写lingo程序，急

model:

sets:

firehouse/1..3/:num

fire/1..7/:

link(firehouse,fire):distance,x

endsets

data:

num=3,2,2

enddata

min=@sum(link:distance*x)

@for(firehouse(i):@sum(fire(j):x(i,j))=num(i))

@for(fire(j):@sum(firehouse(i):x(i,j))=1)

@for(link:@gin(x))

end

结果

X( 1, 1)0.0000000.000000

X( 1, 2)1.0000000.000000

X( 1, 3)0.0000000.000000

X( 1, 4)0.0000000.000000

X( 1, 5)0.0000000.000000

X( 1, 6)1.0000000.000000

X( 1, 7)1.0000000.000000

X( 2, 1)0.0000000.000000

X( 2, 2)0.0000000.000000

X( 2, 3)1.0000000.000000

X( 2, 4)1.0000000.000000

X( 2, 5)0.0000000.000000

X( 2, 6)0.0000000.000000

X( 2, 7)0.0000000.000000

X( 3, 1)1.0000000.000000

X( 3, 2)0.0000000.000000

X( 3, 3)0.0000000.000000

X( 3, 4)0.0000000.000000

X( 3, 5)1.0000000.000000

X( 3, 6)0.0000000.000000

X( 3, 7)0.0000000.000000

就是说1号消防站派车到2 6 7

2号到3 4

3号到1 5

sets:

point/1..12/:x,y,r,c

link(point,point):d

endsets

calc:

@for(link(i,j):d(i,j)=@sqrt((x(i)-x(j))^2+(y(i)-y(j))^2))

endcalc

min=@sum(link(i,j):d(i,j)*c(i)*r(i))

@sum(point:c)=2

@for(point:@bin(c))

function x=linliu(c,A,b)

[n1,n2]=size(A)

A=[A,eye(n1)]c=[-c,zeros(1,n1)]

x1=[zeros(1,n2),b']lk=[n2+1:n1+n2]

fp=fopen('f:\liu.m','wt')

fprintf(fp,'初始单纯形表:\n')

while(1)

x=x1(1:n2)

s1=[lk',b,A]

c

x1

fprintf(fp,' 基b')

for j=1:n2+n1

fprintf(fp,' x%d ',j)

end

fprintf(fp,'\n')

for i=1:n1

fprintf(fp,'%4.1f %4.1f',lk(i),b(i))

for j=1:n2+n1

fprintf(fp,'%4.1f ',A(i,j))

end

fprintf(fp,'\n')

end

fprintf(fp,' ')

for i=1:n1+n2

fprintf(fp,'%4.1f ',c(i))

end

fprintf(fp,'\n解为：\n')

for i=1:n2+n1

fprintf(fp,'x(%d)=%3.2f,',i,x1(i))

end

cc=[]ci=[]

for i=1:n1+n2

if c(i)<0

cc=[cc,c(i)]

ci=[ci,i]

end

end

nc=length(cc)

if nc==0

fprintf('达到最优解')

fprintf(fp,'\n达到最优解')

break

end

cliu=cc(1)

cl=ci(1)

for j=1:nc

if abs(cc(j))>abs(cliu)

cliu=cc(j)

cl=j

end

end

cc1=[]ci1=[]

for i=1:n1

if A(i,cl)>0

cc1=[cc1,A(i,cl)]

ci1=[ci1,i]

end

end

nc1=length(cc1)

if nc1==0

fprintf('无有界最优解')

fprintf(fp,'无有界最优解')

break

end

cliu=b(ci1(1))/cc1(1)

cl1=ci1(1)

for j=1:nc1

if b(ci1(j))/cc1(j)<cliu

cliu=b(ci1(j))/cc1(j)

cl1=ci1(j)

end

end

fprintf(fp,'\nx(%d)入基，x(%d)出基\n',cl,lk(cl1))

fprintf('\nx(%d)入基，x(%d)出基\n',cl,lk(cl1))

b(cl1)=b(cl1)/A(cl1,cl)

A(cl1,:)=A(cl1,:)/A(cl1,cl)

for k=1:n1

if k~=cl1

b(k)=b(k)-b(cl1)*A(k,cl)

A(k,:)=A(k,:)-A(cl1,:).*A(k,cl)

end

end

c=c-c(cl).*A(cl1,:)

x1(lk(cl1))=0

lk(cl1)=cl

for kk=1:n1

x1(lk(kk))=b(kk)

end

x=x1(1:n2)

end

fclose(fp)

检验：p31页 运筹学 清华版

format rat

c=[2 3]A=[1 24 00 4]b=[81612]

x=linliu(c,A,b)

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