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1 MANAGEMENT SCIENCE doi /mnsc ec pp. ec1 ec7 e-companion ONLY AVAILABLE IN ELECTRONIC FORM informs 2008 INFORMS Electronic Companion Customized Bundle Pricing for Information Goods: A Nonlinear Mixed-Integer Programming Approach by Shin-yi Wu, Lorin M. Hitt, Pei-yu Chen, and G. Anandalingam, Management Science, doi /mnsc

2 Customized Bundle Pricing for Information Goods: A Nonlinear Mixed Integer Programming Approach On-line Supplement 1. Solution Approach 1.1. Lagrangian Relaxation By using the Lagrangian relaxation method, we can transform the primal problem (IP) mentioned above into the following Lagrangian relaxation problem (LR) where constraint (5) is dualized: Problem LR φ( a)= ( P B ) X MY + a (1 X ) (11) s.t. Max P, Si, Xi, Y ( ) i = 1,..., J i i i= 1,..., I = 1,..., J i = 1,..., J i= 1,..., I i = 1,..., J i S R - P Y, i = 1,..., I; = 1,..., J (12) i i S = ( R - P) X, i = 1,..., I (13) ( R - P) X 0, i = 1,..., I; = 1,..., J (14) i i X Y, i = 1,..., I; = 1,..., J (15) i S 0, i = 1,..., I (16) i P 0, = 1,..., J (17) X = 0 or 1, i = 1,..., I; = 1,..., J (18) i Y = 0 or 1, = 1,..., J (19) Since we relax only one constraint here, we have one type of corresponding Lagrangian Multiplier a i which must be nonnegative because the constraint we relaxed is an inequality constraint. Note that we can rearrange the terms in obective function of Problem LR and obtain the following obective function: Max φ a P 1,..., 1,..., B ai Xi MY J i I a = = i= 1,..., I i P, Si, Xi, Y ( )= [ ( ) ]+ (20) 1

3 This nonlinear integer programming problem LR is complex and hard to solve. However, it is amenable to simplification. First, given constraint (13) and (14), constraint (16) is redundant, so we can remove it. Second, constraint (13) simply defines the consumer surplus. If we replace S i in constraint (12) with k=1,,j (R ik -P k )X ik, we can remove constraint (13). After making these substitutions, we get an equivalent problem as follows: Problem LR ( a)= [ ( P B a ) X MY ]+ a (21) φ 1,..., 1,..., i i = J i= I i= 1,..., I i P, Xi, Y s.t. k= 1,..., J Max ( ) ( R - P ) X R - P Y, i = 1,..., I; = 1,..., J (22) ik k ik i ( R - P) X 0, i = 1,..., I; = 1,..., J i i (23) X Y, i = 1,..., I; = 1,..., J (24) i P 0, = 1,..., J (25) X = 0 or 1, i = 1,..., I; = 1,..., J (26) i Y = 0 or 1, = 1,..., J (27) Note LR can be decomposed into J different problems except for the complicated constraint (22). Unfortunately (22) cannot be dualized because the resulting Lagrangian obective function will have the nonlinear terms of P X i, P k X ik and P Y as well as the double summation over J, which makes it impossible to decompose the problem further into J different subproblems. As a result, we decide to drop constraint (22) and solve a relaxed version of LR, which we call Problem R. Problem R Va ( )= [ ( P B a ) X MY ]+ a (28) s.t. Max P, Xi, Y = 1,..., J i= 1,..., I i i i= 1,..., I i ( R - P) X 0, i = 1,..., I; = 1,..., J (29) i i X Y, i = 1,..., I; = 1,..., J i (30) P 0, = 1,..., J (31) X = 0 or 1, i = 1,..., I; = 1,..., J (32) i Y = 0 or 1, = 1,..., J (33) 2

4 It is not hard to see that Problem R is actually the summation of J independent subproblems (one for each, as shown in (34)-(39)) plus a constant term i=1,,i a i. Subproblem Max P, Xi, Y s.t. i= 1,..., I ( P B a ) X MY (34) i i ( R - P) X 0, i = 1,..., I (35) i i X Y, i = 1,..., I (36) i P 0, (37) X = 0 or 1, i = 1,..., I (38) Y i = 0 or 1, (39) Each such subproblem is actually equivalent to the following setting. Given each customer s reservation price for the bundle of goods (i.e., R i ), we are trying to maximize vendor s total profit by deciding whether to offer the bundle of goods (i.e., determine Y ) and the price of the bundle P. The fixed cost of offering the bundle is M while the marginal bundle cost to provide this bundle to customer i is B + a i (a i is Lagrangian Multiplier). Again, each customer is assumed to maximize her surplus and she is willing to purchase the bundle (set X i to 1) when the surplus is nonnegative (constraint (35)) Algorithmic Intuition Under this Subproblem, the vendor should offer a bundle of goods if the maximum revenue i=1,,i (P - B - a i )X i achievable by providing this bundle is larger than the fixed cost M. In other words, a single variable Y appears in each Subproblem. If it is 0, then all X i are 0 and the value of the obective function is 0. Therefore we will choose to set Y to 1 only if improvement is possible (the best obective function value for Y =1 is larger than 0). In addition, the vendor is willing to sell the bundle with goods to customer i only when it is profitable: P - B - a i 0, which would make it optimal for X i to be 1. Keeping these in mind, we first try to find the most profitable price P and use it to calculate the maximum revenue i=1,,i (P - B - a i )X i for each subproblem. If the resulting revenue is larger than M, the vendor should offer the bundle by setting Y = 1. We then set X i of those profitable and willing-to-buy 3

5 customers to 1, others to 0, and the optimal obective function value of this subproblem is equal to (maximum revenue - M). On the other hand, if the maximum revenue achievable is smaller than M, the vendor should not offer the bundle. We then simply set all decision variables P and Y to 0, which in turn set all X i to zero, and the optimal obective function value of this subproblem is equal to 0. To derive the most profitable price P, let us first sort the reservation price R i of all potential customers in descending order and call them R 1, R 2,, R I. Obviously, if we set the price above R 1, no customer will be willing to buy and the maximum revenue achievable is 0. So setting the price above R 1 is not profitable. Now assume we set the price between R k and R k+1 (R k > R k+1 ), then at most k customers will be willing to buy. We soon discover that if we raise the price a bit to R k, the total number of willingto-buy customers will not change while the maximum profit achievable will rise for sure. In other words, any price setting between R k and R k+1 is no better than R k, and when considering the best price, we only have to take R 1, R 2,, R I as possible candidates. To this point, we successfully decreased the set of possible optimal pricing P from a continuous set to a discrete and finite one. Since the set is finite (with size I), we can simply choose the price that maximizes revenue from this set. After the above analysis, we can design the following algorithm to solve Problem R, the further relaxed problem of Problem LR where constraint (22) is removed. Step 1. =1,,J { choose the best price P among R 1, R 2,, R I by comparing the maximum revenue achievable; if maximum revenue M, set Y = 1 and set X i of those profitable and willing-to-buy customers to 1, others to 0; else set P, X i, and Y to 0 } Step 2. Add the obective function value of J subproblems together plus a constant term i=1,,i a i. This is the obective function value of Problem R. 4

6 1.3. The Dual Problem and the Subgradient Method According to the algorithm proposed above, although we could not directly solve the Lagrangian relaxation problem Problem LR, we can successfully solve Problem R. According to the weak Lagrangian duality theorem, the optimal obective value of Problem LR is an upper bound of the optimal obective value of Primal Problem IP. And since Problem R is a relaxation of Problem LR, the optimal obective value of Problem R is also an upper bound of the optimal obective value of IP. We then construct the following dual problem to calculate the tightest upper bound and solve the dual problem by using the subgradient method (more details about the subgradient method can be found in the Bertsekas, 1999): min V(a) (D) s.t. a 0. where V(a) is Problem R. Let the vector S be a subgradient of V(a) at (a 1, a 2,, a I ). In iteration k of the subgradient optimization procedure, the multiplier a is updated by a k+1 = a k - α k S k where S k = 1 - =1,,J X i. The step size α k k is determined by α k h V( a ) φ = β where φ h is the primal obective value, which k 2 S we get from a heuristic solution (a lower bound of the optimal primal obective value), and β is a scalar, 0 β 2. We run the subgradient method for 1000 iterations, the value of β is reset every 200 iterations, from 2 to 1.6, 1.2, 0.8 and finally 0.4, and it is divided by 2 after 40 iterations without upper bound improvement. The initial value of the multiplier vector a is chosen to be the 0 vector and φ h is always updated to the best lower bound obtained at the time. After the implementation of the subgradient optimization procedure, while we didn t observe a point with zero subgradient, we have upper bound on the optimal obective value of the primal problem. However, as expected, no primal feasible solution was found in the process due to the structure and complexity of this bundle pricing problem since the subdivision into J independent subproblems means 5

7 that constraint (5) is likely to be violated. In the following subsection, we try to utilize the Lagrangian solutions to develop a heuristic algorithm to get the primal feasible solutions Getting Primal Feasible Solutions For each iteration of the subgradient method, we get a Lagrangian solution which is affected by the updating of the Lagrangian Multiplier a. But because we relaxed constraints (5) and (22) above, the Lagrangian solutions we get for Problem R may not be feasible for the original primal problem. However, from past experiences, we know that the degree of violations is often low and it is possible to obtain a good primal feasible solution using heuristics which start from a Lagrangian solution. We now describe this heuristic in which we use P and Y from the Lagrangian solutions as given and recalculate X i (and also S i ) to get the primal feasible solutions. We can use this to generate a sequence of primal feasible solutions by adusting the sequence of Lagrangian feasible solutions we get from the subgradient method. Algorithm Bundling : Step 1. For each potential customer i, choose the bundle k (only those with Y = 1) with the largest positive surplus. Set X ik to 1 and other X i to 0. (This enforces constraint (22) and constraint (5)). Step 2. For each offered bundle (those with Y = 1), calculate its revenue achieved i=1,,i (P - B )X i. Choose the bundle k with the smallest revenue. If the revenue is smaller than M, we do not offer this bundle by setting the corresponding P, X i, and Y to 0 and go to step 1. (The rationale behind this is that based on our primal obective function (1), it is reasonable for us to offer the bundle only when the revenue from the bundle could cover the fixed cost M.) Step 3. Calculate consumer surplus S i according to constraint (3). More information about Lagrangian relaxation and subgradient methods can be found in Fisher (1981), which was recently chosen as one of the top ten most influential Management Science paper in the past 50 years, and Nowak (2005). 6

8 References: Bertsekas, D. Nonlinear Programming. Nashua, NH: Athena Scientific, pp , Fisher, M. The LaGrangian-Relaxation Method for Solving Integer Programming-Problems, Management Science (27:1), 1981, pp Nowak, I. Lagrangian Decomposition of Block-separable Mixed-Integer All-Quadratic programs, Mathematical Programming 102, 2005, pp