optimization of resources for organization with limited resources

Solve this in matlab for i=5; V=[10 9 12 8 9]; d=[4 8 6 7 5]; B=5; c=0.05

[Dyuman Joshi]

What have you tried yet?

[Torsten]

Use "fmincon".

[Akshay Tanaji Bhosale]

but fmincon is use to solve linear program but my problem is nonlinear in nature.

[Dyuman Joshi]

"but fmincon is use to solve linear program ..."

Wrong. The description of fmincon literally says - "Find minimum of constrained nonlinear multivariable function"

[Akshay Tanaji Bhosale]

% Given data
v = [10, 20, 30, 40, 50];
d = [4, 5, 6, 7, 8];
c = [0.05, 0.05, 0.05, 0.05, 0.05];
B = 5;

% Define the objective function to maximize
fun = @(a) -sum(v .* (a ./ (a + d + c)) - a);

% Number of elements (size of v, d, and c)
n = numel(v);

% Set initial guess for 'a' (starting point for the optimization)
% Make sure the initial guess satisfies the constraints: a_i >= 0
initialGuess = [2,2,2,2,2];

% Define the nonlinear inequality constraint: a_i >= 0
nonlincon = @(a) deal([], a);

% Define the linear inequality constraint: sum(a_i) <= B
A = ones(1, n);
b = B;

% Set options for fmincon (optional)
options = optimoptions('fmincon', 'Display', 'iter');

% Perform the constrained optimization
a_optimal = fmincon(@(a) -sum(fun(a)), initialGuess, A, b, [], [], [], [], nonlincon, options);

% Calculate the optimal objective function value
optimal_objective = -fun(a_optimal);

% Display the optimal values of 'a' and the objective function value
disp("Optimal values of 'a':");
disp(a_optimal);
disp("Optimal objective function value:");
disp(optimal_objective);

I have written this code but the values coming of a_optimal are negative. but I have written the a_i>=0 constraint.

[Walter Roberson]

@Akshay Tanaji Bhosale

Why are you using nonlinear equality constraints for that, when you could simply use lb (lowerbound) ?

Answer 1

[Dyuman Joshi]

I have corrected your code and added comments (double %) for reference -

% Given data
v = [10, 20, 30, 40, 50];
d = [4, 5, 6, 7, 8];
c = 0.05;
B = 5;
% Define the objective function to maximize
fun = @(a) -sum(v .* (a ./ (a + d + c)) - a);

% Number of elements (size of v, d, and c)
n = numel(v);

% Set initial guess for 'a' (starting point for the optimization)
% Make sure the initial guess satisfies the constraints: a_i >= 0
initialGuess = 2*ones(1,n);

%%There is no non-linear inequality, thus nonlincon is []
% Define the nonlinear inequality constraint: a_i >= 0
nonlincon = []; 

%%You need to specify the bounds on the variables - 
%Bounds on variables
lb = zeros(1,n);
ub = Inf(1,n);

% Define the linear inequality constraint: sum(a_i) <= B
A = ones(1, n);
b = B;
% Set options for fmincon (optional)
options = optimoptions('fmincon', 'Display', 'iter');

%%Correced the fmincon call to include bounds
% Perform the constrained optimization
a_optimal = fmincon(@(a) fun(a), initialGuess, A, b, [], [], lb, ub, nonlincon, options);

                                            First-order      Norm of
 Iter F-count            f(x)  Feasibility   optimality         step
    0       6   -2.522299e+01    5.000e+00    1.669e+00
    1      12   -1.970688e+01    6.326e-01    9.245e-01    2.600e+00
    2      18   -1.778736e+01    0.000e+00    1.905e-01    5.597e-01
    3      24   -1.811167e+01    0.000e+00    6.461e-02    2.078e-01
    4      30   -1.815469e+01    0.000e+00    3.416e-03    4.387e-02
    5      36   -1.815754e+01    0.000e+00    1.003e-03    4.258e-03
    6      42   -1.815916e+01    0.000e+00    3.102e-04    3.530e-03
    7      48   -1.815956e+01    0.000e+00    1.872e-05    8.340e-04
    8      54   -1.815956e+01    0.000e+00    2.323e-06    2.369e-05

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

% Calculate the optimal objective function value
optimal_objective = -fun(a_optimal);
% Display the optimal values of 'a' and the objective function value
disp("Optimal values of 'a':");

Optimal values of 'a':

disp(a_optimal);

    0.0000    0.1433    0.9118    1.6277    2.3172

disp("Optimal objective function value:");

Optimal objective function value:

disp(optimal_objective);

   18.1596

[Akshay Tanaji Bhosale]

thanx