Nonlinear Opt.: Auxiliary functions 1

Inhaltsverzeichnis

1 1. Nonlinear Opt.: Auxiliary functions
2 1.1 Theory
3 1.1.1 Example
4 1.2 Balance of dominant function
5 1.3 Balance of Extraversion and Introversion
6 2. Example
7 3. Sources

1. Nonlinear Opt.: Auxiliary functions

1.1 Theory

If someone would use the dominant function the whole time, he would be one-sided, always taking in information or always coming to decisions. Because of that, there is a secondary function the so called auxiliary function. The auxiliary function develops after the dominant function.You have to critically understand that the basis for good type progress is a well-developed auxiliary function that supports the dominant function.

1.1.1 Example

To set this in an example: The auxiliary function can be thought of as the first sailor on the ship whereas the captain is the dominant function. During the youth and puberty, persons come to develop skills in and rely on their dominant and auxiliary functions. They give less attention to the opposite functions.

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1.2 Balance of dominant function

Everyone needs to be able to take in new information and everyone needs to be able to come to decisions about that information. The auxiliary function helps to ensure you doing both of them. On the one hand if someone is always gathering information, he would be blown around like a small boat with an oversized sail and a small keel—driven by any change in wind direction. Such a person would be constantly drawn by new cognitions, but would have difficulty making decisions or coming to conclusions. The auxiliary function brings the person’s focus to decisions. On the other hand, if a person is only making decisions, he would be like a boat with a very large keel and a small sail—very sure and solid, but not open to new wind direction. Such a person would be sure of his decisions, but would be unable to take in new information to change his behavior as conditions changed. The auxiliary function brings the person’s focus to information. A sample for dominant Intuitive models are have Thinking or Feeling as their auxiliary function. So if they prefer Feeling mostly we would find that the Feeling function developed after intuition next in their lives. Therefore they would still concentrate to have the most weight to their Intuitive perceptions, though then they would make use of Feeling to reason and to make decisions about the intuitive information they took in.

1.3 Balance of Extraversion and Introversion

Everyone is used to pay attention to the outer world and lead into action so everyone needs the chance to pause for reflection and to pay back attention to their inner world. Extraverts must turn to their inner world at times and Introverts have to turn to the outer world at times. The auxiliary function solve this balancing act. As an Extravert you would use your dominant function in the outside world. For getting the balance you would use your auxiliary function in the inner world. The outer world must earn more attention to you, but your auxiliary step in if you need to be involved in your inner world. An Extravert would never stop to reflect without using the auxiliary. And as an Introvert, you use your dominant function in your inner world. For balance, you use your auxiliary function in the outer world. The inner world is of more interest to you, but your auxiliary is there when you need to be involved in the outer world. An Introvert might never move into action without using the auxiliary. The auxiliary function provides needed Introversion for Extraverts, and needed Extraversion for Introverts. The dominant and auxiliary are the two middle letters of your four-letter type. They are sometimes called your function pair.

2. Example

Minimize ${x_{1}}^{2}+2{x_{2}}^{2}$ , subject to Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): -x_{1}-x_{2}+1 \leq 0, x\in \mathbb{R}^{2}

For this problem the Lagrangian is given by Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): L(x,u)= {x_{1}}^{2}+2{x_{2}}^{2}+ u(-x_{1}-x_{2}+1) . The KKT conditions yield: Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): \frac{\partial L(x,u)}{\partial x_{1}}=2x_{1}-u=0

and Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): \frac{\partial L(x,u)}{\partial x_{2}}=4x_{2}-u=0; u(-x_{1}-x_{2}+1)=0

. Solving these results in Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): x_{1}^{*}=\frac{2}{3}; x_{2}^{*}=\frac{2}{3}; \bar{u}=\frac{4}{3}

(u=0 yields an infeasible solution).

To consider this example using penalty method, define the penalty function Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): p(x)= \left\{\begin{matrix} (-x_{1}-x_{2}+1)^{2} , if g_{i}(x)> 0\\0 , if g_{i}(x)\leq 0 \end{matrix}\right.

The unconstrained problem is then, minimize ${x_{1}}^{2}+2{x_{2}}^{2}+\mu p(x)$

If p(x) = 0, then the optimal solution is x* = (0, 0) which is infeasible.

Therefore Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): p(x)=(-x_{1}-x_{2}+1)^{2} , \theta (x,\mu )= {x_{1}}^{2}+2{x_{2}}^{2}+ \mu (-x_{1}-x_{2}+1)^{2}

the necessary conditions for the optimal solution yield the following: Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): \frac{\partial \theta (x,\mu )}{\partial x_{1}}=2x_{1}+2\mu (-x_{1}-x_{2}+1)(-1)

and Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): \frac{\partial \theta (x,\mu )}{\partial x_{2}}=2x_{1}+2\mu (-x_{1}-x_{2}+1)(-1)=0

.

$x_{\mu _{1}}={\frac {2\mu }{(1+3\mu )}}$ and $x_{\mu _{2}}={\frac {\mu }{(2+3\mu )}}$ for any fixed µ. When μ → ∞, this converges to the optimum solution of Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): x^*=(\frac{2}{3},\frac{1}{3}) .

Now we define $(u_{\mu })_{i}=2\mu (maximum\left\{0,g_{i}(x_{\mu })\right\})$ , then Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): u_{\mu }= 2\mu (1-\frac{2\mu }{(2+3\mu )}-\frac{\mu }{(2+3\mu )})=2\mu (1-\frac{3\mu }{(2+3\mu)})=\frac{4\mu }{(2+3\mu)} .

Then it is readily seen that Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): \lim_{\mu \to \infty }(\frac{4\mu }{(2+3\mu )})=\frac{4}{3}=\bar{u}