Up- and downgrading the euclidean 1-median problem and knapsack Voronoi diagrams

Abstract

We consider the 1-median problem with euclidean distances with uncertainty in the weights, expressed as possible changes within given bounds and a single budget constraint on the total cost of change. The upgrading (resp. downgrading) problem consists of minimizing (resp. maximizing) the optimal 1-median objective value over these weight changes. The upgrading problem is shown to belong to the family of continuous single facility location-allocation problems, whereas the downgrading problem reduces to a convex but highly non-differentiable optimization problem. Several structural properties of the optimal solution are proven for both problems, using novel planar partitions, the knapsack Voronoi diagrams, and lead to polynomial time solution algorithms.

Appendix 1: Unique optimal allocation at optimal upgrading solution

Consider the reformulation (5) of the upgrading Fermat–Weber problem

$$\begin{aligned} \min _{z\in Z^*} \min _{x\in \mathbb R^2} \sum _{a\in A} z_a\Vert x-a\Vert \end{aligned}$$

(20)

where $Z^*$ is the set of optimal allocations corresponding to cells of the (min) Knapsack Voronoi diagram developed in Sect. 2.4. Defining then the function $F= \min _{z\in Z^*} F_z$, where $F_{z} \ : \mathbb R^2 \rightarrow R\ : x \mapsto \sum _{a\in A} z_a\Vert x-a\Vert$ for each $z\in Z^*$ another inversion of minimizations leads to the reformulation

$$\begin{aligned} \min _{x\in \mathbb R^2} F(x) \end{aligned}$$

(21)

Let also for any $y\in \mathbb R^2$ the active set be $Z^*(y)= \left. \left\{ \ z\in Z^*\ \right\bracevert F_z(y)=F(y)\ \right\}$, which is of course never empty. The following easy lemma was proven in Plastria and Elosmani (2008).

Lemma 5

$y$ is a local minimum of $F$ iff $y$ is a local minimum of $F_z$ for each $z\in Z^*(y)$

Theorem 6

At any local minimum of $F$ not in $A$ there is exactly one active $z\in Z^*$.

Proof

Consider any point $y\notin A$ at which two different optimal allocations in $Z^*$ exist. At any interior point of a cell of the MinKVD there is only a single optimal allocation, so $y$ cannot be of this type and must lie on some MinKVD edge (possibly a vertex) separating two (neighbouring) MinKVD cells. Hence there is a weighted bisector $B_{de}$ of two different points $d\ne e\in A$ containing this edge, and we have $\Vert y-d\Vert /c_d=\Vert y-e\Vert /c_e$. The optimal allocations $z\ne z'\in Z^*(y)$ corresponding to the two cells separated by this edge must differ in their $d$ and $e$ components only. Let $A'$ denote the set of points of $A$ strictly weighted closer to $y$ than both $d$ and $e$; it follows that $z_a=z'_a=u_a$ for all $a\in A'$, and $z_b=z'_b=l_b$ for all $b\in A\setminus (A'\cup \{\ d,e\ \})$. Assuming wlog that $c_d(u_d-l_d)\le c_e(u_e-l_e)$, three situations may arise, similar to Sect. 3.2.3:

• (1) $z_d=u_d$ and $z_e<u_e$, while $z'_e=u_e$ and $z'_d<u_d$

• (2) $z_d=u_d$ and $l_e<z_e\le u_e$, while $z'_d=l_d$ and $l_e<z'_e\le u_e$

• (3) $z_d=l_d$ and $l_e<z_e\le u_e$, while $z'_e=l_e$ and $l_d<z'_d\le u_d$

In any case the functions $F_z$ and $F_{z'}$ differ only in their two terms involving $d$ and $e$, so we have in each of these respective cases

• (1) $(F_z-F_{z'})(x)=(u_d-z'_d)\Vert x-d\Vert - (u_e-z_e)\Vert x-e\Vert$

• (2) $(F_z-F_{z'})(x)=(u_d-l_d)\Vert x-d\Vert - (z'_e-z_e)\Vert x-e\Vert$

• (3) $(F_z-F_{z'})(x)=(l_d-z'_d)\Vert x-d\Vert - (l_e-z_e)\Vert x-e\Vert$

But it also holds that $\sum _{a\in A}c_az_a= B= \sum _{a\in A}c_az'_a$ from which we respectively obtain

• (1) $c_d(u_d-z'_d)= c_e(u_e-z_e)\ne 0$

• (2) $c_d(u_d-l_d) = c_e(z'_e-z_e)\ne 0$

• (3) $c_d(l_d-z'_d)=c_e(l_e-z_e)\ne 0$

and hence in all cases that

$$\begin{aligned} (F_z-F_{z'})(x)= K\left( \frac{\Vert x-d\Vert }{c_d}-\frac{\Vert x-e\Vert }{c_e}\right) \end{aligned}$$

for some $K\ne 0$. Note that this confirms that $F_z(y)=F_{z'}(y)$ since $y\in B_{de}$.

Due to the assumption that $y\notin A$ both $F_z$ and $F_{z'}$ are differentiable at $y$, and so also $F_z-F_{z'}$, the gradient of which at $y$ is

$$\begin{aligned} \nabla (F_z-F_{z'})(y)= K\left( \frac{d-y}{c_d\Vert y-d\Vert }-\frac{e-y}{c_e\Vert y-e\Vert }\right) \end{aligned}$$

In case $c_d\ne c_e$ we have

$$\begin{aligned} \left\| \frac{d-y}{c_d\Vert y-d\Vert } \right\| = \frac{1}{c_d} \ne \frac{1}{c_e} = \left\| \frac{e-y}{c_e\Vert y-e\Vert } \right\| \end{aligned}$$

so $\nabla (F_z-F_{z'})(y)\ne 0$.

But if $c_d=c_e$, then by $y\in B_{de}$ we have $\Vert d-y\Vert =\Vert e-y\Vert \ne 0$ and $\nabla (F_z-F_{z'})(y)=\frac{d-e}{\Vert d-y\Vert c_d}\ne 0$.

It follows that in any case $\nabla F_z(y)\ne \nabla F_{z'}(y)$, so that these gradients cannot vanish both at $y$. This shows that $y$ cannot be simultaneously a local minimum of $F_z$ and $F_{z'}$, which, by lemma 5 implies that $y$ is not a local minimum of $F$. $\square$