2
$\begingroup$

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

$min{\space}F(\bf{X})=[f_1(x),...,f_n(x)]$
$h_k(x)=0{\space\space\space} k=1,...,n_e$
$g_i(x)\leq0{\space\space\space} i=1,...,n$
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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Electricman 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
1
$\begingroup$

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: ...").

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Electricman 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 $\endgroup$ – Electricman 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 $\endgroup$ – Electricman Feb 10 '15 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.