北交大LINGO考試1 轉(zhuǎn)運問題等

北交大LINGO考試1轉(zhuǎn)運問題等北交大LINGO考試1轉(zhuǎn)運問題等北交大LINGO考試1轉(zhuǎn)運問題等V:1.0精細整理，僅供參考北交大LINGO考試1轉(zhuǎn)運問題等日期：20xx年X月《物流軟件》實驗報告實驗編號：學號：序號：姓名：班級：2015年10月一，用lingo解決運輸問題建立數(shù)學模型設(shè)Cij為從工廠Ai銷售到銷售點Bj運費單價，Xij為從Ai到B覺得運輸量，因此總運費為第i個產(chǎn)地的運出量應(yīng)小于等于該地的需求量第j個銷售地的運入量應(yīng)等于該地的需求量因此得到數(shù)學模型如下：Min；s.t.Xij>=0,i=1,2,…,m,j=1,2,…,n.編寫相關(guān)的模型如下MODEL:SETS:warehouse/1,2,3,4/:a;customer/1,2,3,4,5/:b;ROUTES(WAREHOUSE,CUSTOMER):c,x;ENDSETSDATA:a=20,15,25,30;b=25,20,15,20,10;c=3,11,3,10,4,1,9,2,8,3,7,4,10,5,6,5,8,2,6,2;enddata[OBJ]min=@sum(ROUTES:c*x);@FOR(WAREHOUSE(I):[SUP]@SUM(CUSTOMER(J):X(I,J))<=a(i));@for(customer(j):[dem]@sum(warehouse(i):x(i,j))=b(j));End！ROUTES為派生集合名，c,x為屬性，c表示運費，x表決策的運量！customer為銷售點集合名，1,2,3,4是元素；b為屬性，表示需求量！warehouse為庫存集合名，1,2,3,4是元素；a為屬性，表示產(chǎn)量上限在lingo運行Globaloptimalsolutionfound.Objectivevalue:300.0000Totalsolveriterations:0VariableValueReducedCostA(1)20.000000.000000A(2)15.000000.000000A(3)25.000000.000000A(4)30.000000.000000B(1)25.000000.000000B(2)20.000000.000000B(3)15.000000.000000B(4)20.000000.000000B(5)10.000000.000000C(1,1)3.0000000.000000C(1,2)11.000000.000000C(1,3)3.0000000.000000C(1,4)10.000000.000000C(1,5)4.0000000.000000C(2,1)1.0000000.000000C(2,2)9.0000000.000000C(2,3)2.0000000.000000C(2,4)8.0000000.000000C(2,5)3.0000000.000000C(3,1)7.0000000.000000C(3,2)4.0000000.000000C(3,3)10.000000.000000C(3,4)5.0000000.000000C(3,5)6.0000000.000000C(4,1)5.0000000.000000C(4,2)8.0000000.000000C(4,3)2.0000000.000000C(4,4)6.0000000.000000C(4,5)2.0000000.000000X(1,1)10.000000.000000X(1,2)0.0000005.000000X(1,3)10.000000.000000X(1,4)0.0000003.000000X(1,5)0.0000001.000000X(2,1)15.000000.000000X(2,2)0.0000005.000000X(2,3)0.0000001.000000X(2,4)0.0000003.000000X(2,5)0.0000002.000000X(3,1)0.0000006.000000X(3,2)20.000000.000000X(3,3)0.0000009.000000X(3,4)5.0000000.000000X(3,5)0.0000005.000000X(4,1)0.0000003.000000X(4,2)0.0000003.000000X(4,3)5.0000000.000000X(4,4)15.000000.000000X(4,5)10.000000.000000RowSlackorSurplusDualPriceOBJ300.0000-1.000000SUP(1)0.0000000.000000SUP(2)0.0000002.000000SUP(3)0.0000002.000000SUP(4)0.0000001.000000DEM(1)0.000000-3.000000DEM(2)0.000000-6.000000DEM(3)0.000000-3.000000DEM(4)0.000000-7.000000DEM(5)0.000000-3.000000從上面結(jié)果只知，最少運費為300二，用lingo解決轉(zhuǎn)運問題（1）建立數(shù)學模型轉(zhuǎn)運問題中有2個倉庫，4個顧客，3個中間環(huán)節(jié)。用ai表示第i個工廠的產(chǎn)量，bk表示第k個顧客的需求量，Cij1表示倉庫到物流中心的運費單價，Cjk2表示物流中心到顧客的運費單價，Xij1表示倉庫到物流中心的運量，Xij2表示物流中心到顧客的運量，則轉(zhuǎn)運問題的數(shù)學表達式為Min+s.t.，i=1,2;(運出量不大于儲存量)=,j=1,2,3;（運入量等于運出量）,k=1,2,3,4;（運入量應(yīng)等于需求量）X1>=0X2>=0（2）用lingo編程MODEL:SETS:WAREHOUSE/W1,W2/:PRODUCE;CENTER/P1,P2,P3/;CUSTOMER/D1,D2,D3,D4/:REQUIRE;LINKI(WAREHOUSE,CENTER):CI,XI;LINKII(CENTER,CUSTOMER):CII,XII;ENDSETSDATA:PRODUCE=11,9;REQUIRE=3,6,5,6;CI=13,12,15,11,13,22;CII=3,11,3,10,1,9,2,8,7,4,10,5;ENDDATA[obj]min=@sum(linki:ci*xi)+@sum(linkii:cii*xii);@for(WAREHOUSE(i):[sup]@sum(center(j):xi(i,j))<=produce(i));@for(center(j):[mid]@sum(WAREHOUSE(i):xi(i,j))=@sum(customer(k):xii(j,k)));@for(customer(k):[dem]@sum(center(j):xii(j,k))=require(k));End！LINKI和LINKII都是派生集合名，CI,CII是屬性，表對應(yīng)的運費，XI,XII也是屬性，表對應(yīng)的決策運量。!CUSTOMER是顧客需求集合名，D1，D2，D3，D4是元素，require是屬性！CENTER是物流中心集合名，P1，P2，P3是元素。！WAREHOUSE是倉庫集合名W1，W2是元素，PRODUCE是屬性運行結(jié)果如下Globaloptimalsolutionfound.Objectivevalue:347.0000Totalsolveriterations:0VariableValueReducedCostPRODUCE(W1)11.000000.000000PRODUCE(W2)9.0000000.000000REQUIRE(D1)3.0000000.000000REQUIRE(D2)6.0000000.000000REQUIRE(D3)5.0000000.000000REQUIRE(D4)6.0000000.000000CI(W1,P1)13.000000.000000CI(W1,P2)12.000000.000000CI(W1,P3)15.000000.000000CI(W2,P1)11.000000.000000CI(W2,P2)13.000000.000000CI(W2,P3)22.000000.000000XI(W1,P1)0.0000003.000000XI(W1,P2)5.0000000.000000XI(W1,P3)6.0000000.000000XI(W2,P1)9.0000000.000000XI(W2,P2)0.0000000.000000XI(W2,P3)0.0000006.000000CII(P1,D1)3.0000000.000000CII(P1,D2)11.000000.000000CII(P1,D3)3.0000000.000000CII(P1,D4)10.000000.000000CII(P2,D1)1.0000000.000000CII(P2,D2)9.0000000.000000CII(P2,D3)2.0000000.000000CII(P2,D4)8.0000000.000000CII(P3,D1)7.0000000.000000CII(P3,D2)4.0000000.000000CII(P3,D3)10.000000.000000CII(P3,D4)5.0000000.000000XII(P1,D1)3.0000000.000000XII(P1,D2)0.0000002.000000XII(P1,D3)5.0000000.000000XII(P1,D4)1.0000000.000000XII(P2,D1)0.0000000.000000XII(P2,D2)0.0000002.000000XII(P2,D3)0.0000001.000000XII(P2,D4)5.0000000.000000XII(P3,D1)0.0000009.000000XII(P3,D2)6.0000000.000000XII(P3,D3)0.00000012.00000XII(P3,D4)0.0000000.000000RowSlackorSurplusDualPriceOBJ347.0000-1.000000SUP(W1)0.0000001.000000SUP(W2)0.0000000.000000MID(P1)0.000000-11.00000MID(P2)0.000000-13.00000MID(P3)0.000000-16.00000DEM(D1)0.000000-14.00000DEM(D2)0.000000-20.00000DEM(D3)0.000000-14.00000DEM(D4)0.000000-21.00000從結(jié)果可以看出，W1向P1，P2，P3分別運送0，5，6；W2向P1，P2，P3分別運送9，0，0。P1向D1，D2，D3，D4,分別配送3，0，5，1；P2向D1，D2，D3，D4分別配送0，0，0，5；P3向D1，D2，D3，D4分別配送0，6，0，0。最少運費為347元/單位改變以上的擴建規(guī)模，經(jīng)過分析，每個倉庫最多可以擴建2單位。有9種可能性W1+1，W2+0（無解）W1+1，W2+1（無解）W1+1，W2+2（388）W1+2，W2+0（無解）W1+2，W2+1（387）W1+2，W2+2（387）W1+0，W2+0（無解）W1+0，W2+1（無解）W1+0，W2+2（無解）應(yīng)該選擇方案5，即擴建W1兩單位，擴建W2一單位。相應(yīng)程序如下MODEL:SETS:WAREHOUSE/W1,W2/:PRODUCE;CENTER/P1,P2,P3/;CUSTOMER/D1,D2,D3,D4/:REQUIRE;LINKI(WAREHOUSE,CENTER):CI,XI;LINKII(CENTER,CUSTOMER):CII,XII;ENDSETSDATA:PRODUCE=13,11;REQUIRE=6,6,5,6;CI=13,12,15,11,13,22;CII=3,11,3,10,1,9,2,8,7,4,10,5;ENDDATA[obj]min=@sum(linki:ci*xi)+@sum(linkii:cii*xii);@for(WAREHOUSE(i):[sup]@sum(center(j):xi(i,j))<=produce(i));@for(center(j):[mid]@sum(WAREHOUSE(i):xi(i,j))=@sum(customer(k):xii(j,k)));@for(customer(k):[dem]@sum(center(j):xii(j,k))=require(k));end結(jié)果運行Globaloptimalsolutionfound.Objectivevalue:387.0000Totalsolveriterations:11VariableValueReducedCostPRODUCE(W1)13.000000.000000PRODUCE(W2)11.000000.000000REQUIRE(D1)6.0000000.000000REQUIRE(D2)6.0000000.000000REQUIRE(D3)5.0000000.000000REQUIRE(D4)6.0000000.000000CI(W1,P1)13.000000.000000CI(W1,P2)12.000000.000000CI(W1,P3)15.000000.000000CI(W2,P1)11.000000.000000CI(W2,P2)13.000000.000000CI(W2,P3)22.000000.000000XI(W1,P1)0.0000003.000000XI(W1,P2)7.0000000.000000XI(W1,P3)6.0000000.000000XI(W2,P1)10.000000.000000XI(W2,P2)0.0000000.000000XI(W2,P3)0.0000006.000000CII(P1,D1)3.0000000.000000CII(P1,D2)11.000000.000000CII(P1,D3)3.0000000.000000CII(P1,D4)10.000000.000000CII(P2,D1)1.0000000.000000CII(P2,D2)9.0000000.000000CII(P2,D3)2.0000000.000000CII(P2,D4)8.0000000.000000CII(P3,D1)7.0000000.000000CII(P3,D2)4.0000000.000000CII(P3,D3)10.000000.000000CII(P3,D4)5.0000000.000000XII(P1,D1)5.0000000.000000XII(