Mixed Integer Quadratic Programming with linear constraints in Matlab calling Gurobi

Question

I have some troubles to understand how to implement the following MIQP (Mixed Integer Quadratic Programming) with linear constraints in Matlab calling Gurobi.

Let me explain in a schematic way my setting.

(1) x is the unknown and it is a column vector with size 225x1.

(2) The objective function (which should be minimised wrto x) looks like

which can be rewritten as

I have a Matlab script computing alpha, Q,c (Q,c sparse) when some_known_parameters1 are given:

function [alpha, Q,c]=matrix_objective_function(some_known_parameters1)

%...

end

(3) The constraints are linear in x, include equalities and inequalities, and are written in the form

I have a Matlab script computing Aeq,beq,Aineq,bineq (Aeq,Aineq sparse) when some_known_parameters2 is given:

function [Aeq,beq,Aineq,bineq]=constraints(some_known_parameters2)

%...

end

(4) Some components of x are restricted to be in {0,1}. I have a Matlab script producing a string of letters B (binary), C (continous) when some_known_parameters3 is given:

function type=binary_continuous(some_known_parameters3)

%...

end

Now, I need to put together (1)-(4) using Gurobi. I am struggling to understand how. I found this example but it looks very cryptic to me. Below I report some lines I have attempted to write, but they are incomplete and I would like your help to complete them.

clear 
rng default

%Define some_known_parameters1, 
 some_known_parameters2,some_known_parameters3 [...]

%1) generate alpha,Q,c,Aeq,beq,Aineq,bineq,type with Q,c,Aeq, Aineq sparse
[alpha, Q,c]=matrix_objective_function(some_known_parameters1)
[Aeq,beq,Aineq,bineq]=constraints(some_known_parameters2)
type=binary_continuous(some_known_parameters3)



%2) Set up Gurobi
clear model;
model.A=[Aineq; Aeq];
model.rhs=full([bineq(:); beq(:)]); 
model.sense=[repmat('<', size(Aineq,1),1); repmat('=', size(Aeq,1),1)];
model.Q=Q; %not sure?
model.alpha=alpha; %not sure?
model.c=c; %not sure?
model.vtype=type;
result=gurobi(model); %how do I get just the objective function here without the minimiser?

Questions:

(1) I'm not sure about

model.Q=Q; 
model.alpha=alpha; 
model.c=c;

I'm just trying to set the matrices of the objective function using the letters provided here but it gives me error. The example here seems to me doing

model.Q=Q; 
model.obj=c;

But then how do I set alpha? Is it ignoring it because it does not change the set of solutions?

(2) How do I get as output stored in a matrix just the minimum value of the objective function without the corresponding x?

You should not expect to be able to give gurobi some black-box objective. This is something for NLP territory (where some form of differentiation is happening inside; e.g. Ipopt/Bonmin). Gurobi needs this objective in it's own native form. What form that is depends on your lib/wrapper. In low-level form usually something like 0.5 * x'Qx + q'x with Q psd (convex QP; which is probably the only one gurobi supports; ignoring SOCP generalizations). If that's giving you headaches, look for some more high-level wrapper. Gurobi's Python-API for example supports expr = QuadExpr(x*x + y+y). — sascha
– sascha, Commented Nov 13, 2018 at 19:49
@sascha thanks: my problem is very basic I guess: (a) I think my objective function can be rewritten as Q+x'Hx (I have added this to my question); (b) I still don't understand how to complete steps 3) and 4) above. — Star
– Star, Commented Nov 13, 2018 at 19:55
This is all explained in the docs for this low-level view. Bring your objective in this form and set obj = some_vec, objcon = some_vec and Q = some_matrix. Then (4) is just a string it seems like BBC (binary, binary, continuous). The bounds are vectors lb and ub. — sascha
– sascha, Commented Nov 13, 2018 at 20:02
vtypes is string. lb ub are vectors. 3 vars between (-1,1), then (1,3) and (2,3) will be lb=[-1,1,2] and ub=[1,3,3]. — sascha
– sascha, Commented Nov 14, 2018 at 5:35
That's a modelling thing. I recommend grabbing some integer-programming book. You can use one binary variable and replace all occurences with the term x = 1-2*binVar. x will be in {-1,1} then. Yes, this is annoying in low-level form (but there is no way out without wrappers / lib-support). But nobody should do prototyping on this level imho. — sascha
– sascha, Commented Nov 14, 2018 at 8:15

joni · Accepted Answer · 2018-11-15 09:36:58Z

2

(1) You're right, there's no need to pass the constant alpha since it doesn't affect the optimal solution. Gurobi's MATLAB API only accepts sparse matrices. Furthermore model.obj is always the c vector in the problem statement:

model.Q = sparse(Q); 
model.obj = c;

(2) To get the optimal objective value, you first need to pass your model to gurobi and solve it. Then you can access it via the objval attribute:

results = gurobi(model);
val = results.objval + alpha

answered Nov 15, 2018 at 9:36

joni

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