# Study notes: constraints in Quipu Nelder-Mead solver

In my previous post, I went over the recent changes I made to my F# Nelder-Mead solver, Quipu. In this post, I want to explore how I could go about handling constraints in Quipu.

First, what do I mean by constraints? In its basic form, the solver takes a function, and attempts to find the set of inputs that minimizes that function. Lifting the example from the previous post, you may want to know what values of $(x,y)$ produce the smallest value for $f(x,y)=(x-10)^2+(y+5)^2$. The solution happens to be $(10,-5)$, and Quipu solves that without issues:

#r "nuget: Quipu, 0.2.0"

let f (x, y) = (x - 10.0) ** 2.0 + (y + 5.0) ** 2.0

val it: Solution = Optimal (2.467079917e-07, [|9.999611886; -4.999690039|])


However, in many situations, not every value will do. There might be restrictions on what values are valid, such as “x must be positive”, or “y must be less than 2”. These are known as constraints, and typically result in an inequality constraint, in our case something like $g(x,y) \leq 0$. How could we go about handling such constraints in our solver?

In this post, I will look into some possible approaches to minimize a function under a set of constraints. I will do it the hard way, manually, hoping that the exercise will provide some direction on how to modify the library to make that easy in the future.

## First take: a crude penalty

Let’s keep the original function, $f(x,y)=(x-10)^2+(y+5)^2$, but imagine that we have one constraint: $x \leq 5$. Our original solution $(10,-5)$ is not valid in that case: $x > 5$, or, in technical terms, the constraint is not satisfied.

For any value of $x > 5$, we have a problem. One approach here is to add a penalty to our function, such that any value that does not satisfy a constraint will cause the objective to increase. In that case, the Nelder-Mead solver will avoid movements towards these values. Let’s try that:

let g (x, y) =
f (x, y)
+
// we incorporate the constraint in the objective
(if x <= 5.0 then 0.0 else infinity)


We keep the original objective function, $f$, but we add in a penalty term as well, creating a modified objective function, $g$. If the constraint is not satisfied, apply a penalty of $+\infty$, otherwise ignore the constraint and simply return $0$. Does this work?

let solution1 =
g

val it: Solution = Optimal (25.00000157, [|4.999999857; -5.000363939|])


It does! The optimal solution is now $(5,-5)$, which does indeed satisfy the constraint. We could easily expand on this idea, adding a penalty function for any other constraint we might have.

Are we done, then? Well, not quite. By default, Quipu will start its search around $0$, here $(0,0)$. However, what happens if we started around, say, $(20,0)$?

let problem =
g
|> NelderMead.startFrom (StartingPoint.fromValue [ 20.0; 0.0 ])


Sadly, the solver goes into an endless loop that never terminates. The problem here is that when we start around $(20,0)$, we are in a region where the constraint is not satisfied. Every search direction results in $+\infty$, and the solver has nowhere to go - every direction looks equally bad.

In other words, if we happen to start from a position where constraints are not satisfied, we are going to run into trouble.

Another related problem: this approach will not work well to handle equality constraints. Besides inequality constraints, a common constraint type is an equality. For instance, we might want something like $x+y=20$.

A common trick in optimization to handle equality constraints is to turn them into 2 inequalities. A perhaps counter-intuitive way to state $x+y=20$ is the following: $x+y \leq 20$ and $x+y \geq 20$. While this might appear weird, it is convenient. If we can handle inequality constraints, we get equality constraints for free, by converting equalities into pairs of inequalities.

However, this will be causing our penalty function issues. By definition, unless we are exactly on the values that satisfy our constraint, one of the two inequalities will result in a penalty of $+\infty$. Like in the previous situation we discussed, every move by the solver will look equally terrible.

## Second take: better penalties

So what can we do? One approach is to use a progressive penalty. We want a function that will return $0$ when the constraint is satisfied, and a value that becomes increasingly larger as we move further away from it being satisfied.

We could for instance do something like this:

let simplePenalty (x, y) =
f (x, y)
+
(if x <= 5.0 then 0.0 else ((x - 5.0) ** 2.0))


Now the penalty term will return $0$ if the constraint is satisfied, and $(x-5)^2$ otherwise. For values close to $5$, the penalty will be small, but as $x$ moves further away from $5$, the penalty will become steeper and steeper.

Let’s try this out, starting our search from the point that was giving us trouble before, $(20,0)$:

simplePenalty
|> NelderMead.startFrom (StartingPoint.fromValue [ 20.0; 0.0 ])

val it: Solution = Optimal (12.50000014, [|7.500152792; -4.999694746|])


Does this work? Well, sort of, but not really. On the one hand, the solver does not get stuck, and returns an optimal solution, $(7.5,-5)$. On the other hand, the solution is neither optimal (it should be $(5,-5)$), nor the constraint satisfied either ($7.5 > 5$). Instead of no answer, we get a pretty bad answer.

The issue here is that the constraint is “soft”. A minor violation of the constraint will result in a small penalty. In other words, if our solution does not satisfy the constraint, but is close to the limit, the penalty is small enough to be acceptable, so to speak.

What we could do then is make the penalty steeper. Let’s make it 100 times steeper:

let steeperPenalty (x, y) =
f (x, y)
+
(if x <= 5.0 then 0.0 else 100.0 * ((x - 5.0) ** 2.0))

val it: Solution = Optimal (24.75247568, [|5.049540943; -5.000552666|])


Still not quite right, but much better. How about 10,000 steeper?

Optimal (24.99750049, [|5.000501938; -4.999554925|])


As we crank up the aggressiveness of the penalty, we get solutions that are closer and closer to the correct answer. This suggests a possible strategy: solve iteratively, starting with a soft constraint, and make it progressively steeper, until we are close enough.

Here is a quick sketch of how this might look like. First, we create a penalty that takes in a coefficient, describing how aggressive the penalty is:

let penalty coeff =
let f (x, y) =
f (x, y)
+
(if x <= 5.0 then 0.0 else coeff * ((x - 5.0) ** 2.0))
f


Then, we solve our problem, starting with a coefficient of 1, and increasing it by a factor 10 each iteration, starting from the solution identified during the previous pass:

let rec solve (i: int, startingPoint: seq<float>) =
printfn $"Iteration {i}" let coeff = 10.0 ** i let solution = penalty coeff |> NelderMead.minimize |> NelderMead.startFrom (StartingPoint.fromValue startingPoint) |> NelderMead.solve if i >= 5 then solution else let nextStart = match solution with | Optimal (_, x) -> x | _ -> failwith "Ooops" printfn$"{List.ofArray nextStart}"
solve (i + 1, nextStart)

solve (0, [ 20.0; 0.0 ])

Iteration 0
[7.500152791643328; -4.999694746077196]
Iteration 1
[5.454595839229248; -4.999507451495937]
Iteration 2
[5.049548981640678; -5.000000777860209]
Iteration 3
[5.00499712795882; -5.000337947630191]
Iteration 4
[5.000501780625028; -5.000266170674201]
Iteration 5


This is a sketch, and would need some refinements. In particular, stopping after 5 iterations is totally arbitrary. We should probably stop once the constraints are all within certain bounds (and figure out pesky details like what to do if we never manage to satisfy the constraints…).

With that caveat, things do appear to work as expected. As we make the constraint iteratively stiffer, $x$ gets progressively closer and closer to $5$. I also tried out an equality constraint, and the results were what I expected.

For the sake of completeness, I also need to point out an odd result. If you run the algorithm for a little longer, you might observe that the $y$ values oscillate between $5$ and $4.06$. I am not sure what is going on at that point.

## Parting words

As I was doing some reading on constrained optimization, I came across another approach, barrier functions. Where penalty functions add a penalty to the objective when a constraint is not satisfied, barrier functions create a penalty inside the feasible domain. The closer you approach a penalty, the steeper the penalty.

This is an interesting approach, but after some thinking, I believe it won’t work for Nelder-Mead. A barrier should work well if the search algorithm relies on gradients, because the step size depends on the gradient. However, Nelder-Mead does not rely on gradients (which is one of its advantages). While the step direction depends on the function, the step size depends only on the geometry of the current simplex. As a result, a barrier would have no direct impact: the standard algorithm could still take steps leading outside of the feasible domain, where all constraints are satisfied, and we would encounter the same exact issue we had with our original, crude penalty function.

I imagine you could modify the algorithm to perhaps take more adaptive steps, but in the meantime, the penalty direction seems more promising. I will probably take a stab at incorporating constraints in the current solver using penalties in the next few weeks - we’ll see how that goes!

In the meantime, you can find the current code here:

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