I know L-BFGS-B only supports simple box constrains of the form: $l_i \leq x_i \leq u_i$, where $l_i$ and $u_i$ are constants. For my specific optimization problem, I need to specify some simple linear constraints of the form:

$x_{i-1} + k \leq x_i \leq x_{i+1}-k $ where $k>0$ is a fixed constant. Essentially, I need each variable $x_i$ to stay in-between (and within a certain distance away from) the two variables at index $i-1$ and $i+1$. Is it possible to achieve this with L-BFGS-B?

Can the $l$ and $u$ constraint vectors be modified after each iteration? One silly idea I have tried (and seems to work, although I have not extensively tested it) is to modify $l$ and $u$ after each iteration with this rule:

$$l_i = x_{i-1} + k$$ $$u_i = x_{i+1} - k$$

Thanks in advance!

EDIT: I forgot to mention that the first variable ($x_0$) is also subject to a constant lower bound and the last variable ($x_{N-1}$) is subject to an constant upper bound:

$$x_0 \geq k$$

$$x_{N-1} \leq k_{max}$$

This makes recasting the problem into a basis where the unknowns are the differences of two $x$ variables not feasible since these two extra bounds (which in the current system are actual box constraints) would then become linear constraints.


1 Answer 1


One approach to this problem is to reparameterize your problem in terms of



$z_{i}=x_{i}-x_{i-1}$, $i=2, \ldots, n-1$.

You can then rewrite your objective function in terms of the the $z_{i}$ variables by substituting

$x_{i}=x_{1}+z_{1}+\ldots + z_{i}$, $i=2, 3, \ldots, n$.

Your separation constraints become

$k \leq z_{i}$

  • 1
    $\begingroup$ ... and in case your gradient routine is something you'd rather not touch, you can use the chain rule to get the gradient in the new coordinates $\partial f / \partial z = (\partial f / \partial x) (\partial z / \partial x)^{-1}$ then feed that into your LBFGS-B routine. $\endgroup$
    – GoHokies
    Commented Mar 20, 2018 at 18:29
  • $\begingroup$ I actually had this idea originally, except it will not work because I have actual lower and upper bound constraints for the first and last $x$ variables respectively: $k \leq x_0$ and $x_{N-1} \leq k_{max}$. I will update my question to reflect this. $\endgroup$
    – Costis
    Commented Mar 21, 2018 at 3:44
  • $\begingroup$ The lower bound on the first variable isn't a problem. The constraint that the last variable is less than or equal to an upper bound isn't something you can address with this approach. $\endgroup$ Commented Mar 21, 2018 at 4:21

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