As you know, a generic Multiobjective optimization problem can be stated as follows:

$h_k(x)=0{\space\space\space} k=1,...,n_e$
$g_i(x)\leq0{\space\space\space} i=1,...,n$
where $\bf{X}=[x_1, x_2, ... ,x_j]$

definitions : Objective Space is a vector space including objective functions,i.e.$[f_1(x),...,f_n(x)]$ , of the Multiobjective Optimization problem as its dimensions. It is different from solution space, which is a vector space with decision variables,i.e.$[x_1, x_2, ... ,x_j]$, of the Multiobjective Optimization problem as the dimensions.

It is obvious that no one can plot feasible solution space when number of decision variables are more than three, i.e., $j>3$. Also, It is not possible to plot feasible objective space when number of objectives are more than three, i.e., $n>3$.
I want to pull your attention to the case that we have 5 decision variables so we cannot plot the solution space, and we have three objective functions. Having three objective functions enables us to plot feasible objective space. Objective space for a MO problem including three objective functions of $f_1(.)$ , $f_2(.)$ and $f_(3)$ is shown in the figure:
enter image description here
where $\mu_1,\mu_2,\mu_3$ are three objective functions of the Multiobjective Optimization problem.

Now my question is:
How to plot feasible objective space of a Generic Multiobjective Optimization problem?
For example, imagine the problem bellow with the given constraints and tell me how can I obtain the feasible objective space similar to the one in the figure.

$f_1(X)= norm(x)^2$
$f_2(X)= 3x_1+2x_2 - x_3/3 + 0.01(x_4 - x_5)^3$ $f_3(X)= x_1^2 + 3x_2^2 + 0.2(x_3 - x_5)^3 + log(x_4^2 + x_1^2 + x_2^2 + 1)$

Subject to:
$h_1(X) = x_1 + 2x_2 - x_3 - 0.5x_4 + x_5 - 2$
$h_2(X) = 4x_1 - 2x_2 + 0.8x_3 + 0.6x_4 + 0.5x_5^2$
$g_1(X)= norm(x)^2 - 10$

Please note that, I don't expect the solution of the given problem. Please give me some applicable insights about obtaining the graphing of feasible objective space.

  • $\begingroup$ I haven't got what the objective space is. If it is a finite dimensional real vector space (as is seems from what you wrote), well then it's the $\mathbb{R}^n$… If "feasible objective space" is the set of values that the objective can attain, then it is just the image of the feasible domain under the objective function. $\endgroup$ – Dirk Feb 9 '15 at 11:18
  • $\begingroup$ That is correct, the definition of feasible objective space is what you've stated. But, the question is how can I plot such space, In my actual problem I have more than 20 constraints with 30 decision varibles and 3 objective functions. so I want to know what systematic apprach should I use to plot the feasible objective space @Dirk $\endgroup$ – SAH Feb 9 '15 at 14:23
  • $\begingroup$ I know it's not the same, as it doesn't give you exactly the solution space or a plot of the surface, but I have had success in demonstrating multi parameter design spaces with radar charts. $\endgroup$ – user7257 Apr 10 '15 at 16:16

You can't. Our mind is not equipped to visualize things in spaces with more than 3 dimensions. The best you can do is to visualize projections onto lower-dimensional spaces, or to find analoga (e.g., "the solution space is a polyhedron in 7 dimensions; if it were in 3 dimensions, it would look like this: ...").

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  • $\begingroup$ The objective space of the given problem is stil 3. meaning that a similar figure like the given one can obtained by cutting the space with the constraints. I think you have mistaken with decision varibles space with are more than 3. @wolfgang-bangerth $\endgroup$ – SAH Feb 8 '15 at 21:05
  • $\begingroup$ I'm confused. The example you show specifically has variables $x_1\ldots x_5$, i.e., your design space has 5 dimensions. You can't visualize this. $\endgroup$ – Wolfgang Bangerth Feb 9 '15 at 14:00
  • $\begingroup$ Yes, I cannot visualize the feasible design space(so-called solution space ). But as I have 3 objective functions so I can still visualize the feasible objective space, like the given figure. Please read my question again, I've updated and made it more clear today. @wolfgang-bangerth $\endgroup$ – SAH Feb 9 '15 at 14:19
  • $\begingroup$ Oh, I think I finally understand what you mean. What you're saying is this: let $\Omega\subset{\mathbb R}^j$ the feasible set with respect to your equality and inequality constraints. Then you want to plot the set $\Lambda\subset{\mathbb R}^n$ (where $n$ is the number of objective functions so that $\Lambda=\{(f_1(x),\ldots,f_j(x))^T: \; x\in\Omega\}$. In other words, $\Lambda$ is the set of possible values you could get for your objective functions. Is this right? $\endgroup$ – Wolfgang Bangerth Feb 10 '15 at 4:40
  • $\begingroup$ Yes That's exactly what I am saying. I am an engineer, so I don't use to precise mathematical definition as you mathematicians use to. Also I think it should be $\Lambda=\{(f_1(x),\ldots,f_n(x))^T: \; x\in\Omega\}$ , Because there are $n$ objective functions not $j$, right?. and about the answer, what should I do, Thanks @wolfgang-bangerth $\endgroup$ – SAH Feb 10 '15 at 7:21

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