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CEC 2006 g24

The problem

The CEC 2006 special session on constrained optimization (Liang et al., 2006) collected 24 test problems, g01 to g24, with their best known solutions and rules for comparing algorithms. g24 is the last, the report's equations 49 and 50 (page 15). The report takes it from Floudas et al. (1999, Handbook of Test Problems in Local and Global Optimization, Kluwer). It has two variables, x1 in [0, 3] and x2 in [0, 4].

Minimize

f(x) = −x1 − x2

subject to two inequalities, each g(x) ≤ 0:

g1 = −2 x1⁴ + 8 x1³ − 8 x1² + x2 − 2
g2 = −4 x1⁴ + 32 x1³ − 88 x1² + 96 x1 + x2 − 36

Each is an upper bound on x2 that depends on x1. Written that way, they are simpler than they look:

g1:  x2 ≤ 2 (x1 (x1 − 2))² + 2
g2:  x2 ≤ 4 ((x1 − 1)(x1 − 3))²

The first bound is at least 2 everywhere. The second is 0 at x1 = 1 and at x1 = 3, where only x2 = 0 is feasible. The report says the feasible region consists of two disconnected sub-regions: they are the parts left and right of x1 = 1, and they meet at the single point (1, 0).

The minimum is f = −5.50801327159536, at x = (2.32952019747762, 3.17849307411774), where both constraints are active. The report prints x without the comma, as "2.329520197477623.17849307411774"; the two numbers add up to −f. genoxide's docs mark the minimum as proven, from the shape above: at each x1, the best x2 is the least of the two bounds and 4, so f is a function of x1 alone, and its least value is where the two bounds cross, at a root of x1⁴ − 12 x1³ + 40 x1² − 48 x1 + 17. The report's x1 is that root to 14 digits.

What makes it hard

Little, for most algorithms. The report's table 3 gives its feasible share as 79.6556 %, but that is the share where g1 alone holds. With both constraints, the share is 44.21 %, computed for this page by integrating the bounds above. Either way, random points are often feasible.

What remains is a trap for a local search. Along x1, f has three local minima, each where the two bounds on x2 cross:

x1 = 0.6116, x2 = 3.4421, f = −4.0537   (the left part)
x1 = 1.5996, x2 = 2.8204, f = −4.4200
x1 = 2.3295, x2 = 3.1785, f = −5.5080   (the global minimum)

Each sits in a corner between the two curved bounds. A search that settles into one of the other two corners has to cross a region where f gets worse to reach the third.

Representation

A Real genome of 2 genes, x1 in [0, 3] and x2 in [0, 4]. genoxide's problems::cec2006::G24 is the fitness: the value f(x) and the total constraint violation, the sum of max(0, g(x)) over the two constraints, 0 for a feasible solution.

genoxide compares fitnesses with Deb's feasibility rules (Deb, 2000, Computer Methods in Applied Mechanics and Engineering 186: 311-338): a feasible solution beats an infeasible one, two feasible ones compare by value, and two infeasible ones by violation. The rules need no penalty weights.

Algorithm

CMA-ES (Hansen and Ostermeier, 2001, Evolutionary Computation 9(2): 159-195) samples a population from a normal distribution, and adapts its mean, step size and covariance matrix. It uses genoxide's defaults, a population of 4 + ⌊3 ln 2⌋ = 6, a step size of 0.3 of each gene's range and a random start, with IPOP restarts (Auger and Hansen, 2005, IEEE CEC 2005: 1769-1776): when a run converges, the next starts from a random point with twice the population. A sample outside the bounds is drawn again, up to 100 times, and then clipped to them. Deb's rules rank the samples.

The run has the report's budget of 500,000 evaluations, and stops once its best solution is feasible with an absolute error f(x) − f* of at most 1e-8. The report counts a run as successful with an error of at most 1e-4; the example asks for more.

Why the restarts: without them, CMA-ES met the target on 22 of 25 seeds, after a median of 1,104 evaluations. The other three converged to the two other corners, twice at f = −4.4200 and once at −4.0537, and stayed there. With IPOP, all 25 met the target, after a median of 1,104 evaluations (from 930 to 3,306). SHADE (Tanabe and Fukunaga, 2013, IEEE CEC 2013: 71-78), genoxide's default differential evolution, met it on all 25 too, but after a median of 15,600; L-SHADE after a median of 6,948.

Output

The first line names the run. The second gives what stopped it, after how many evaluations, the error f(x) − f and whether the best solution is feasible: "< 1e-8" means the run met its target. The third gives the evaluations to the first feasible solution, and to an error of 1e-4, the report's criterion of success. The fourth gives the restarts and the population of each run. The fifth compares f(x) with f, to 6 significant digits. The sixth gives the solution, and the last the two constraints: "active" for a constraint on its boundary (|g| ≤ 1e-6), else the value of g, negative when it's satisfied. In Python, run evaluates the problem in Rust, so both versions print the same.

The page's plot shows each variable on its range, and each constraint's state: violated, active or satisfied. Its curve shows the error f − f* of the best feasible solution, and of the population's median, on a log scale. The best's curve begins at the first feasible solution, and the median's once half the population is feasible.

The project page plays this run back.

Good results

A good run is feasible and ends within 1e-4 of f*, the report's success. CMA-ES with restarts meets the target of 1e-8 with every seed tried.

Seed 1 needs no restart. Its first generation of 6 samples has a feasible one. The run meets the report's criterion after 522 evaluations and the target after 954. The solution is the report's x* to 6 digits, with both constraints active.

Reference: Liang, J. J., Runarsson, T. P., Mezura-Montes, E., Clerc, M., Suganthan, P. N., Coello Coello, C. A. and Deb, K. (2006). Problem Definitions and Evaluation Criteria for the CEC 2006 Special Session on Constrained Real-Parameter Optimization. Technical report, Nanyang Technological University, Singapore.

Known optimum: −5.50801327159536 (proven)

Source: examples/cec2006_g24

Interactive run: tachsin.gr/projects/genoxide/examples/cec2006-g24

cargo run --release --example cec2006_g24
//! CEC 2006 g24: the linear function −x1 − x2 of 2 variables under 2 quartic inequality
//! constraints, from the CEC 2006 special session on constrained optimization (Liang et al.,
//! 2006). The minimum is −5.50801327159536, proven, with both constraints active.
//!
//! genoxide's `G24` gives the value of a solution and its constraint violation, which Deb's
//! feasibility rules compare: a feasible solution beats an infeasible one. CMA-ES, restarted with
//! a growing population (IPOP) when it converges, searches the 2 variables within the report's
//! budget of 500,000 evaluations, and stops once the error f(x) − f* is at most 1e-8. The example
//! prints the best solution and its constraints.
//!
//! With `GENOXIDE_TRACE=<file>`, it also writes a trace of its run for the plot on the example's
//! page, with `trace.rs`.
//!
//! ```text
//! cargo run --release --example cec2006_g24
//! ```

mod trace;

use genoxide::prelude::*;
use genoxide::problems::Problem;
use genoxide::problems::cec2006::G24;

// the CEC 2006 report's budget of evaluations per run
const BUDGET: u64 = 500_000;
// the run stops once its best is feasible with an error f(x) - f* at most this
const ERROR: f64 = 1e-8;
// the report counts a run as successful once its error is at most this
const SUCCESS: f64 = 1e-4;
// a constraint within this of its boundary is active
const ACTIVE: f64 = 1e-6;

fn main() -> Result<()> {
    let problem = G24;
    let optimum = problem.optimum().expect("known");
    let f_star = optimum.value();
    let cmaes = Cmaes::builder(problem.representation())
        .restarts(cmaes::Restarts::Ipop)
        .minimize()
        .seed(1)
        .build()?;
    // with GENOXIDE_TRACE=<file>, a trace of the run for the plot on the example's page
    let mut trace = trace::Trace::from_env();
    // the evaluations when the best is first feasible, and when its error first meets the
    // report's criterion of success
    let (mut feasible, mut success) = (None, None);
    // the population sizes of the runs: IPOP doubles it at each restart
    let mut sizes: Vec<usize> = Vec::new();
    let outcome = Engine::new(cmaes, problem)
        .stop_when(Stop::target(f_star + ERROR).or(Stop::evaluations(BUDGET)))
        .on_generation(|snapshot| {
            let progress = snapshot.progress();
            let best = progress.best().filter(|best| best.is_feasible());
            let error = best.and_then(Fitness::score).map(|value| value - f_star);
            if error.is_some() {
                feasible.get_or_insert(progress.evaluations());
            }
            if error.is_some_and(|error| error <= SUCCESS) {
                success.get_or_insert(progress.evaluations());
            }
            let size = snapshot.population().len();
            if sizes.last() != Some(&size) {
                sizes.push(size);
            }
            trace.record(snapshot);
        })
        .run()?;

    let best = outcome.best_fitness();
    let value = best.score().expect("valid");
    let x = outcome.best_genome();
    println!("CMA-ES with IPOP restarts and Deb's feasibility rules on g24, seed 1");
    let (stop, error) = if outcome.stop_reason() == StopReason::Target {
        ("stopped by the target", format!("< {ERROR:.0e}"))
    } else {
        ("stopped", format!("{:.1e}", value - f_star))
    };
    let evaluations = outcome.evaluations();
    let feasibility = if best.is_feasible() {
        "feasible"
    } else {
        "infeasible"
    };
    println!("{stop} after {evaluations} evaluations: f(x) - f* {error}, {feasibility}");
    println!(
        "first feasible after {} evaluations, f(x) - f* <= 1e-4 after {}",
        count(feasible),
        count(success)
    );
    println!("{}", restarts(&sizes));
    println!(
        "f(x) {}, f* {} ({})",
        significant(value, 6),
        significant(f_star, 6),
        if optimum.is_proven() {
            "proven"
        } else {
            "best known"
        }
    );
    let genes: Vec<String> = (1..)
        .zip(&x[..])
        .map(|(i, xi)| format!("x{i} {}", significant(*xi, 6)))
        .collect();
    println!("{}", genes.join(", "));
    let constraints: Vec<String> = (1..)
        .zip(problem.constraints(x).inequalities())
        .map(|(i, &g)| format!("g{i} {}", state(g)))
        .collect();
    println!("{}", constraints.join(", "));
    trace.write();
    Ok(())
}

// a constraint g(x) <= 0: "active" on its boundary, else its value
fn state(g: f64) -> String {
    if g.abs() <= ACTIVE {
        "active".to_string()
    } else {
        significant(g, 4)
    }
}

// the restarts, from the population sizes of the runs
fn restarts(sizes: &[usize]) -> String {
    let sizes: Vec<String> = sizes.iter().map(usize::to_string).collect();
    match sizes.len() {
        1 => format!("no restart, a population of {}", sizes[0]),
        2 => format!("1 restart, populations {}", sizes.join(", ")),
        n => format!("{} restarts, populations {}", n - 1, sizes.join(", ")),
    }
}

// the evaluations, or "never"
fn count(evaluations: Option<u64>) -> String {
    evaluations.map_or("never".to_string(), |evaluations| evaluations.to_string())
}

// `digits` significant digits, e.g. 29.9953 or -30665.5 for 6
fn significant(value: f64, digits: i32) -> String {
    let magnitude = value.abs().log10().floor() as i32;
    let decimals = (digits - 1 - magnitude).max(0) as usize;
    format!("{value:.decimals$}")
}
python examples/cec2006_g24/main.py
"""CEC 2006 g24: the linear function −x1 − x2 of 2 variables under 2 quartic inequality
constraints, from the CEC 2006 special session on constrained optimization (Liang et al., 2006).
The minimum is −5.50801327159536, proven, with both constraints active.

genoxide's ``G24`` gives the value of a solution and its constraint violation, which Deb's
feasibility rules compare: a feasible solution beats an infeasible one. CMA-ES, restarted with a
growing population (IPOP) when it converges, searches the 2 variables within the report's budget
of 500,000 evaluations, and stops once the error f(x) − f* is at most 1e-8. The example prints the
best solution and its constraints. ``run`` evaluates the problem in Rust.

With ``GENOXIDE_TRACE=<file>``, it also writes a trace of its run for the plot on the example's
page, with trace.py.

    python examples/cec2006_g24/main.py
"""

import math

import genoxide as gx
import numpy as np

from trace import Trace

# the CEC 2006 report's budget of evaluations per run
BUDGET = 500_000
# the run stops once its best is feasible with an error f(x) - f* at most this
ERROR = 1e-8
# the report counts a run as successful once its error is at most this
SUCCESS = 1e-4
# a constraint within this of its boundary is active
ACTIVE = 1e-6


def significant(value, digits):
    """``digits`` significant digits, e.g. 29.9953 or -30665.5 for 6."""
    magnitude = math.floor(math.log10(abs(value)))
    return f"{value:.{max(digits - 1 - magnitude, 0)}f}"


def state(g):
    """A constraint g(x) <= 0: "active" on its boundary, else its value."""
    return "active" if abs(g) <= ACTIVE else significant(g, 4)


def scientific(value, decimals):
    """Scientific notation as Rust writes it, e.g. 1.2e-5 for 1 decimal."""
    mantissa, exponent = f"{value:.{decimals}e}".split("e")
    return f"{mantissa}e{int(exponent)}"


def count(evaluations):
    """The evaluations, or "never"."""
    return "never" if evaluations is None else str(evaluations)


def restarts(sizes):
    """The restarts, from the population sizes of the runs."""
    joined = ", ".join(map(str, sizes))
    if len(sizes) == 1:
        return f"no restart, a population of {sizes[0]}"
    if len(sizes) == 2:
        return f"1 restart, populations {joined}"
    return f"{len(sizes) - 1} restarts, populations {joined}"


problem = gx.problems.cec2006.G24()
optimum = problem.optimum
f_star = optimum.value
cmaes = gx.Cmaes(problem.genome, objective=problem.objective, restarts="ipop", seed=1)
# with GENOXIDE_TRACE=<file>, a trace of the run for the plot on the example's page
trace = Trace(problem)
# the evaluations when the best is first feasible, and when its error first meets the report's
# criterion of success
first = {"feasible": None, "success": None}
# the population sizes of the runs: IPOP doubles it at each restart
sizes = []


def on_generation(progress):
    _, violations = problem.evaluate(progress.best_genome[np.newaxis])
    if violations[0] == 0.0:
        error = progress.best_fitness - f_star
        if first["feasible"] is None:
            first["feasible"] = progress.evaluations
        if first["success"] is None and error <= SUCCESS:
            first["success"] = progress.evaluations
    size = len(progress.population)
    if not sizes or sizes[-1] != size:
        sizes.append(size)
    trace.record(progress)


result = cmaes.run(
    problem, target=f_star + ERROR, evaluations=BUDGET, on_generation=on_generation
)

value = result.best_fitness
print("CMA-ES with IPOP restarts and Deb's feasibility rules on g24, seed 1")
if result.stop_reason == "target":
    stop, error = "stopped by the target", f"< {scientific(ERROR, 0)}"
else:
    stop, error = "stopped", scientific(value - f_star, 1)
feasibility = "feasible" if result.violation == 0.0 else "infeasible"
print(f"{stop} after {result.evaluations} evaluations: f(x) - f* {error}, {feasibility}")
print(
    f"first feasible after {count(first['feasible'])} evaluations, f(x) - f* <= 1e-4 after "
    f"{count(first['success'])}"
)
print(restarts(sizes))
proven = "proven" if optimum.proven else "best known"
print(f"f(x) {significant(value, 6)}, f* {significant(f_star, 6)} ({proven})")
x = result.best_genome.tolist()
print(", ".join(f"x{i} {significant(xi, 6)}" for i, xi in enumerate(x, 1)))
constraints = problem.constraints(result.best_genome).tolist()
print(", ".join(f"g{i} {state(g)}" for i, g in enumerate(constraints, 1)))
trace.write()

What it prints, from a seeded run:

CMA-ES with IPOP restarts and Deb's feasibility rules on g24, seed 1
stopped by the target after 954 evaluations: f(x) - f* < 1e-8, feasible
first feasible after 6 evaluations, f(x) - f* <= 1e-4 after 522
no restart, a population of 6
f(x) -5.50801, f* -5.50801 (proven)
x1 2.32952, x2 3.17849
g1 active, g2 active