On the Power of Static Assignment Policies for Robust Facility Location Problems
aa r X i v : . [ m a t h . O C ] N ov On the Power of Static Assignment Policies forRobust Facility Location Problems
Omar El Housni , Vineet Goyal , and David Shmoys ORIE, Cornell Tech, New York, USA [email protected] IEOR, Columbia University , New York, USA [email protected] ORIE, Cornell University, Ithaca, USA [email protected]
Abstract.
We consider a two-stage robust facility location problem ona metric under an uncertain demand. The decision-maker needs to de-cide on the (integral) units of supply for each facility in the first stageto satisfy an uncertain second-stage demand, such that the sum of firststage supply cost and the worst-case cost of satisfying the second-stagedemand over all scenarios is minimized. The second-stage decisions areonly assignment decisions without the possibility of adding recourse sup-ply capacity. This makes our model different from existing work on two-stage robust facility location and set covering problems. We consider animplicit model of uncertainty with an exponential number of demandscenarios specified by an upper bound k on the number of second-stageclients. In an optimal solution, the second-stage assignment decisionsdepend on the scenario; surprisingly, we show that restricting to a fixed(static) fractional assignment for each potential client irrespective of thescenario gives us an O (log k/ log log k )-approximation for the problem.Moreover, the best such static assignment can be computed efficientlygiving us the desired guarantee. We consider two-stage robust facility location problems under demand uncer-tainty where we are given a set of clients and a set of facilities in a commonmetric space. In the first stage, the decision-maker needs to select the (integral)units supply at each facility. The uncertain demand is then selected adversariallyand needs to be satisfied by the existing supply with the minimum assignmentcost in the second stage. The goal is to determine the first-stage supply suchthat the sum of first-stage supply cost and the worst-case assignment cost overall demand scenarios is minimized. Our problem is motivated by settings wherethe lead time to procure supply is large and obtaining additional units of supplyin the second stage is not feasible. The uncertain second-stage demand mustthen be satisfied by supply units from the first stage, a common constraint inmany applications. This is a departure from existing work on two-stage robust
El Housni et al. facility location, network design and, more generally, robust covering problemsthat have been studied extensively in the literature [6,7,1,9] where the second-stage decisions allow for “adding more supply” (more specifically adding moresets/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 k on the number of second-stage demand clients. Since the number of scenarios is exponentially-many (spec-ified compactly), we can not efficiently solve even the LP relaxation for the prob-lem. In contrast, if the set of second-stage scenarios are explicitly specified (forthe explicit scenario model, see for instance [6,1]), we can write a polynomially-sized LP relaxation with assignment decisions for each scenario. The main chal-lenge then is related to obtaining an integral solution, which for the case of setcovering and several network design problems can be reduced to deterministicversions (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 approximatelysolve the linear relaxation of the problem efficiently. The implicit model of un-certainty with an upper bound on number of uncertain second-stage clients orelements has been studied extensively in the literature. Feige et al. [7] show undera reasonable complexity assumption that it is hard to solve the LP relaxation of atwo-stage set covering problem within a factor better than Ω (log n/ log log n ) un-der this implicit model of uncertainty. They also give an O (log n )-approximationfor the 0 − O (log n )-approximation for the set-covering problem, thereby matchingthe deterministic approximation guarantee. El Housni and Goyal [12] show that astatic policy (that is, linear in the set of second-stage elements, also referred to asan affine policy) gives an O (log n/ log log n )-approximation for the two-stage LP,thereby matching the hardness lower bound for the fractional problem. Gupta etal. [10] and Khandekar et al. [14] present approximations for several network de-sign problems under the implicit model of uncertainty. Although there is a largebody of work in this direction, as we mentioned earlier, our problem is differentfrom set covering, since there is no possibility of adding recourse capacity; priorresults do not imply an approximation for our model. Other related work.
Many variants of facility location problems have beenstudied extensively in the literature, including both deterministic versions aswell as variants that address demand uncertainty. We refer the reader to [17] fora review of the many variants of deterministic facility location problems. Amongthe models that address demand uncertainty in the facility location problems, inaddition to robust [2,4], there are also stochastic [11,13,20,16] and distribution-ally 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 scenar-ios and the goal is to minimize the total expected cost. A distributionally robustmodel can be thought of as a hybrid between stochastic and robust where thesecond-stage distribution is selected adversarially from a pre-specified set andthe goal is to minimize the worst-case expected cost. We refer the reader to the obust Facility Location Problems 3 survey [19] for an extensive review of facility location problems under uncer-tainty.
Let us begin with a formal problem definition. We are given a set of n facilities F and m clients C in a common metric d where d ij denotes the distance between i and j . For each facility i ∈ F , there is a cost c i per unit of supply at i . Thedemand uncertainty is modeled by an implicit set of scenarios C k that includesall subsets of clients C of size at most k . The decision-maker needs to select an(integral) number of units of supply x i for each facility i ∈ F in the first-stage.An adversary observes the first stage decisions and selects a worst-case demandscenario S ∈ C k that must be satisfied with the first-stage supply, where eachclient in the realized scenario needs one unit of supply. The goal is to minimizethe sum of the first-stage supply cost and the worst-case assignment cost overall second-stage demand scenarios. We refer to this problem as soft-capacitatedrobust facility location ( SCRFL ). Typically in the literature, soft-capacitated referto settings where violations of capacity upper bounds are allowed. The analoguehere is that we can add any amount of supply in a facility without upper bounds( x i ∈ Z + ) but we pay a per unit cost of supply.We consider a class of static assignment policies, where each of the m clientshas a static fractional assignment to facilities that is independent of the scenario,leading to a feasible second-stage solution for each demand scenario, while re-specting supply capacities. Note this is a restriction, since the optimal second-stage assignment decisions are scenario-dependent in general. As a warm-up, weshow that static assignment policies are optimal for the uncapacitated case withunlimited supply at each open facility (i.e., there is a cost c i to open facility i with unlimited supply). We refer to this problem as uncapacitated robust facilitylocation ( URFL ). This is based on the intuition that each client can be assignedto the closest open facilities in an optimal solution in any scenario; this leads tooptimality of a static assignment policy for the LP relaxation (Theorem 1).
Theorem 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 atfacilities is constrained (or equivalently, there is a cost per unit of supply). Themain contribution in this paper is to show that static assignment policies give an O (log k/ log log k )-approximation for the LP relaxation of ( SCRFL ) (Theorem 2).We show this by constructing such a solution, starting from an optimal first-stagesupply. The optimal static assignment policies can be computed efficiently bysolving a compact LP.
Theorem 2.
A static assignment policy gives O (log k/ log log k ) -approximationfor the linear relaxation of ( SCRFL ) . Furthermore, the fractional supply in the first stage can be rounded to anintegral supply using ideas similar to rounding algorithms for the deterministic
El Housni et al. facility location [18]. In particular, the static assignment solution for the unca-pacitated case can be rounded to give a 4-approximation algorithm for (
URFL ).The static assignment solution for the soft-capacitated case can be roundedwithin a constant factor, which results in an O (log k/ log log k )-approximationalgorithm for ( SCRFL ). We would like to note that while the fractional assign-ment is static in our approximate LP solution, our integral assignment for anyclient in the second-stage depends on the other demand clients in the scenario;thereby, making our static assignment policy adaptive in implementation.
In this section, we consider the uncapacitated robust facility location problem ( URFL ) where for each i ∈ F , there is a cost c i to open facility i with unlimitedsupply. The problem can be stated as the following integer program, where eachbinary variable x i , i ∈ F indicates if facility i is opened and each y Sij , i ∈ F , j ∈ S, S ∈ C k indicates the assignment of client j to facility i in scenario S .min X i ∈F c i x i + max S ∈C k X i ∈F X j ∈ S d ij y Sij s.t. X i ∈F y Sij ≥ , ∀ S ∈ C k , ∀ j ∈ S,x i ≥ y Sij , ∀ i ∈ F , ∀ S ∈ C k , ∀ j ∈ S,x i ∈ { , } , y Sij ≥ , ∀ i ∈ F , ∀ S ∈ C k , ∀ j ∈ S. ( URFL)
Note that the second-stage problem is a transportation problem and sincethe demand of a client is integral (0 or 1), the optimal solution y Sij is integralas well. The special case of (
URFL ) where the uncertainty set contains only asingle scenario corresponds to the NP-hard classical and well-studied uncapaci-tated facility location problem, which is hard to approximate within a constantbetter than 1.463 unless NP has an O ( n O (log log n ) )-time algorithm [8]. We let( LP-URFL ) denote the linear relaxation of (
URFL ), where we replace x i ∈ { , } by x i ≥ i ∈ F . We would like to note that, in our model, the numberof scenarios could be exponential in the dimension of the problem. Hence, ingeneral, even the linear relaxation of such a problem could be challenging. How-ever, we show that ( LP-URFL ) can be solved in polynomial time using a
StaticAssignment Policy for the second-stage variables. Moreover, we can round thefractional solution losing only a constant factor, thereby getting a constant ap-proximation for (
URFL ). This section serves as a warm-up for introducing andmotivating static assignment policies before addressing the class of capacitatedrobust facility location problems. obust Facility Location Problems 5
Consider an optimal solution of (
URFL ). Since each open facility can have anunlimited amount of supply, each client in the realized scenario is assigned to theclosest facility among the opened ones. Therefore, a client j is always assignedto the same open facility in all scenarios S where j ∈ S . The same observationholds as well for ( LP-URFL ) where each client is assigned to the same fractionallyopened facilities independent of the realized scenario. Thus, the assignment of aclient is static. This can be captured by the following policy.
Static assignment policy.
There exists y ij ≥ i ∈ F , j ∈ C such that ∀ S ∈ C k , ∀ i ∈ F , ∀ j ∈ S, y
Sij = y ij . (1) Proof of Theorem 1 . Let ( x ∗ , y ∗ S , S ∈ C k ) be an optimal solution to (LP-URFL) .Since there are no capacities on facilities, each client j is assigned to the clos-est fractionally opened facilities. In particular, for each j ∈ C , let π j be apermutation of F = { , . . . , n } such that d π j (1) j ≤ d π j (2) j ≤ . . . ≤ d π j ( n ) j ,and let ℓ = min { p | x ∗ π j (1) + x ∗ π j (2) + . . . + x ∗ π j ( p ) ≥ } . Denote ˆ x π j ( ℓ ) =1 − ( x ∗ π j (1) + . . . + x ∗ π j ( ℓ − ) . The optimal solution can be written in the form (1)as follows: for S ∈ C k and j ∈ S ; y Sij = x ∗ i for i ∈ { π j (1) , . . . , π j ( ℓ − } , y Sij = ˆ x i for i = π j ( ℓ ), and y Sij = 0, otherwise. ⊓⊔ Let (
Static-URFL ) denote the problem after restricting the second-stage vari-ables y Sij in (LP-URFL) to a policy (1), which can then be reformulated as follows:min X i ∈F c i x i + max S ∈C k X i ∈F X j ∈C ( j ∈ S ) · d ij y ij s.t. X i ∈F y ij ≥ , ∀ j ∈ C ,x i ≥ y ij ≥ , ∀ i ∈ F , ∀ j ∈ C . ( Static-URFL )From Theorem 1, (
Static-URFL ) is equivalent to (LP-URFL) . The number ofvariables in (
Static-URFL ) is reduced to a polynomial number since the y ij nolonger depend on the scenario S . The inner maximization problem is still takenover an exponential number of scenarios; however we can separate efficiently overthese scenarios and write an efficient compact LP formulation for ( Static-URFL ):max S ∈C k { X i ∈F X j ∈C ( j ∈ S ) · d ij y ij } = max h ∈ [0 , |C| { X i ∈F X j ∈C d ij y ij h j | X j ∈C h j ≤ k } = min µ, ω ≥ { kµ + X j ∈C ω j | µ + ω j ≥ X i ∈F d ij y ij , ∀ j ∈ C} , (2)where the first equality holds because the optimal solution of the right maxi-mization problem occurs at the extreme points of the k -ones polytope, whichcorresponds to the worst-case scenarios of C k and the second equality followsfrom strong duality. Therefore, by dropping the min and introducing µ and all ω j as variables, we reformulate ( Static-URFL ) as the following linear program:
El Housni et al. min X i ∈F c i x i + kµ + X j ∈C ω j s.t. µ + ω j ≥ X i ∈F d ij y ij , ∀ j ∈ C , X i ∈F y ij ≥ , ∀ j ∈ C ,x i ≥ y ij , ∀ i ∈ F , ∀ j ∈ C ,x i ≥ , y ij ≥ , ω j ≥ , µ ≥ , ∀ i ∈ F , ∀ j ∈ C . (3)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 workon rounding techniques from the literature of deterministic facility location prob-lems. In fact, the LP rounding technique in Shmoys et al. [18], which gives a 4-approximation algorithm to the deterministic uncapacitated problem also gives a4-approximation algorithm for (
Static-URFL ). The idea is to define a ball aroundeach client of radius equal to the fractional assignment cost of the client (whichis independent of any scenario for our static policy). Then, we open facilities innon-intersecting balls of ascending radius. The result is given in the followingtheorem and for completeness, the details of the rounding are in Appendix A.
Theorem 3. ( 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 thebest constant approximation for (URFL) , but we introduce it as a warm-up formotivating the static assignment policy before presenting our main result in thenext section.
In this section, we consider the soft-capacitated robust facility location ( SCRFL )which is similar to (
URFL ) except that each facility i incurs a linear supply cost,where c i is the cost per unit of supply. We refer to x i as the supply (or capacity)in facility i . Each client in the realized scenario needs to be satisfied by one unitof supply. The problem is called soft-capacitated since there is no upper boundon x i , ( x i ∈ Z + ). The problem is given by the following integer program:min X i ∈F c i x i + max S ∈C k X i ∈F X j ∈ S d ij y Sij s.t. X i ∈F y Sij ≥ , ∀ S ∈ C k , ∀ j ∈ S,x i ≥ X j ∈ S y Sij , ∀ i ∈ F , ∀ S ∈ C k , ∀ j ∈ S,x i ∈ N , y Sij ≥ , ∀ i ∈ F , ∀ S ∈ C k , ∀ j ∈ S. ( SCRFL) obust Facility Location Problems 7
We let (
LP-SCRFL ) denote the linear relaxation of (
SCRFL ), where we replace x i ∈ N by x i ≥
0, for each i ∈ F . We would like to note that even the linearrelaxation ( LP-SCRFL ) is challenging to solve since it has exponentially-manyvariables (scenarios). Unlike the uncapacitated case, the static assignment policy(1) is not optimal for (
LP-SCRFL ) and the optimal assignment for each clientdepends, in general, on the realized scenario. In particular, the same client couldbe assigned to different facilities in different scenarios. In contrast, we showthe surprising result that a static assignment policy gives O (log k/ log log k )-approximation to ( LP-SCRFL ). Moreover, we can round the solution of the staticassignment policy to an integral solution for (
SCRFL ) and only lose an additionalconstant factor. We let (
Static-SCRFL ) denote the problem when we restrict thesecond-stage variables y Sij in (
LP-SCRFL ) to static assignment policies (1). Theproblem can then be reformulated as follows:min X i ∈F c i x i + max S ∈C k X i ∈F X j ∈C ( j ∈ S ) · d ij y ij s.t. X i ∈F y ij ≥ , ∀ j ∈ C ,x i ≥ max S ∈C k X j ∈C ( j ∈ S ) · y ij , ∀ i ∈ F ,x i ≥ , y ij ≥ , ∀ i ∈ F , ∀ j ∈ C . ( Static-SCRFL) O ( log k log log k ) -approximation algorithm Our main contribution in this section is to show that a static assignment pol-icy (1) gives O (log k/ log log k )-approximation for (LP-SCRFL) (Theorem 2). Toprove this theorem, we consider an optimal solution of ( LP-SCRFL ) and massageit to construct a solution of the form (1) while losing O (log k/ log log k ) factor.We first present our construction and several structural lemmas and then givethe proof of Theorem 2. Our construction.
Let x ∗ : ( x ∗ i ) i ∈F be an optimal first-stage solution of ( LP-SCRFL ), let OPT be the corresponding optimal first-stage cost and let OPT be the corresponding optimal second-stage cost. We will classify the clients C into three subsets C , C , C using Procedure 1 (below) and then specify a staticassignment policy for each subset. We use the following notation in the proce-dure. Let α > r = 5 · OPT /k . For ℓ ≥ j ∈ C , we let B ℓj denote theball centered at client j of radius ℓr . We initialize the sets F ← F and C ← C and update them at iteration, as explained in the procedure, until C becomesempty. We let Cl ( B ) denote the set of clients in C that are inside the ball B and let Sp ( B ) denote the total optimal supply of facilities F that are inside theball B , i.e., Sp ( B ) = X i ∈ F ( i ∈ B ) · x ∗ i and Cl ( B ) = { j ∈ C | j ∈ B } . El Housni et al.
Procedure 1
1: Initialize C ← C , C ← ∅ , C ← ∅ , C ← ∅ , F ← F .2: while C = ∅ do
3: Pick a client j ∈ C . Initialize ℓ = 14: if | Cl ( B ℓ − j ) | ≥ k then C ← C ∪ Cl ( B ℓ − j ), C ← C \ Cl ( B ℓ − j )6: Stop, return to line 27: if Sp ( B ℓj ) < · | Cl ( B ℓ − j ) | then C ← C ∪ Cl ( B ℓ − j ), C ← C \ Cl ( B ℓ − j )9: Stop, return to line 210: if Sp ( B ℓj ) ≥ α · | Cl ( B ℓ +1 j ) | then C ← C ∪ Cl ( B ℓ +1 j ), C ← C \ Cl ( B ℓ +1 j ),12: F ← F \ { i ∈ F | i ∈ B ℓj }
13: Stop, return to line 214: else ℓ ← ℓ + 1, return to line 4. Note that both Sp ( B ) and Cl ( B ) depend on the current sets of facilities F andclients C , 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 F and C .In the procedure, while the set C is not empty, we pick a client j ∈ C andgrow three balls around it: B ℓ − j ( internal ball), B ℓj ( medium ball) and B ℓ +1 j ( external ball) starting with ℓ = 1. For each ℓ , we check if the number of clientsin the internal ball B ℓ − j is greater than k (line 4); if this is the case, we removethem from C , put them in C and restart in line 2. If not, we check if the supplyin the medium ball B ℓj is sufficient to satisfy half of the clients in the internalball B ℓ − j (line 7); if that is not the case, we remove those clients from C , putthem in C , and restart in line 2. Otherwise, we finally check if the supply in thethe medium ball B ℓj is sufficient to satisfy a fraction 1 / α of the clients in theexternal ball B ℓ +1 j (line 10); if that is the case we remove all the clients in B ℓ +1 j and put them in C , we also remove all the facilities in B ℓj and restart in line2. If none of these three conditions holds, we increase ℓ to ℓ + 1. First, we showthat after at most log α k increments (i.e., ℓ ≤ log α k ), one of three conditionsmust hold and therefore we will remove some clients from C and restart in line 2.Which implies that after a finite number of iterations, the set C becomes empty.In particular, We have the following lemma. Lemma 1.
In Procedure 1, after a finite number of iterations, the set C becomesempty and C ∪ C ∪ C is equal to C . Moreover, ℓ is always less than log α k .Proof. Fix a client j and let ℓ ≥
1. If none of the three conditions (“if” state-ments) holds then α · | Cl ( B ℓ − j ) | ≤ α · Sp ( B ℓj ) < | Cl ( B ℓ +1 j ) | . obust Facility Location Problems 9 Therefore, the number of clients grows geometrically when we increase the radiusof the balls and by induction, we have that α ℓ ≤ α ℓ · | Cl ( B j ) | < | Cl ( B ℓ +1 j ) | , where | Cl ( B j ) | ≥
1, since Cl ( B j ) contains at least the client j . Hence, afterat most log α k increments, we will reach k clients, and must stop by the firstcondition, and return to line 2. Hence, we always have ℓ ≤ log α k . Finally since,we remove at least one client at each iteration of the while loop, the set C becomes empty after at most |C| iterations, and finally C ∪ C ∪ C = C . ⊓⊔ Now we are ready to present our static assignment policy for (LP-SCRFL) .The following three lemmas show our constructed static assignment for eachclient in the three subsets C , C , C . Moreover, we specify the supply used tosatisfy each subset of these clients and present the analysis for the assignmentcost.For each client in the set C , we know that it belongs to a ball with at least k clients. By the feasibility of the optimal solution, this implies that there exists k units of supply in x ∗ “close” to this ball. Hence, we satisfy this client by usingthe same fraction 1 /k of x ∗ (static assignment) while paying a small assignmentcost (roughly a constant times the radius of the ball). Since, there are at most k clients in each scenario, and each one is using at most x ∗ /k , we need to dedicateonly one x ∗ for all clients in C . Formally, we have the following lemma. Lemma 2.
There exists a static assignment policy for C such that each clientin C is using at most the supply x ∗ /k and has an assignment cost less than O (log α k/k ) · OPT , i.e., there exists (˜ y ij ) i ∈F ,j ∈ C such that for each j ∈ C : X i ∈F ˜ y ij ≥ , x ∗ i k ≥ ˜ y ij ≥ , ∀ i ∈ F , and X i ∈F d ij ˜ y ij = O (log α k ) · OPT k . Proof.
Let j be a client of C . It is sufficient to show that the following mini-mization problem is feasible and its optimal cost is O (log α k/k ) · OPT . Considermin (X i ∈F d ij y ij (cid:12)(cid:12)(cid:12)(cid:12) X i ∈F y ij ≥ , x ∗ i k ≥ y ij ≥ , ∀ i ∈ F ) . (4)Problem (4) must be feasible since the total supply in x ∗ is greater than thetotal demand in any scenario, i.e., P i ∈F x ∗ i ≥ k . Recall that a client j in C belongs to one of the sets Cl ( B ℓ − t ) for some t ∈ C and ℓ ≤ log α k (Lemma1) such that | Cl ( B ℓ − t ) | ≥ k . Consider a scenario S formed by k clients from Cl ( B ℓ − t ). Let denote y S the assignment of scenario S in the optimal solution.Consider the following candidate solution for (4): y ij = 1 k · X p ∈ S y Sip , ∀ i ∈ F . We have, by the feasibility of the optimal solution, for each i ∈ F , 0 ≤ y ij ≤ k x ∗ i and X i ∈F y ij = 1 k · X i ∈F X p ∈ S y Sip ≥ k X p ∈ S . Therefore, our solution is feasible for (4). Moreover, we have X i ∈F d ij y ij = 1 k · X i ∈F X p ∈ S d ij y Sip ≤ k · X i ∈F X p ∈ S d ip y Sip + 1 k · X p ∈ S d pj ≤ k · OPT + 2(2 ℓ − r ≤ OPT k + 2(2 log α k − · · OPT k = O (log α k ) · OPT k , where the first inequality follows from the triangle inequality and the fact that P i ∈F y Sip = 1 for all p ∈ S in the optimal solution. For the second inequality,we use the definition of OPT to bound the first term, the second term d pj isbounded by the diameter of the ball B ℓ − j which contains client j and all clients p ∈ S . ⊓⊔ Now, consider the set C . By construction, these are clients such that thereis not enough supply within a distance r = 5OPT /k to satisfy half of them.Therefore, intuitively they need to pay “large” distances in the optimal assign-ment cost if they show up all together in the same scenario. In the followinglemma, we show that we can have no more than k of these clients. As we wouldshow later, this would imply that we can dedicate a supply x ∗ to C and makea static assignment of all the clients C to this x ∗ . Lemma 3.
The set C has at most k clients.Proof. Suppose, for the sake of contradiction, that | C | > k . Let G , G , . . . , G T be the disjoint subsets of clients added at each iteration in the construction of C in Procedure 1. In particular, C = G ∪ G ∪ . . . ∪ G T for some T where:(i) for t = 1 , , . . . , T, G t = Cl ( B ℓ t − j t ) for some client j t and 1 ≤ ℓ t ≤ log α k .(ii) the supply Sp ( B ℓ t j t ) is less than half of the clients in G t .(iii) each set G t has strictly less than k clients, since the procedure has to failthe first “if” statement before adding G t into C .Recall that Sp ( B ℓ t j t ) = X i ∈ F ( i ∈ B ℓ t j t ) · x ∗ i , where F is the current set of facilities in the procedure (and is not all F sincesome facilities have been removed in line 12 of the procedure). However, wewould like to emphasize that when a facility has been removed (in line 12 of the obust Facility Location Problems 11 procedure), all clients within distance r from this facility were removed as well(line 11). This is true, since when we remove the facilities in a medium ball, say B ℓj , (line 12), we remove all clients in the corresponding external ball B ℓ +1 j (line 11). Hence, the remaining clients in C are at least (2 ℓ + 1) r − (2 ℓ ) r = r away from the removed facilities. In particular, for the clients G t , the supply thathas been removed before they were added to C is at least r away from them.Therefore, all the facility in F that are within a distance r from a client in G t belong to the set F that verifies P i ∈ F ( i ∈ B ℓ t j t ) · x ∗ i ≤ ·| G t | . This implies thatthe supply of all facilities within a distance r from G t in the optimal solution,is less than half of the clients G t . Hence, if all of the clients G t show up in ascenario, the optimal second-stage solution needs to pay an assignment cost ofat least r · | G t | / G t according to their cardinalities: wlog assume that | G | ≥| G | ≥ . . . ≥ | G T | . We construct a scenario ˆ S by taking clients from the sets G , G , . . . until we hit k . This is possible since by assumption | C | > k . Assumethat | G | + | G | + . . . | G p − | + | ¯ G p | = k, for some p , where 2 ≤ p ≤ T . Note that ¯ G p is a subset of G p , since we can reach k before taking all the clients of the last set G p . For each t = 1 , . . . , p −
1, theoptimal second-stage decision needs to pay at least r · | G t | /
2. Therefore,OPT ≥ · r · ( | G | + | G | + . . . | G p − | ) . We did not include G p , since not all these clients are necessary in the scenarioˆ S , but only | ¯ G p | of them. Since ¯ G p has the smallest cardinality | G | + | G | + . . . | G p − | ≥
12 ( | G | + | G | + . . . | G p − | + | ¯ G p | ) = k . Therefore, OPT ≥ r · k · OPT , which is a contradiction. Therefore, | C | ≤ k . ⊓⊔ Finally, for clients C , we show that there exists | C | / α units of supply“close” to them. In particular, we can multiply these units by 2 α , dedicate themto C and make a static assignment for C . We have the following lemma. Lemma 4.
There exists a supply ˆ x that has a cost at most α · OPT and thereexists a static assignment policy such that all clients in C are assigned to supply ˆ x and each client in C has an assignment cost that is O (log α k/k ) · OPT , i.e.,there exists (ˆ y ij ) i ∈F ,j ∈ C and (ˆ x i ) i ∈F such that for all j ∈ C : X i ∈F ˆ y ij ≥ , ˆ x i ≥ X j ∈ C ˆ y ij , ∀ i ∈ F , ˆ x i ≥ , ˆ y ij ≥ , ∀ i ∈ F , X i ∈F c i ˆ x i ≤ α · OPT and X i ∈F d ij ˆ y ij = O (log α k ) · OPT k . Proof.
Let G , G , . . . , G T be the disjoint subsets of clients added at each itera-tion to construct C in Procedure 1. In particular, C = G ∪ G ∪ . . . ∪ G T forsome T such that: for all t = 1 , , . . . , T, G t = Cl ( B ℓ t +1 j t ) for some client j t andsome integer ℓ t with 1 ≤ ℓ t ≤ log α k . Moreover, the supply Sp ( B ℓ t j t ) is greaterthan a 1 / α fraction of the clients in G t . Hence, for each ball B ℓ t j t , we multiplythe supply by 2 α , move it to the cheapest facility in this ball and make a staticassignment of all clients G t to this cheapest facility. Since the supply in B ℓ t j t isremoved along with clients G t , it will not be used by the other clients in C .Formally, let i t be the cheapest facility in the ball B ℓ t j t . We define, for eachfacility i in B ℓ t j t , ˆ x i = 2 α P i ′ ∈ F ( i ′ ∈ B ℓ t j t ) · x ∗ i ′ if i = i t and ˆ x i = 0 otherwise.For each client j ∈ C , we let ˆ y ij = 1 for i = i t and j ∈ G t , and let ˆ y ij = 0,otherwise. Therefore, the first desired constraints in the lemma are verified. Letus check the last one. The distance between a client and its assigned facility inour solution is at most r plus the diameter of the ball B ℓ t j t , i.e., r + 4 ℓ t r ≤ (4 log α k + 1) · · OPT /k = O (log α k/k ) · OPT . ⊓⊔ Proof of Theorem 2.
Let (˜ y ij ) i ∈F ,j ∈ C be the solution given in Lemma 2 for sat-isfying the clients in C . We dedicate a supply x ∗ to clients C . Let (ˆ y ij ) i ∈F ,j ∈ C and (ˆ x i ) i ∈F be the solution given in Lemma 4 for satisfying the clients in C .Finally, we know from Lemma 3 that C has at most k clients, and therefore C is a scenario. So we dedicate a supply x ∗ to C and let the optimal assignment y C ij be our static assignment solution for C . In particular, we give the followingsolution to (LP-SCRFL) , where the first stage solution is 2 x ∗ + ˆ x and the staticassignment policy is for all i ∈ F : y ij = ˜ y ij for j ∈ C , y ij = y C ij for j ∈ C , and y ij = ˆ y ij for j ∈ C . It is clear that P i ∈F y ij ≥
1, for each j in C ∪ C ∪ C .Moreover, for any scenario S ∈ C k and i ∈ F , X j ∈ S y ij = X j ∈ S ∩ C ˜ y ij + X j ∈ S ∩ C y C ij + X j ∈ S ∩ C ˆ y ij ≤ X j ∈ S ∩ C x ∗ i k + x ∗ i + ˆ x i ≤ x ∗ i + ˆ x i . Therefore, our solution is feasible for (LP-SCRFL) . Let us evaluate its cost. Thecost of the first stage is at most 2
OP T + 2 α OPT = O ( α ) · OPT . For thesecond-stage cost, consider any scenario S ∈ C k , We have X i ∈F X j ∈ S d ij y ij = X j ∈ S ∩ C X i ∈F d ij ˜ y ij + X j ∈ S ∩ C X i ∈F d ij y C ij + X j ∈ S ∩ C X i ∈F d ij ˆ y ij ≤ X j ∈ S ∩ C O (log α k ) · OPT k + OPT + X j ∈ S ∩ C O (log α k ) · OPT k ≤ O (log α k ) · OPT + OPT + O (log α k ) · OPT = O (log α k ) · OPT . obust Facility Location Problems 13 By balancing the terms α and log α k , we choose α = log k/ log log k which gives O (log k/ log log k )-approximation to (LP-SCRFL) . ⊓⊔ Similar to the uncapacitated problem, we can solve (
Static-SCRFL ) efficientlyusing a compact linear program. In fact, we dualize the inner maximizationproblem in the objective function of (
Static-SCRFL ) in the same way as (2). Inaddition to that, we reformulate the second constraint in (
Static-SCRFL ) usingthe same dualization technique as follows: for each i ∈ F ,max S ∈C k { X j ∈ S y ij } = max h ∈ [0 , | C | { X j ∈C y ij h j | X j ∈C h j ≤ k } = min η i ,λ ij ≥ { kη i + X j ∈C λ ij | η i + λ ij ≥ y ij , ∀ j ∈ C} . The linear program is given bymin X i ∈F c i x i + kµ + X j ∈C ω j s.t. µ + ω j ≥ X i ∈F d ij y ij , ∀ j ∈ C , X i ∈F y ij ≥ , ∀ j ∈ C ,x i ≥ kη i + X j ∈C λ ij , ∀ i ∈ F ,η i + λ ij ≥ y ij , ∀ i ∈ F , ∀ j ∈ C ,x i , y ij , λ ij , η i , ω j , µ ≥ , ∀ i ∈ F , ∀ j ∈ C , (5)which can be reduced, after removing the variables y ij , tomin X i ∈F c i x i + kµ + X j ∈C ω j s.t. µ + ω j ≥ X i ∈F d ij ( η i + λ ij ) , ∀ j ∈ C , X i ∈F η i + λ ij ≥ , ∀ j ∈ C ,x i ≥ kη i + X j ∈C λ ij , ∀ i ∈ F ,x i , λ ij , η i , ω j , µ ≥ , ∀ i ∈ F , ∀ j ∈ C . (6)Finally, we round the optimal solution of ( Static-SCRFL ) to an integral so-lution using the filtering and rounding techniques from Shmoys et al. [18] whilelosing only a factor of 12. Again, this rounding technique was designed forthe deterministic facility location problem, but the same argument works as well for (
Static-SCRFL ). Finally, since (
Static-SCRFL ) gives O (log k/ log log k )-approximation to ( LP-SCRFL ) and we only loose a constant factor in the round-ing, this results in O (log k/ log log k )- approximation algorithm for ( SCRFL ). Westate the result in the following theorem and, for completeness, we present thedetails of the rounding in Appendix B.
Theorem 4. ( Static-SCRFL ) can be rounded to give O ( log k log log k ) -approximationalgorithm to ( SCRFL ) . Note that after rounding the supply in our solution of (
Static-SCRFL ), the inte-gral second-stage assignment for each realized scenario is a transportation prob-lem and therefore its optimal solution is integral. We would like to emphasizethat while the fractional assignment in our solution is static, our integral assign-ment is not necessarily static. In fact, a policy with an integral static assignmentcould even be bad our model.
In this paper, we give a O (log k/ log log k )-approximation for soft-capacitated ro-bust facility location problems with an implicit model of demand uncertainty. Itis an interesting open question to study whether there exists a constant approx-imation algorithm for the problem, even in special cases such as the Euclideanmetric. Our solution approach relies on static fractional assignment policies,which we show are optimal for the uncapacitated problem and give a strongtheoretical guarantee for soft-capacitated case. Static assignment policies, whilereasonable for the case of soft-capacities can be shown to be arbitrarily bad forthe case of hard-capacities, where in addition to cost per unit, there is also anupper bound on supply at each facility. It is another interesting open directionto study any non-trivial approximation in this setting. obust Facility Location Problems 15 References
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A Rounding and proof of Theorem 3
Rounding (Shmoys et al.[18]). Recall that n is the number of facilities and m is the number of clients. Let ( x ∗ , y ∗ ) = (cid:0) ( x ∗ i ) i ∈F , ( y ∗ ij ) i ∈F ,j ∈C (cid:1) be an opti-mal solution for ( Static-SCRFL ). Let OPT and OPT respectively be the cor-responding optimal first-stage and second-stage cost. In particular, OPT =max S ∈C k P i ∈F P j ∈C ( j ∈ S ) · d ij y ∗ ij . For each client j ∈ C , let L j = X i ∈F d ij y ∗ ij . We sort L j in ascending order. Suppose without loss of generality, that L ≤ L ≤ . . . ≤ L m . For each client j ∈ C , we consider B j the ball centered at theclient j with radius αL j (where α ≥ L j and open the cheapest facilityin each selected ball. That is, we consider every client in this sorted order, andonly include its ball if it does not intersect any previously selected ball. Eachclient is assigned to its closest opened facility. For each client j ∈ C , L j ≥ X i/ ∈ B j d ij y ∗ ij ≥ X i/ ∈ B j αL j y ∗ ij . Hence, 1 α ≥ X i/ ∈ B j y ∗ ij , which implies X i ∈ B j y ∗ ij ≥ − α . Therefore, X i ∈ B j x ∗ i ≥ − α . By summing over the non-intersecting balls, the cost of our opened facilities isless than − α OPT . Now let us consider the assignment cost. For client j , if itsball B j is chosen by the greedy procedure above, then the client is assigned tothe opened facility i in B j and d ij ≤ αL j . If not, the client j is assigned to a obust Facility Location Problems 17 facility i no further away than the one chosen in the ball B k that intersected B j .By the triangle inequality: d ij ≤ d iv + d jv ≤ αL j + 2 αL k ≤ αL j , where v ∈ B j ∩ B k and L k ≤ L j . Hence, for any scenario S ∈ C k , the second-stage cost is at most 3 α · P j ∈ S L j , which is at most 3 α · OPT . By optimizingover α , we take α = , which results in the 4-approximation with respect toOPT + OPT . Moreover, since ( Static-URFL ) is equivalent to (LP-URFL) and (LP-URFL) is a relaxation of (
URFL ), the rounding above gives 4-approximationalgorithm for (
URFL ). B Rounding and proof of Theorem 4
Rounding (Shmoys et al.[18]). Let ( x ∗ , y ∗ ) = (cid:0) ( x ∗ i ) i ∈F , ( y ∗ ij ) i ∈F ,j ∈C (cid:1) be anoptimal solution of ( Static-SCRFL ). Let z = max S ∈C k P i ∈F ,j ∈ S d ij y ∗ ij and set α ∈ (0 , j ∈ C , let π be a permutation of facilities that serve j in the optimal solution of ( Static-SCRFL ) such that d π (1) j ≤ d π (2) j ≤ . . . ≤ d π ( n ) j . Define the radius d j ( α ) = d π ( i ∗ ) j where i ∗ = min { i ′ | P i ′ i =1 y ∗ π ( i ) j ≥ α } .Following the same definitions in Shmoys el al. [18], we say that a solution ( x , y )is g -close if for any j ∈ C , y ij > d ij ≤ g j . Claim.
Given a feasible solution ( x , y ), we can construct a g -close feasible so-lution (¯ x , ¯ y ) such that ¯ x = α x , ¯ z = α z and g j ≤ d j ( α ) , ∀ j ∈ C where¯ z = max S ∈C k P i ∈F ,j ∈ S d ij ¯ y ij . Proof.
In fact, for each j , we set ¯ y π ( i ) j = y π ( i ) j /α for 1 ≤ i ≤ i ∗ and ¯ y π ( i ) j = 0otherwise. We have X i ∈F ¯ y ij = i ∗ X i =1 α y π ( i ) j ≥ . The second constraint in (
Static-SCRFL ) is trivially verified since x is multipliedby α and y ij are either multiplied by α or set to 0. Moreover, all edges in oursolution with a positive flow have a cost at most d j ( α ). Hence, (¯ x , ¯ y ) is g -closewith g j ≤ d j ( α ) for all j . ⊓⊔ Let (¯ x , ¯ y ) be the g -close solution corresponding to the optimal solution( x ∗ , y ∗ ). For the ease of notation, we just use the notation ( x , y ) instead of(¯ x , ¯ y ) in the rest of the proof. We round-up all x i ≥ to ⌈ x i ⌉ . We pay at most afactor 2 in the first-stage cost. Let us focus on the remaining fractional facilities,say ˆ F , i.e., ˆ F = { i ∈ F | < x i < / } . We let ˆ C denote the set of clients such that half of their supply is coming fromˆ F , i.e., ˆ C = { j ∈ C | X i ∈ ˆ F y ij ≥ } . We sort the clients in ˆ C in ascending order of g j and do the following until ˆ C becomes empty. We take the client j with the smallest g j in ˆ C , let say j ′ . Let V = { i ∈ ˆ F : y ij ′ > } , and T = { j ∈ C : ∃ i ∈ V s.t. y ij > } . We put the ⌈ P i ∈ V x i ⌉ units of supply in the cheapest facility in V which wedenote f c . We set each x i in V \ { f c } to 0. The clients in T are the ones affectedby this change. We will route all of their demand from V to f c and make thisassignment static. Note that, at this step, we have routed only their demandfrom V and not their entire demand. This is feasible because ⌈ X i ∈ V x i ⌉ ≥ X i ∈ V x i ≥ X j ∈C ( j ∈ S ) · X i ∈ V y ij for any scenario S . Since we choose the clients in ˆ C in ascending order of g j , thetriangle inequality ensures that the solution is 3 g -close. We keep doing this untilˆ C becomes empty.Let us analyze the final cost. We lost a factor α in both first-stage andsecond-stage cost. Then, we lost a factor 2 in the first-stage cost by rounding upthe solution and moving the supply to the cheapest facilities after each rounding.We lost another factor 2 in the first stage to satisfy C \ ˆ C . For the second stage,we have a solution that is 3 g -close with g j ≤ d j ( α ). Note that d j ( α ) ≤ − α X i ∈F d ij y ij . In particular, for any scenario S , X j ∈ S g j ≤ − α X j ∈ S X i ∈F d ij y ij . Hence, we loose − α in the second stage. Overall, the factor is4 α OPT + 3 α (1 − α ) OPT . We set α = , which gives 12 approximation to ( LP-SCRFL ). Finally, since(