问题
I need to maximize the equation 3x+y in matlab with the following constraints:
2x+y<=6, x+3y<=9, and x,y>=0
I am having a really hard time figuring out how to put in the constraints in a way that I can relate them back to the original equation. I am new to matlab and am having trouble figuring this out.
Thanks in advance!
回答1:
As @Franck mentioned, you can in general use fmincon
to solve optimization problems. However, as your problem is simply a linear programming problem, the solution is much simpler (and guaranteed to be optimal) :
f = -[3 1]; % Note the minus as we want maximization
A = [2 1; 1 3];
b = [6; 9];
LB = [0 0];
[X, FVAL] = linprog(f,A,b,[],[],LB)
Will give:
X =
3.0000
0.0000
FVAL =
-9.0000
Hence the optimum is found at point (3,0) and the resulting value is 9.
Try help linprog
to read more about this very usefull function.
回答2:
Create the following files and run maximize_stuff:
maximize_stuff.m:
function [] = maximize_stuff()
x0 = [2 2]; % fmincon starts at X0 and finds a minimum X
[x,fval] = fmincon('objfun',x0,[],[],[],[],[0;0],[Inf;Inf],'constraint');
fval = -fval; % Because we want to find the maximum, not the minimum
x
fval
end
objfun.m
function f=objfun(x)
f = 3*x(1) + x(2);
f = -f; % Because we want to find the maximum, not the minimum
end
constraint.m :
function [c,ceq]=constraint(x)
c1 = 2 * x(1) + x(2) - 6;
c2= x(1) + 3*x(2) - 9;
c = [c1;c2];
ceq = [];
end
It should return:
>> maximize_stuff
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in
feasible directions, to within the default value of the function tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
Active inequalities (to within options.TolCon = 1e-06):
lower upper ineqlin ineqnonlin
2 1
x =
3.0000 0
fval =
9.0000
You can verify the results http://www.wolframalpha.com/input/?i=2x%2By%3C%3D6%3B+x%2B3y%3C%3D9%3B+x%3E%3D0%3By%3E%3D0%3B :
A very good tutorial: http://www.math.colostate.edu/~gerhard/classes/331/lab/fmincon.html
fmincon is called as follows:
with linear inequality constraints Ax£b only (as in linprog): [x,fval]=fmincon('objfun',x0,A,b)
with linear inequality constraints and linear equality constraints Aeq·x=beq only: [x,fval]=fmincon('objfun',x0,A,b,Aeq,beq)
with linear inequality and equality constraints, and in addition a lower bound of the form x³lb only: [x,fval]=fmincon('objfun',x0,A,b,Aeq,beq,lb) If only a subset of the variables has a lower bound, the components of lb corresponding to variables without lower bound are -Inf. For example, if the variables are (x,y), and x³1 but y has no lower bound, then lb=[1;-Inf].
with linear inequality and equality constraints and lower as well as an upper bound of the form x£ub only: [x,fval]=fmincon('objfun',x0,A,b,Aeq,beq,lb,ub) If only a subset of the variables has an upper bound, the components of ub corresponding to variables without upper bound are Inf. For example, if the variables are (x,y) and x£1 but y has no lower bound, then lb=[1;Inf].
with linear inequality and equality constraints, lower and upper bounds, and nonlinear inequality and equality constraints: [x,fval]=fmincon('objfun',x0,A,b,Aeq,beq,lb,ub,'constraint') The last input argument in this call is the name of a function file (denoted constraint in these notes and saved as constraint.m in the working directory), in which the nonlinear constraints are coded.
来源:https://stackoverflow.com/questions/19085938/maximize-3xy-with-constraints-in-matlab