# optimization function: sum of root squares of sum of two quadratic

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Hello

I am a graduate student in computer science, making a thesis on uncertainty geometry. During my thesis I came across the following optimization problem, which I was not able solve: Given k Independent variables {$q_i$}, and 2n linear values that depend on theme. I need to solve the following optimization problem (finding its maximum and minimum values):

Target: $\sum \limits_{e=1}^{n-1} \sqrt{(n_e.x+\sum \limits_{i=1}^{k}a_{ei} q_i)^2+(n_e.y+\sum \limits_{i=1}^{k}b_{ei} q_i)^2}$

Subject to constrains: $\forall k \geq i \geq 1 \hspace {10 mm} c_{i}^{max} \geq q_{i} \geq c_{i}^{min}$

Each Independent variable $q_i$ has a symmetric domain around zero. with its unique min $c_{i}^{min}$ and max $c_{i}^{max}$ values, such that $-|c_{i}^{max}| = - c_{i}^{max} = c_{i}^{min}$

Explanation: This is a geometric model of k independent variables and n points in the plane. The location of each point is linearly depending on the k independent variables. That is the location of a point v plane determined by his two dimensions values, v.x and v.y each of which is a linear sum of k independent variables, their coefficients of linear equation, and a unique fixed point. Example:

v.x = $n_v.x$ + $\sum \limits_{i=1}^{k} a_{vi}q_i$

v.y = $n_v.y$ + $\sum \limits_{i=1}^{k} b_{vi}q_i$

When all the k independent variables receive a zero value, the point is in location ($n_v.x, n_v.y$).

Given n points in the plane, we define the graph tree T, where each of his n vertices represents a different point in the plane. And the weight of each arc is equal to the distance between the two points that his vertices represent. For example if v and w are two point in the plane, that there vertices in T as an arc between theme, thene the arc weight is equal to the distance between them. And d (v, w) is equal to:

v.x = $n_v.x$ + $\sum \limits_{i=1}^{k} a_{vi}q_i$

v.y = $n_v.y$ + $\sum \limits_{i=1}^{k} b_{vi}q_i$

w.x = $n_v.x$ + $\sum \limits_{i=1}^{k} a_{wi}q_i$

w.y = $n_v.y$ + $\sum \limits_{i=1}^{k} b_{wi}q_i$

$n_e.x = n_w.x - n_v.x$

$n_e.y = n_w.y - n_v.y$

$\forall 1≤i≤k \hspace {10 mm} a_{ei} = a_{wi} - a_{vi}$

$\forall 1≤i≤k \hspace {10 mm} b_{ei} = b_{wi} - b_{vi}$

d(w,v) = $\sqrt{(w.x-v.x)^2+(w.y-v.y)^2}$ = $\sqrt{(e.x)^2+(e.y)^2}$ = $\sqrt{(n_e.x+\sum \limits_{i=1}^{k}a_{ei} q_i)^2+(n_e.y+\sum \limits_{i=1}^{k}b_{ei} q_i)^2}$

And total weight of the entire tree is equal to:

Target: $\sum \limits_{e=1}^{n-1} \sqrt{(n_e.x+\sum \limits_{i=1}^{k}a_{ei} q_i)^2+(n_e.y+\sum \limits_{i=1}^{k}b_{ei} q_i)^2}$

Subject to constrains: $\forall 1≤i≤k \hspace {10 mm} c_{i}^{max}≤q_{i}≤c_{i}^{min}$

When $c_{i}^{min}$ and $c_{i}^{max}$ are the constraints on the variable $q_i$. So this is an optimization problem with constrains box.

I found a simple solution to the problem, when instead of k independent variables {$q_i$}, there are n pairs of independent variables ($v.x, v.y$), with constrains box.

target: $\sum \limits_{e=1}^{n-1} \sqrt{(e.x)^2+(e.y)^2}$ = $\sum \limits_{e=1}^{n-1} \sqrt{(w.x-v.x)^2+(w.y-v.y)^2}$

Subject to constrains: $\forall 1≤i≤n-1\hspace {10 mm} Ae_i≤b$

The following article explains the solution: A Newton Barrier method for Minimizing a Sum of Euclidean Norms subject to linear equality constraints But I could not find a solution for my more difficult problem, as it presented above. Solutions and help from others are welcome.

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