min F(S_1,S_2,):1.25 S_1^2+0.4S_2^2 Min ʄ (S_1,S_2,S_3 ): S_1^2+0.35S_2^2 2S_1^2+0.6S_2^2≤5100000 5S_1^2+3S_2^2 ≤14250000

Can you help me solve a bi-level programming problem that contains two levels and two constraints? How do I get the solution? I need the code, the solution, and the algorithm. Can you help me?

minF(s1,s2)=1.25S1^2 +0.4S2^2

minf(s1,s2)=s1^2+035s2^2

s.to 2s1^2+0.6s2^2 <=5100000

5s1^2+3s_2^2<=14250000

s1,s2>=0

[Torsten]

Could you state your problem properly ? It's not readable for me.

[othman warde]

minF(s1,s2)=1.25S1^2 +0.4S2^2

minf(s1,s2)=s1^2+035s2^2

s.to 2s1^2+0.6s2^2 <=5100000

5s1^2+3s_2^2<=14250000

s1,s2>=0

[Torsten]

You can't minimize two objective functions simultaneously. Maybe you want to determine a Pareto-Optima curve:

https://de.mathworks.com/help/gads/gamultiobj.html

?

[othman warde]

I just want to find the optimal values ​​for both

:s1,s2,minF(s1,s2) .minf(s1,s2)

[Torsten]

Say you have the optimization problem:

min: s1
min: s2

s.c.

s1 + s2 = 1
s1, s2 >=0

What would be the solution ?

s1 = 1 and s2 = 0 

or

s1 = 0 and s2 = 1 

or

s1 = 0.5 and s2 = 0.5 

or

...

?

Same conflict for your problem, I guess.

Or do you want to have the solution for both problems separately - one solution for

minF(s1,s2)=1.25s1^2 +0.4s2^2 
s.to        
2s1^2+0.6s2^2 <=5100000
5s1^2+3s_2^2<=14250000
s1,s2>=0

second solution for

minf(s1,s2)=s1^2+035s2^2 
s.to        
2s1^2+0.6s2^2 <=5100000
5s1^2+3s_2^2<=14250000
s1,s2>=0

[Steven Lord]

Suppose you had a set S1 of (s1, s2) values that made minF be 1 and minf be 0.4, another set S2 that made minF be 0.75 and minf be 0.75, and a third set S3 that made minF be 100 and minf be 0. Just considering those three sets, which would you consider to be optimal and why?

S1 = [1 0.4];
S2 = [0.75 0.75];
S3 = [100, 0];

Is it S1 since that has the lowest mean of minF and minf?

[mean(S1), mean(S2), mean(S3)]

ans = 1×3

0.7000 0.7500 50.0000

Is it S2 because that has the smallest maximum value?

[max(S1), max(S2), max(S3)]

ans = 1×3

1.0000 0.7500 100.0000

Is it S3 because that has the smallest minimum value?

[min(S1), min(S2), min(S3)]

ans = 1×3

0.4000 0.7500 0

Without knowing what you mean by optimal in this scenario we can't really recommend an approach.

[othman warde]

I have two types of medication s1 ,s2 The optimal amount of medication is represented by cartons = s1

The optimal amount of second medication is represented by carton = s2

minF(S1,S2)= The lowest cost achieved at the upper level

minf(S1,S2)= The lowest cost achieved at the lower level

[Torsten]

So you have two minimizations that can be solved independently from each other ? Or what do you want to tell us with your description of the problem ?

[othman warde]

% Define the upper level objective function

upperLevelObj = @(s) 1.25*s(1)^2 + 0.4*s(2)^2;

% Define the lower level objective function

lowerLevelObj = @(s) s(1)^2 + 0.35*s(2)^2;

% Define the upper level constraint function

upperLevelConstr = @(s) [2*s(1)^2 + 0.6*s(2)^2 - 5100000;

5*s(1)^2 + 3*s(2)^2 - 14250000];

% Set up the optimization problem for the upper level

upperLevelProblem.objective = upperLevelObj;

upperLevelProblem.x0 = [0, 0]; % Initial guess for (s_1, s_2)

upperLevelProblem.Aineq = [];

upperLevelProblem.bineq = [];

upperLevelProblem.Aeq = [];

upperLevelProblem.beq = [];

upperLevelProblem.lb = [];

upperLevelProblem.ub = [];

upperLevelProblem.nonlcon = [];

upperLevelProblem.options = optimoptions('fmincon', 'Display', 'off');

% Solve the upper level problem

[sUpperOptimal, upperOptimalCost] = fmincon(upperLevelProblem);

% Set up the optimization problem for the lower level

lowerLevelProblem.objective = lowerLevelObj;

lowerLevelProblem.x0 = sUpperOptimal; % Use upper level optimal solution as initial guess

lowerLevelProblem.Aineq = [];

lowerLevelProblem.bineq = [];

lowerLevelProblem.Aeq = [];

lowerLevelProblem.beq = [];

lowerLevelProblem.lb = [];

lowerLevelProblem.ub = [];

lowerLevelProblem.nonlcon = upperLevelConstr;

lowerLevelProblem.options = optimoptions('fmincon', 'Display', 'off');

% Solve the lower level problem

[sLowerOptimal, lowerOptimalCost] = fmincon(lowerLevelProblem);

% Display the optimal values and costs

disp('Upper Level Optimal Solution:');

disp(['s_1 = ', num2str(sUpperOptimal(1))]);

disp(['s_2 = ', num2str(sUpperOptimal(2))]);

disp(['Upper Level Optimal Cost: ', num2str(upperOptimalCost)]);

disp('Lower Level Optimal Solution:');

disp(['s_1 = ', num2str(sLowerOptimal(1))]);

disp(['s_2 = ', num2str(sLowerOptimal(2))]);

disp(['Lower Level Optimal Cost: ', num2str(lowerOptimalCost)]);

Is that correct.........................؟؟؟؟؟؟؟؟؟؟؟

Answer 1

[Torsten]

My guess is you want to maximize, not minimize the objective function. The result of mimimizing it is obviously s_1 = s_2 = 0 in both cases.

% Define the upper level objective function
upperLevelObj = @(s) 1.25*s(1)^2 + 0.4*s(2)^2;

% Define the lower level objective function
lowerLevelObj = @(s) s(1)^2 + 0.35*s(2)^2;

% Define the upper level constraint function
upperLevelConstr = @(s) deal([2*s(1)^2 + 0.6*s(2)^2 - 5100000;
                              5*s(1)^2 + 3*s(2)^2 - 14250000],[]);

% Set up the optimization problem for the upper level
objective = upperLevelObj;
x0 = [0; 0]; % Initial guess for (s_1, s_2)
Aineq = [];
bineq = [];
Aeq = [];
beq = [];
lb = [0;0];
ub = [Inf ;Inf];
nonlcon = upperLevelConstr;
options = optimoptions('fmincon', 'Display', 'off', 'TolX',1e-16,'TolFun',1e-16);

% Solve the upper level problem
[sUpperOptimal, upperOptimalCost] = fmincon(objective,x0,Aineq,bineq,Aeq,beq,lb,ub,nonlcon,options);

% Set up the optimization problem for the lower level
objective = lowerLevelObj;
x0 = sUpperOptimal; % Use upper level optimal solution as initial guess
Aineq = [];
bineq = [];
Aeq = [];
beq = [];
lb = [0;0];
ub = [Inf;Inf];
nonlcon = upperLevelConstr;
options = optimoptions('fmincon', 'Display', 'off', 'TolX',1e-16,'TolFun',1e-16);

% Solve the lower level problem
[sLowerOptimal, lowerOptimalCost] = fmincon(objective,x0,Aineq,bineq,Aeq,beq,lb,ub,nonlcon,options);

% Display the optimal values and costs
disp('Upper Level Optimal Solution:');

Upper Level Optimal Solution:

disp(['s_1 = ', num2str(sUpperOptimal(1))]);

s_1 = 7.037e-08

disp(['s_2 = ', num2str(sUpperOptimal(2))]);

s_2 = 1.2715e-07

disp(['Upper Level Optimal Cost: ', num2str(upperOptimalCost)]);

Upper Level Optimal Cost: 1.2657e-14

disp('Lower Level Optimal Solution:');

Lower Level Optimal Solution:

disp(['s_1 = ', num2str(sLowerOptimal(1))]);

s_1 = 8.6229e-08

disp(['s_2 = ', num2str(sLowerOptimal(2))]);

s_2 = 1.4824e-07

disp(['Lower Level Optimal Cost: ', num2str(lowerOptimalCost)]);

Lower Level Optimal Cost: 1.5127e-14

[othman warde]

How can I extract the output ...s1,s2

Should I put the full code in MATLAB or how?

I entered it in MATLAB but it shows an error ؟

[Torsten]

I always use StrgC and StrgV to copy-paste code.

And if there are outputs in the copied code, I delete these lines.

And if it shows an error, I can't tell where this error comes from because as you can see: the above code runs without problems (at least under MATLAB R2023a). In case the error persists, include code and full error message.

[othman warde]

Can the bat algorithm be used?

[Torsten]

I've never heard of this algorithm, but you might be interested to study the Wikipedia article

https://en.wikipedia.org/wiki/Bat_algorithm

where also a link to MATLAB/Octave code is given.