Also suppose problem (1) has a minimum and the minimum can be achieved by an unique minimizer .
Now if I have a point such that is very small, then how small is the distance . We might expect that will imply that . This is true if is compact and is continuous. But this does not tell what is the quantitative relationship between the optimal gap, i.e., and the distance to the solution, i.e., .
In this post, I am going to show that for linear programming (LP) problem, the optimal gap and distance to solutions are the same up to a multiplicative constant which only depend on the problem data.
To start, consider an LP in the standard form, i.e.,
Denote the solution set of problem (1) to be , and the distance to the solution set to be . Note that the norm here is arbitrary, not necessarily the Euclidean norm.
We are now ready to state the theorem.
Theorem 1 (Sandwich inequality of optimal gap and distance to solutions of LP) For problem (1), theres exist constants which depends only on and such that for all feasible , i.e., ,
The above theorem shows that the role of optimal gap, i.e., , and the distance to the solution set, i.e. , are the same up to a multiplicative constant. The right inequality of the theorem is usually referred as linear growth in the optimization literature.
The proof below is constructive and we can in fact take where and for and . Here is the dual norm and are extreme points and extreme rays. We assume there are many extreme points and many extreme rays with first s and first are in the optimal set . See the proof for more detail.
The idea of the proof mainly relies on the extreme point and extreme rays representation of the feasible region and the optimal set,i.e., and .
Proof: The feasible region can be written as
Here s are extreme points of the feasible region and s are extreme rays. By scaling the extreme rays, we can assume that for all .
The optimal set can also be written as
We assume here the first many and many are in the optimal set and the rest of and s are not for notation simplicity.
We denote and where . Note since the s not in the optimal set should have inner product with to be positive.
We first prove the second inquality, i.e., .
Now take an arbitrary feasible , it can be written as
for some and .
The objective value of is then
We use the fact that for all here.
Subtract the above by . We have
The second equality is due to and the inequality is because of the definition of and the s are positive.
The distance between and is the infimum of
By taking with and and apply triangular inequality to the above quantity, we have
The first inequality is the triangular inequality and . The second inequality is applying the definition of and . The first equality is due to and the second equality is due to .
Thus the distance between and is bounded above by
Since by our previous argument, we see that setting
should give us
We now prove the inequality
Note that the infimum in is actually achieved by some . The reason is that we can first pick a , then
But the set is actually bounded and closed ( is closed as it is a convex combination of finite points plus a conic combination of extreme vectors), thus Weierstrass theorems tells us that the infimum is actually achieved by some .
Now take such that . We have
where is the dual norm of . Thus letting finishes the proof.
From the proof, we see that two possible choice of and are where and for and . These are not optimal and can be sharpened. I probably will give a sharper constant in a future post.