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11institutetext: ORIE, Cornell Tech, New York, USA
11email: [email protected]
22institutetext: IEOR, Columbia University , New York, USA
22email: [email protected]
33institutetext: ORIE, Cornell University, Ithaca, USA
33email: [email protected]

On the Power of Static Assignment Policies for Robust Facility Location Problems

Omar El Housni 11    Vineet Goyal 22    David Shmoys 33
Abstract

We consider a two-stage robust facility location problem on a metric under an uncertain demand. The decision-maker needs to decide on the (integral) units of supply for each facility in the first stage to satisfy an uncertain second-stage demand, such that the sum of first stage supply cost and the worst-case cost of satisfying the second-stage demand over all scenarios is minimized. The second-stage decisions are only assignment decisions without the possibility of adding recourse supply capacity. This makes our model different from existing work on two-stage robust facility location and set covering problems. We consider an implicit model of uncertainty with an exponential number of demand scenarios specified by an upper bound kk on the number of second-stage clients. In an optimal solution, the second-stage assignment decisions depend on the scenario; surprisingly, we show that restricting to a fixed (static) fractional assignment for each potential client irrespective of the scenario gives us an O(logk/loglogk)O(\log k/\log\log k)-approximation for the problem. Moreover, the best such static assignment can be computed efficiently giving us the desired guarantee.

1 Introduction

We consider two-stage robust facility location problems under demand uncertainty where we are given a set of clients and a set of facilities in a common metric space. In the first stage, the decision-maker needs to select the (integral) units supply at each facility. The uncertain demand is then selected adversarially and needs to be satisfied by the existing supply with the minimum assignment cost in the second stage. The goal is to determine the first-stage supply such that the sum of first-stage supply cost and the worst-case assignment cost over all demand scenarios is minimized. Our problem is motivated by settings where the lead time to procure supply is large and obtaining additional units of supply in the second stage is not feasible. The uncertain second-stage demand must then be satisfied by supply units from the first stage, a common constraint in many applications. This is a departure from existing work on two-stage robust facility location, network design and, more generally, robust covering problems that have been studied extensively in the literature [6, 7, 1, 9] where the second-stage decisions allow for “adding more supply” (more specifically adding more sets/facilities to satisfy the requirement).

In this paper, we consider an implicit model of uncertainty with exponentially-many demand scenarios specified by an upper bound kk on the number of second-stage demand clients. Since the number of scenarios is exponentially-many (specified compactly), we can not efficiently solve even the LP relaxation for the problem. In contrast, if the set of second-stage scenarios are explicitly specified (for the explicit scenario model, see for instance [6, 1]), we can write a polynomially-sized LP relaxation with assignment decisions for each scenario. The main challenge then is related to obtaining an integral solution, which for the case of set covering and several network design problems can be reduced to deterministic versions (see, for instance, Dhamdhere et al. [6]).

In contrast, in an implicit model of uncertainty (with possibly exponentially-many scenarios), one of the fundamental challenges is to even approximately solve the linear relaxation of the problem efficiently. The implicit model of uncertainty with an upper bound on number of uncertain second-stage clients or elements has been studied extensively in the literature. Feige et al. [7] show under a reasonable complexity assumption that it is hard to solve the LP relaxation of a two-stage set covering problem within a factor better than Ω(logn/loglogn)\Omega(\log n/\log\log n) under this implicit model of uncertainty. They also give an O(log2n)O(\log^{2}n)-approximation for the 010-1 two-stage robust set-covering problem. Gupta et al. [9] give an improved O(logn)O(\log n)-approximation for the set-covering problem, thereby matching the deterministic approximation guarantee. El Housni and Goyal [12] show that a static policy (that is, linear in the set of second-stage elements, also referred to as an affine policy) gives an O(logn/loglogn)O(\log n/\log\log n)-approximation for the two-stage LP, thereby matching the hardness lower bound for the fractional problem. Gupta et al. [10] and Khandekar et al. [14] present approximations for several network design problems under the implicit model of uncertainty. Although there is a large body of work in this direction, as we mentioned earlier, our problem is different from set covering, since there is no possibility of adding recourse capacity; prior results do not imply an approximation for our model.

Other related work. Many variants of facility location problems have been studied extensively in the literature, including both deterministic versions as well as variants that address demand uncertainty. We refer the reader to [17] for a review of the many variants of deterministic facility location problems. Among the models that address demand uncertainty in the facility location problems, in addition to robust [2, 4], there are also stochastic [11, 13, 20, 16] and distributionally robust [15, 3, 5] models that have been studied extensively in the literature. In a stochastic model, there is a distribution on the second-stage demand scenarios and the goal is to minimize the total expected cost. A distributionally robust model can be thought of as a hybrid between stochastic and robust where the second-stage distribution is selected adversarially from a pre-specified set and the goal is to minimize the worst-case expected cost. We refer the reader to the survey [19] for an extensive review of facility location problems under uncertainty.

1.1 Our Contributions

Let us begin with a formal problem definition. We are given a set of nn facilities \cal F and mm clients 𝒞\cal C in a common metric dd where dijd_{ij} denotes the distance between ii and jj. For each facility ii\in{\cal F}, there is a cost cic_{i} per unit of supply at ii. The demand uncertainty is modeled by an implicit set of scenarios 𝒞k{\cal C}^{k} that includes all subsets of clients 𝒞\cal C of size at most kk. The decision-maker needs to select an (integral) number of units of supply xix_{i} for each facility ii\in\cal F in the first-stage. An adversary observes the first stage decisions and selects a worst-case demand scenario S𝒞kS\in{\cal C}^{k} that must be satisfied with the first-stage supply, where each client in the realized scenario needs one unit of supply. The goal is to minimize the sum of the first-stage supply cost and the worst-case assignment cost over all second-stage demand scenarios. We refer to this problem as soft-capacitated robust facility location (SCRFL). Typically in the literature, soft-capacitated refer to settings where violations of capacity upper bounds are allowed. The analogue here is that we can add any amount of supply in a facility without upper bounds (xi+)x_{i}\in\mathbb{Z_{+}}) but we pay a per unit cost of supply.

We consider a class of static assignment policies, where each of the mm clients has a static fractional assignment to facilities that is independent of the scenario, leading to a feasible second-stage solution for each demand scenario, while respecting supply capacities. Note this is a restriction, since the optimal second-stage assignment decisions are scenario-dependent in general. As a warm-up, we show that static assignment policies are optimal for the uncapacitated case with unlimited supply at each open facility (i.e., there is a cost cic_{i} to open facility ii with unlimited supply). We refer to this problem as uncapacitated robust facility location (URFL). This is based on the intuition that each client can be assigned to the closest open facilities in an optimal solution in any scenario; this leads to optimality of a static assignment policy for the LP relaxation (Theorem 1.1).

Theorem 1.1

A static assignment policy is optimal for the linear relaxation of (URFL).

The optimality of static assignment is not true in general when the supply at facilities is constrained (or equivalently, there is a cost per unit of supply). The main contribution in this paper is to show that static assignment policies give an O(logk/loglogk)O(\log k/\log\log k)-approximation for the LP relaxation of (SCRFL) (Theorem 1.2). We show this by constructing such a solution, starting from an optimal first-stage supply. The optimal static assignment policies can be computed efficiently by solving a compact LP.

Theorem 1.2

A static assignment policy gives O(logk/loglogk)O(\log k/\log\log k)-approximation for the linear relaxation of (SCRFL).

Furthermore, the fractional supply in the first stage can be rounded to an integral supply using ideas similar to rounding algorithms for the deterministic facility location [18]. In particular, the static assignment solution for the uncapacitated case can be rounded to give a 44-approximation algorithm for (URFL). The static assignment solution for the soft-capacitated case can be rounded within a constant factor, which results in an O(logk/loglogk)O(\log k/\log\log k)-approximation algorithm for (SCRFL). We would like to note that while the fractional assignment is static in our approximate LP solution, our integral assignment for any client in the second-stage depends on the other demand clients in the scenario; thereby, making our static assignment policy adaptive in implementation.

2 Warm-up: uncapacitated robust facility location

2.1 Problem formulation

In this section, we consider the uncapacitated robust facility location problem (URFL) where for each ii\in{\cal F}, there is a cost cic_{i} to open facility ii with unlimited supply. The problem can be stated as the following integer program, where each binary variable xi,ix_{i},i\in{\cal F} indicates if facility ii is opened and each yijS,i,jS,S𝒞ky_{ij}^{S},i\in{\cal F},j\in S,S\in{\cal C}^{k} indicates the assignment of client jj to facility ii in scenario SS.

min\displaystyle\min icixi+maxS𝒞kijSdijyijS\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F}}\sum_{j\in{S}}d_{ij}y_{ij}^{S} (URFL)
s.t. iyijS1,\displaystyle\sum_{i\in{\cal F}}y_{ij}^{S}\geq 1,\qquad S𝒞k,jS,\displaystyle\forall S\in{\cal C}^{k},\forall j\in{S},
xiyijS,\displaystyle x_{i}\geq y_{ij}^{S},\qquad i,S𝒞k,jS,\displaystyle\forall i\in{\cal F},\forall S\in{\cal C}^{k},\forall j\in{S},
xi{0,1},yijS0,\displaystyle x_{i}\in\{0,1\},\;y_{ij}^{S}\geq 0,\qquad i,S𝒞k,jS.\displaystyle\forall i\in{\cal F},\forall S\in{\cal C}^{k},\forall j\in{S}.

Note that the second-stage problem is a transportation problem and since the demand of a client is integral (0 or 1), the optimal solution yijSy^{S}_{ij} is integral as well. The special case of (URFL) where the uncertainty set contains only a single scenario corresponds to the NP-hard classical and well-studied uncapacitated facility location problem, which is hard to approximate within a constant better than 1.463 unless NP has an O(nO(loglogn))O(n^{O(\log\log n)})-time algorithm [8]. We let (LP-URFL) denote the linear relaxation of (URFL), where we replace xi{0,1}x_{i}\in\{0,1\} by xi0x_{i}\geq 0 for each ii\in{\cal F}. We would like to note that, in our model, the number of scenarios could be exponential in the dimension of the problem. Hence, in general, even the linear relaxation of such a problem could be challenging. However, we show that (LP-URFL) can be solved in polynomial time using a Static Assignment Policy for the second-stage variables. Moreover, we can round the fractional solution losing only a constant factor, thereby getting a constant approximation for (URFL). This section serves as a warm-up for introducing and motivating static assignment policies before addressing the class of capacitated robust facility location problems.

2.2 Static assignment policy

Consider an optimal solution of (URFL). Since each open facility can have an unlimited amount of supply, each client in the realized scenario is assigned to the closest facility among the opened ones. Therefore, a client jj is always assigned to the same open facility in all scenarios SS where jSj\in S. The same observation holds as well for (LP-URFL) where each client is assigned to the same fractionally opened facilities independent of the realized scenario. Thus, the assignment of a client is static. This can be captured by the following policy.

Static assignment policy. There exists yij0y_{ij}\geq 0 for each i,j𝒞i\in{\cal F},j\in{\cal C} such that

S𝒞k,i,jS,yijS=yij.\forall{S}\in{\cal C}^{k},\forall i\in{\cal F},\forall j\in S,\qquad\qquad y_{ij}^{S}=y_{ij}. (1)

Proof of Theorem 1.1. Let (𝒙,𝒚S,S𝒞k)(\boldsymbol{x}^{*},\boldsymbol{y}^{*S},S\in{\cal C}^{k}) be an optimal solution to (LP-URFL). Since there are no capacities on facilities, each client jj is assigned to the closest fractionally opened facilities. In particular, for each j𝒞j\in{\cal C}, let πj\pi_{j} be a permutation of ={1,,n}{\cal F}=\{1,\ldots,n\} such that dπj(1)jdπj(2)jdπj(n)jd_{\pi_{j}(1)j}\leq d_{\pi_{j}(2)j}\leq\ldots\leq d_{\pi_{j}(n)j}, and let =min{p|xπj(1)+xπj(2)++xπj(p)1}\ell=\min\{p\;|\;x^{*}_{\pi_{j}(1)}+x^{*}_{\pi_{j}(2)}+\ldots+x^{*}_{\pi_{j}(p)}\geq 1\}. Denote x^πj()=1(xπj(1)++xπj(1)).\hat{x}_{\pi_{j}(\ell)}=1-(x^{*}_{\pi_{j}(1)}+\ldots+x^{*}_{\pi_{j}({\ell-1)}}). The optimal solution can be written in the form (1) as follows: for S𝒞kS\in{\cal C}^{k} and jSj\in S; yijS=xi{y}_{ij}^{S}=x^{*}_{i} for i{πj(1),,πj(1)}i\in\{\pi_{j}(1),\ldots,\pi_{j}(\ell-1)\}, yijS=x^i{y}_{ij}^{S}=\hat{x}_{i} for i=πj()i=\pi_{j}(\ell), and yijS=0{y}_{ij}^{S}=0, otherwise. ∎

Let (Static-URFL) denote the problem after restricting the second-stage variables yijSy_{ij}^{S} in (LP-URFL) to a policy (1), which can then be reformulated as follows:

min\displaystyle\min icixi+maxS𝒞kij𝒞𝟏(jS)dijyij\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F}}\sum_{j\in{\cal C}}{\bf 1}(j\in S)\cdot d_{ij}y_{ij} (Static-URFL)
s.t. iyij1,\displaystyle\sum_{i\in{\cal F}}y_{ij}\geq 1,\qquad j𝒞,\displaystyle\forall j\in{\cal C},
xiyij0,\displaystyle x_{i}\geq y_{ij}\geq 0,\qquad i,j𝒞.\displaystyle\forall i\in{\cal F},\forall j\in{\cal C}.

From Theorem 1.1, (Static-URFL) is equivalent to (LP-URFL). The number of variables in (Static-URFL) is reduced to a polynomial number since the yijy_{ij} no longer depend on the scenario SS. The inner maximization problem is still taken over an exponential number of scenarios; however we can separate efficiently over these scenarios and write an efficient compact LP formulation for (Static-URFL):

maxS𝒞k{ij𝒞𝟏(jS)dijyij}=max𝒉[0,1]|𝒞|{ij𝒞dijyijhj|j𝒞hjk}\displaystyle\max_{S\in{\cal C}^{k}}\;\{\sum_{i\in{\cal F}}\sum_{j\in{\cal C}}{\bf 1}(j\in S)\cdot d_{ij}y_{ij}\}=\max_{\boldsymbol{h}\in[0,1]^{|{\cal C}|}}\;\{\sum_{i\in{\cal F}}\sum_{j\in{\cal C}}d_{ij}y_{ij}h_{j}\;|\;\sum_{j\in{\cal C}}h_{j}\leq k\} (2)
=minμ,𝝎0{kμ+j𝒞ωj|μ+ωjidijyij,j𝒞},\displaystyle=\min_{\mu,\boldsymbol{\omega}\geq 0}\;\{k\mu+\sum_{j\in{\cal C}}\omega_{j}\;|\;\mu+\omega_{j}\geq\sum_{i\in{\cal F}}d_{ij}y_{ij},\;\forall j\in{\cal C}\},

where the first equality holds because the optimal solution of the right maximization problem occurs at the extreme points of the kk-ones polytope, which corresponds to the worst-case scenarios of 𝒞k{\cal C}^{k} and the second equality follows from strong duality. Therefore, by dropping the min and introducing μ\mu and all ωj\omega_{j} as variables, we reformulate (Static-URFL) as the following linear program:

min\displaystyle\min icixi+kμ+j𝒞ωj\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+k\mu+\sum_{j\in{\cal C}}\omega_{j} (3)
s.t. μ+ωjidijyij,\displaystyle\mu+\omega_{j}\geq\sum_{i\in{\cal F}}d_{ij}y_{ij},\qquad j𝒞,\displaystyle\forall j\in{\cal C},
iyij1,\displaystyle\sum_{i\in{\cal F}}y_{ij}\geq 1,\qquad j𝒞,\displaystyle\forall j\in{\cal C},
xiyij,\displaystyle x_{i}\geq y_{ij},\qquad i,j𝒞,\displaystyle\forall i\in{\cal F},\forall j\in{\cal C},
xi0,yij0,ωj0,μ0,\displaystyle x_{i}\geq 0,\;y_{ij}\geq 0,\;\omega_{j}\geq 0,\;\mu\geq 0,\qquad i,j𝒞.\displaystyle\forall i\in{\cal F},\forall j\in{\cal C}.

Finally, we round the solution of (Static-URFL) to an integral solution for (URFL) while losing only a constant factor. This can be done using prior work on rounding techniques from the literature of deterministic facility location problems. In fact, the LP rounding technique in Shmoys et al. [18], which gives a 44-approximation algorithm to the deterministic uncapacitated problem also gives a 44-approximation algorithm for (Static-URFL). The idea is to define a ball around each client of radius equal to the fractional assignment cost of the client (which is independent of any scenario for our static policy). Then, we open facilities in non-intersecting balls of ascending radius. The result is given in the following theorem and for completeness, the details of the rounding are in Appendix 0.A.

Theorem 2.1

(Static-URFL) can be rounded to give a 4-approximation to (URFL).

We would like to note that the focus in this section is not about finding the best constant approximation for (URFL), but we introduce it as a warm-up for motivating the static assignment policy before presenting our main result in the next section.

3 Soft-capacitated robust facility location

3.1 Problem formulation

In this section, we consider the soft-capacitated robust facility location (SCRFL) which is similar to (URFL) except that each facility ii incurs a linear supply cost, where cic_{i} is the cost per unit of supply. We refer to xix_{i} as the supply (or capacity) in facility ii. Each client in the realized scenario needs to be satisfied by one unit of supply. The problem is called soft-capacitated since there is no upper bound on xix_{i}, (xi+)x_{i}\in\mathbb{Z_{+}}). The problem is given by the following integer program:

min\displaystyle\min icixi+maxS𝒞kijSdijyijS\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F}}\sum_{j\in{S}}d_{ij}y_{ij}^{S} (SCRFL)
s.t. iyijS1,\displaystyle\sum_{i\in{\cal F}}y_{ij}^{S}\geq 1,\qquad S𝒞k,jS,\displaystyle\forall S\in{\cal C}^{k},\forall j\in{S},
xijSyijS,\displaystyle x_{i}\geq\sum_{j\in S}y_{ij}^{S},\qquad i,S𝒞k,jS,\displaystyle\forall i\in{\cal F},\forall S\in{\cal C}^{k},\forall j\in{S},
xi,yijS0,\displaystyle x_{i}\in\mathbb{N},\;y_{ij}^{S}\geq 0,\qquad i,S𝒞k,jS.\displaystyle\forall i\in{\cal F},\forall S\in{\cal C}^{k},\forall j\in{S}.

We let (LP-SCRFL) denote the linear relaxation of (SCRFL), where we replace xix_{i}\in\mathbb{N} by xi0x_{i}\geq 0, for each ii\in{\cal F}. We would like to note that even the linear relaxation (LP-SCRFL) is challenging to solve since it has exponentially-many variables (scenarios). Unlike the uncapacitated case, the static assignment policy (1) is not optimal for (LP-SCRFL) and the optimal assignment for each client depends, in general, on the realized scenario. In particular, the same client could be assigned to different facilities in different scenarios. In contrast, we show the surprising result that a static assignment policy gives O(logk/loglogk)O(\log k/\log\log k)-approximation to (LP-SCRFL). Moreover, we can round the solution of the static assignment policy to an integral solution for (SCRFL) and only lose an additional constant factor. We let (Static-SCRFL) denote the problem when we restrict the second-stage variables yijSy_{ij}^{S} in (LP-SCRFL) to static assignment policies (1). The problem can then be reformulated as follows:

min\displaystyle\min icixi+maxS𝒞kij𝒞𝟏(jS)dijyij\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F}}\sum_{j\in{\cal C}}{\bf 1}(j\in S)\cdot d_{ij}y_{ij} (Static-SCRFL)
s.t. iyij1,\displaystyle\sum_{i\in{\cal F}}y_{ij}\geq 1,\qquad j𝒞,\displaystyle\forall j\in{\cal C},
ximaxS𝒞kj𝒞𝟏(jS)yij,\displaystyle x_{i}\geq\max_{S\in{\cal C}^{k}}\;\;\sum_{j\in{\cal C}}{\bf 1}(j\in S)\cdot y_{ij},\qquad i,\displaystyle\forall i\in{\cal F},
xi0,yij0,\displaystyle x_{i}\geq 0,\;y_{ij}\geq 0,\qquad i,j𝒞.\displaystyle\forall i\in{\cal F},\forall j\in{\cal C}.

3.2 An O(logkloglogk)O(\frac{\log k}{\log\log k})-approximation algorithm

Our main contribution in this section is to show that a static assignment policy (1) gives O(logk/loglogk)O(\log k/\log\log k)-approximation for (LP-SCRFL) (Theorem 1.2). To prove this theorem, we consider an optimal solution of (LP-SCRFL) and massage it to construct a solution of the form (1) while losing O(logk/loglogk)O(\log k/\log\log k) factor. We first present our construction and several structural lemmas and then give the proof of Theorem 1.2.

Our construction. Let 𝒙:(xi)i\boldsymbol{x^{*}}:(x^{*}_{i})_{i\in{\cal F}} be an optimal first-stage solution of (LP-SCRFL), let OPT1\mathrm{OPT}_{1} be the corresponding optimal first-stage cost and let OPT2\mathrm{OPT}_{2} be the corresponding optimal second-stage cost. We will classify the clients 𝒞{\cal C} into three subsets C1,C2,C3C_{1},C_{2},C_{3} using Procedure 1 (below) and then specify a static assignment policy for each subset. We use the following notation in the procedure. Let α>1\alpha>1 and r=5OPT2/kr=5\cdot\mathrm{OPT}_{2}/k. For 1\ell\geq 1 and j𝒞j\in{\cal C}, we let BjB_{j}^{\ell} denote the ball centered at client jj of radius r\ell r. We initialize the sets FF\leftarrow{\cal F} and C𝒞C\leftarrow{\cal C} and update them at iteration, as explained in the procedure, until CC becomes empty. We let Cl(B)Cl(B) denote the set of clients in CC that are inside the ball BB and let Sp(B)Sp(B) denote the total optimal supply of facilities FF that are inside the ball BB, i.e.,

Sp(B)=iF𝟏(iB)xiandCl(B)={jC|jB}.Sp(B)=\sum_{i\in{F}}{\bf 1}(i\in B)\cdot x_{i}^{*}\qquad\text{and}\qquad Cl(B)=\{j\in{C}\;|\;j\in B\}.

Note that both Sp(B)Sp(B) and Cl(B)Cl(B) depend on the current sets of facilities FF and clients CC, which we update at each iteration of the while loop in the procedure. But for the ease of notation, we do not refer to them with the indices FF and CC.

In the procedure, while the set CC is not empty, we pick a client jCj\in C and grow three balls around it: Bj21B_{j}^{2\ell-1} (internal ball), Bj2B_{j}^{2\ell} (medium ball) and Bj2+1B_{j}^{2\ell+1} (external ball) starting with =1\ell=1. For each \ell, we check if the number of clients in the internal ball Bj21B_{j}^{2\ell-1} is greater than kk (line 4); if this is the case, we remove them from CC, put them in C1C_{1} and restart in line 2. If not, we check if the supply in the medium ball Bj2B_{j}^{2\ell} is sufficient to satisfy half of the clients in the internal ball Bj21B_{j}^{2\ell-1} (line 7); if that is not the case, we remove those clients from CC, put them in C2C_{2}, and restart in line 2. Otherwise, we finally check if the supply in the the medium ball Bj2B_{j}^{2\ell} is sufficient to satisfy a fraction 1/2α1/2\alpha of the clients in the external ball Bj2+1B_{j}^{2\ell+1} (line 10); if that is the case we remove all the clients in Bj2+1B_{j}^{2\ell+1} and put them in C3C_{3}, we also remove all the facilities in Bj2B_{j}^{2\ell} and restart in line 2. If none of these three conditions holds, we increase \ell to +1\ell+1. First, we show that after at most logαk\log_{\alpha}k increments (i.e., logαk\ell\leq\log_{\alpha}k), one of three conditions must hold and therefore we will remove some clients from CC and restart in line 2. Which implies that after a finite number of iterations, the set CC becomes empty. In particular, We have the following lemma.

Procedure 1
1:Initialize C𝒞,C1,C2,C3,FC\leftarrow{\cal C},C_{1}\leftarrow\emptyset,C_{2}\leftarrow\emptyset,C_{3}\leftarrow\emptyset,F\leftarrow{\cal F}.
2:while C{C}\neq\emptyset  do
3:     Pick a client jCj\in C. Initialize =1\ell=1
4:     if  |Cl(Bj21)|k|Cl(B_{j}^{2\ell-1})|\geq k  then
5:         C1C1Cl(Bj21)C_{1}\leftarrow C_{1}\cup Cl(B_{j}^{2\ell-1}),   CCCl(Bj21){C}\leftarrow{C}\setminus Cl(B_{j}^{2\ell-1})
6:         Stop, return to line 2      
7:     if  Sp(Bj2)<12|Cl(Bj21)|Sp(B_{j}^{2\ell})<\frac{1}{2}\cdot|Cl(B_{j}^{2\ell-1})|  then
8:         C2C2Cl(Bj21)C_{2}\leftarrow C_{2}\cup Cl(B_{j}^{2\ell-1}),   CCCl(Bj21){C}\leftarrow{C}\setminus Cl(B_{j}^{2\ell-1})
9:         Stop, return to line 2      
10:     if  Sp(Bj2)12α|Cl(Bj2+1)|Sp(B_{j}^{2\ell})\geq\frac{1}{2\alpha}\cdot|Cl(B_{j}^{2\ell+1})|  then
11:         C3C3Cl(Bj2+1)C_{3}\leftarrow C_{3}\cup Cl(B_{j}^{2\ell+1}),   CCCl(Bj2+1){C}\leftarrow{C}\setminus Cl(B_{j}^{2\ell+1}),
12:         FF{iF|iBj2}{F}\leftarrow{F}\setminus\{i\in{F}\;|\;i\in B_{j}^{2\ell}\}
13:         Stop, return to line 2
14:     else+1\ell\leftarrow\ell+1, return to line 4.      
Lemma 1

In Procedure 1, after a finite number of iterations, the set CC becomes empty and C1C2C3C_{1}\cup C_{2}\cup C_{3} is equal to 𝒞{\cal C}. Moreover, \ell is always less than logαk\log_{\alpha}k.

Proof

Fix a client jj and let 1\ell\geq 1. If none of the three conditions (“if” statements) holds then

α|Cl(Bj21)|2αSp(Bj2)<|Cl(Bj2+1)|.\alpha\cdot|Cl(B_{j}^{2\ell-1})|\leq 2\alpha\cdot Sp(B_{j}^{2\ell})<|Cl(B_{j}^{2\ell+1})|.

Therefore, the number of clients grows geometrically when we increase the radius of the balls and by induction, we have that

αα|Cl(Bj1)|<|Cl(Bj2+1)|,\alpha^{\ell}\leq\alpha^{\ell}\cdot|Cl(B_{j}^{1})|<|Cl(B_{j}^{2\ell+1})|,

where |Cl(Bj1)|1|Cl(B_{j}^{1})|\geq 1, since Cl(Bj1)Cl(B_{j}^{1}) contains at least the client jj. Hence, after at most logαk\log_{\alpha}k increments, we will reach kk clients, and must stop by the first condition, and return to line 2. Hence, we always have logαk\ell\leq\log_{\alpha}k. Finally since, we remove at least one client at each iteration of the while loop, the set CC becomes empty after at most |𝒞||{\cal C}| iterations, and finally C1C2C3=𝒞C_{1}\cup C_{2}\cup C_{3}={\cal C}. ∎

Now we are ready to present our static assignment policy for (LP-SCRFL). The following three lemmas show our constructed static assignment for each client in the three subsets C1,C2,C3C_{1},C_{2},C_{3}. Moreover, we specify the supply used to satisfy each subset of these clients and present the analysis for the assignment cost.

For each client in the set C1C_{1}, we know that it belongs to a ball with at least kk clients. By the feasibility of the optimal solution, this implies that there exists kk units of supply in 𝒙\boldsymbol{x}^{*} “close” to this ball. Hence, we satisfy this client by using the same fraction 1/k1/k of 𝒙\boldsymbol{x}^{*} (static assignment) while paying a small assignment cost (roughly a constant times the radius of the ball). Since, there are at most kk clients in each scenario, and each one is using at most 𝒙/k\boldsymbol{x}^{*}/k, we need to dedicate only one 𝒙\boldsymbol{x}^{*} for all clients in C1C_{1}. Formally, we have the following lemma.

Lemma 2

There exists a static assignment policy for C1C_{1} such that each client in C1C_{1} is using at most the supply 𝐱/k\boldsymbol{x^{*}}/k and has an assignment cost less than O(logαk/k)OPT2O(\log_{\alpha}k/k)\cdot\mathrm{OPT}_{2}, i.e., there exists (y~ij)i,jC1(\tilde{y}_{ij})_{i\in{\cal F},j\in{C_{1}}} such that for each jC1j\in C_{1} :

iy~ij1,xiky~ij0,i,andidijy~ij=O(logαk)OPT2k.\sum_{i\in{\cal F}}\tilde{y}_{ij}\geq 1,\qquad\frac{x_{i}^{*}}{k}\geq\tilde{y}_{ij}\geq 0,\;\forall i\in{\cal F},\;\;\;\;\text{and}\;\;\;\;\sum_{i\in{\cal F}}d_{ij}\tilde{y}_{ij}=O(\log_{\alpha}k)\cdot\frac{\mathrm{OPT}_{2}}{k}.
Proof

Let jj be a client of C1C_{1}. It is sufficient to show that the following minimization problem is feasible and its optimal cost is O(logαk/k)OPT2O(\log_{\alpha}k/k)\cdot\mathrm{OPT}_{2}. Consider

min{idijyij|iyij1,xikyij0,i}.\min\;\left\{\sum_{i\in{\cal F}}d_{ij}y_{ij}\;\bigg{|}\;\sum_{i\in{\cal F}}y_{ij}\geq 1,\;\;\frac{x_{i}^{*}}{k}\geq y_{ij}\geq 0,\;\;\forall i\in{\cal F}\right\}. (4)

Problem (4) must be feasible since the total supply in 𝒙\boldsymbol{x}^{*} is greater than the total demand in any scenario, i.e., ixik\sum_{i\in{\cal F}}x_{i}^{*}\geq k. Recall that a client jj in C1C_{1} belongs to one of the sets Cl(Bt21)Cl(B_{t}^{2\ell-1}) for some t𝒞t\in{\cal C} and logαk\ell\leq\log_{\alpha}k (Lemma 1) such that |Cl(Bt21)|k|Cl(B_{t}^{2\ell-1})|\geq k. Consider a scenario SS formed by kk clients from Cl(Bt21)Cl(B_{t}^{2\ell-1}). Let denote 𝒚S\boldsymbol{y}^{S} the assignment of scenario SS in the optimal solution. Consider the following candidate solution for (4):

yij=1kpSyipS,i.y_{ij}=\frac{1}{k}\cdot\sum_{p\in S}y^{S}_{ip},\;\forall i\in{\cal F}.

We have, by the feasibility of the optimal solution, for each ii\in{\cal F}, 0yij1kxi0\leq y_{ij}\leq\frac{1}{k}x_{i}^{*} and

iyij=1kipSyipS1kpS1=1.\sum_{i\in{\cal F}}y_{ij}=\frac{1}{k}\cdot\sum_{i\in{\cal F}}\sum_{p\in S}y^{S}_{ip}\geq\frac{1}{k}\sum_{p\in S}1=1.

Therefore, our solution is feasible for (4). Moreover, we have

idijyij\displaystyle\sum_{i\in{\cal F}}d_{ij}y_{ij} =1kipSdijyipS\displaystyle=\frac{1}{k}\cdot\sum_{i\in{\cal F}}\sum_{p\in S}d_{ij}y^{S}_{ip}
1kipSdipyipS+1kpSdpj\displaystyle\leq\frac{1}{k}\cdot\sum_{i\in{\cal F}}\sum_{p\in S}d_{ip}y^{S}_{ip}+\frac{1}{k}\cdot\sum_{p\in S}d_{pj}
1kOPT2+2(21)r\displaystyle\leq\frac{1}{k}\cdot\mathrm{OPT}_{2}+2(2\ell-1)r
OPT2k+2(2logαk1)5OPT2k=O(logαk)OPT2k,\displaystyle\leq\frac{\mathrm{OPT}_{2}}{k}+2(2\log_{\alpha}k-1)\cdot 5\cdot\frac{\mathrm{OPT}_{2}}{k}=O(\log_{\alpha}k)\cdot\frac{\mathrm{OPT}_{2}}{k},

where the first inequality follows from the triangle inequality and the fact that iyipS=1\sum_{i\in{\cal F}}y_{ip}^{S}=1 for all pSp\in S in the optimal solution. For the second inequality, we use the definition of OPT2\mathrm{OPT}_{2} to bound the first term, the second term dpjd_{pj} is bounded by the diameter of the ball Bj21B_{j}^{2\ell-1} which contains client jj and all clients pSp\in S. ∎

Now, consider the set C2C_{2}. By construction, these are clients such that there is not enough supply within a distance r=5OPT2/kr=5\mathrm{OPT}_{2}/k to satisfy half of them. Therefore, intuitively they need to pay “large” distances in the optimal assignment cost if they show up all together in the same scenario. In the following lemma, we show that we can have no more than kk of these clients. As we would show later, this would imply that we can dedicate a supply 𝒙\boldsymbol{x}^{*} to C2C_{2} and make a static assignment of all the clients C2C_{2} to this 𝒙\boldsymbol{x}^{*}.

Lemma 3

The set C2C_{2} has at most kk clients.

Proof

Suppose, for the sake of contradiction, that |C2|>k|C_{2}|>k. Let G1,G2,,GTG_{1},G_{2},\ldots,G_{T} be the disjoint subsets of clients added at each iteration in the construction of C2C_{2} in Procedure 1. In particular, C2=G1G2GTC_{2}=G_{1}\cup G_{2}\cup\ldots\cup G_{T} for some TT where:

  • (i)

    for t=1,2,,T,t=1,2,\ldots,T, Gt=Cl(Bjt2t1)G_{t}=Cl(B_{j_{t}}^{2\ell_{t}-1}) for some client jtj_{t} and 1tlogαk1\leq\ell_{t}\leq\log_{\alpha}k.

  • (ii)

    the supply Sp(Bjt2t)Sp(B_{j_{t}}^{2\ell_{t}}) is less than half of the clients in GtG_{t}.

  • (iii)

    each set GtG_{t} has strictly less than kk clients, since the procedure has to fail the first “if” statement before adding GtG_{t} into C2C_{2}.

Recall that

Sp(Bjt2t)=iF𝟏(iBjt2t)xi,Sp(B_{j_{t}}^{2\ell_{t}})=\sum_{i\in{F}}{\bf 1}(i\in B_{j_{t}}^{2\ell_{t}})\cdot x_{i}^{*},

where FF is the current set of facilities in the procedure (and is not all \cal F since some facilities have been removed in line 12 of the procedure). However, we would like to emphasize that when a facility has been removed (in line 12 of the procedure), all clients within distance rr from this facility were removed as well (line 11). This is true, since when we remove the facilities in a medium ball, say Bj2B_{j}^{2\ell}, (line 12), we remove all clients in the corresponding external ball Bj2+1B_{j}^{2\ell+1} (line 11). Hence, the remaining clients in CC are at least (2+1)r(2)r=r(2\ell+1)r-(2\ell)r=r away from the removed facilities. In particular, for the clients GtG_{t}, the supply that has been removed before they were added to C2C_{2} is at least rr away from them. Therefore, all the facility in \cal F that are within a distance rr from a client in GtG_{t} belong to the set FF that verifies iF𝟏(iBjt2t)xi2|Gt|\sum_{i\in{F}}{\bf 1}(i\in B_{j_{t}}^{2\ell_{t}})\cdot x_{i}^{*}\leq 2\cdot|G_{t}|. This implies that the supply of all facilities within a distance rr from GtG_{t} in the optimal solution, is less than half of the clients GtG_{t}. Hence, if all of the clients GtG_{t} show up in a scenario, the optimal second-stage solution needs to pay an assignment cost of at least r|Gt|/2r\cdot|G_{t}|/2.

Order the sets GtG_{t} according to their cardinalities: wlog assume that |G1||G2||GT||G_{1}|\geq|G_{2}|\geq\ldots\geq|G_{T}|. We construct a scenario S^\hat{S} by taking clients from the sets G1G_{1}, G2,G_{2},\ldots until we hit kk. This is possible since by assumption |C2|>k|C_{2}|>k. Assume that

|G1|+|G2|+|Gp1|+|G¯p|=k,|G_{1}|+|G_{2}|+\ldots|G_{p-1}|+|\bar{G}_{p}|=k,

for some pp, where 2pT2\leq p\leq T. Note that G¯p\bar{G}_{p} is a subset of Gp{G}_{p}, since we can reach kk before taking all the clients of the last set Gp{G}_{p}. For each t=1,,p1t=1,\ldots,p-1, the optimal second-stage decision needs to pay at least r|Gt|/2r\cdot|G_{t}|/2. Therefore,

OPT212r(|G1|+|G2|+|Gp1|).\mathrm{OPT}_{2}\geq\frac{1}{2}\cdot r\cdot(|G_{1}|+|G_{2}|+\ldots|G_{p-1}|).

We did not include GpG_{p}, since not all these clients are necessary in the scenario S^\hat{S}, but only |G¯p||\bar{G}_{p}| of them. Since G¯p\bar{G}_{p} has the smallest cardinality

|G1|+|G2|+|Gp1|12(|G1|+|G2|+|Gp1|+|G¯p|)=k2.|G_{1}|+|G_{2}|+\ldots|G_{p-1}|\geq\frac{1}{2}(|G_{1}|+|G_{2}|+\ldots|G_{p-1}|+|\bar{G}_{p}|)=\frac{k}{2}.

Therefore,

OPT2rk4=5OPT24,\mathrm{OPT}_{2}\geq r\cdot\frac{k}{4}=5\cdot\frac{\mathrm{OPT}_{2}}{4},

which is a contradiction. Therefore, |C2|k|C_{2}|\leq k. ∎

Finally, for clients C3C_{3}, we show that there exists |C3|/2α|C_{3}|/2\alpha units of supply “close” to them. In particular, we can multiply these units by 2α2\alpha, dedicate them to C3C_{3} and make a static assignment for C3C_{3}. We have the following lemma.

Lemma 4

There exists a supply 𝐱^\boldsymbol{\hat{x}} that has a cost at most 2αOPT12\alpha\cdot\mathrm{OPT}_{1} and there exists a static assignment policy such that all clients in C3C_{3} are assigned to supply 𝐱^\boldsymbol{\hat{x}} and each client in C3C_{3} has an assignment cost that is O(logαk/k)OPT2O(\log_{\alpha}k/k)\cdot\mathrm{OPT}_{2}, i.e., there exists (y^ij)i,jC3(\hat{y}_{ij})_{i\in{\cal F},j\in{C_{3}}} and (x^i)i(\hat{x}_{i})_{i\in{\cal F}} such that for all jC3j\in C_{3} :

iy^ij1,x^ijC3y^ij,i,x^i0,y^ij0,i,\sum_{i\in{\cal F}}\hat{y}_{ij}\geq 1,\qquad\hat{x}_{i}\geq\sum_{j\in C_{3}}\hat{y}_{ij},\;\forall i\in{\cal F},\qquad\hat{x}_{i}\geq 0,\;\hat{y}_{ij}\geq 0,\;\forall i\in{\cal F},
icix^i2αOPT1andidijy^ij=O(logαk)OPT2k.\sum_{i\in{\cal F}}c_{i}\hat{x}_{i}\leq 2\alpha\cdot\mathrm{OPT}_{1}\qquad\text{and}\qquad\sum_{i\in{\cal F}}d_{ij}\hat{y}_{ij}=O(\log_{\alpha}k)\cdot\frac{\mathrm{OPT}_{2}}{k}.
Proof

Let G1,G2,,GTG_{1},G_{2},\ldots,G_{T} be the disjoint subsets of clients added at each iteration to construct C3C_{3} in Procedure 1. In particular, C3=G1G2GTC_{3}=G_{1}\cup G_{2}\cup\ldots\cup G_{T} for some TT such that: for all t=1,2,,T,t=1,2,\ldots,T, Gt=Cl(Bjt2t+1)G_{t}=Cl(B_{j_{t}}^{2\ell_{t}+1}) for some client jtj_{t} and some integer t\ell_{t} with 1tlogαk1\leq\ell_{t}\leq\log_{\alpha}k. Moreover, the supply Sp(Bjt2t)Sp(B_{j_{t}}^{2\ell_{t}}) is greater than a 1/2α1/2\alpha fraction of the clients in GtG_{t}. Hence, for each ball Bjt2tB_{j_{t}}^{2\ell_{t}}, we multiply the supply by 2α2\alpha, move it to the cheapest facility in this ball and make a static assignment of all clients GtG_{t} to this cheapest facility. Since the supply in Bjt2tB_{j_{t}}^{2\ell_{t}} is removed along with clients GtG_{t}, it will not be used by the other clients in C3C_{3}.

Formally, let iti_{t} be the cheapest facility in the ball Bjt2tB_{j_{t}}^{2\ell_{t}}. We define, for each facility ii in Bjt2tB_{j_{t}}^{2\ell_{t}}, x^i=2αiF𝟏(iBjt2t)xi\hat{x}_{i}=2\alpha\sum_{i^{\prime}\in{F}}{\bf 1}(i^{\prime}\in B_{j_{t}}^{2\ell_{t}})\cdot x_{i^{\prime}}^{*} if i=iti=i_{t} and x^i=0\hat{x}_{i}=0 otherwise. For each client jC3j\in C_{3}, we let y^ij=1\hat{y}_{ij}=1 for i=iti=i_{t} and jGtj\in G_{t}, and let y^ij=0\hat{y}_{ij}=0, otherwise. Therefore, the first desired constraints in the lemma are verified. Let us check the last one. The distance between a client and its assigned facility in our solution is at most rr plus the diameter of the ball Bjt2tB_{j_{t}}^{2\ell_{t}}, i.e.,

r+4tr(4logαk+1)5OPT2/k=O(logαk/k)OPT2.r+4{\ell}_{t}r\leq(4\log_{\alpha}k+1)\cdot 5\cdot\mathrm{OPT}_{2}/k=O(\log_{\alpha}k/k)\cdot\mathrm{OPT}_{2}.

Proof of Theorem 1.2. Let (y~ij)i,jC1(\tilde{y}_{ij})_{i\in{\cal F},j\in{C_{1}}} be the solution given in Lemma 2 for satisfying the clients in C1C_{1}. We dedicate a supply 𝒙\boldsymbol{x}^{*} to clients C1C_{1}. Let (y^ij)i,jC3(\hat{y}_{ij})_{i\in{\cal F},j\in{C_{3}}} and (x^i)i(\hat{x}_{i})_{i\in{\cal F}} be the solution given in Lemma 4 for satisfying the clients in C3C_{3}. Finally, we know from Lemma 3 that C2C_{2} has at most kk clients, and therefore C2C_{2} is a scenario. So we dedicate a supply 𝒙\boldsymbol{x}^{*} to C2C_{2} and let the optimal assignment yijC2y_{ij}^{C_{2}} be our static assignment solution for C2C_{2}. In particular, we give the following solution to (LP-SCRFL), where the first stage solution is 2𝒙+𝒙^2\boldsymbol{x}^{*}+\boldsymbol{\hat{x}} and the static assignment policy is for all ii\in{\cal F}: yij=y~ijy_{ij}=\tilde{y}_{ij} for jC1j\in C_{1}, yij=yijC2y_{ij}={y}^{C_{2}}_{ij} for jC2j\in C_{2}, and yij=y^ijy_{ij}=\hat{y}_{ij} for jC3j\in C_{3}. It is clear that iyij1\sum_{i\in{\cal F}}y_{ij}\geq 1, for each jj in C1C2C3C_{1}\cup C_{2}\cup C_{3}. Moreover, for any scenario S𝒞kS\in{\cal C}^{k} and ii\in{\cal F},

jSyij\displaystyle\sum_{j\in S}y_{ij} =jSC1y~ij+jSC2yijC2+jSC3y^ij\displaystyle=\sum_{j\in S\cap C_{1}}\tilde{y}_{ij}+\sum_{j\in S\cap C_{2}}y^{C_{2}}_{ij}+\sum_{j\in S\cap C_{3}}\hat{y}_{ij}
jSC1xik+xi+x^i2xi+x^i.\displaystyle\leq\sum_{j\in S\cap C_{1}}\frac{x_{i}^{*}}{k}+x_{i}^{*}+\hat{x}_{i}\leq 2x_{i}^{*}+\hat{x}_{i}.

Therefore, our solution is feasible for (LP-SCRFL). Let us evaluate its cost. The cost of the first stage is at most 2OPT1+2αOPT1=O(α)OPT12OPT_{1}+2\alpha\mathrm{OPT}_{1}=O(\alpha)\cdot\mathrm{OPT}_{1}. For the second-stage cost, consider any scenario S𝒞kS\in{\cal C}^{k}, We have

ijSdijyij\displaystyle\sum_{i\in{\cal F}}\sum_{j\in S}d_{ij}y_{ij} =jSC1idijy~ij+jSC2idijyijC2+jSC3idijy^ij\displaystyle=\sum_{j\in S\cap C_{1}}\sum_{i\in{\cal F}}d_{ij}\tilde{y}_{ij}+\sum_{j\in S\cap C_{2}}\sum_{i\in{\cal F}}d_{ij}y^{C_{2}}_{ij}+\sum_{j\in S\cap C_{3}}\sum_{i\in{\cal F}}d_{ij}\hat{y}_{ij}
jSC1O(logαk)OPT2k+OPT2+jSC3O(logαk)OPT2k\displaystyle\leq\sum_{j\in S\cap C_{1}}O(\log_{\alpha}k)\cdot\frac{\mathrm{OPT}_{2}}{k}+\mathrm{OPT}_{2}+\sum_{j\in S\cap C_{3}}O(\log_{\alpha}k)\cdot\frac{\mathrm{OPT}_{2}}{k}
O(logαk)OPT2+OPT2+O(logαk)OPT2\displaystyle\leq O(\log_{\alpha}k)\cdot\mathrm{OPT}_{2}+\mathrm{OPT}_{2}+O(\log_{\alpha}k)\cdot\mathrm{OPT}_{2}
=O(logαk)OPT2.\displaystyle=O(\log_{\alpha}k)\cdot\mathrm{OPT}_{2}.

By balancing the terms α\alpha and logαk\log_{\alpha}k, we choose α=logk/loglogk\alpha=\log k/\log\log k which gives O(logk/loglogk)O(\log k/\log\log k)-approximation to (LP-SCRFL). ∎

Similar to the uncapacitated problem, we can solve (Static-SCRFL) efficiently using a compact linear program. In fact, we dualize the inner maximization problem in the objective function of (Static-SCRFL) in the same way as (2). In addition to that, we reformulate the second constraint in (Static-SCRFL) using the same dualization technique as follows: for each ii\in{\cal F},

maxS𝒞k{jSyij}\displaystyle\max_{S\in{\cal C}^{k}}\;\{\sum_{{j\in S}}y_{ij}\} =max𝒉[0,1]|C|{j𝒞yijhj|j𝒞hjk}\displaystyle=\max_{\boldsymbol{h}\in[0,1]^{|C|}}\;\{\sum_{{j\in{\cal C}}}y_{ij}h_{j}\;\;|\;\;\sum_{j\in{\cal C}}h_{j}\leq k\}
=minηi,λij0{kηi+j𝒞λij|ηi+λijyij,j𝒞}.\displaystyle=\min_{\eta_{i},\lambda_{ij}\geq 0}\;\{k\eta_{i}+\sum_{j\in{\cal C}}\lambda_{ij}\;|\;\eta_{i}+\lambda_{ij}\geq y_{ij},\;\forall j\in{\cal C}\}.

The linear program is given by

min\displaystyle\min icixi+kμ+j𝒞ωj\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+k\mu+\sum_{j\in{\cal C}}\omega_{j} (5)
s.t. μ+ωjidijyij,\displaystyle\mu+\omega_{j}\geq\sum_{i\in{\cal F}}d_{ij}y_{ij},\qquad j𝒞,\displaystyle\forall j\in{\cal C},
iyij1,\displaystyle\sum_{i\in{\cal F}}y_{ij}\geq 1,\qquad j𝒞,\displaystyle\forall j\in{\cal C},
xikηi+j𝒞λij,\displaystyle x_{i}\geq k\eta_{i}+\sum_{j\in{\cal C}}\lambda_{ij}, i,\displaystyle\forall i\in{\cal F},
ηi+λijyij,\displaystyle\eta_{i}+\lambda_{ij}\geq y_{ij}, i,j𝒞,\displaystyle\forall i\in{\cal F},\forall j\in{\cal C},
xi,yij,λij,ηi,ωj,μ0,\displaystyle x_{i},y_{ij},\lambda_{ij},\eta_{i},\omega_{j},\mu\geq 0,\qquad i,j𝒞,\displaystyle\forall i\in{\cal F},\forall j\in{\cal C},

which can be reduced, after removing the variables yijy_{ij}, to

min\displaystyle\min icixi+kμ+j𝒞ωj\displaystyle\sum_{i\in{\cal F}}c_{i}x_{i}+k\mu+\sum_{j\in{\cal C}}\omega_{j} (6)
s.t. μ+ωjidij(ηi+λij),\displaystyle\mu+\omega_{j}\geq\sum_{i\in{\cal F}}d_{ij}(\eta_{i}+\lambda_{ij}),\qquad j𝒞,\displaystyle\forall j\in{\cal C},
iηi+λij1,\displaystyle\sum_{i\in{\cal F}}\eta_{i}+\lambda_{ij}\geq 1,\qquad j𝒞,\displaystyle\forall j\in{\cal C},
xikηi+j𝒞λij,\displaystyle x_{i}\geq k\eta_{i}+\sum_{j\in{\cal C}}\lambda_{ij}, i,\displaystyle\forall i\in{\cal F},
xi,λij,ηi,ωj,μ0,\displaystyle x_{i},\lambda_{ij},\eta_{i},\omega_{j},\mu\geq 0,\qquad i,j𝒞.\displaystyle\forall i\in{\cal F},\forall j\in{\cal C}.

Finally, we round the optimal solution of (Static-SCRFL) to an integral solution using the filtering and rounding techniques from Shmoys et al. [18] while losing only a factor of 12. Again, this rounding technique was designed for the deterministic facility location problem, but the same argument works as well for (Static-SCRFL). Finally, since (Static-SCRFL) gives O(logk/loglogk)O(\log k/\log\log k)-approximation to (LP-SCRFL) and we only loose a constant factor in the rounding, this results in O(logk/loglogk)O(\log k/\log\log k)- approximation algorithm for (SCRFL). We state the result in the following theorem and, for completeness, we present the details of the rounding in Appendix 0.B.

Theorem 3.1

(Static-SCRFL) can be rounded to give O(logkloglogk)O(\frac{\log k}{\log\log k})-approximation algorithm to (SCRFL).

Note that after rounding the supply in our solution of (Static-SCRFL), the integral second-stage assignment for each realized scenario is a transportation problem and therefore its optimal solution is integral. We would like to emphasize that while the fractional assignment in our solution is static, our integral assignment is not necessarily static. In fact, a policy with an integral static assignment could even be bad our model.

4 Conclusion.

In this paper, we give a O(logk/loglogk)O(\log k/\log\log k)-approximation for soft-capacitated robust facility location problems with an implicit model of demand uncertainty. It is an interesting open question to study whether there exists a constant approximation algorithm for the problem, even in special cases such as the Euclidean metric. Our solution approach relies on static fractional assignment policies, which we show are optimal for the uncapacitated problem and give a strong theoretical guarantee for soft-capacitated case. Static assignment policies, while reasonable for the case of soft-capacities can be shown to be arbitrarily bad for the case of hard-capacities, where in addition to cost per unit, there is also an upper bound on supply at each facility. It is another interesting open direction to study any non-trivial approximation in this setting.

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Appendix 0.A Rounding and proof of Theorem 2.1

Rounding (Shmoys et al.[18]). Recall that nn is the number of facilities and mm is the number of clients. Let (𝒙,𝒚)=((xi)i,(yij)i,j𝒞)(\boldsymbol{x}^{*},\boldsymbol{y}^{*})=\left((x^{*}_{i})_{i\in{\cal F}},(y^{*}_{ij})_{i\in{\cal F},j\in{\cal C}}\right) be an optimal solution for (Static-SCRFL). Let OPT1\mathrm{OPT}_{1} and OPT2\mathrm{OPT}_{2} respectively be the corresponding optimal first-stage and second-stage cost. In particular, OPT2=maxS𝒞kij𝒞𝟏(jS)dijyij\mathrm{OPT}_{2}=\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F}}\sum_{j\in{\cal C}}{\bf 1}(j\in S)\cdot d_{ij}y^{*}_{ij}. For each client j𝒞j\in{\cal C}, let

Lj=idijyij.L_{j}=\sum_{i\in{\cal F}}d_{ij}y^{*}_{ij}.

We sort LjL_{j} in ascending order. Suppose without loss of generality, that L1L2LmL_{1}\leq L_{2}\leq\ldots\leq L_{m}. For each client j𝒞j\in{\cal C}, we consider BjB_{j} the ball centered at the client jj with radius αLj\alpha L_{j} (where α1\alpha\geq 1 will be defined later). We choose greedily the non-intersecting balls in ascending order of LjL_{j} and open the cheapest facility in each selected ball. That is, we consider every client in this sorted order, and only include its ball if it does not intersect any previously selected ball. Each client is assigned to its closest opened facility. For each client j𝒞j\in{\cal C},

LjiBjdijyijiBjαLjyij.L_{j}\geq\sum_{i\notin B_{j}}d_{ij}y^{*}_{ij}\geq\sum_{i\notin B_{j}}\alpha L_{j}y^{*}_{ij}.

Hence,

1αiBjyij,\frac{1}{\alpha}\geq\sum_{i\notin B_{j}}y^{*}_{ij},

which implies

iBjyij11α.\sum_{i\in B_{j}}y^{*}_{ij}\geq 1-\frac{1}{\alpha}.

Therefore,

iBjxi11α.\sum_{i\in B_{j}}x^{*}_{i}\geq 1-\frac{1}{\alpha}.

By summing over the non-intersecting balls, the cost of our opened facilities is less than 111αOPT1\frac{1}{1-\frac{1}{\alpha}}\mathrm{OPT}_{1}. Now let us consider the assignment cost. For client jj, if its ball BjB_{j} is chosen by the greedy procedure above, then the client is assigned to the opened facility ii in BjB_{j} and dijαLjd_{ij}\leq\alpha L_{j}. If not, the client jj is assigned to a facility ii no further away than the one chosen in the ball BkB_{k} that intersected BjB_{j}. By the triangle inequality:

dijdiv+djvαLj+2αLk3αLj,d_{ij}\leq d_{iv}+d_{jv}\leq\alpha L_{j}+2\alpha L_{k}\leq 3\alpha L_{j},

where vBjBkv\in B_{j}\cap B_{k} and LkLjL_{k}\leq L_{j}. Hence, for any scenario S𝒞kS\in{\cal C}^{k}, the second-stage cost is at most 3αjSLj3\alpha\cdot\sum_{j\in S}L_{j}, which is at most 3αOPT23\alpha\cdot\mathrm{OPT}_{2} . By optimizing over α\alpha, we take α=43\alpha=\frac{4}{3}, which results in the 4-approximation with respect to OPT1+OPT2\mathrm{OPT}_{1}+\mathrm{OPT}_{2}. Moreover, since (Static-URFL) is equivalent to (LP-URFL) and (LP-URFL) is a relaxation of (URFL), the rounding above gives 4-approximation algorithm for (URFL).

Appendix 0.B Rounding and proof of Theorem 3.1

Rounding (Shmoys et al.[18]). Let (𝒙,𝒚)=((xi)i,(yij)i,j𝒞)(\boldsymbol{x}^{*},\boldsymbol{y}^{*})=\left((x^{*}_{i})_{i\in{\cal F}},(y^{*}_{ij})_{i\in{\cal F},j\in{\cal C}}\right) be an optimal solution of (Static-SCRFL). Let z=maxS𝒞ki,jSdijyijz=\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F},j\in{S}}d_{ij}y^{*}_{ij} and set α(0,1)\alpha\in(0,1). For each client j𝒞j\in{\cal C}, let π\pi be a permutation of facilities that serve jj in the optimal solution of (Static-SCRFL) such that dπ(1)jdπ(2)jdπ(n)jd_{\pi(1)j}\leq d_{\pi(2)j}\leq\ldots\leq d_{\pi(n)j}. Define the radius dj(α)=dπ(i)jd_{j}(\alpha)=d_{\pi(i^{*})j} where i=min{i|i=1iyπ(i)jα}i^{*}=\min\{i^{\prime}\;|\;\sum_{i=1}^{i^{\prime}}y^{*}_{\pi(i)j}\geq\alpha\}. Following the same definitions in Shmoys el al. [18], we say that a solution (𝒙,𝒚)(\boldsymbol{x},\boldsymbol{y}) is gg-close if for any j𝒞j\in{\cal C}, yij>0y_{ij}>0 implies that dijgjd_{ij}\leq g_{j}.

Claim. Given a feasible solution (𝒙,𝒚)(\boldsymbol{x},\boldsymbol{y}), we can construct a gg-close feasible solution (𝒙¯,𝒚¯)(\bar{\boldsymbol{x}},\bar{\boldsymbol{y}}) such that 𝒙¯=1α𝒙\bar{\boldsymbol{x}}=\frac{1}{\alpha}\boldsymbol{x}, z¯=1αz\bar{z}=\frac{1}{\alpha}z and gjdj(α),j𝒞g_{j}\leq d_{j}(\alpha),\;\forall j\in{\cal C} where z¯=maxS𝒞ki,jSdijy¯ij\bar{z}=\max_{S\in{\cal C}^{k}}\sum_{i\in{\cal F},j\in{S}}d_{ij}\bar{y}_{ij}.

Proof

In fact, for each jj, we set y¯π(i)j=yπ(i)j/α\bar{y}_{\pi(i)j}=y_{\pi(i)j}/\alpha for 1ii1\leq i\leq i^{*} and y¯π(i)j=0\bar{y}_{\pi(i)j}=0 otherwise. We have

iy¯ij=i=1i1αyπ(i)j1.\sum_{i\in{\cal F}}\bar{y}_{ij}=\sum_{i=1}^{i^{*}}\frac{1}{\alpha}y_{\pi(i)j}\geq 1.

The second constraint in (Static-SCRFL) is trivially verified since xx is multiplied by 1α\frac{1}{\alpha} and yijy_{ij} are either multiplied by 1α\frac{1}{\alpha} or set to 0. Moreover, all edges in our solution with a positive flow have a cost at most dj(α)d_{j}(\alpha). Hence, (𝒙¯,𝒚¯)(\bar{\boldsymbol{x}},\bar{\boldsymbol{y}}) is gg-close with gjdj(α)g_{j}\leq d_{j}(\alpha) for all jj. ∎

Let (𝒙¯,𝒚¯)(\bar{\boldsymbol{x}},\bar{\boldsymbol{y}}) be the gg-close solution corresponding to the optimal solution (𝒙,𝒚)(\boldsymbol{x}^{*},\boldsymbol{y}^{*}). For the ease of notation, we just use the notation (𝒙,𝒚)(\boldsymbol{x},\boldsymbol{y}) instead of (𝒙¯,𝒚¯)(\bar{\boldsymbol{x}},\bar{\boldsymbol{y}}) in the rest of the proof. We round-up all xi12x_{i}\geq\frac{1}{2} to xi\lceil x_{i}\rceil. We pay at most a factor 22 in the first-stage cost. Let us focus on the remaining fractional facilities, say ^\hat{{\cal F}}, i.e.,

^={i|  0<xi<1/2}.\hat{{\cal F}}=\{i\in{\cal F}\;\;|\;\;0<x_{i}<1/2\}.

We let 𝒞^\hat{{\cal C}} denote the set of clients such that half of their supply is coming from ^\hat{{\cal F}}, i.e.,

𝒞^={j𝒞|i^yij12}.\hat{{\cal C}}=\{j\in{\cal C}\;\;|\;\;\sum_{i\in\hat{{\cal F}}}y_{ij}\geq\frac{1}{2}\}.

We sort the clients in 𝒞^\hat{{\cal C}} in ascending order of gjg_{j} and do the following until 𝒞^\hat{{\cal C}} becomes empty. We take the client jj with the smallest gjg_{j} in 𝒞^\hat{{\cal C}}, let say jj^{\prime}. Let

V={i^:yij>0},V=\{i\in\hat{{\cal F}}\;:\;y_{ij^{\prime}}>0\},

and

T={j𝒞:iV s.t. yij>0}.T=\{j\in{\cal C}\;:\;\exists i\in V\text{ s.t. }y_{ij}>0\}.

We put the iVxi\lceil\sum_{i\in V}x_{i}\rceil units of supply in the cheapest facility in VV which we denote fcf_{c}. We set each xix_{i} in V{fc}V\setminus\{f_{c}\} to 0. The clients in TT are the ones affected by this change. We will route all of their demand from VV to fcf_{c} and make this assignment static. Note that, at this step, we have routed only their demand from VV and not their entire demand. This is feasible because

iVxiiVxij𝒞𝟏(jS)iVyij\lceil\sum_{i\in V}x_{i}\rceil\geq\sum_{i\in V}x_{i}\geq\sum_{j\in{\cal C}}{\bf 1}(j\in S)\cdot\sum_{i\in V}y_{ij}

for any scenario SS. Since we choose the clients in 𝒞^\hat{\cal C} in ascending order of gjg_{j}, the triangle inequality ensures that the solution is 3g3g-close. We keep doing this until 𝒞^\hat{\cal C} becomes empty.

Let us analyze the final cost. We lost a factor 1α\frac{1}{\alpha} in both first-stage and second-stage cost. Then, we lost a factor 22 in the first-stage cost by rounding up the solution and moving the supply to the cheapest facilities after each rounding. We lost another factor 22 in the first stage to satisfy 𝒞𝒞^{\cal C}\setminus\hat{\cal C}. For the second stage, we have a solution that is 3g3g-close with gjdj(α)g_{j}\leq d_{j}(\alpha). Note that

dj(α)11αidijyij.d_{j}(\alpha)\leq\frac{1}{1-\alpha}\sum_{i\in{\cal F}}d_{ij}y_{ij}.

In particular, for any scenario SS,

jSgj11αjSidijyij.\sum_{j\in S}g_{j}\leq\frac{1}{1-\alpha}\sum_{j\in S}\sum_{i\in{\cal F}}d_{ij}y_{ij}.

Hence, we loose 31α\frac{3}{1-\alpha} in the second stage. Overall, the factor is

4αOPT1+3α(1α)OPT2.\frac{4}{\alpha}\mathrm{OPT}_{1}+\frac{3}{\alpha(1-\alpha)}\mathrm{OPT}_{2}.

We set α=12\alpha=\frac{1}{2}, which gives 12 approximation to (LP-SCRFL). Finally, since (Static-SCRFL) gives O(logk/loglogk)O(\log k/\log\log k)-approximation to (LP-SCRFL) and (LP-SCRFL) is a relaxation of (SCRFL), this results in O(logk/loglogk)O(\log k/\log\log k)-approximation algorithm for (SCRFL).