Make or Buy: Revenue Maximization in Stackelberg Scheduling Games

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1 Make or Buy: Revenue Maximization in Stackelberg Scheduling Games Toni Böhnlein, Oliver Schaudt, and Joachim Schauer 3 Universität zu Köln, Institut für Informatik, Weyertal 80, 5093 Köln, boehnlein@zpr.uni-koeln.de RWTH Aachen, Institut für Mathematik, Pontdriesch 0, 506 Aachen, schaudt@mathc.rwth-aachen.de 3 University of Graz, Department of Statistics and Operations Research, Universitätsstr. 5, 800 Graz, joachim.schauer@uni-graz.at Abstract In a Stackelberg pricing game a distinguished player, the leader, chooses prices for a set of items, and the other player, the follower, seeks to buy a minimal cost feasible subset of the items. The goal of the leader is to maximize her revenue, which is determined by the sold items and their prices. Typically, the follower is given by a combinatorial covering problem, e.g., his feasible subsets are the edges of a spanning tree or the edges of an s-t-path in a network. We initiate the study of Stackelberg pricing games where the follower solves a maximization problem. In this model, the leader offers a payment to include her items in the follower s solution. Our motivation stems from the following situation: assume the leader has a set of jobs,..., k to complete. A job i may either (a) be completed for a given cost b(i) using her own resources or (b) offered to the follower at a variable price p(i) to complete it for her. The objective function to be maximized by the leader is the sum of the margins b(i) p(i) over those jobs i that are completed by the follower. Informally, the question is which jobs should be outsourced and what profit the leader has to offer. Our main result says that the problem can be solved to optimality in polynomial-time when the jobs have fixed starting and terminating times and the follower is computes a maximum weight scheduling on a single machine. To show that the situation changes when the follower is given by other optimization problems, we prove APX-hardness for a scheduling problem that can be modeled as a bipartite maximum weight matching problem. Moreover, we show APX-hardness in the case of the maximum weight spanning tree problem. On a more general note, we prove Σ p -completeness if the follower has a general combinatorial optimization problem given in the form of a finite ground set and a feasibility oracle. This shows that while the follower s problem is NP-complete, the leader s problem is hard even if she has an NP-oracle at hand. 998 ACM Subject Classification G..0 Discrete Mathematics General Keywords and phrases Algorithmic pricing, Stackelberg games, Revenue maximization. Introduction Suppose an agent seeks to complete a set of jobs,..., k. Job i may either (a) be executed for a given fixed cost b(i) using the agent s own resource or (b) offered to a manufacturer at a variable price p(i) to carry it out for him. If the manufacturer finishes an offered job i, the agent pays the price p(i). The agent s objective is to maximize the sum of the margins b(i) p(i) over those jobs i that are finished by the manufacturer. Typically, this is called a make or buy decision. Toni Böhnlein, Oliver Schaudt, Joachim Schauer; licensed under Creative Commons License CC-BY Leibniz International Proceedings in Informatics Schloss Dagstuhl Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany

2 Stackelberg Pricing Games Whether it is profitable to outsource a job or not, depends on the manufacturer s offer situation. The agent might have competitors who also offer a payment to the manufacturer to carry out their jobs. Moreover, the manufacturer s schedule has to obey a number of constraints. For instance, it might be impossible to execute two jobs in the same time window since the manufacturer only has one machine available. When setting the prices, the agent is aware of the competitors jobs, the constraints they imply and the offered payments. After prices are set, the manufacturer selects a feasible subset of all jobs offered by the agent and her competitors. His objective is to maximize the income. We study the problem of computing prices that are optimal for the leader, for different types of constraints of the manufacturer s schedule. This class of pricing problems features a hierarchical dependency. First, the agent sets prices; then the manufacturer selects a set of jobs. In the literature, such problems are captured by a game-theoretic model called Stackelberg Pricing Games. Originally, in a Stackelberg Pricing Game one player chooses prices for a number of items. After that, one or several other players are interested in buying these items. Following the standard terminology, the player to choose the prices is called the leader while the other players are called followers. The goal of the leader is to maximize her revenue while followers want to minimize their costs. Depending on the follower s preferences, computing optimal prices can be a highly non-trivial problem. A major line of research studies Stackelberg Pricing Games where the follower s preferences are given by a combinatorial optimization problem. Labbé et al. [4] model road-toll setting problems by a Stackelberg Pricing Game based on the shortest path problem. In this game, the leader sets prices for a subset of priceable edges of a network graph while the remaining edges have fixed costs. Each follower has a pair of vertices (s, t) and buys a minimum cost path from s to t. The cost of a path depends on both the fixed cost and the prices set by the leader. Roche et al. [6] show that the problem is NP-hard, even if there is only one follower, and it has later been shown to be APX-hard [6, 3]. More recently, other combinatorial optimization problems were studied in their Stackelberg Pricing Game version. For example, Cardinal et al. [0, ] investigate the Stackelberg Minimum Spanning Tree Game, proving APX-hardness and giving some approximation results. Our contribution is a model to capture scenarios where the follower solves a maximization problem. To model the make or buy problem sketched above, we introduce a Stackelberg Pricing Game that is based on the well-known Interval Scheduling Problem. In this problem, there is one machine and a set of weighted jobs I. Each job i has a fixed starting time s i R and termination time t i R. Hence, a job can be represented by an interval [s i, t i ] on the line. We say that two intervals overlap if their intersection is non-empty. The objective is to find a subset of non-overlapping intervals of maximum total weight. On the left-hand side in Figure we have an instance of an interval scheduling problem if we only consider the solid intervals a, b, c, d. An optimal solution is the set {a, b} with a total weight of 7. According to every algorithms textbook, this problem can be solved efficiently via dynamic programming. We call the agent leader and the manufacturer follower. In our Stackelberg Pricing Game, the solid intervals a, b, c, d represent the jobs of the competitors. The dashed lines x, z, y are the jobs of the leader with their respective costs 4, 3, 5. First, the leader has to set prices p x, p y and p z. Higher prices are more appealing to the follower. However, lower prices are more profitable for the leader. On the right hand side of Figure the prices p x = 3, p y = 0, p z = 4 are set. The follower selects the jobs c, x, z with total weight 9. Note that the prices p x and p z are optimal in the following sense: if either of p x or p z is decreased by some ε, the intervals x or z are not selected by the follower. The leader obtains a margin

3 T. Böhnlein, O. Schaudt, J. Schauer 3 of under these prices. The optimal prices p x =, p y =, p z = 0 yield margin 4. Under these prices the solutions {b, c, x, y} and {a, b} are optimal for the follower; both have a weight of 7. A common assumption for Stackelberg Pricing Games is that the follower is cooperative: he always chooses the optimal solution which is most profitable for the leader. a c x 5 d p x /4 y p y /3 4 z b p z /5 a c x 5 3/4 d y 0/3 4 z b 4/5 Figure Stackelberg Interval Scheduling Another scheduling problem that fits into our scenario can be formalized as a matching problem. Here, the follower has m machines, and there are n jobs. Each machine can execute at most one job in total. This situation can be modeled by a bipartite graph G = (U V, E). The vertices in U correspond to the m machines and the vertices in V to the n jobs. If the follower receives a payment to execute job i on machine j, there is an edge connecting the respective vertices with the payment as its weight. Hence, the follower solves a maximum weight matching problem to maximize his income. We extend this to a Stackelberg Pricing Game, as follows. Say k of the n jobs belong to the leader. An edge e which is incident to the corresponding vertices is a priceable edge and has cost b(e). The leader s objective is to set prices p(e) that maximize the sum of the margins p(e) b(e) over the priceable edges that are part of a maximum weight matching computed by the follower. Our results. To make things more formally, we use a slightly different notion as in the introductory scenario. We say the leader receives a benefit b(i) if job i is executed by the follower. The leader s objective is to maximize her revenue, the sum of the margins b(i) p(i) over the jobs i executed by the follower. Our main result is a polynomial time algorithm that solves the Stackelberg Interval Scheduling Game. Since it will be more handy later on, we restate this game in terms of an independent set problem in an interval graph. Associated with an instance of the interval scheduling problem I we construct the corresponding interval graph G = (V, E) with vertex weights. For each interval i I there is a corresponding vertex v i V with weight w(v i ) = w(i). If two intervals i, j overlap, there is an edge (v i, v j ) E. Recall that an independent set in a graph is a subset of mutually non-adjacent vertices. The interval scheduling problem is equivalent to finding a maximum weight independent set of G. Stackelberg Interval Scheduling (SIS) Input: An interval graph G = (V, E) with priceable vertices P V and P = k. For every v P there is a benefit b(v) R and for every u F = V \ P there is a fixed weight w(u) R. Objective: Find a price function p : P R 0 maximizing max{b(s P ) p(s P ) S is a maximum weight independent set}, where the weight of an independent set S is defined as w(s F ) + p(s P ).

4 4 Stackelberg Pricing Games The assumption that the follower picks a solution maximizing the leader s revenue is common for such pricing games and made to avoid technicalities (cf. [5]). The main result of our paper reads as follows. Theorem. The SIS problem can be solved in time O(k 3 ( V (G) + E(G) )) given an instance (G, P, w, b) with P = k. The proof builds on a careful analysis of the structure of an optimal price function. It analyses on the linear programming formulation of a restricted version of the SIS problem. Based on these insights we use dynamic programming to solve the full problem. We remark that our theorem is one of the few cases in which one can solve a Stackelberg Pricing Problem based on a non-trivial optimization problem to optimality. As a complementary result, we show that the second scheduling problem mentioned above the one that is based on a matching problem is APX-hard. We formulate the corresponding Stackelberg Pricing Game for general graphs. Stackelberg Matching Game Input: A graph G = (V, E) with priceable edges P E where k = P. For every e P there is a benefit b(e) R and for every e F = E \ P there are fixed weights w(e) R. Objective: Find a price function p : P R 0 maximizing max{b(m P ) p(m P ) M is a maximum weight matching}, where the weight of a matching M is defined as w(m F ) + p(m P ). The main result of Section 3 is that this problem is hard to approximate, unless P=NP. Theorem. The Stackelberg Matching Game is APX-hard even when the graph G is bipartite, w(f) {, } for all f F, and b(e) = 4 for all e P. As we will see in the proof of the theorem, the graph used for the reduction has another special property. For a priceable edge e = (u, v), the edges with fixed weights that are adjacent to e are either incident to u or v. Hence, the pricing problem for the second scheduling problem which we mentioned in the introduction belongs to the class of APX-hard problems. Moreover, the theorem shows that the Stackelberg Interval Scheduling Game on perfect graphs instead of interval graphs is APX-hard. The matching problem on a graph is equivalent to the independent set problem on its line graph. Line graphs of bipartite graphs are perfect graphs. Additionally, in Section 4 we show that a maximization version of the Stackelberg Pricing Game for spanning trees is APX-hard. This follows readily with the proof by Cardinal et al. [0] for the minimization version. Yet, it shows that a Stackelberg Pricing Game based on a maximization problem is hard if the follower optimizes over a matroid. In Section 5 we take a step back from our example problems and study the complexity of the Stackelberg Pricing Game in its own right. It turns out that there are combinatorial optimization problems in NP such that the corresponding Stackelberg Pricing Problem is Σ p complete. In other words, such a pricing problem is computationally difficult even if an NP oracle is provided, unless the polynomial hierarchy collapses to the second level.

5 T. Böhnlein, O. Schaudt, J. Schauer 5 Related work. A substential body of work focuses on specific network problems in their Stackelberg Pricing Game version. As mentioned before, Labbé et al. [4] use the Stackelberg Shortest Path Game to model road-toll problems. They establish NP-hardness and use LP bilevel formulations to solve smaller instances of the problem. Roche et al. [6] formulate a combinatorial approximation algorithm with logarithmic approximation guarantee. Moreover, a lower bound on the approximability is due to Briest et al. [6]: they show that the Stackelberg Shortest Path Game is NP-hard to approximate within a factor of less than. This is an improvement above previous results by Joret [3] showing APX-hardness of this pricing problem. For a comprehensive survey on the Stackelberg Shortest Path Game and bi-level programming see Hoesel [8] or Labbé and Violin [5]. Bilo et al. [4] show that maximizing the revenue in the Stackelberg Shortest Path Tree Game is NP-hard, and give an efficient algorithm assuming that the number of priceable edges is constant. Later their algorithm was improved by Cabello [9]. Briest et al. [8] develop a polynomial time algorithm for a special case of the Stackelberg Bipartite Vertex Cover Game. Their algorithm is based on max-flow computations. An improved algorithm, building on the preflow-push algorithm, was later given by Baïou and Barahona []. Cardinal et al. [0] prove several positive approximation results for the Stackelberg Minimum Spanning Tree Game. In the same paper, they prove that revenue maximization for this game is APX-hard. Cardinal et al. [] prove that the Stackelberg Minimum Spanning Tree Game remains NP-hard if the instances are planar graphs. Moreover, the problem becomes polynomial-time solvable on graphs of bounded treewidth. Bilo et al. [3] consider further variations. Briest et al. [8] analyze Stackelberg Pricing Games where the follower solves a combinatorial optimization problem. They show that if there is a polynomial time algorithm to solve the follower s problem, then there is a log(k) approximation for the pricing problem. Interestingly, this approximation algorithm uses a single price strategy. Independently, a slightly more general result was obtained by Balcan et al. []. Böhnlein et al. [5] study the single price strategy in a non-discrete setting. Another interesting scenario arises when the optimization problem of the follower is not solved to optimality by the follower. Briest et al. [7] make the first step in this direction. They study Stackelberg Pricing Games were the follower has an NP-hard optimization problem and runs a known approximation algorithm to solve it. For example, the Stackelberg Knapsack Game is shown to be NP-hard, if the follower uses a greedy -approximate algorithm, and a + ɛ approximation algorithm is derived. For the Stackelberg Vertex Cover Game, where the follower implements a primal-dual approach, the revenue maximization problem allows an exact polynomial-time algorithm. Stackelberg Interval Scheduling Games The main feat of this section is the proof of Theorem. Recall that an instance of the interval scheduling problem is given by a set I of intervals. Each interval is specified by a tuple (s i, e i ) R with s i < e i and a weight w(i) R. As mentioned in the introduction, associated to I there is an interval graph G = (V, E). It is well-known that the interval representation of an interval graph is not unique. We say that (v,..., v k ) is an interval order of {v,..., v k } V if there exists an interval representation such that s v... s vk. Before we can turn to the general problem, we need to analyze a special case. For this special case of the Stackelberg Interval Scheduling Game we set aside the benefits on the priceable vertices P. Instead, we require that P S where S is the follower s solution. The

6 6 Stackelberg Pricing Games goal of the leader is then to minimize the total price under which all her vertices are bought. If all priceable vertices have to be part of S, P itself has to be an independent set. Translated back to an interval representation, this means that all priceable intervals must pairwise be non-overlapping. Therefore, we call this special case Stackelberg Interval Scheduling with non-overlapping intervals. In the next section we treat the special case of non-overlapping intervals in order to attend the general case in Section... Stackelberg Interval Scheduling with non-overlapping intervals Let us first state the problem formally. Stackelberg Interval Scheduling with non-overlapping intervals (NOSIS) Input: An interval graph G = (V, E) with priceable vertices P V and fixed weights w(u) for u F = V \ P. The set P is an independent set of G and P = k. Objective: Find a price function p : P R 0 such that there is a maximum weight independent set S with P S and min{p(p ) S is a maximum weight independent set} is minimum. The weight of an independent set S is defined as w(s F ) + p(s P )... A Linear Programming formulation of NOSIS Briest et al. [8] formulate a linear program for so-called Stackelberg Network Pricing Games which generalize our interval scheduling game. Their linear program has exponentially many constraints. However, it can be solved in polynomial time using a separation oracle. In this section, we formulate a linear program for NOSIS where the number of constraints is of order O(k ). More importantly, this formulation simplifies the proofs of results we give later. Given a NOSIS instance (G, P, w) and a price function p R k 0, let IS G(p) be the set of maximum weight independent sets of G under p. As all elements of IS G (p) have the same weight, define opt(is G (p)) := w(f S) + p(p S), for some S IS G (p). We call a price function p feasible if there exists an S IS G (p) such that P S. To set up the linear program, several independent sets of subgraphs of G have to be computed. These subgraphs do not contain any priceable vertices. Thus, for a subset U F, let IS G (U ) be a maximum weight independent set of the induced subgraph G[U ]. Note that in contrast to the previous definition IS G (U ) is an independent set. If there are several maximum weight independent sets of G[U ], select an arbitrary one. Let opt(is G (U )) := w(is G (U )). If p is feasible, then opt(is G (p)) must be at least opt(is G (F )), the maximum weight of an independent set avoiding all priceable vertices. Let U = {u F (u, v) E, v P } be the set of vertices with fixed weights that are not adjacent to any priceable vertex. Hence, opt(is G (F )) opt(is G (U)) is a lower bound for the total price of a feasible price function.

7 T. Böhnlein, O. Schaudt, J. Schauer 7 This lower bound yields the first constraint of our linear program: every feasible price function p necessarily satisfies p(p ) opt(is G (F )) opt(is G (U)). However, this constraint is not sufficient to ensure that p is feasible. For example, the price vector p = (opt(is G (F )) opt(is G (U)), 0,..., 0) satisfies the constraint. But it is unlikely that under p any priceable vertex with price 0 is an element of a maximum weight independent set. In fact, analogous constraints for every subset of priceable vertices are necessary. Luckily it is sufficient to consider constraints for subsets of vertices that are consecutive in an interval order, see Proposition. Let (s, e) be an interval representation of G and I = (v, v,..., v k ) a corresponding interval order of the priceable vertices P. Let v i,..., v j be a consecutive sub-series of I for i j k. We derive a lower bound for p(v i ) p(v j ), which will yield an additional constraint for the linear program. The constraint is set up under the condition that the vertices v,..., v i and v j+,..., v k are included in every optimal solution of the follower. We need some additional notation. Let M i,j := {u F e vi < s u and e u < s vj+ }, where we put e v0 = and s vk+ = to avoid further technicalities. For a maximum weight independent set of G[M i,j ], define IS(G[i, j]) := IS G (M i,j ) As above, we want to subtract the weight of an independent set of vertices that are not adjacent to vertices of P. Therefore, let and OP := {u F (u, v) E, v P } IS(Ḡ[i, j]) := IS G(M i,j \ OP ). Shorthand, we write opt(g[i, j]) and opt(ḡ[i, j]) for opt(is(g[i, j])) and opt(is(ḡ[i, j])), respectively. We will consider constraints of the form p(v i ) p(v j ) opt(g[i, j]) opt(ḡ[i, j]). Such a constraint yields a lower bound on the total price of the vertices v i,..., v j under p. Definition 3 (NOSIS Linear Program (NOSIS-LP)). For a NOSIS instance (G, P, w) with the interval order I = (v,..., v k ) of P, define the following linear program. min p(v ) + p(v ) p(v k ) s.t. p(v i ) p(v j ) opt(g[i, j]) opt(ḡ[i, j]) i j k p(v ), p(v ),..., p(v k ) R + 0 Note that these are O(k ) constraints in total. Any interval representation is suitable for our purpose, but we need to fix one.

8 8 Stackelberg Pricing Games Proposition. Given a NOSIS instance (G, P, w) with the interval order I = (v,..., v k ) of P. A price vector p R k 0 is feasible if and only if it satisfies the constraints of the corresponding NOSIS-LP. Proof. To show sufficiency, let p be a price vector satisfying all constraints of the NOSIS- LP. Choose a maximum weight independent set S IS G (p) such that Q := P \ S has minimum cardinality. If Q = 0, P S and we are done. Thus, suppose that Q for a contradiction. The vertices of Q can be split up into consecutive parts according to the interval order I. Fix a maximum cardinality subset Q Q of consecutive vertices according to I. Let v i, v j Q be the vertices with the, respectively, smallest and largest index in I. That is, v i S and v j+ S if existent. In particular, w(m i,j S) = opt(g[i, j]). () Moreover, by assumption, p satisfies the following constraint: p(v i ) p(v j ) opt(g[i, j]) opt(ḡ[i, j]). () Consider the set S = (S\M i,j ) {v i,..., v j } IS(Ḡ[i, j]). The independent set IS(Ḡ[i, j]) contains no vertices that are adjacent to a priceable vertex or a vertex in S. Hence, S is an independent set of G. Due to () and (), S has at least the same weight as S. Since S has maximum weight, S and S must have the same weight. However, P \ S < Q, in contradiction to the choice of S. Consequently, p is feasible. To show necessity, let p be a feasible price vector. That is, there exists an independent set S IS G (p) and P S. But suppose a constraint of the NOSIS-LP is violated. Thus, for some i j k we have p(v i ) p(v j ) + opt(ḡ[i, j]) < opt(g[i, j]). (3) The violated constraint states that the vertices confined by v i and v j+ of S can be exchanged to form an independent set with larger weight than S. To see this, consider the set S = (S \ (M i,j {v i,..., v j })) IS(G[i, j]). It is straightforward that S is an independent set of G. Due to (3), S has a larger weight than S. Consequently, S is not a maximum weight independent set, a contradiction to the choice of S. Proposition implies that an optimal price vector can be computed using the NOSIS-LP. Recall that an optimal price vector p is feasible and p(p ) is minimum. We close this section with a technical lemma for later use. Lemma. Given a NOSIS instance (G, P, w) and an interval order I = (v,..., v k ) of P. For l < m < r k, it holds that opt(ḡ[l, m ]) + opt(ḡ[m +, r]) = opt(ḡ[l, r]). Proof. Recall that IS(Ḡ[l, r]) is a maximum weight independent set of G[M l,r \ OP ]. The graph G[M l,r \ OP ] is the disjoint union of the graphs G[M i,i M i+,i+ ], i = l,... r.

9 T. Böhnlein, O. Schaudt, J. Schauer 9 Hence, r opt(ḡ[l, r]) = = i=l m i=l opt(is G (M i,i M i+,i+ )) opt(is G (M i,i M i+,i+ )) + = opt(ḡ[l, m ]) + opt(ḡ[m +, r]). r opt(is G (M i,i M i+,i+ )) i=m.. A Combinatorial Algorithm In this section, we introduce a combinatorial algorithm for NOSIS. The algorithm computes a price vector based on an interval order (v,..., v k ) of P in k iterations. In iteration i [k] the price for v i is computed based on one constraint of the NOSIS-LP. We need one more bit of notation. Let p = (p(v l ),..., p(v r )) be a price vector for v l,..., v r P where l r k. Suppose vertices v l,..., v r with prices of p are added to G[M l,r ] to set up an independent set instance. We write and IS(G[l, r](p)) := IS G[Ml,r {v l,...,v r}](p) opt(g[l, r](p)) := opt(is G[Ml,r {v l,...,v r}(p)). Algorithm Stackelberg Interval Scheduling with non-overlapping intervals : Input: a NOSIS instance (G, P, w) and an interval order (v,..., v k ) of P : for i :=,..., k do 3: p(v i ) = opt(g[, i](p(v ),..., p(v i ), 0)) opt(ḡ[, i]) i l= p(v l) 4: return (p(v ),..., p(v k )) Lemma. The running time of Algorithm on a NOSIS instance (G, P, w) is of order O(k ( V + E )). Proof. In each iteration two induced subgraphs of G are constructed. Given an interval representation of G, this can be done in time of order O( V + E ). To evaluate opt(), an independent set instance on an interval graph has to be solved. This takes O( V + E ) time []. Consequently, the running time of Algorithm is of order O(k ( V + E )). Proposition. Given a NOSIS instance (G, P, w) and an interval order I = (v,..., v k ) of P. If p R k 0 is a price vector computed by Algorithm, then p is feasible. Proof. Let p R k 0 be a price vector computed by Algorithm for (G, P, w) and I. To show that p is feasible, we show that p satisfies all constraints of the NOSIS-LP. For j i k we need to show p(v j ) p(v i ) opt(g[j, i]) opt(ḡ[j, i]).

10 0 Stackelberg Pricing Games In iteration i of Algorithm the price vector p i = (p(v ),..., p(v i ), 0) is used to compute p(v i ) as i p(v i ) = opt(g[, i](p i )) opt(ḡ[, i]) p(v l ). (4) Put p j = (p(v ),..., p(v j ), 0,..., 0) R i and observe that opt(g[, i](p i )) opt(g[, i](p j )). Note that j opt(g[, i](p j )) p(v l ) + opt(ḡ[, j ]) + opt(g[j, i]), l= l= since we compare the optimum solution of the follower to the best solution that includes v,..., v j. The two inequalities yield j opt(g[, i](p i )) p(v l ) + opt(ḡ[, j ]) + opt(g[j, i]). (5) l= Together, (5) and (4) yield i l=j p(v l ) opt(g[j, i]) + opt(ḡ[, j ]) opt(ḡ[, i]). Applying Lemma opt(ḡ[, j ]) opt(ḡ[, i]) = opt(ḡ[j, i]) finishes the proof. Proposition 3. Given a NOSIS instance (G, P, w) and an interval order I = (v,..., v k ) of P. Algorithm computes a feasible price vector p with minimum total weight p(p ). Proof. Let p R k 0 be a price vector computed by Algorithm for (G, P, w) and I. The vector p is feasible due to Proposition. It is left to show that p also has minimum total weight. For (G, P, w) let O be the set of feasible price vectors with minimum total weight. We compare p to a price vector q = arg max o O { l p(v l ) = o(v l ) for l < l }, breaking ties arbitrarily. That is, q is an optimal price vector that agrees with p on as many positions as possible, starting at position. If q and p agree in all entries, we are done. So, suppose l [k] is the smallest index where p and q differ. That is, p(v l ) q(v l ) and p(v l ) = q(v l ) for all l < l. Claim. p(v l ) < q(v l ) Proof. Since q is feasible, l opt(g[, l](q)) = q(v l ) + opt(ḡ[, l]) + q(v l). l = We set up an new price vector q l = (q(v ),..., q(v l ), 0). As the prices of vertices do not increase, it follows that opt(g[, l](q l )) opt(g[, l](q)).

11 T. Böhnlein, O. Schaudt, J. Schauer By definition, p l = (p(v ),..., p(v l ), 0) = q l and opt(g[, l](p l )) = opt(g[, l](q l )). As a result of the inequations above and the identity l l = q(v l ) = l l = p(v l ), l opt(g[, l](p l )) p(v l ) + opt(ḡ[, l]) + q(v l). (6) l = holds. In iteration l Algorithm computes l p(v l ) = opt(g[, l](p l )) opt(ḡ[, l]) p(v l ) (7) l = Applying (6) in (7) yields p(v l ) q(v l ). Since p(v l ) q(v l ), p(v l ) < q(v l ) follows. To continue with the proof, we construct a new price vector q from q. Set q = q and change two entries: q (v l ) = p(v l ) and q (v l+ ) = q(v l+ ) + (q(v l ) p(v l )). Thus, the surplus which q(v l ) has compared to p(v l ) is shifted over to q(v l+ ). Both q and q are of the same (minimum) total weight since q (v l ) + q (v l+ ) = p(v l ) + (q(v l+ ) + q(v l ) p(v l )) = q(v l ) + q(v l+ ). (8) To finish the proof, we show that q satisfies the constraints of the NOSIS-LP. Remember that q is feasible. Since only q (v l ) was reduced in the construction of q from q, it suffices to check constraints that involve q (v l ). To do this, consider the constraint q (v i ) q (v j ) opt(g[i, j]) opt(ḡ[i, j]), for i l j. Case : l = j. According to the construction of q, we have that p(v l ) = q (v l ), for i l l. Since p is a feasible price vector, q satisfies the constraint. Case : l < j. Recall that the constraints of the NOSIS-LP were constructed for consecutive vertices of I. Therefore, the constraint also involves q (v l+ ). With (8) it follows that q satisfies the constraint. Consequently, q is a feasible price vector with minimum total weight. As q agrees with p on more consecutive indices than q, we reach a contradiction, and p = q, which shows that p has minimum total weight. As a result, Algorithm computes an optimal price vector for a NOSIS instance in polynomial time. The proof of Proposition 3 can also be interpreted as a transformation of an arbitrary, optimal price vector into the price vector computed by Algorithm. The degrees of freedom of the constraints are shifted to higher indices of the given interval order. The first entry of a price vector p computed by Algorithm is special. It is computed independently of the other prices. With the following lemma, we show that p without the first entry is feasible for the remaining priceable vertices. This property is essential for the next section. Lemma 3. Given a NOSIS instance (G, P, w) and an interval order I = (v,..., v k ) of P. If p R k 0 is a price vector computed by Algorithm, then there exists an S IS G (0, p(v ),..., p(v k )) such that {v,..., v k } S.

12 Stackelberg Pricing Games Proof. Let p R k 0 be a price vector computed by Algorithm for (G, P, w) and I. The price vector p is feasible due to Proposition. With Lemma it follows that opt(g[, k](p)) = Algorithm computes k l= p(v l ) + opt(ḡ[3, k]) + p(v ) + opt(ḡ[, ]). (9) p(v ) + opt(ḡ[, ]) = opt(g[, ]). (0) Applying (0) in (9) yields opt(g[, k](p)) = k l= p(v l ) + opt(ḡ[3, k]) + opt(g[, ]). () Now, we set the price of v to zero to form p = (0, p(v ),..., p(v k )). Hence, opt(g[, k](p )) opt(g[, k](p)). () We remark that k p(v l ) + opt(ḡ[3, k]) + opt(g[, ]) opt(g[, k](p ), (3) l= since we compare the optimum solution of the follower to the best solution that includes v,..., v k. Together, () and (3) yield opt(g[, k](p)) opt(g[, k](p )). (4) Inequalities () and (4) sandwich opt(g[, k](p ). Therefore, opt(g[, k](p )) = opt(g[, k](p)) = k l= p(v l ) + opt(ḡ[3, k]) + opt(g[, ]). The last equation states that there exists an S IS G (p ) such that {v,..., v k } S.. Stackelberg Interval Scheduling In the general Stackelberg Interval Scheduling problem it is not required that the priceable vertices form an independent set. We begin this section by stating the definition of the general game. Stackelberg Interval Scheduling (SIS) Input: An interval graph G = (V, E) with priceable vertices P V and P = k. For every v P there is a benefit b(v) R and for every u F = V \ P there are fixed weights w(u) R. Objective: Find a price vector p : P R 0 such that (i) there exists a maximum weight independent set S of G, and (ii) the revenue b(s) p(s) is maximum. The weight of S is defined as w(f S) + p(p S). A SIS instance is characterized by a 4-tuple (G, P, w, b). In this section, we formulate a polynomial time algorithm that solves SIS, building upon results from the previous sections. In particular, the structure of a price vector computed by Algorithm is crucial.

13 T. Böhnlein, O. Schaudt, J. Schauer 3 Outline of the SIS algorithm The input of the algorithm is a SIS instance (G, P, w, b) and an interval order I = (v,..., v k ) of P. Two techniques characterize the algorithm. The first ingredient uses the NOSIS machinery to compute optimal prices for an independent set P of G where P P. For this, we compute the induced subgraph G = G[F P ]. Now (G, P, w) is an instance of NOSIS. Algorithm allows us to compute an optimal price vector p. In turn, the price vector p can be extended to a price vector p for the original SIS instance by setting the price of vertices in the set P \ P to zero. For the fixed set P, the benefits are constant, and p has a minus sign in the leader s objective function of the SIS instance. Consequently, the price vector p yields maximum revenue under the given assumption. We denote this maximum revenue rev G (P ). On this account, the algorithm selects several independent sets of P and determines their maximum revenue. However, it cannot explicitly enumerate all independent sets of P since it is supposed to run in polynomial time. The second ingredient is dynamic programming: the algorithm solves an SIS instance from a series of sub-instances. The first sub-instance contains only the priceable vertex v. In each iteration, one priceable vertex is added according to the interval order I. Define P i := {v, v,..., v i } and G i := G[F P i ] for i [k]. Note that in each induced subgraph all vertices with fixed weights are present. The sub-instance of iteration i is (G i, P i, w, b Pi ). In iteration i our algorithm computes an independent set S i {v, v,..., v i } with v i S i. The set S i is the optimal selection to form an independent set of P i and v i. Formally, for any independent set Q P i we have rev Gi (Q v i ) rev Gi (S i ). In Proposition 4 we show that the optimal solution in iteration i can be computed with the following recursion: S i = arg max rev Gi (S j {v i }) j<i, (v i,v j) E Note that S j {v i } is an independent set since we require that (v i, v j ) E. To allow our algorithm to test sets where S j =, a priceable vertex v 0 with S 0 = is added. The vertex v 0 is isolated in G and b(v 0 ) = 0. The interval order I is extended to (v 0, v,..., v k ) and we put P i = P i {v 0 } for all i [k]. After computing optimal independent sets for v 0, v, v,..., v k we can select the globally optimal set and thus solve the instance to optimality. Algorithm Stackelberg Interval Scheduling : Input: A SIS instance (G, P, w, b) and an interval order (v 0, v,..., v k ) of P {v 0 } : S 0 =, globalmax = 0, p = 0 3: for i =,..., k do 4: localmax = 0 5: for j < i and (v i, v j ) E do 6: G := G[F S j {v i }] 7: compute vector p that yields maximum revenue r for (G, S j {v i }, w) 8: if r > localmax then 9: localmax = r 0: S i = S j {v i } : if localmax > globalmax then : globalm ax = localm ax 3: p = p where the price for every v P \ S i is 0 4: return p

14 4 Stackelberg Pricing Games Lemma 4. Algorithm runs in time of order O(k 3 ( V + E )). Proof. Due to Lemma, the running time of line 7 is of order O(k ( V + E )). In the worst the inner for-loop (line 6-3) is executed k l= l = k(k+) times. Thus, the two nested for-loops yield a running time of order O(k ). In total, the running time of Algorithm is of order O(k 3 ( V + E )). Proposition 4. Given a SIS instance (G, P, w, b) and an interval order I = (v 0, v,..., v k ) of P {v 0 }. Algorithm computes an optimal price vector. Proof. To prove the proposition, we show that it is not necessary to explicitly enumerate all independent sets to determine an optimal selection of priceable vertices. If a new priceable vertex is considered, it suffices to test the pre-computed independent sets in line 6 of Algorithm. We show that a different selection of vertices cannot yield more revenue. Let i, j [k] such that j < i and (v i, v j ) E. Suppose that we are in the i-th iteration of line 3. Let us compare the set S j to a different selection of vertices. For this, let Q {v, v,... v j } be another independent set with v j Q. Furthermore, let p and q be optimal price vectors for the NOSIS instances (G[F S j ], S j, w) and (G[F Q], Q, w), respectively. By induction we may assume that b(s j ) p(s j ) b(q) q(q). (5) Now, we show that S j is an optimal selection of an independent set if v j and v i are fixed. For ease of notation let G S = G[F S j {v i }] and G Q = G[F Q {v i }]. Let ˆp and ˆq be optimal price vectors for the NOSIS instances (G S, S j {v i }, w) and (G Q, Q {v i }, w), respectively. Suppose that b(s j ) ˆp(S j ) + b(v i ) ˆp(v i ) < b(q) ˆq(Q) + b(v i ) ˆq(v i ). (6) From (5) and (6) we derive a contradiction. More precisely, we show that ˆq (without ˆq(v i )) yields more revenue than q for (G[F Q], Q, w). Let Ī = (v k, v k,..., v ) be the reversed interval order of I. This interval order induces interval orders ĪS = Ī[S j {v i }] and ĪQ = Ī[Q {v i}] for the priceable vertices of S j {v i } and Q {v i }, respectively. We choose the two price vectors ˆp and ˆq as the price vectors computed by Algorithm for the interval orders ĪS and ĪQ, respectively. We claim the following. Claim. ˆp(v i ) = ˆq(v i ) Proof. As v i is the first entry of ĪS and ĪQ, Algorithm computes and ˆp(v i ) = opt(g S [i, i]) opt(ḡs[i, i]). (7) ˆq(v i ) = opt(g Q [i, i]) opt(ḡq[i, i]). (8) Recall that opt(g[i, i]) is the weight of a maximum weight independent set of G[M i,i ] where M i,i F is confined by v i and v i+. The definition of M i,i depends only on the predecessor and successor of v i of the current interval order. For both ĪS and ĪQ, v i has no predecessor and v j as successor. Hence, opt(g S [i, i]) = opt(g Q [i, i]) and, similarly, opt(ḡs[i, i]) = opt(ḡq[i, i]). With (7) and (8) it follows that ˆp(v i ) = ˆq(v i ).

15 T. Böhnlein, O. Schaudt, J. Schauer 5 Consider the NOSIS instance (G S, S j {v i }, w). With p and ˆq(v i ) we form the price vector p = (p(v ),..., p(v j ), ˆq(v i )). Claim 3. The price vector p is feasible for (G S, S j {v i }, w). Proof. The vector p is feasible for (G[F S j ], S j, w). Note that, G S and G[F S j ] contain the same set of vertices with fixed weights. Extend p to the price vector p 0 = (p(v ),..., p(v j ), 0) for the instance (G S, S j {v i }, w). With respect to the induced interval order I[S j {v i }], opt(g S [, i](p 0 )) = j p 0 (v l ) + opt(ḡs[, j ]) + opt(g S [i, i]) (9) l= holds. As the prices of vertices are reduced, opt(g S [, i](p 0 )) opt(g S [, i](p )). Note that only the price of v i is increased when comparing p 0 to p. If the weight of a maximum weight independent set of G S is strictly larger under p than under p 0, then a maximum weight independent set under p must contain v i. Thus, opt(g S [, i](p )) = j p (v l ) + opt(ḡs[, i]) + ˆq(v i ) l= Applying Lemma yields opt(g S [, i](p )) = Due to Claim, j p (v l ) + opt(ḡs[, j ]) + ˆq(v i ) + opt(ḡs[i, i]). (0) l= ˆq(v i ) + opt(ḡs[i, i]) = opt(g S [i, i]). () Using () in (0) yields opt(g S [, i](p )) = j p (v l ) + opt(ḡs[, j ]) + opt(g S [i, i]). l= Hence, together with (9), the last equation yields opt(g S [, i](p 0 )) = opt(g S [, i](p )). This is a contradiction to our assumption and opt(g S [, i](p 0 )) = opt(g S [, i](p )) follows. With (9) it follows that opt(g S [, i](p )) = Substituting () yields opt(g S [, i](p )) = j p 0 (v l ) + opt(ḡs[, j ]) + opt(g S [i, i]) l= i p (v l ) + opt(ḡs[, i]). l= The last equation states that p is feasible for (G S, S j v i, w).

16 6 Stackelberg Pricing Games Since ˆp is an optimal price vector for (G S, S j v i, w), b(s j ) ˆp(S j ) + b(v i ) ˆp(v i ) b(s j ) p(s j ) + b(v i ) ˆq(v i ). With (6) it follows that b(q) ˆq(Q) + b(v i ) ˆq(v i ) > b(s j ) p(s j ) + b(v i ) ˆq(v i ) which simplifies to b(q) ˆq(Q) > b(s j ) p(s j ) With (5) it follows that b(q) ˆq(Q) > b(q) q(q). () Claim 4. Price vector ˆq (without ˆp(v i )) is feasible for (G[F Q], Q, w). Proof. As ˆq was computed with ĪQ, the claim follows from Lemma 3. Equation states that vector ˆq (without ˆp(v i )) yields more revenue than q for (G[F Q], Q, w). This contradicts that q is an optimal price vector. As a consequence, there can be no independent set like Q that yields strictly more revenue than S j when v j and v i are fixed. Therefore, it is valid to test only the pre-computed independent sets in line 6 of Algorithm. After the computation, our algorithm chooses the optimal selection of priceable vertices with the variable globalm ax. The corresponding price vector is returned. Toghether, Lemma 4 and Proposition 4 prove Theorem, which we restate. Theorem. The SIS problem can be solved in time O(k 3 ( V (G) + E(G) )) given an instance (G, P, w, b) with P = k. 3 Stackelberg Matching Game One of the most prominent optimization problems on graphs is the matching problem. Based on its weighted version we formulate a corresponding Stackelberg Pricing Game. Formally, the game is defined as follows. Stackelberg Matching Game Input: A graph G = (V, E) with priceable edges P E where k = P. For every e P there is a benefit b(e) R and for every e F = E \ P there are fixed weights w(e) R. Objective: Find a price function p : P R 0 maximizing max{b(m P ) p(m P ) M is a maximum weight matching}, where the weight of a matching M is defined as w(m F ) + p(m P ). The following theorem states that the Stackelberg Matching Game is NP-hard even if (i) the number of distinct fixed weights is restricted and (ii) only bipartite graphs are considered. We need the theorem to prepare for the main result of this section: Theorem. Theorem 4. The Stackelberg Matching Game is NP-hard even when the graph G is bipartite, w(f) {, } for all f F, and b(e) = 4 for all e P.

17 T. Böhnlein, O. Schaudt, J. Schauer 7 Proof. We provide a reduction from the NP-complete Set Cover problem. Shorthand, we write C for c C c where C is a family of subsets. Set Cover Input: A ground set B = {û,..., û n } and a family S = {S,..., S m } of subsets of B where S = B. Objective: Find a subfamily C S with minimum cardinality such that C = B. Given a Set Cover instance (B, S) with n = B and m = S we construct a Stackelberg Matching Game instance (G, P, b, w). The graph G = (V, E = P F ) is set up as follows. There is the set B V that contains a vertex u j B for each û j B. A subset-gadget as depicted in Figure (a) is constructed for each subset of S. The set of priceable edges consists of two disjoint subsets P = P sub P cov. Let S i S and û j S i, then there is a vertex u i,j V \ B and a priceable edge σ i,j = (u i,j, u j ) P cov with benefit b(σ i,j ) = 4. An edge σ i,j is called a covering-edge of u j. The subset P cov P contains all covering-edges. there are two vertices v i, w i V \ B as well as the priceable edge s i = (v i, w i ) P sub with benefit b(s i ) = 4. The edge s i is called the subset-edge of S i. The subset P sub P contains all subset-edges. there are paths of length two connecting v i to each u i,j. Such a path consists of a vertex ū i,j V \ B as well as the edge (u i,j, ū i,j ) F with weight w((u i,j, ū i,j )) = and the edge (ū i,j, v i ) F with weight w((ū i,j, v i )) =. Moreover, we define σ i = {σ i,j û j S i } for i =,..., m. ū, u, σ, u s w v ū, u, σ, u u j ū,3 u,3 σ,3 u 3 (a) The subset-gadget of S = {u, u, u 3}. Priceable edges are dashed and have benefit 4 (not displayed). (b) One of n extended star components obtained after removing weight edges. Figure Illustration of the Stackelberg Matching Game construction. Let us quickly verify some structural properties of G that we need to prove the theorem. Note that w(f) {, } for all f F and b(e) = 4 for all e P. Claim 5. The graph G is bipartite.

18 8 Stackelberg Pricing Games Proof. Let V = {v i i =,..., m} {u i,j i =,..., m, j =,..., n} and V = B {w i i =,..., m} {ū i,j i =,..., m, j =,..., n}. The two subsets V and V partition V. Since no pair of vertices in either V or V is adjacent, G is bipartite. Given a price function p there can be several maximum weight matchings. We call the maximum weight matchings that yield maximum revenue for the leader maximum revenue matchings under p. The idea of the construction is (i) to enforce certain matchings as maximum revenue matchings by having large enough benefits on all priceable edges. The leader has to ensure that a maximum weight matching contains a maximum-cardinality matching of the priceable edges. This means all subset-edges and one covering-edge for each vertex u B must be part of a maximum weight matching. (ii) With Claim 7 we will see that under an optimal price function all covering-edges in a maximum revenue matching have price. If a subset-gadget has a covering-edge with price, then its subset-edge has to have price. Without such a covering-edge it suffices to set price for the subset-edge. Consequently, a subset-gadget is more expensive if it is covering some ground elements and finding a minimum set cover coincides with computing an optimal price function. Example. Figure 3(a) displays an instance of the Stackelberg Matching Game constructed from a Set Cover instance. On the right-hand side (b) a minimum set cover {S, S } is encoded as their subset-edges have price. For the subset-edges of S 3 and S 4, which are not part of the set cover, price suffices to be part of a maximum weight matching. They are less expensive for the leader. Note that both S and S cover u. There are two equivalent matchings: one contains σ, and the other σ,. The two matchings have the same weight and yield the same revenue for the leader. Our proof relies on establishing bounds on the prices of edges. An upper bound can be obtained from the weight of adjacent edges, as follows. Assume we have a priceable edge e = (u, v) and there is no other priceable edge adjacent to e. Then a price of U e = max{w(f) f = (u, w) F } + max{w(f) f = (v, w) F } guarantees that e is an element of a maximum weight matching. This follows from a simple exchange argument. If we assume that b(e) > U e, then U e guarantees that e is an element of every maximum revenue matching since there are no other priceable edges adjacent to e. Furthermore, setting a price higher than U e on the edge e does not result in an optimal price function. Reducing its price to U e yields more revenue and e is still guaranteed to be an element of every maximum revenue matching. We deduce that U e is an upper bound on the price of edge e in an optimal price function. In particular, for a subset-edge s P sub we have U s =. For a covering-edge σ P cov we obtain U σ =, but there are other priceable edges adjacent to σ. Therefore, setting a price of U σ does not guarantee that σ is an element of a maximum weight matching. In Claim 7 we show that U σ is an upper bound of p(σ) in an optimal price function p. Before we prove the main proposition, we present three claims that formalize the ideas given above. The first claim identifies the structure of maximum revenue matchings. Claim 6. A maximum revenue matching contains m subset-edges and n covering-edges. Proof. Let p be an optimal price function and let M be a maximum revenue matching under p. Suppose there is a subset-edge s M. Consider the subset-gadget of s. There is an edge e with fixed weight which is adjacent to s and e M. To see this, consider the edge e with fixed weight adjacent to e. Suppose that e M. Since M has maximum weight,

19 T. Böhnlein, O. Schaudt, J. Schauer 9 s σ, σ,3 σ,4 u u u 3 u 4 u u u 3 u 4 σ, σ, σ 3,3 σ 4,4 0 0 s s 3 s 4 (a) The Stackelberg Matching Game instance constructed from the Set Cover instance. (b) Optimal prices on the priceable edges are added and a maximum revenue matching is displayed (bold). Figure 3 Illustration of the reduction for the Set Cover instance ({u, u, u 3, u 4}, {S = {u, u 3, u 4}, S = {u, u }, S 3 = {u 3}, S 4 = {u 4}}). e M. But now exchanging e for e yields a matching of more weight. We conclude that e M. Furthermore, we must have p(s) <. If p(s) =, M does not yields maximum revenue. If p(s) >, M is not a maximum weight matching. Adjacent to e there is a covering-edge σ P cov. Case : σ M. We set up a new price function p from p where we change p (s) =. Let M = (M \ {e}) {s}. The matchings M and M have the same weight under p. Hence, matching M has maximum weight under p and yields more revenue than M under p. This contradicts that p yields maximum revenue. Case : σ M. Since e M, we could have that p(σ) = 0. This is the worst case for our purpose as σ yields 4 revenue. We set up a new price function p from p where we change p (s) = and p (σ) =. Let M = (M \ {e}) {s}. The matchings M and M have the same weight under p. Hence, matching M has maximum weight under p. The edges s and σ yield a revenue of 5 under p and a revenue of 4 under p (assuming that p(σ) = 0). As p yields a larger revenue than p, we reach a contradiction. In conclusion, we have s M for all s P sub. Suppose there is a vertex u j B and for all covering-edges σ P cov incident to u j we have σ M. For each such edge σ there is an edge e with fixed weight adjacent to σ. We must have that all such edges e M. Otherwise we could add a corresponding covering-edge and form a matching that yields more revenue than M. Choose one edge σ incident to u j. Let e be the edge with fixed weight that is adjacent to σ. We set up a