Nonlinear variant of a knapsack problem

In [1] a problem is posed:

Original Problem
\[\begin{align}\max & \sum_i \log_{100}(\color{darkblue}w_i) \cdot \color{darkred}x_i \\ & \frac{\sum_i \color{darkblue}w_i \color{darkred}x_i}{\sqrt{\color{darkred}k}} \le 10,000\\ & \color{darkred} k = \sum_i \color{darkred}x_i \\ & \color{darkred}x_i \in \{0,1\} \end{align}\]

The data vector \(w_i\) is assumed to integer valued with \(w_i\ge 1\). Hence the logarithm can be evaluated without a problem. Also we can assume that the number of items can be around 1,000.

I don't think I have ever seen \(\log_{100}()\) being used in a model. Most programming (and modeling) languages only support natural logarithms \(\ln()\) and may be \(\log_{10}()\). We can convert things by: \[\log_{100}(x) = \frac{\ln(x)}{\ln(100)}\] This means the objective can be written as \[\max \frac{1}{\ln(100)} \sum_i \ln(w_i) x_i\] Essentially, the \(\log_{100}()\) function just adds a scaling factor. We can simplify the objective to \[\max \sum_i \ln(w_i)x_i\] (The objective value will be different, but the optimal solution will be the same).

The constraint can be rewritten as: \[\sum_i w_i x_i \le 10,000 \sqrt{k}\] If we ignore the all zero solution, we can assume \(k\ge 1\). This bound will make sure the square root function is always differentiable. With this, we have a standard MINLP (Mixed Integer Nonlinear Programming) model.

MINLP Model
\[\begin{align}\max & \sum_i \ln(\color{darkblue}w_i) \color{darkred}x_i \\ & \sum_i \color{darkblue}w_i \color{darkred}x_i \le 10,000\sqrt{\color{darkred}k}\\ & \color{darkred} k = \sum_i \color{darkred}x_i \\ & \color{darkred}x_i \in \{0,1\} \\ & \color{darkred} k \ge 1 \end{align}\]

This model has only a single, well behaved, non-linearity. Literally, there is only one nonlinear nonzero element.  In addition, the model is convex. So we don't expect any problems. For \(w_i\),  I generated 1000 random integer values from the interval \([1,10000]\).

MINLP Results
Solver	Obj	Time	Notes
Dicopt	1057.1355	0.6	3 NLP, 2 MIP subproblems
SBB	1056.9361	95	Node limit exceeded
Bonmin	1056.9472	600	Time limit exceeded
Bonmin	1057.1355	16	Option: bonmin.algorithm B-OA

The outer-approximation based algorithms (Dicopt, Bonmin with B-OA option) do much better than the branch & bound algorithms (SBB, default Bonmin). Even the global solvers do better:

Global Solver Results
Solver	Obj	Time
Baron	1057.1355	2
Antigone	1057.1355	1
Couenne	1057.1355	9

The problem can also be formulated as a convex MIQCP (Mixed Integer Quadratically Constrained Programming) model:

MIQCP Model
\[\begin{align}\max & \sum_i \ln(\color{darkblue}w_i) \color{darkred}x_i \\ & \color{darkred}y^2 \le \color{darkred}k\\ & \color{darkred} k = \sum_i \color{darkred}x_i \\ & \color{darkred}y = \frac{\sum_i \color{darkblue}w_i \color{darkred}x_i}{10,000} \\ & \color{darkred}x_i \in \{0,1\} \\ & \color{darkred} k \ge 1, \color{darkred}y \ge 0 \end{align}\]

Solvers like Cplex may convert this into a Cone problem (MISOCP).

Finally, we can also linearize this model by observing that \(k\in \{1,\dots,1000\}\). So we don't really have a continuous function \(f(k)=\sqrt{k}\), but rather only need function values at the integer points. We can exploit this by making this explicit. We can write: \[\begin{align} & k = \sum_i i\cdot \delta_i \\ & \sqrt{k} = \sum_i \sqrt{i}\cdot \delta_i \\ & \sum_i \delta_i = 1 \\ & \delta_i \in \{0,1\}\end{align}\] This is essentially a SOS1 (Special Ordered Set of Type 1) structure implementing a table lookup. The MIP model looks like:

MIP Model

\[\begin{align}\max & \sum_i \ln(\color{darkblue}w_i) \color{darkred}x_i \\ & \color{darkred} k = \sum_i \color{darkred}x_i \\ & \sum_i \color{darkblue}w_i \color{darkred} x_i \le 10,000 \color{darkred}q \\ & \color{darkred} k = \sum_i i \cdot \color{darkred}\delta_i \\ & \color{darkred} q = \sum_i \sqrt{i} \cdot \color{darkred}\delta_i \\ & \sum_i \color{darkred} \delta_i = 1 \\ & \color{darkred}x_i, \color{darkred}\delta_i \in \{0,1\} \\ & \color{darkred} k, \color{darkred} q \ge 1 \end{align}\]

When we solve this problem we see:

MIQCP and MIP Results
Model	Solver	Obj	Time
MIQCP	Cplex	1057.1355	0.3
MIP	Cplex	1057.1355	0.6

Conclusion: the question in the original post was: how to solve this problem? Here we proposed three different models: an MINLP, MIQCP and MIP model. All these models can be solved quickly. It is noted that the quadratic and linear models are not approximations: they give the same solution as the original MINLP model. Pure non-linear branch & bound methods are having a bit of a problem with the MINLP model, but Outer-Approximation works very well.

References

1. How do we solve a variant of the knapsack problem in which the capacity of the knapsack keeps increasing as we add more items into the knapsack?, https://stackoverflow.com/questions/59242370/how-do-we-solve-a-variant-of-the-knapsack-problem-in-which-the-capacity-of-the-k