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2.4 Computational Models

3.1.2 The Radii

Radius Associated with a Point. We extend the original definition of a radius associated with a point, given in Section 2.3, to powers of metric spaces. More precisely, for each point xi ∈ X, we define the value ri to be the radius of the ball with center xi that satisfies

X

x∈X∩B(xi,ri)

r`i − D(xi, x)`= f . (3.1)

Observe that there still exists only one solution to the radius ri since the left hand side of Equation (3.1) is continuous and strictly monotonically increasing with ri. For any i ∈ {1, . . . , n}, we have (f /n)1/` ≤ ri ≤ f1/`.

In case of uniform opening cost f = 1 and a metric exponent ` = 1, Bădoiu et al. [14]

discovered a useful relation between the value weight(B(xi, ri)) and the radius ri. Their result can be generalized to any uniform opening cost f ≥ 0 and any metric exponent

` ≥ 1. We obtain the following lemma:

Lemma 3.1.1. For each xi ∈ X, we have

weight (B(xi, ri)) ≥ f ri` .

3.1 Preliminaries 33

Proof. Due to the definition of ri, we have

X

x∈X∩B(xi,ri)

(ri`− D(xi, x)`) = f ,

which implies

X

x∈X∩B(xi,ri)

ri` ≥ f .

Since weight(B(xi, ri)) = |{x ∈ X | x ∈ B(xi, ri)}|, we obtain weight(B(xi, ri)) ≥ f /r`i. Sum of the Radii. Bădoiu et al. [14] proved that the sum of the radii associated with the points in X is a good approximation of the optimal facility location cost for X. Again, their result can be generalized to any uniform opening cost f ≥ 0 and any metric exponent

` ≥ 1.

In the proof of the generalized result, we use a modified version of the Mettu-Plaxton algorithm. More precisely, this version works exactly as Algorithm 2.3.1 except that, in the first step, it computes, for each point xi ∈ X, the radius ri that satisfies Equation (3.1), instead of the original radius proposed by Mettu and Plaxton [87]. We will first show that this modified Mettu-Plaxton algorithm is still a constant-factor approximation. Based on this result, we will then prove that the sum of the exponentiated radii approximates the optimal cost FacLoc*(X, f, `) within a constant factor.

Let FMPbe the set of open facilities computed by the modified Mettu-Plaxton algorithm.

In the following, we will show that FacLoc(X, FMP, f, `) ≤ 3` · FacLoc*(X, f, `). The argumentation is basically the same as in [87]. Only a few minor adaptations to our scenario have been made.

Claim 3.1.2. For any point xi ∈ X, there exists an open facility xj ∈ FMP such that rj ≤ ri and D(xi, xj) ≤ 2 · ri.

Proof. If there is no such open facility xj with rj ≤ ri in B(xi, 2 · ri), then we open a facility at xi and xi belongs to FMP.

Claim 3.1.3. Let xi and xj be distinct open facilities in FMP. Then, we have D(xi, xj) >

2 · max{ri, rj}.

Proof. Without loss of generality, we assume that rj ≤ ri. It follows that xj ∈ B(x/ i, 2 · ri).

Otherwise, the point xi would not be an open facility. Thus, we have D(xi, xj) > 2 · ri ≥ 2 · rj .

For any point xj ∈ X and an arbitrary set of open facilities F0 ⊆ X, let charge(xj, F0) := D(xj, F0)`+ X

xi∈F0

max{0, r`i − D(xi, xj)`} .

Claim 3.1.4. For an arbitrary set of open facilities F0 ⊆ X, we have

X

xj∈X

charge(xj, F0) = FacLoc(X, F0, f, `) . Proof. Due to the definition of charge(·, ·) and Equation (3.1), we get

X

xj∈X

charge(xj, F0)

= X

xj∈X

D(xj, F0)`+ X

xj∈X

X

xi∈F0

max{0, ri`− D(xi, xj)`}

= X

xj∈X

D(xj, F0)`+ X

xi∈F0

X

xj∈X∩B(xi,ri)

(ri`− D(xi, xj)`)

= X

xj∈X

D(xj, F0)`+ X

xi∈F0

f

= FacLoc(X, F0, f, `) .

Claim 3.1.5. Let xj ∈ X be any point, let F0 ⊆ X be an arbitrary set of open facilities, and let xi ∈ F0 be any open facility. If we have D(xj, xi) = D(xj, F0), then charge(xj, F0) ≥ max{r`i, D(xj, xi)`}.

Proof. If xj ∈ B(x/ i, ri), then

charge(xj, F0) ≥ D(xj, F0)`

= D(xj, xi)`

> ri` . Otherwise, we have

charge (xj, F0) ≥ D (xj, F0)`+ri`− D(xj, xi)`

= D (xj, xi)`+ri`− D(xj, xi)`

= ri`

≥ D(xj, xi)` .

Claim 3.1.6. Let xj ∈ X be any point, and let xi be any open facility in FMP. If xjB(xi, ri), then charge(xj, FMP) ≤ r`i.

Proof. By Claim 3.1.3, there is no open point xm ∈ FMP such that we have i 6= m and xj ∈ B(xm, rm). Since D(xj, FMP) ≤ D(xj, xi), we obtain

charge (xj, FMP) = D (xj, FMP)`+r`i − D (xj, xi)`

≤ D (xj, xi)`+r`i − D (xj, xi)`

= r`i .

3.1 Preliminaries 35

Claim 3.1.7. Let xj ∈ be any point, and let xibe any open facility in FMP. If xj ∈ B(x/ i, ri), then we have charge(xj, FMP) ≤ D(xj, xi)`.

Proof. The correctness of the claim follows immediately, unless there is an open facility xm ∈ FMP such that xj ∈ B(xm, rm). If such an open facility xm exists, then Claims 3.1.3 and 3.1.6 imply D(xi, xm) > 2 · max{ri, rm} and charge(xj, FMP) ≤ r`m. Furthermore, by triangle inequality, we obtain

D(xj, xi) ≥ D(xi, xm) − D(xj, xm)

> 2rm− rm

= rm , which proves charge(xj, FMP) ≤ rm` ≤ D(xj, xi)`.

Claim 3.1.8. For any point xj ∈ X and an arbitrary set of open facilities F0 ⊆ X, we have charge(xj, FMP) ≤ 3`· charge(xj, F0).

Proof. Let xi be some open facility in F0 such that we have D(xj, xi) = D(xj, F0). By Claim 3.1.2, there exists a facility xm ∈ FMP such that we have rm ≤ ri and D(xi, xm) ≤ 2 · ri.

If xj ∈ B(xm, rm), then we get charge(xj, FMP) ≤ rm` by Claim 3.1.6. Since Claim 3.1.5 implies charge(xj, F0) ≥ r`i, we can conclude

charge(xj, FMP) ≤ r`m

≤ r`i

≤ charge(xj, F0) . This proves the assertion in case that we have xj ∈ B(xm, rm).

If xj ∈ B(x/ m, rm), then charge(xj, FMP) ≤ D(xj, xm)` by Claim 3.1.7. Thus, by triangle inequality, we get

charge(xj, FMP) ≤ D(xj, xm)`

≤ (D(xj, xi) + D(xi, xm))`

≤ (D(xj, xi) + 2 · ri)`

≤ 3`· max{D(xj, xi)`, r`i} . Now, the assertion follows by Claim 3.1.5.

Lemma 3.1.9. FacLoc(X, FMP, f, `) ≤ 3`· FacLoc*(X, f, `) Proof. The assertion follows from Lemmas 3.1.4 and 3.1.8.

Based on the results above, we can prove the following lemma:

Lemma 3.1.10.

1

2`+1 · FacLoc*(X, f, `) ≤ X

xi∈X

ri` ≤ 6`· FacLoc*(X, f, `)

Proof. We first prove the lower bound and then the upper bound. The argumentation is basically the same as in [14]. Only a few minor adaptations to our scenario have been made.

Lower bound: Let FMP be the set of open facilities computed by the modified Mettu-Plaxton algorithm. Then, it follows from Claim 3.1.2 that

2`· X

xi∈X

r`iX

xi∈X

D(xi, FMP)` . (3.2)

Next, we show that we also have

2`· X

and the modified Mettu-Plaxton algorithm would not open a facility at xj, which is a contradiction. Hence, we obtain

which proves Inequality (3.3). Due to Inequalities (3.2) and (3.3), we get 2`+1· X

3.1 Preliminaries 37

Upper bound: Due to Lemma 3.1.9, we know that

FacLoc(X, FMP, f, `) ≤ 3`· FacLoc*(X, f, `) . Thus, to prove the upper bound, it remains to show that

X

xi∈X

ri` ≤ 2`· FacLoc(X, FMP, f, `) .

Due to Claim 3.1.4, we have

2`· FacLoc(X, FMP, f, `) = 2`· X

xi∈X

charge(xi, FMP)

≥ 2`·

X

xi∈FMP

r`i + X

xj∈X\FMP

max{rδ(j)` , D(xj, xδ(j))`}

,

where δ(j) denotes the index of the facility in FMP that is closest to xj. Thus, if we can show that

2`·

X

xi∈FMP

r`i + X

xj∈X\FMP

max{rδ(j)` , D(xj, xδ(j))`}

X

xi∈X

ri` , (3.4)

then we are done. It is sufficient to prove

rj` ≤ 2`−1 ·D(xj, xδ(j))`+ r`δ(j) (3.5) because this implies max{rδ(j)` , D(xj, xδ(j))`} ≥ rj`/2` and Inequality (3.4) follows. We prove the correctness of Inequality (3.5) by contradiction. Hence, we assume that

r`j > 2`−1 ·D(xj, xδ(j))`+ r`δ(j)) .

We can easily prove by induction that 2`−1· (a`+ b`) ≥ (a + b)` for any a, b ≥ 0. Thus, we obtain

rj` >D(xj, xδ(j)) + rδ(j))` ,

which, in turn, would imply B(xδ(j), rδ(j)) ⊆ B(xj, rj). Furthermore, by applying triangle inequality and 2`−1 · (a`+ b`) ≥ (a + b)` for an a, b ≥ 0, we get

D(xj, xm)`D(xj, xδ(j)) + D(xδ(j), xm)`

≤ 2`−1·D(xj, xδ(j))`+ D(xδ(j), xm)`

as upper bound on the exponentiated distance between xj and any point xm ∈ B(xδ(j), rδ(j)).

Now, we obtain

X

xm∈X∩B(xj,rj)

rj`− D(xj, xm)`

X

xm∈X∩B(xδ(j),rδ(j))

rj`− D(xj, xm)`

> X

xm∈X∩B(xδ(j),rδ(j))

2`−1·D(xj, xδ(j))`+ r`δ(j)− 2`−1·D(xj, xδ(j))`+ D(xδ(j), xm)`

= 2`−1 · X

xm∈X∩B(xδ(j),rδ(j))

rδ(j)` − D(xδ(j), xm)`

= 2`−1 · f

≥ f ,

which is a contradiction because the definition of rj requires

X

xm∈X∩B(xj,rj)

r`j− D(xj, xm)` = f .

It follows that Inequality (3.5) is true, which was the only thing left to prove the assertion of the lemma.