%Solution to Problem 3.4 in Chapter 3 (Linear Programming).
%  In this alternative, we do not use the contract amounts specifically
%  in the program.

%Build the constraint matrix:
A=zeros(20,12);
b=zeros(20,1);
Aeq=zeros(3,12);  %Contract amounts
beq=[10000;8000;8000];  %Contract Amounts

Cattle=1:4; Sheep=5:8; Chicken=9:12;
Aeq(1,Cattle)=[1 1 1 1];
Aeq(2,Sheep)=[1 1 1 1];
Aeq(3,Chicken)=[1  1 1 1];

%Shortages:
I=eye(4);
A(1:4,Cattle)=I; A(1:4,Sheep)=I; A(1:4,Chicken)=I;
b(1:4)=[6000;10000;5000;5000];
clear I

%Vitamins:
A(5,Cattle)=[-2 0 -4 2];
A(6,Sheep)=[-2 0 -4 2];
A(7,Chicken)=[-4 -2 -6 0];
A(8,Chicken)=[2 0 4 -2];

%Protein:
Temp=[-4 1 -6 -2];
A(9,Cattle)=Temp;
A(10,Sheep)=Temp;
A(11,Chicken)=Temp;

%Calcium
A(12,Cattle)=[1 -3 1 1];
A(13,Sheep)=[0 -4 0 0];
A(14,Chicken)=[0 -4 0 0];

%Fat:
Temp=[-4 -2 -2 -5];
A(15,Cattle)=Temp;
A(16,Sheep)=Temp;
A(17,Chicken)=Temp;
clear Temp

%Fat Max:
A(18,Cattle)=[0 -2 -2 1];
A(19,Sheep)=[2 0 0 3];
A(20,Chicken)=[2 0 0 3];

%Delete the max fat:
A(19:20,:)=[];
b(19:20)=[];


c=repmat([.2 .12 .24 .12],1,3);
lb=zeros(12,1);

[f fopt]=linprog(c,A,b,Aeq,beq,lb)

%Verify the numbers:
TotalCorn=sum(f([1,5,9]))/1000
TotalLime=sum(f([2,6,10]))/1000
TotalSoy=sum(f([3,7,11]))/1000
TotalFish=sum(f([4,8,12]))/1000

TotalCattle=sum(f(Cattle))/1000
TotalSheep=sum(f(Sheep))/1000
TotalChicken=sum(f(Chicken))/1000