Hardness of an Asymmetric Stackelberg Network Pricing Game Davide Bilò, Luciano Gualà, Guido Proietti

Pricing problems: a wide class of optimization problems pricing problems for Unit-Demand Consumers: Min-buying model Max-buying model Rank-buying model With/without price ladder constraint … pricing problems for Single-Minded Consumers: general on a network Network Pricing games Bilevel optimization problems with a game-theoretic flavour

Network pricing game 2-player game we have a (di)graph G=(V,E=RB) each eR has a fixed cost c(e) leader sets a price p(e) for each eB the follower computes (buys) a (feasible) subgraph H(p) of G which minimizes (H,c,p) leader’s revenue: rev(H(p)) goal: find prices in order to maximize the revenue

Example: StackMST The follower computes an MST T(p) of G 1 6 6 (T,c,p) =  c(e)+ p(e) 7 eT eT 4 3 8 2 2 1 1 1 rev(T(p)) = p(e) 4 eT(p) 1 1

Example: StackMST The follower computes an MST T(p) of G 6 6 1 (T,c,p) =  c(e)+ p(e) 7 eT eT 4 3 8 2 2 1 1 1 rev(T(p)) = p(e) 4 eT(p) 1 1 The revenue is 13

Related work: StackSP [Roch et al., Networks’05] strongly NP-hard O(log |B|)-apx [Joret, tech. report’08] APX-hard [Briest,Khanna, tech. report’09] Not approximable within 2-

Related work: StackMST [Cardinal et al., WADS’07] APX-hard O(log |B|)-apx (single-price algorithm) [Cardinal et al., WINE’09] NP-hard for planar graphs Polynomial-time solvable for bounded treewidth graphs

Related work [Briest et al., STACS’08]: O(log |B|+log f)-apx for binary demand Stackelberg games with f followers binary demand: Single-price algorithm (H,c,p) =  c(e)+ p(e) eH eH rev(H(p)) = p(e) eH(p)

Related work [Bilò,Gualà,Proietti,Widmayer, WINE’08] (symmetric) StackSPT NP-hard Efficient poly-time algorithm when |B| is constant (T,c,p) =  dT(r,v) =  loadT(e)c(e)+ loadT(e)p(e) v eT eT rev(T(p)) = loadT(e)p(e) eT(p)

Related work [Briest,Hoefer,Gualà,Ventre, WINE’09]: What when follower’s problem is NP-hard? Follower uses a 2-apx algorithm for Min-Knapsack Strong NP-hard (2+)-apx Follower uses a primal-dual algorithm for Set Cover APX-hard when element frequency f>2 exact poly-time algorithm when f=2

Asymmetric Stackelberg Shortest Path Tree (ASSPT) game The follower computes an SPT T(p) rooted at r r (T,c,p) =  dT(r,v) 1 4 v 4 ? =  loadT(e)c(e)+ loadT(e)p(e) 8 eT eT ? 6 ? rev(T(p)) = p(e) 1 eT(p)

Our results Not approximable within n1/2- (n-1)-apx algorithm unweighted star Equivalence with the Rooted Maximum Leaf Outbranching (RMLO) problem APX-hard in general, NP-hard for DAGs [Alon et al., Discrete Math., 2009] 92-apx algorithm [Daligault, Thomassé,IWPEC’09] APX-hard for DAGs unweighted chain Exact poly-time algorithm when G=(V, RB) is a DAG NP-hard otherwise

Theorem For every >0, the ASSPT game is not approximable within a factor n1/2-, even on DAGs and when c(e){0,1}, for each eR reduction from Maximum Independent Set (MIS) problem MIS not approximable within n1-, for every >0, unless P=NP

G G’ of n nodes of N=(n2) nodes v v n copies of v 1 u u r n copies of u 1 1 v’ v’ n copies of v’ Claim 1: There exists an optimal pricing p* with the form p*:B  {0,1} Claim 2: Any pricing p:B  {0,1} defines an independent set Ip and the revenue of p is at most |Ip|n+n-|Ip| Claim 3: rev(p*) is at least |I*|n + n-|I*|, where I* is a MIS

G G’ ≥ ≥ > > rev(p*) n|I*| + n - |I*| n|I*| n1-’ of n nodes of N=(n2) nodes v v n copies of v 1 u u r n copies of u 1 1 v’ v’ n copies of v’ rev(p*) n|I*| + n - |I*| n|I*| ≥ ≥ > n1-’ n|Ip| + n - |Ip| rev(p) n|Ip| + n |I*| > n1- |Ip|

Theorem The exists an (n-1)-approximation algorithm for the ASSPT game price every edge e=(u,v) as p(e)=max{0,d(v)-d(u)} d(x): red distance, i.e. distance from r to x in G=(V,R) Notice: every blue edge e with p(e)>0 is selected by the follower

T(p*) r : the path in T(p*) with maximum revenue rev()≥ 1/(n-1) r* we’ll show that rev(T(p)) ≥ rev()

since every ei with p(ei)>0 is selected k-1   d(vi)+wi+1 ≥ d(ui+1) i=1,…,k-1 r i=1 w1 = d(u1) w1 d(uk) = d(uk) u1 e1 k k v1   (d(vi)-d(ui)) ≥ d(vk) - wi w2 i=1 i=1 u2 e2 v2 k  w3 rev()  d(vk)- wi i=1 . since every ei with p(ei)>0 is selected uk k  ek rev(T(p)) ≥ (d(vi)-d(ui)) vk i=1

Open problems Close the gap between lower and upper bounds Is the single-price algorithm the best algorithm one can design?

Positioning MST problem leader places non-parallel blue edges and price them Then, the follower computes an MST Leader collects the revenue Constraints: Leader has a budget B a placement cost (u,v), for each feasible position must place edges with total placement cost at most B T 4 4 8 1 1 3 6 7 2

Positioning MST problem 3 variants: without placement costs Uniform placement costs Unrestricted placement costs a modification of the single-price algorithm always guarantees a O(log n)-apx Subcase of StackMST ??? can we say more? generalizes StackMST

Thank you