Worst-case performance analysis

In this section, we will prove two worst-case performance guarantees when ARSA is applied to two special cases of our facility location problem. To describe the two special cases, we state three assumptions below.

Assumption 1. g(·) is the kink function g^K(·) in (4.1) for some a > 0 and B > 0.

Note that under this assumption, it is without loss of generality to assume that there is no location whose s_i > ^B_a. If such a location exists, an optimal solution will either contain only this location or does not contain it. We may thus safely remove this location for a while, solve the remaining problem, and at the end check whether selecting only this location is actually the best option.

Another assumption we need for ARSA to have a worst-case performance guarantee is the following.

Assumption 2. The net stand-alone benefits s_i− h_i, i ∈ I, satisfy max{s_i− h_i} > 0.

Algorithm 1 approximation-relaxation-sorting algorithm (ARSA)

1: Find the kink function g^K(·) that approximate g(·). Replace g(·) by g^K(·) to obtain the problem (P ).

2: Split the problem to n subproblems, one with an additional constraint P

i∈Ix_i = k, k = 1, ..., n. Let the subproblems be (P₁), ..., and (P_n).

3: for k from 1 to n do

4: Relax the integer constraints in (P_k).

5: Solve the relaxation of (P_k). Let x^k_i be the value of x_i in the optimal solution.

6: Sort locations so that x^k₍₁₎ ≥ · · · x^k_(n), where ties are broken arbitrarily.

7: Construct a solution ¯x such that ¯x₍₁₎ = · · · = ¯x_(k)= 1 and 0 otherwise.

8: Construct a solution ˆx such that ˆx₍₁₎= · · · = ˆx_(k−1)= ˆx_(k+1) = 1 and 0 otherwise.

9: if z(¯x) > z(ˆx) then

10: Report ¯x as the proposed solution for (P_k).

11: else

12: Report ˆx as the proposed solution for (Pk).

13: end if

14: end for

15: Report the best solution among the proposed solutions for the n subproblems.

This assumption says that the most positive net stand-alone benefit is greater than zero. In other words, there must be some facility which can bring profit itself. With this assumption, we know that the maximum benefit of building only one facility is positive.

Under Assumption 2, we define a ratio that directly affects our worst-case performance guarantee.

Definition 2. For an instance of our facility location problem in (3.1), we define its critical ratio as

r = −min_i∈I{s_i− h_i} max_i∈I{s_i− h_i}.

Note that r is readily available when an instance of our facility location problem is given. Also note that according to our definition, r is negative if s_i− h_i > 0 for all i ∈ I or positive if there is at least one location whose net stand-alone benefit s_i − h_i < 0.

Finally, we have r ≥ −1. r = −1 if and only if s_i− h_i are identical for all i ∈ I.

Finally, we need all network benefits to be equal. One example in which the assump-tion may be valid is to build car/bike sharing staassump-tions where a drive/ride between any two locations are equally possible. The next assumption formalize this requirement.

Assumption 3. There is a constant t ≥ 0 such that the network benefits t_ij, [i, j] ∈ E, satisfy

t_ij = t ∀[i, j] ∈ E.

Before proving that ARSA has a worst-case performance guarantee under our three assumptions, we first prove Lemma 1. It states that, under Assumption 3, an optimal solution to the relaxation of subproblem (P_k) has either zero or two x^k_is that are fractional.

Lemma 1. Suppose that Assumption 3 is satisfied. An optimal solution to the relax-ation of subproblem (P_k) has either zero or two fractional components. If there are two fractional components, we have

Proof. First, note that Assumption 3 and the additional constraint P

i∈Ix_i = k together make the relaxation of subproblem (P_k) become

maxxi,p p −X

where the variable yij is not needed anymore, and the total network benefit is exactly (^k(k−1)₂ )t regardless of which k locations are selected. We then follow the idea of Caprara et al. (2000) to prove this result. For our linear program with n + 1 variables, at least n + 1 constraints are binding at an optimal extreme point solution. If at least three xis are fractional, we will have at most n − 3 constraints binding in the set of constraints x_i ∈ [0, 1]. The maximum number of binding constraints is thus n, which is not enough.

If exactly two xis are fractional, the first three constraints must all be binding, and the right-hand-side values of the first two constraints must be identical. Because P

i∈Ix_i must be an integer, there is no solution with only one fractional x_i. Finally, no fractional variable is also a possible outcome.

With Lemma 1, we are now ready to prove that ARSA is a _2+r¹ -approximation algo-rithm for our facility location problem when our three assumptions hold.

Proposition 2. Suppose that Assumptions 1, 2, and 3 hold. For the problem defined in (3.1), let z^∗ and z⁰ be the objective values of an optimal solution and the solution reported by ARSA, respectively. We then have _z^z∗⁰ ≥ _2+r¹ .

Proof. Let z_k^∗ denotes the objective value of an optimal solution to subproblem (P_k) (with constraint P

i∈Ixi = k) and z_k^LP denote that to its relaxation. Note that

k=1,...,nmax {z_k^LP} ≥ max

k=1,...,n{z_k^∗} = z^∗. (4.2)

Our plan is to prove that, for each subproblem (P_k), at least one of ARSA’s n solutions will be at least one-third as good as an optimal solution to the relaxation. More precisely, let x^{LP −k} and x^ARSA−k be an optimal solution to the relaxation of (Pk) and the solution reported by ARSA, we will show that

max Combining (4.2) and (4.3), we will be able to complete the proof.

We now prove (4.3). From lemma 1, we know x^{LP −k} has either two or no fractional value. If it has no fractional value, ARSA obviously selects the optimal solution to make x^ARSA−k = x^{LP −k} and achieves z_k^LP. Now suppose that there are two fractional variables, say, x^{LP −k}₁ and x^{LP −k}₂ . Without loss of generality, let x^{LP −k}₁ = c = 1 − x^{LP −k}₂ for some c ∈ (0, 1) and s1 ≥ s2. Let L0 be the set of k − 1 locations with x^{LP −k}_i = 1. For the relaxation, we know

z_k^LP =X

(s_i− h_i) + c(s₁− h₁) + (1 − c)(s₂− h₂).

ARSA will select the k − 1 locations in L₀ and either location 1 or location 2 to form a

i.e., the total benefit obtained in the solution L₂ does not exceed B. Therefore, we have

z(x^ARSA−k) = X

2+r > ¹₂. In this case, ARSA is guaranteed to perform pretty well. If there is at least one location such that si− hi < 0, we have r > 0. If that negative si− hi is far below 0, our r will be large and the guarantee would be small. Finally, note that r = −1 if and only if all s_i− h_i are identical. In this case, indeed ARSA will obtain the optimal solution (and that is why the performance guarantee is 1).

We next prove that ARSA is a ¹₂-approximation algorithm for our facility location problem in another special case. We first state another assumption, which is a stronger one than Assumption 3.

Assumption 4. The network benefits t_ij, [i, j] ∈ E, satisfy

tij = 0 ∀[i, j] ∈ E.

We now start proving ARSA has performance greater than max{¹₂,_2+r¹ } under Assumptions 1 and 4. Note that now Assumption 2 is not needed.

Proposition 3. Suppose that Assumptions 1 and 4 hold. For the problem defined in (3.1), let z^∗ and z⁰ be the objective values of an optimal solution and the solution reported by ARSA, respectively. We then have _z^z∗⁰ ≥ max{¹₂,_2+r¹ }.

Proof. Due to the similarity between this proof and the previous one, we only describe the different part. Let F_N = {i : s_i− h_i < 0} while N is the abbreviation of “negative.”

Since t = 0, the optimal solution of the integer problem does not contain any facility in FN.

Suppose the optimal solution contains k facilities. For the subproblem (P_k), we know that z_k^LP > z^∗. And the optimal solution x^k also contains no facility in F_N.

We now have a new bound of the difference between z_k^LP − z(x^ARSA−k), which is

z_k^LP − z(x^ARSA−k) = c((s₁− h₁) − (s₂− h₂))

≤ (s₁− h₁) + max{0, r} max

i∈I {x_i− h_i}

≤ (1 + max{0, r}) max{x_i− h_i},

where the first inequality comes from the fact that s₂− h₂ ≥ 0; otherwise, it will not be set to be a positive value when solving the relaxation of (P_k). It then follows that

z(x^ARSA−k) + (1 + max{0, r})z(x^ARSA−1) ≥ z_k^LP

and thus max{z(x^ARSA−k), z(x^ARSA−1)} ≥ max{¹₂,_2+r¹ }z_k^LP. This completes the proof.