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```  using LinearAlgebra
function rijesi_simplex(a,b,c)
d=[]
if(size(c)[2]==size(a)[2])
baza=zeros(1,size(b)[2])
for i in 1:(size(b)[2])
baza[i]=i+size(c)[2]
end
for i in 1:size(b)[2]
t=b[i]
for j in 1:size(a)[2]
t=[t a[i,j]]
end
if(isempty(d))
d=t;
else
d=[d;t]
end
end
d=[d;0 c]
r=Matrix(1.0I,size(b)[2],size(b)[2])
t1=zeros(1,size(b)[2])
t=[r; t1]
d=[d t]
while true
display(d)
w=d[size(d)[1],2];
er=2;
for p in 2:size(d)[2]
if(w<d[size(d)[1],p])
w=d[size(d)[1],p]
er=p;
end
end
print("er=");
println(er)
#w je najveci element
#er je kolona tog elementa
if(w<=0)
break;
end
#pivotiranje
#provjera parametara
qw=1
for p in 1:size(d)[1]
if(d[p,er]>0)
qw=0
end
end
if(qw==1)
return "rjesenje je neograniceno"
end
#djenljenje za mini
op=1
op1=d[1,1]/d[1,er]
print("op1=")
println(op1)
for p in 2:size(d)[1]-1
if(d[p,er]>0)
if(op1>d[p,1]/d[p,er])
print("pp=")
println(p)
op1=d[1,p]/d[p,er]
op=p
print("op2=")
println(op1)
end
end
end
print("P=")
println(op)
#imamo pivot elemtn na er,op
baza[op]=er-1
#izmjena reda sa pivotom
d1=copy(d)
for i in 1:size(d)[2]
d[op,i]=d[op,i]/d1[op,er]
end
for i in 1:size(d)[1]
if(i!=op&&d[i,er]!=0)
for j in 1:size(d)[2]
d[i,j]=d[i,j]-d[op,j]*d1[i,er]
end
end
end
end
#ispis z
z=zeros(1,size(c)[2])
z=[z -d[size(d)[1],1]]
for i in 1:size(c)[2]
if(findfirst(x->x==i,baza)!=nothing)
z[i]=d[Int(baza[1,findfirst(x->x==i,baza)[2]]),1]
end
end
display(baza)
return z
else
return "neispravni parametri";
end
end

a=[30 16; 14 19; 11 26; 0 1]
size(a)[2]
b=[22800 14100 15959 550]
c=[800 1000]
rijesi_simplex(a,b,c)
size(c)[2]
size(a)[2]
a=[0.5 0.3; 0.1 0.2]
b=[150 60]
c=[3 1]
rijesi_simplex(a,b,c)

```