# DTLZ¶

## DTLZ1¶

The only difficulty provided by this problem is the convergence to the Pareto-optimal hyper-plane.The search space contains $$(11^k-1)$$ local Pareto-optimal fronts, where an MOEA can get attracted before reaching the global Pareto-optimal front.

Definition

$$\newcommand{\boldx}{\mathbf{x}} \begin{array} \mbox{Minimize} & f_1(\boldx) = \frac{1}{2}x_1x_2\cdots x_{M-1}(1+g(\boldx_M)), \\ \mbox{Minimize} & f_2(\boldx) = \frac{1}{2}x_1x_2\cdots (1-x_{M-1})(1+g(\boldx_M)), \\ \vdots & \vdots \\ \mbox{Minimize} & f_{M-1}(\boldx) = \frac{1}{2}x_1 (1-x_2)(1+g(\boldx_M)), \\ \mbox{Minimize} & f_M(\boldx) = \frac{1}{2}(1-x_1) (1+g(\boldx_M)), \\[2mm] \mbox{subject to} & 0 \leq x_i \leq 1, \quad \mbox{for i=1,2,\ldots,n}. \end{array}$$

The last $$k=(n-M+1)$$ variables are represented as $$\boldx_M$$. The functional $$g(\boldx_M)$$ requires $$|\boldx_M|=k$$ variables and must take any function with $$g\geq 0$$.

We suggest the following: $$g(\boldx_M) = 100\left[|\boldx_M| + \sum_{x_i\in \boldx_M} (x_i-0.5)^2 - \cos(20 \pi (x_i-0.5)) \right]$$

Optimum

The Pareto-optimal solution corresponds to $$x_i=0.5$$ (for all $$x_i\in \boldx_M$$) and the objective function values lie on the linear hyper-plane: $$\sum_{m=1}^M f_m^{\ast} = 0.5$$.

Plot

[1]:

from pymoo.factory import get_problem, get_reference_directions, get_visualization
from pymoo.util.plotting import plot

ref_dirs = get_reference_directions("das-dennis", 3, n_partitions=12)

pf = get_problem("dtlz1").pareto_front(ref_dirs)


## DTLZ2¶

This function can also be used to investigate an MOEA’s ability to scale up its performance in large number of objectives. Like in DTLZ1, for $$M>3$$, the Pareto-optimal solutions must lie inside the first octant of the unit sphere in a three-objective plot with $$f_M$$ as one of the axes. Since all Pareto-optimal solutions require to satisfy $$\sum_{m=1}^M f_m^2 = 1$$, the difference between the left term with the obtained solutions and one can be used as a metric for convergence as well. Besides the suggestions given in DTLZ1, the problem can be made more difficult by replacing each variable $$x_i$$ (for $$i=1$$ to $$(M-1)$$) with the mean value of $$p$$ variables: $$x_i = \frac{1}{p}\sum_{k=(i-1)p+1}^{ip} x_k$$.

Definition

$$\begin{array} \mbox{Min.} & f_1(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \cdots \cos (x_{M-2}\pi/2) \cos (x_{M-1}\pi/2), \\ \mbox{Min.} & f_2(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \cdots \cos (x_{M-2}\pi/2) \sin (x_{M-1}\pi/2), \\ \mbox{Min.} & f_3(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \cdots \sin (x_{M-2}\pi/2), \\ \vdots & \vdots \\ %\mbox{Min.} & f_{M-1}(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \sin (x_2\pi/2), \\ \mbox{Min.} & f_{M}(\boldx) = (1+g(\boldx_M)) \sin (x_1\pi/2), \\[2mm] \mbox{with} & g(\boldx_M) = \sum_{x_i \in \boldx_M} (x_i-0.5)^2, \\[2mm] & 0 \leq x_i \leq 1, \quad \mbox{for i=1,2,\ldots,n}. \end{array}$$

Once again, the $$\boldx$$ vector is constructed with $$k=n-M+1$$ variables.

Optimum

The Pareto-optimal solutions corresponds to $$x_i=0.5$$ for all $$x_i\in \boldx_M$$.

Plot

[2]:

pf = get_problem("dtlz2").pareto_front(ref_dirs)


## DTLZ3¶

In order to investigate an MOEA’s ability to converge to the global Pareto-optimal front, we suggest using the above problem with the $$g$$ function equal to Rastrigin. The $$g$$ function introduces $$(3^k-1)$$ local Pareto-optimal fronts and one global Pareto-optimal front. All local Pareto-optimal fronts are parallel to the global Pareto-optimal front and an MOEA can get stuck at any of these local Pareto-optimal fronts, before converging to the global Pareto-optimal front (at $$g^{\ast}=0$$).

Definition

$$\begin{array} \mbox{Min.} & f_1(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \cdots \cos (x_{M-2}\pi/2) \cos (x_{M-1}\pi/2), \\ \mbox{Min.} & f_2(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \cdots \cos (x_{M-2}\pi/2) \sin (x_{M-1}\pi/2), \\ \mbox{Min.} & f_3(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \cdots \sin (x_{M-2}\pi/2), \\ \vdots & \vdots \\ %\mbox{Minimize} & f_{M-1}(\boldx) = (1+g(\boldx_M)) \cos (x_1\pi/2) \sin (x_2\pi/2), \\ \mbox{Min.} & f_{M}(\boldx) = (1+g(\boldx_M)) \sin (x_1\pi/2), \\ \mbox{with} & g(\boldx_M) = 100\left[|\boldx_M| + \sum_{x_i\in \boldx_M} (x_i-0.5)^2 - \cos(20 \pi (x_i-0.5))\right], \\ & 0 \leq x_i \leq 1, \quad \mbox{for i=1,2,\ldots,n}. \end{array}$$

Optimum

The global Pareto-optimal front corresponds to $$x_i = 0.5$$ for $$x_i\in \boldx_M$$.

Plot

[3]:

pf = get_problem("dtlz3").pareto_front(ref_dirs)


## DTLZ4¶

In order to investigate an MOEA’s ability to maintain a good distribution of solutions, we modify problem DTLZ2 with a different parametric variable mapping:

Definition

$$\begin{array} \mbox{Min.} & f_1(\boldx) = (1+g(\boldx_M)) \cos (x^{\alpha}_1\pi/2) \cdots \cos (x^{\alpha}_{M-2}\pi/2) \cos (x^{\alpha}_{M-1}\pi/2), \\ \mbox{Min.} & f_2(\boldx) = (1+g(\boldx_M)) \cos (x^{\alpha}_1\pi/2) \cdots \cos (x^{\alpha}_{M-2}\pi/2) \sin (x^{\alpha}_{M-1}\pi/2), \\ \mbox{Min.} & f_3(\boldx) = (1+g(\boldx_M)) \cos (x^{\alpha}_1\pi/2) \cdots \sin (x^{\alpha}_{M-2}\pi/2), \\ \vdots & \vdots \\ %\mbox{Minimize} & f_{M-1}(\boldx) = (1+g(\boldx_M)) \cos (x^{\alpha}_1\pi/2) \sin (x^{\alpha}_2\pi/2), \\ \mbox{Min.} & f_{M}(\boldx) = (1+g(\boldx_M)) \sin (x^{\alpha}_1\pi/2), \\ \mbox{with} & g(\boldx_M) = \sum_{x_i \in \boldx_M} (x_i-0.5)^2, \\[2mm] & 0 \leq x_i \leq 1, \quad \mbox{for i=1,2,\ldots,n}. \end{array}$$

The parameter $$\alpha=100$$ is suggested here. This modification allows a dense set of solutions to exist near the $$f_M$$-$$f_1$$ plane.

Optimum

The global Pareto-optimal front corresponds to $$x_i = 0.5$$ for $$x_i\in \boldx_M$$.

Plot

[4]:

pf = get_problem("dtlz4").pareto_front(ref_dirs)


## DTLZ5¶

This problem will test an MOEA’s ability to converge to a curve and will also allow an easier way to visually demonstrate (just by plotting $$f_M$$ with any other objective function) the performance of an MOEA. Since there is a natural bias for solutions close to this Pareto-optimal curve, this problem may be easy for an algorithm to solve.

Definition

$$\begin{array} \mbox{Min.} & f_1(\boldx) = (1+g(\boldx_M)) \cos (\theta_1\pi/2) \cdots \cos (\theta_{M-2}\pi/2) \cos (\theta_{M-1}\pi/2), \\ \mbox{Min.} & f_2(\boldx) = (1+g(\boldx_M)) \cos (\theta_1\pi/2) \cdots \cos (\theta_{M-2}\pi/2) \sin (\theta_{M-1}\pi/2), \\ \mbox{Min.} & f_3(\boldx) = (1+g(\boldx_M)) \cos (\theta_1\pi/2) \cdots \sin (\theta_{M-2}\pi/2), \\ \vdots & \vdots \\ %\mbox{Min.} & f_{M-1}(\boldx) = (1+g(\boldx_M)) \cos (\theta_1\pi/2) \sin (\theta_2\pi/2), \\ \mbox{Min.} & f_{M}(\boldx) = (1+g(\boldx_M)) \sin (\theta_1\pi/2), \\ \mbox{with} & \theta_i = \frac{\pi}{4(1+g(\boldx_M))}\left(1+2g(\boldx_M) x_i\right), \quad \mbox{for i=2,3,\ldots,(M-1)}, \\ & g(\boldx_M) = \sum_{x_i \in \boldx_M} (x_i-0.5)^2, \\ & 0 \leq x_i \leq 1, \quad \mbox{for i=1,2,\ldots,n}. \end{array}$$

Optimum

The Pareto-optimal front corresponds to $$x_i=0.5$$ for all $$x_i\in \boldx_M$$ and function values satisfy $$\sum_{m=1}^M f_m^2 = 1$$.

Plot

[5]:

pf = get_problem("dtlz5").pareto_front()


## DTLZ6¶

The above test problem can be made harder by making a similar modification to the $$g$$ function in DTLZ5, as done in DTLZ3.

Definition

However, in DTLZ6, we use a different $$g$$ function:

$$g(\boldx_M) = \sum_{x_i\in \boldx_M} x_i^{0.1}.$$

The size of $$\boldx_M$$ vector is chosen as $$10$$ and the total number of variables is identical as in DTLZ5. The above change in the problem makes NSGA-II and SPEA2 difficult to converge to the true Pareto-optimal front as in DTLZ5.

Optimum

Here, the Pareto-optimal front corresponds to $$x_i=0$$ for all $$x_i\in \boldx_M$$.

Plot

[6]:

pf = get_problem("dtlz6").pareto_front()


## DTLZ7¶

This test problem has $$2^{M-1}$$ disconnected Pareto-optimal regions in the search space. The functional $$g$$ requires $$k=|\boldx_M|=n-M+1$$ decision variables. This problem will test an algorithm’s ability to maintain subpopulation in different Pareto-optimal regions.

Definition

$$\begin{array} \mbox{Minimize} & f_1(\boldx_1)=x_1, \\ \mbox{Minimize} & f_2(\boldx_2)=x_2, \\ \vdots & \vdots \\ \mbox{Minimize} & f_{M-1}(\boldx_{M-1})=x_{M-1}, \\ \mbox{Minimize} & f_M(\boldx) = (1+g(\boldx_M))h(f_1,f_2,\ldots,f_{M-1},g), \\ \mbox{where} & g(\boldx_M) = 1 + \frac{9}{|\boldx_M|}\sum_{x_i\in \boldx_M} x_i, \\ & h(f_1,f_2,\ldots,f_{M-1},g) = M-\sum_{i=1}^{M-1} \left[ \frac{f_i}{1+g}\left(1 + \sin (3\pi f_i)\right)\right], \\ \mbox{subject to} & 0 \leq x_i \leq 1, \quad \mbox{for i=1,2,\ldots,n}. \end{array}$$

Optimum

The Pareto-optimal solutions corresponds to $$\boldx_M=\vec{0}$$.

Plot

[7]:

pf = get_problem("dtlz7").pareto_front()