貨物總數(shù)分別為60、55、51、43、41、52，現(xiàn)有8個客戶各要一批貨，數(shù)量分別為35，37，22，32，41，32，43，38。各供貨倉庫到8個客戶處的單位貨物運輸價見表V1V2V3V4V5V6V7V8W162674258W249538582W352197433W476739271W523957265W655228143試確定各倉庫到各客戶處的貨物調(diào)運數(shù)量，使總的運輸費用最小。解：設(shè)表示從第個倉庫到第個客戶的貨物運量。表示從第個倉庫到第個客戶的單位貨物運價，表示第個倉庫的最大供貨量，表示第個客戶的訂貨量。目標函數(shù)是使總運輸費用

2、最少，約束條件有三個：1、各倉庫運出的貨物總量不超過其庫存數(shù)2、各客戶收到的貨物總量等于其訂貨數(shù)量3、非負約束數(shù)學(xué)模型為：編程如下：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.由以上結(jié)果可以清楚的看到由各倉庫到各客戶處的貨物調(diào)運數(shù)量，由此得出的符合條件的最佳運貨方案，而使運

11、費最低，最低為664。二、運輸規(guī)劃重慶有三家電子廠分別是新普,隆宇和恒華,生產(chǎn)的筆記本電腦將要運向北京,天津,廣東,上海四個城市銷售,其產(chǎn)量和銷售量見下表：（單位：萬臺）表：1-1北京天津廣東上海產(chǎn)量新普626730隆宇495325恒華881521銷量15172212-問：哪種銷售方案將會取得最少的運輸費用,費用為多少？針對該運輸問題,為了方便計算,可以設(shè)新普（A1）,隆宇（A2）和恒華（A3）分別銷往北京（B1）、天津（B2）、廣東（B3）和上海（B4）四個城市銷售量為.建立以下模型：表：1-2B1B2B3B4產(chǎn)量A1626730A2495325A3881521銷量15172212-目標（T

12、he objective）最少費用：約束條件：供應(yīng)限制（The supply constrains） 指標約束（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運行結(jié)果為：Feasible solution found. Extended solver steps: 12 Total solver iterations: 346 Model Title: 鋼管下料；Variable Value Row Slack or Sur