數(shù)學(xué)建模第五次優(yōu)秀論文研讀報告

1、Matlab創(chuàng)新獎第一問（1）A區(qū)交巡警平臺管轄范圍Matlab程序方法一：clcclearA=load(全區(qū)線路節(jié)點連接.txt);B=load(全區(qū)數(shù)據(jù).txt);C=load(巡警平臺.txt);qidian=A(:,1);zhongdian=A(:,2);jiedian=B(:,1);jiedianx=B(:,2);jiediany=B(:,3);z=inf*ones(length(jiedian),length(jiedian);for i=1:length(jiedian) z(i,i)=0;endfor i=1:length(qidian) z(qidian(i),zhongdi

2、an(i)=sqrt(jiedianx(qidian(i)-jiedianx(zhongdian(i).2+(jiediany(qidian(i)-jiediany(zhongdian(i).2); z(zhongdian(i),qidian(i)=z(qidian(i),zhongdian(i);endD,path=floyd(z);guanxia=;for i=1:length(jiedian) t=inf; for j=1:length(C) if tD(i,C(j) t=D(i,C(j); m=C(j); end end guanxia=guanxia;m,i;endAguanxia=

3、;for j=1:20 for i=1:length(jiedian) if guanxia(i,1)=j Aguanxia=Aguanxia;guanxia(i,:); end endendAguanxia方法二：clearA=xlsread(A10.xls,全市交通路口節(jié)點數(shù)據(jù),A2:E93);B=xlsread(A10.xls,全市交通路口的路線,A2:B144);d=ones(92,92)*inf;x=A(:,2);y=A(:,3);for i=1:92 for k=1:143 if B(k,1)=i&B(k,2)93 j=B(k,2); d(i,j)=sqrt(x(i)-x(j).2

4、+(y(i)-y(j).2); end endendfor i=1:92 for j=1:92 if i=j d(i,j)=0; end if d(i,j)=x*d);for(yaodao(i):sum(pingtai(j):x(i,j)=1);for(pingtai(j):sum(yaodao(i):x(i,j)=1);for(fangan(i,j):bin(x(i,j);EndGlobal optimal solution found. Objective value: 80.15457 Objective bound: 80.15457 Infeasibilities: 0.000000

5、 Extended solver steps: 0 Total solver iterations: 1308 求出最遠(yuǎn)距離為平臺7至平臺9，距離為80.15，需要的時間為8.0154分。再以距離為約束，以總調(diào)度時間最短為目標(biāo)。model:sets:yaodao/1.13/;pingtai/1.20/;fangan(yaodao,pingtai):x,d;endsetsdata:d=file(w.txt);text()=table(x);text()=table(d);enddatamin=sum(fangan:x*d);for(fangan:x*d=80.15457);for(yaodao(

6、i):sum(pingtai(j):x(i,j)=1);for(pingtai(j):sum(yaodao(i):x(i,j)=1);for(fangan(i,j):bin(x(i,j);end Global optimal solution found. Objective value: 461.8849 Objective bound: 461.8849 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 0故總調(diào)度時間是461.88。（3）增加平臺后管轄范圍的分配方案： 先在不增加平臺的根

7、據(jù)工作量均衡情況下重新分配管理范圍，這里引進(jìn)了變異系數(shù)（與標(biāo)準(zhǔn)差和方差的功能一下，但適用范圍更廣，在量綱和平均值不同的情況下變異系數(shù)能代替標(biāo)準(zhǔn)差求數(shù)據(jù)的波動情況）； 作者用交巡警平臺的工作量的變異系數(shù)來表示均衡性，其值越小均衡性越好。平臺工作量的平均值為目標(biāo)函數(shù)1：變異系數(shù)最小目標(biāo)函數(shù)2：最長出警時間最短；約束條件：平臺的約束：使每個平臺不至于閑置，至少管理自己所在的路口。每個路口都被平臺管轄的約束。出警時間少于3min。第五篇模型一：clearclcT=load(1.txt);S=load(a.txt);M=10000;for a=1:92 for b=1:92 if a=b A(a,b)=

8、0; else A(a,b)=inf; end for i=1:140 if (a=S(i,1)&b=S(i,2)|(b=S(i,1)&a=S(i,2) A(a,b)=(T(a,1)-T(b,1)2+(T(a,2)-T(b,2)2)0.5; end end endendAn=92;D=A; %賦初值for(i=1:n)for(j=1:n)R(i,j)=j;end;end %賦路徑初值for(k=1:n)for(i=1:n)for(j=1:n)if(D(i,k)+D(k,j)D(i,j)D(i,j)=D(i,k)+D(k,j); %更新dijR(i,j)=k;end;end;end %更新rij

9、k %顯示迭代步數(shù)D %顯示每步迭代后的路長R %顯示每步迭代后的路徑pd=0;for i=1:n %含有負(fù)權(quán)時if(D(i,i)0)pd=1;break;end;end %存在一條含有頂點vi 的負(fù)回路if(pd)break;end %存在一條負(fù)回路, 終止程序end %程序結(jié)束v=;for i=1:20 for j=21:92 if D(i,j)30 E(i,j)=0; else E(i,j)=D(i,j); end endendEu=v(3,:);u(find(u=0)=;程序運行結(jié)果為：服務(wù)臺在3min中內(nèi)可以到達(dá)的路口11 42 43 44 64 65 66 67 68 69 70

10、71 72 73 74 75 76 77 78 79 8022 40 42 43 44 66 67 68 69 70 71 72 73 74 75 76 7833 43 44 54 55 64 65 66 67 68 70 7644 57 58 60 62 63 64 65 6655 47 48 49 50 51 52 53 56 58 5966 47 48 50 51 52 56 58 5977 30 31 32 33 34 47 4888 31 32 33 34 35 36 37 45 46 4799 31 32 33 34 35 36 37 45 4610101111 25 26 271

11、212 251313 21 22 23 2414141515 311616 33 34 35 36 37 45 461717 40 41 42 43 70 721818 71 72 73 74 77 78 79 80 81 82 83 84 85 87 88 89 90 911919 64 65 66 67 68 69 70 71 73 74 75 76 77 78 79 80 81 82 832020 81 82 83 84 85 86 87 88 89 90 91服務(wù)臺管轄路口11 67 68 69 71 73 74 75 76 7822 40 43 44 70 72 3933 54 55

12、 65 66 44 57 60 62 63 64 55 49 50 51 52 53 56 58 596677 30 32 47 48 6188 33 46 99 31 34 35 4510101111 26 271212 251313 21 22 23 2414141515 28 291616 36 37 381717 41 42 1818 80 81 82 831919 77 79 2020 84 85 86 87 88 89 90 91 92模型二：利用Lingo求解如下：model:sets:yaodao/1.13/;pingtai/1.20/;fangan(yaodao,pingta

13、i):x,d;endsetsdata:d=file(data.txt);enddatamin=max(fangan:x*d);for(yaodao(i):sum(pingtai(j):x(i,j)=1);for(pingtai(j):sum(yaodao(i):x(i,j)=2; sum(pt(i)|i#ge#21:a(i)=5; !增加2-5個平臺;for(pt(i)|i#le#20:a(i)=1);!原本的20個平臺;for(link(i,j):x(i,j)=a(i);!每個非平臺的站點只能被一個平臺管轄;for(jd(j):sum(pt(i):x(i,j)=1);for(pt(i):x(

14、i,i)=a(i);!每個平臺自己管轄自己;for(link(i,j):x(i,j)*d(i,j)=57.0053);!管轄范圍要在57.0053之內(nèi);for(link:bin(x); for(pt(i):bin(a); !2個0-1變量;end 服務(wù)臺重新分配的管轄路口2929 283232 30 31 35 556666 757272 65 69 70 71 73 78 79 809292 85 86 87 88 90 9111 42 43 44 642233 44 565566 57 58 60 6277 52 59 61 88 49 51 5399 36 37 50 541010111

15、11212 271313 23 24 25 261414 21 221515 33 341616 45 46 47 481717 38 39 40 411818 63 671919 68 74 76 77 81 892020 82 83 84模型四：model: sets: pt/1.154/:a,f; jd/1.154/:c; link(pt,jd):x,d;end setsdata: c=file(anfalvc.txt);d=file(dc.txt); enddatamin=max(pt(i):(sum(jd(j):x(i,j)*c(j);!min=max(pt(i):(sum(jd(j

16、):x(i,j)*d(i,j);!min=(max(pt(i):(sum(jd(j):x(i,j)*a(j)*c(j)-min(pt(i):(sum(jd(j):x(i,j)*a(j)*c(j)/max(pt(i):(sum(jd(j):x(i,j)*a(j)*c(j); !工作量均衡目標(biāo);!max(pt(i):(sum(jd(j):x(i,j)*c(j)=2; sum(pt(i)|i#ge#18:a(i)=5; !sum(pt(i)|i#ge#18:a(i)=0; !增加平臺;for(pt(i)|i#le#17:a(i)=1);!原本的17個平臺;for(link(i,j):x(i,j)=a

17、(i);!每個非平臺的站點只能被一個平臺管轄;for(jd(j):sum(pt(i):x(i,j)=1);for(pt(i):x(i,i)=a(i);!每個平臺自己管轄自己;for(link(i,j):x(i,j)*d(i,j)=78.4);!管轄范圍要在78.4之內(nèi);for(link:bin(x); for(pt(i):bin(a); !2個0-1變量;for(pt(i):f(i)=sum(jd(j):x(i,j)*c(j);!每個平臺的總案發(fā)量；end城區(qū)第一個平臺第二平臺第三個平臺第四個平臺第五個平臺189270271284296337341342364367402410411437456503506540546566模型五：model:sets:yaodao/1.17/;pingtai/1.80/;fangan(yaodao,pingtai):x,d;endsetsdata:d=file(data1.txt);enddatamin=max(fangan:x*d);for(yaodao(i):sum(pingtai(j):x(i,j)