13信算3班张超第1次作业要点

练习1铁路平板车装货问题（用Lingo求解）有七种规格的包装箱要装到两节铁路平板车上去。 包装箱的宽和高是一样的，厚度（t ,cm计） 及重量（w，kg计）不同。表1给出了包装箱的厚度、重量以及数量。每节平板车有10.2m长的地方可装包装箱，载重为40t。由于当地货运的限制，对于 C5, C6, C7类包装箱的总数有一个特别限制：箱子所占的空间（厚度）不能超过302.7cm。试把包装箱装到平板车上，使得浪费空间最小。种类C1C2C3C4C5C6C7t/cm48.753.061.372.048.752.064.0w/kg200030001000500400020001000n/件8796648决策变量：设包装箱种类为A（i）（i=1,2,.,7）,平板车为B（ j）（j=1,2）,设有x（ i，j）个A（ i）类包装箱装到平板车 B （j）上。目标函数：z=48.7*x(1,1)+48.7*x(1,2)+53.0*x(2,1)+53.0*x(2,2)+61.3*x(3,1)+61.3*x(3,2)+72.0*x(4,1)+72.0*x(4,2)+48.7*x(5,1)+48.7*x(5,2)+52.0*x(6,1)+52.0*x(6,2)+64.0*x(7,1)+64.0*x(7,2).约束条件：数量约束：各类包装箱数量有限，即x(1,1)+x(1,2)=8;x(2,1)+x(2,2)=7;x(3,1)+x(3,2)=9;x(4,1)+x(4,2)=6;x(5,1)+x(5,2)=6; x(6,1)+x(6,2)=4;x (7,1)+x(8,1)=8;空间约束：每节平板车有10.2m长的地方可装包装箱，即48.7*x(1,1)+53.0*x(2,1)+61.3*x(3,1)+72.0*x(4,1)+48.7*x(5,1)+52.0*x(6,1)+64.0*x(7,1)=1020;48.7*x(1,2)+53.0*x(2,2)+61.3*x(3,2)+72.0*x(4,2)+48.7*x(5,2)+52.0*x(6,2)+64.0*x(7,2)=1020;载重约束：每节平板车载重为 40t，即2000*x(1,1)+3000*x(2,1)+1000*x(3,1)+500*x(4,1)+4000*x(5,1)+2000*x(6,1)+1000*x (7,1)=40000;2000*x(1,2)+3000*x(2,2)+1000*x(3,2)+500*x(4,2)+4000*x(5,2)+2000*x(6,2)+1000*x (7,2)=40 000;特别约束：箱子所占的空间(厚度)不能超过 302.7cm,即48.7*x(5,1)+52.0*x(6,1)+64.0*x(7,1)+48.7*x(5,2)+52.0*x(6,2)+64.0*x(7,2)=0;综上可得Max z=48.7*x(1,1)+48.7*x(1,2)+53.0*x(2,1)+53.0*x(2,2)+61.3*x(3,1)+61.3*x(3,2)+72.0*x(4,1) +72.0*x(4,2)+48.7*x(5,1)+48.7*x(5,2)+52.0*x(6,1)+52.0*x(6,2)+64.0*x(7,1)+64.0*x(7,2).S.t. x(1,1)+x(1,2)=8;x(2,1)+x(2,2)=7;x(3,1)+x(3,2)=9;x(4,1)+x(4,2)=6;x(5,1)+x(5,2)=6; x(6,1)+x(6,2)=4;x (7,1)+x(8,1)=8;48.7*x(1,1)+53.0*x(2,1)+61.3*x(3,1)+72.0*x(4,1)+48.7*x(5,1)+52.0*x(6,1)+64.0*x(7,1)=1020;48.7*x(1,2)+53.0*x(2,2)+61.3*x(3,2)+72.0*x(4,2)+48.7*x(5,2)+52.0*x(6,2)+64.0*x(7,2)=1020; 2000*x(1,1)+3000*x(2,1)+1000*x(3,1)+500*x(4,1)+4000*x(5,1)+2000*x(6,1)+1000*x (7,1)=40 000;2000*x(1,2)+3000*x(2,2)+1000*x(3,2)+500*x(4,2)+4000*x(5,2)+2000*x(6,2)+1000*x (7,2)=40 000;48.7*x(5,1)+52.0*x(6,1)+64.0*x(7,1)+48.7*x(5,2)+52.0*x(6,2)+64.0*x(7,2)=0;模型求解：软件实现!T表示包装箱的厚度，W表示包装箱的重量，N表示包装箱的数量；model:!铁路平板车装货问题；sets:A/n1.n7/:T,W,N;!包装箱；B/1,2/;!平板车；lin ks(A,B):X;en dsets!目标函数；max=sum(A(i):sum(B(j):X(i,j)*T(i);!长度约束；for(B(j):sum(A(i):X(i,j)*T(i)=1020);!载重约束；sum(B(j):sum(A(i)|i#ge#5:X(i,j)*T(i)=302.7;!厚度约束；for(B(j):sum(A(i):X(i,j)*W(i)=40000);!件数约束；for(A(i):sum(B(j):X(i,j)=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;非负约束：x(i)=0;综上可得：min Z=x1+x2+x3+x4+x5+x6+x7+x8+x9+xio+x11+x12+x13; S.t. x1+x4+x6=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;模型求解：lingo实现Model : !模型 1;min Z=x1+x2+x3+x4+x5+x6+x7+x8+x9+xio+x11+x12+x13;x1+x4+x6=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x9gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8););gin(x10); gin(x11); gin(x12); gin(x13);end运行结果:Global optimal soluti on found.4.0000004.0000000.000000Objective value:Objective bou nd:In feasibilities:Exte nded solver steps:Total solver iterati ons:VariableValueReduced CostX10.0000001.000000X20.0000001.000000X31.0000001.000000X40.0000001.000000X50.0000001.000000X61.0000001.000000X70.0000001.000000X80.0000001.000000X90.0000001.000000X101.0000001.000000X111.0000001.000000X120.0000001.000000X130.0000001.000000Row Slack or Surplus Dual Price-1.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000000.0000001 4.0000002 0.0000003 0.0000004 0.0000005 0.0000006 0.0000007 0.0000008 0.0000009 0.000000Model : !模型 2;min Z=x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12+x13;x1+x4+x6=1；x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x9gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8););gin(x10); gin(x11); gin(x12); gin(x13);x3+x6+x10+x11=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x9gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8););gin(x10); gin(x11); gin(x12); gin(x13);x3+x6+x10+x11=3;x3+x6+x9+x13=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8); gin(x9 )；gin(x10); gin(x11); gin(x12); gin(x13);x3+x6+x10+x11=3;x3+x6+x9+x13=3;x1+x7+x11+x12=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x9gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8););gin(x10); gin(x11); gin(x12); gin(x13);x3+x6+x10+x11=3;x3+x6+x9+x13=3;x1+x7+x11+x12=3;x1+x5+x8+x13=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x9gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8););gin(x10); gin(x11); gin(x12); gin(x13);x3+x6+x10+x11=3;x3+x6+x9+x13=3;x1+x7+x11+x12=3;x1+x5+x8+x13=3;x2+x4+x8+x13=1;x6+x8+x12=1;x1+x2+x3=1;x3+x4+x5+x7=1;x7+x8+x9+x10=1;x10+x12+x13=1;x2+x5+x9+x11=1;x11+x13=1;gin(x9gin(x1); gin(x2); gin(x3); gin(x4); gin(x5); gin(x6); gin(x7); gin(x8););gin(x10); gin(x11); gin(x12); gin(x13);x3+x6+x10+x11=3;x3+x6+x9+x13=3;x1+x7+x11+x12=3;x1+x5+x8+x13=3;x2+x4+x8+x13=3;x2+x6+x7+x13=3;end运行结果：Global optimal soluti on found.Objective value:5.000000Objective bou nd:5.000000In feasibilities:0.000000Exte nded solver steps:0Total solver iterati ons:0VariableValueReduced CostX11.0000001.000000X20.0000001.000000X30.0000001.000000X40.0000001.000000X50.0000001.000000X60.0000001.000000X71.0000001.000000X81.0000001.000000X90.0000001.000000X100.0000001.000000X111.0000001.000000X120.0000001.000000X131.0000001.000000RowSlack or SurplusDual Price15.000000-1.00000020.0000000.00000030.0000000.00000040.0000000.00000050.0000000.00000061.0000000.00000070.0000000.00000080.0000000.00000091.0000000.000000102.0000000.000000112.0000000.000000120.0000000.000000130.0000000.000000141.0000000.000000151.0000000.000000Matlab 实现：编写M1文件如下：c=1 1 1 1 1 1 1 1 1 1 11 1；A=-1 0 0 -1 0 -1 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 -1 0 0 0 -1 0;-1 -1 -1 0 0 0 0 0 0 0 0 0 0;0 0 -1 -1 -1 0 -1 0 0 0 0 0 0;0 0 0 0 0 0 -1 -1 -1 -1 0 0 0;0 0 0 0 0 0 0 0 0 -1 0 -1 -1;0 -1 0 0 -1 0 0 0 -1 0 -1 0 0;0 0 0 0 0 0 0 0 0 0 -1 0 -1;b=-1;-1;-1;-1;-1;-1;-1;-1;Aeq=;beq=;vlb=0;0;0;0;0;0;0;0;0;0;0;0;0;vub=;x,fval=b in tprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =1000001000110 fval =4编写M2文件如下：c=1 1 1 1 1 1 1 1 1 1 1 1 1;A=-1 00-1 0 -1 0 00 0 0 0 0;0 0 0 0 0 -1 0 -10 0 0 -1 0;-1 -1-1 00 00 0000 0 0;0 0-1 -1 -1 0 -1 0 0 0 0 0 0;00 0 0 0 0 -1-1 -1-10 00;0000 0 0 0 00 -1 0 -1 -1;0 -1 0 0 -1 0 00 -1 0 -1 00;0000 00 0000 -1 0 -1;1 0 0 0 0 0 1 0 0 0 1 1 0;b=-1;-1;-1;-1;-1;-1;-1;-1; 3;Aeq=;beq=; vlb=0;0;0;0;0;0;0;0;0;0;0;0;0;vub=;x,fval=b in tprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =100010010000 fval =1编写M3文件如下：c=1 1 1 1 1 1 1 1 1 1 1 1 1;A=-1 00-1 0 -1 0 00 0 0 0 0;0 0 0 0 0 -1 0 -1 0 0 0 -1 0;-1 -1-1 00 00 0000 0 0;0 0-1 -1 -1 0 -1 0 0 0 0 0 0;0 0 0 0 0 0 -1 -1 -1 -10 00;0000 0 0 0 00 -1 0 -1 -1;0 -1 0 0 -1 0 0 0 -1 0 -1 0 0;00 00 00 0000 -1 0 -1;1 0 0 0 0 0 1 0 0 0 1 1 0;1 0 0 0 1 0 0 1 00 0 01；b=-1;-1;-1;-1;-1;-1;-1;-1;3;3;Aeq=;beq=;vlb=0;0;0;0;0;0;0;0;0;0;0;0;0;vub=;x,fval=b in tprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =0100011000001 fval =4编写M4文件如下：c=1 1 1 1 1 1 1 1 1 1 1 1 1;A=-1 00-1 0 -1 0 00 0 0 0 0;0 0 0 0 0 -1 0 -1 0 0 0 -1 0;-1 -1-1 00 0 0 0000 0 0;0 0-1 -1 -1 0 -1 0 0 0 0 0 0;0 0 0 0 0 0 -1 -1 -1 -10 0 0;0000 0 0 0 00 -1 0 -1 -1;0 -1 0 0 -1 0 0 0 -1 0 -1 0 0;00 00 0 0 0000 -1 0 -1;1 0 0 0 0 0 1 0 0 0 1 1 0;1 0 0 0 1 0 0 1 00 0 01;0 1 0001 1 0 0 00 0 1;b=-1;-1;-1;-1;-1;-1;-1;-1;3;3;3;Aeq=;beq=;vlb=0;0;0;0;0;0;0;0;0;0;0;0;0;vub=;x,fval=b in tprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =0101000100001fval =4编写M5文件如下：c=1 1 1 1 1 1 1 1 1 1 1 1 1;A=-1 0 0 -1 0 -1 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 -1 0 0 0 -1 0;-1 -1 -1 0 0 0 00 00 00 0;0 0-1 -1 -1 0 -1 0 0 0 00 0;0 0 0 0 00 -1-1 -1-10 0 0;0 00 00 0 0 00 -1 0 -1 -1;0 -1 0 0-1 0 0 0 -1 0-1 00;0000 0 00 00 0-1 0 -1;1 0 0 0 0 0 1 0 0 0 11 0;1 0 0 0 10 01 000 01;0 10 00 11 0 0 00 0 1;0 1 0 1 0 0 0 10 0 0 0 1;b=-1;-1;-1;-1;-1;-1;-1;-1;3;3;3;3;Aeq=;beq=;vlb=O;O;O;O;O;O;O;O;O;O;O;O;O;vub=;x,fval=bi ntprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =0010010001100fval =4编写M6文件如下：c=1 1 1 1 1 1 1 1 1 1 1 1 1;A=-1 0 0 -1 0 -1 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 -1 0 0 0 -1 0;-1 -1 -1 0 0 00 0000 00;00 -1 -1 -1 0 -1 0 0 0 00 0;0 0 0 0 0 0 -1-1 -1-10 00;0000 00 00 0 -1 0 -1 -1;0 -1 0 0-1 0 0 0 -1 0 -1 00;0 000 00 0000 -10 -1;1 0 0 0 0 0 1 0 0 0 11 0;1 0 0 0 1 0 01 0 0001;01 0001 10 00 0 0 1;0 1 0 1 0 0 0 10 0 0 0 1;0 0 1 00 1 0001 1 0 0;b=-1;-1;-1;-1;-1;-1;-1;-1;3;3;3;3;3;Aeq=;beq=; vlb=0;0;0;0;0;0;0;0;0;0;0;0;0;vub=;x,fval=b in tprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =010010010001fval =4编写M7文件如下：c=1 1 1 1 1 1 1 1 1 1 1 1 1;A=-1 0 0 -1 0 -1 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 -1 0 0 0 -1 0;-1 -1 -1 0 0 00 0000 00;0 0-1 -1 -1 0 -1 0 0 0 00 0;0 0 0 0 0 0-1-1 -1-10 00;0000 00 0 00 -1 0 -1-1;0 -1 0 0-1 0 0 0 -1 0 -100;0 000 00 0000 -10 -1;1 0 0 0 00 1 0 0 0 11 0;1 0 0 0 1 001 0 00 01;01 0001 10 0 00 0 1;0 10 1 0 0 0 10 0 0 0 1;0 0 100 1 00 01 1 0 0;0 0 1 0 0 1 0 0 1 0 0 0 1;b=-1;-1;-1;-1;-1;-1;-1;-1;3;3;3;3;3;3;Aeq=;beq=; vlb=0;0;0;0;0;0;0;0;0;0;0;0;0;vub=;x,fval=b in tprog(c,A,b,Aeq,beq,vlb,vub)运行结果：x =0011100100001 fval =5结果分析：8间展厅的可能有：(3,6,10,11)，( 3,6,9,13)由运行结果可知，管理层用最少数量的双向摄像机覆盖所有的(1,7,11,12)，( 1,5,8,13,)，( 2,6,7,13)，( 2,4,8,13), 6种可能。