Quadratic programming
x = quadprog(H,f)
x = quadprog(H,f,A,b)
x = quadprog(H,f,A,b,Aeq,beq)
x = quadprog(H,f,A,b,Aeq,beq,lb,ub)
x = quadprog(H,f,A,b,Aeq,beq,lb,ub,x0)
x = quadprog(H,f,A,b,Aeq,beq,lb,ub,x0,options)
x = quadprog(problem)
[x,fval]
= quadprog(H,f,...)
[x,fval,exitflag]
= quadprog(H,f,...)
[x,fval,exitflag,output]
= quadprog(H,f,...)
[x,fval,exitflag,output,lambda]
= quadprog(H,f,...)
Finds a minimum for a problem specified by
$$\underset{x}{\mathrm{min}}\frac{1}{2}{x}^{T}Hx+{f}^{T}x\text{suchthat}\{\begin{array}{c}A\cdot x\le b,\\ Aeq\cdot x=beq,\\ lb\le x\le ub.\end{array}$$
H, A, and Aeq are matrices, and f, b, beq, lb, ub, and x are vectors.
f, lb, and ub can be passed as vectors or matrices; see Matrix Arguments.
returns
a vector x
= quadprog(H
,f
)x
that minimizes 1/2*x'*H*x + f'*x
. H
must
be positive definite for the problem to have a finite minimum.
minimizes x
= quadprog(H
,f
,A
,b
)1/2*x'*H*x + f'*x
subject to the restrictions A*x
≤ b
. A
is
a matrix of doubles, and b
is a vector of doubles.
solves
the preceding problem subject to the additional restrictions x
= quadprog(H
,f
,A
,b
,Aeq
,beq
)Aeq*x = beq
. Aeq
is
a matrix of doubles, and beq
is a vector of doubles.
If no inequalities exist, set A = []
and b = []
.
solves
the preceding problem subject to the additional restrictions x
= quadprog(H
,f
,A
,b
,Aeq
,beq
,lb
,ub
)lb
≤ x
≤ ub
. lb
and ub
are
vectors of doubles, and the restrictions hold for each x
component.
If no equalities exist, set Aeq = []
and beq = []
.
Note:
If the specified input bounds for a problem are inconsistent,
the output

solves
the preceding problem starting from the vector x
= quadprog(H
,f
,A
,b
,Aeq
,beq
,lb
,ub
,x0
)x0
.
If no bounds exist, set lb = []
and ub = []
. Some quadprog
algorithms
ignore x0
, see Input Arguments.
solves
the preceding problem using the optimization options specified in x
= quadprog(H
,f
,A
,b
,Aeq
,beq
,lb
,ub
,x0
,options
)options
.
Use optimoptions
to create options
.
If you do not want to give an initial point, set x0 = []
.
returns
the minimum for x
= quadprog(problem
)problem
, where problem
is
a structure described in Input Arguments.
Create problem
by exporting a problem using the
Optimization app; see Exporting Your Work.
[
returns
the value of the objective function at x
,fval
]
= quadprog(H
,f
,...)x
:
fval = 0.5*x'*H*x + f'*x
[
x
,fval
,exitflag
]
= quadprog(H
,f
,...)exitflag
,
a scalar that describes the exit condition of quadprog
.
[
x
,fval
,exitflag
,output
]
= quadprog(H
,f
,...)output
,
a structure that contains information about the optimization.
[
x
,fval
,exitflag
,output
,lambda
]
= quadprog(H
,f
,...)lambda
,
a structure whose fields contain the Lagrange multipliers at the solution x
.

Symmetric matrix of doubles. Represents the quadratic in the
expression  

Vector of doubles. Represents the linear term in the expression  

Matrix of doubles. Represents the linear coefficients in the
constraints  

Vector of doubles. Represents the constant vector in the constraints  

Matrix of doubles. Represents the linear coefficients in the
constraints  

Vector of doubles. Represents the constant vector in the constraints  

Vector of doubles. Represents the lower bounds elementwise in  

Vector of doubles. Represents the upper bounds elementwise in  

Vector of doubles. Optional. The initial point for some
If you do not give  

Options created using Some options are absent from the All Algorithms
All Algorithms Except activeset
trustregionreflective Algorithm Only
interiorpointconvex Algorithm Only
 

Structure encapsulating the


Vector that minimizes  

Value of  

Integer identifying the reason the algorithm terminated. The
following lists the values of
 

Structure containing information about the optimization. The fields are:
 

Structure containing the Lagrange multipliers at the solution
For details, see Lagrange Multiplier Structures. 
Solve a simple quadratic programming problem: find values of x
that
minimize
$$f(x)=\frac{1}{2}{x}_{1}^{2}+{x}_{2}^{2}{x}_{1}{x}_{2}2{x}_{1}6{x}_{2},$$
subject to
x_{1} + x_{2} ≤
2
–x_{1} +
2x_{2} ≤ 2
2x_{1} + x_{2} ≤
3
0 ≤ x_{1},
0 ≤ x_{2}.
In matrix notation this is
$$f(x)=\frac{1}{2}{x}^{T}Hx+{f}^{T}x,$$
where
$$H=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right],\text{}f=\left[\begin{array}{c}2\\ 6\end{array}\right],\text{}x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right].$$
Enter the coefficient matrices:
H = [1 1; 1 2]; f = [2; 6]; A = [1 1; 1 2; 2 1]; b = [2; 2; 3]; lb = zeros(2,1);
Set the options to use the 'interiorpointconvex'
algorithm
with no display:
options = optimoptions('quadprog',... 'Algorithm','interiorpointconvex','Display','off');
Call quadprog
:
[x,fval,exitflag,output,lambda] = ... quadprog(H,f,A,b,[],[],lb,[],[],options);
Examine the final point, function value, and exit flag:
x,fval,exitflag x = 0.6667 1.3333 fval = 8.2222 exitflag = 1
An exit flag of 1
means the result
is a local minimum. Because H
is a positive definite
matrix, this problem is convex, so the minimum is a global minimum.
You can see H
is positive definite by noting all
its eigenvalues are positive:
eig(H) ans = 0.3820 2.6180
Use the 'interiorpointconvex'
algorithm to solve
a sparse quadratic program.
Generate a sparse symmetric matrix for the quadratic form:
v = sparse([1,.25,0,0,0,0,0,.25]); H = gallery('circul',v);
Include the linear term for the problem:
f = 4:3;
Include the constraint that the sum of the terms in
the solution x
must be less than 2
:
A = ones(1,8);b = 2;
Set options to use the 'interiorpointconvex'
algorithm
and iterative display:
opts = optimoptions('quadprog',... 'Algorithm','interiorpointconvex','Display','iter');
Run the quadprog
solver and observe
the iterations:
[x fval eflag output lambda] = quadprog(H,f,A,b,[],[],[],[],[],opts); Firstorder Total relative Iter f(x) Feasibility optimality error 0 2.000000e+000 1.000e+001 4.500e+000 1.200e+001 1 2.630486e+001 0.000e+000 9.465e002 9.465e002 2 2.639877e+001 0.000e+000 3.914e005 3.914e005 3 2.639881e+001 0.000e+000 3.069e015 6.883e015 Minimum found that satisfies the constraints. Optimization completed because the objective function is nondecreasing 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.
Examine the solution:
fval,eflag fval = 26.3988 eflag = 1
For the 'interiorpointconvex'
algorithm, an exit
flag of 1
means the result is a global minimum.
You can use the Optimization app for quadratic programming.
Enter optimtool
at the MATLAB^{®} command line,
and choose the quadprog  Quadratic programming
solver.
For more information, see Optimization App.
[1] Coleman, T.F. and Y. Li, "A Reflective Newton Method for Minimizing a Quadratic Function Subject to Bounds on Some of the Variables," SIAM Journal on Optimization, Vol. 6, Number 4, pp. 1040–1058, 1996.
[2] Gill, P. E., W. Murray, and M. H. Wright, Practical Optimization, Academic Press, London, UK, 1981.
[3] Gould, N. and P. L. Toint. "Preprocessing for quadratic programming." Math. Programming, Series B, Vol. 100, pp. 95–132, 2004.