MATLAB Optimization Problem: Maximizing Objective Function with Constraints

This MATLAB code defines an optimization problem with an objective function and constraints to be solved using the 'optimproblem' function. The code demonstrates how to set up constraints based on the variables 'y' and the matrices 'a' and 'c'.

a=[5.26	5.19	4.21	4.00	2.95	2.74	2.46	1.90	0.64	1.27;
1.90	0.99	1.90	1.13	1.27	2.25	1.48	2.04	3.09	3.51;
5.89	5.61	5.61	4.56	3.51	3.65	2.46	2.46	1.06	0.57;
0.64	1.76	1.27	1.83	2.74	2.60	4.21	3.72	5.05	6.10;
4.42	3.86	3.72	3.16	2.25	2.81	0.78	1.62	1.27	0.50];
c=[0.95	1.05	1.00	1.05	1.10	1.25	1.05	1.30	1.35	1.25;
1.25	1.10	1.35	1.05	1.15	1.35	1.05	1.15	1.35	1.25;
0.3	0.28	0.29	0.32	0.31	0.33	0.32	0.31	0.33	0.31];
prob=optimproblem;
y=optimvar('y',50,'Type','integer','LowerBound',0);
sum=0;
for i=1:5
    for j=1:10
        b=(i-1)*10+j;
        sum=sum+y(b)*y(b)*a(i,j);
    end
end
prob.Objective = 154*sum;
con=optimconstr(50);
cum=0;
for i=1:10
    cum=cum+y(i);
end
con(1)=154*cum>=12000;

tum=0;
for i=10:20
    tum=tum+y(i);
end
con(2)=154*tum>=13000;

bum=0;
for i=20:30
    bum=bum+y(i);
end
con(3)=154*bum>=13000;

dum=0;
for i=30:40
    dum=dum+y(i);
end
con(4)=154*dum>=19000;

lum=0;
for i=40:50
    lum=lum+y(i);
end
con(5)=154*lum>=13000;
con(6)=[
(y(1)+y(11)+y(41))*154<=9500;
(y(2)+y(12)+y(42))*154<=10500;
(y(3)+y(13)+y(43))*154<=10000;
(y(4)+y(14)+y(44))*154<=10500;
(y(5)+y(15)+y(45))*154<=11000;
(y(6)+y(16)+y(46))*154<=12500;
(y(7)+y(17)+y(47))*154<=10500;
(y(8)+y(18)+y(48))*154<=13000;
(y(9)+y(19)+y(49))*154<=13500;
(y(10)+y(20)+y(50))*154<=12500;
(y(31)+y(21))*154<=12500;
(y(32)+y(22))*154<=11000;
(y(33)+y(23))*154<=13500;
(y(34)+y(24))*154<=10500;
(y(35)+y(25))*154<=11500;
(y(36)+y(26))*154<=13500;
(y(37)+y(27))*154<=10500;
(y(38)+y(28))*154<=11500;
(y(39)+y(29))*154<=13500;
(y(40)+y(30))*154<=12500;
];

fee=0;
tee=0;
for i=1:10
    fee=fee+y(i)*c(3,i);
end
for i=1:10
    tee=tee+y(i);
end
ft=fee/tee;
con(26)= ft<=0.305;
con(27)= ft>=0.285;

hee=0;
pee=0;
for i=11:20
    hee=hee+y(i)*c(3,i);
end
for i=11:20
    pee=pee+y(i);
end
gt=hee/pee;
con(28)= gt<=0.305;
con(29)= gt>=0.285;

qee=0;
aee=0;
for i=41:50
    qee=qee+y(i)*c(3,i);
end
for i=41:50
    aee=aee+y(i);
end
mt=qee/aee;
con(30)= mt<=0.305;
con(31)= mt>=0.285;

cube=0;
for i=1:5
    for j=1:10
        l=(i-1)*10+j;
        cube=cube+y(l)*a(i,j);
    end
end

leyan=0;
for i=1:50
    leyan=leyan+y(i);
end

con(32)=cube/14+leyan*8/60<=8;
problem.Constraints = [con(1), con(2),con(3), con(4),con(5), con(6),con(7), con(8),con(9), con(10),con(11), con(12),con(13), con(14),con(15), con(16),con(17), con(18),con(19), con(20),con(21), con(22),con(23), con(24),con(25), con(26),con(27), con(28),con(29), con(30),con(31), con(32)];
[sol,fval,flag]=solve(prob),sol.y

Note: In the original code, there was a typo in the last line where problem should be prob.

Explanation:

1. Objective Function: The code defines an objective function that aims to maximize the value of 'sum'. The variable 'sum' is calculated by multiplying each element of matrix 'a' with the corresponding value of 'y' squared and summing up the results.
2. Constraints: The code defines 32 constraints, which limit the possible values of 'y' to ensure the solution meets specific requirements. These constraints are expressed using logical operators and inequalities.
3. Solving the Problem: The solve function is used to solve the optimization problem defined in the code. The output includes the solution vector 'sol.y', the objective function value 'fval', and a flag indicating the success of the optimization process.

The code addresses a real-world optimization problem where you want to find the best values for a vector 'y' (representing a decision variable) that maximizes a given objective function while satisfying a set of constraints. This is a typical example of an integer programming problem that MATLAB can effectively solve.

Please note that this code assumes you have the Optimization Toolbox installed in MATLAB. The optimproblem and solve functions are part of this toolbox.