Polynomial time algorithms for the Minimax Regret Uncapacitated Lot Sizing Model

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1 Polynomial time algorithms for the Minimax Regret Uncapacitated Lot Sizing Model Dong Li a,, Dolores Romero Morales a a Säid Business School, University of Oxford, Park End Street, Oxford, OX1 1HP, United Kingdom Abstract We study the Minimax Regret Uncapacitated Lot Sizing (MRULS) model, where the production cost function and the demand are subject to uncertainty. We propose a polynomial time algorithm which solves the MRULS model in O(n 6 ) time. We improve this running time to O(n 5 ) when only the demand is uncertain, and to O(n 4 ) when only the production cost function is uncertain. Keywords: robust optimization, minimax regret, lot sizing, production cost and demand uncertainties 1. Introduction In the Uncapacitated Lot Sizing (ULS) model [1], the goal is to find a production plan that minimizes the total production and inventory holding cost, and at the same time satisfies the demands. In an n-period planning horizon, we assume that the demand in period i is given by d i > 0 for i = 1,..., n. For each period i, there is a unit production cost p i, an unit inventory cost h i and a setup cost f i if production takes place. The ULS model can be formulated as follows, ULS : min ( f i y i p i x i h i I i ) s.t. I i1 x i = I i d i, i = 1,..., n x i My i, i = 1,..., n y i {0, 1}, i = 1,..., n x i, I i 0, i = 1,..., n, where M is a large constant, x i the amount produced in period i, I i the inventory level at the end of period i, and y i equals to 1 if x i > 0 and 0 otherwise. Throughout this paper, we use bold characters for vectors, e.g., y = (y i ). Wagner and Whitin [1] proposed an O(n 2 ) time algorithm to solve the ULS model based on the so-called the Zero Inventory Ordering (ZIO) property, i.e., I i1 x i = 0. This property implies that x and I are uniquely determined by y and we only need to optimize over y. Corresponding author. Tel.: 44 (0) addresses: dong.li@sbs.ox.ac.uk (Dong Li), dolores.romero-morales@sbs.ox.ac.uk (Dolores Romero Morales) Realizing the uncertain nature of the input data, we study a robust version of the ULS model, the Minimax Regret ULS model (MRULS). Unlike the absolute robust models, see, e.g., [2, 3] for an overview, the Minimax Regret model minimizes the worst-case regret, or opportunity loss. Under demand uncertainty, Zhang [4] proposed an O(n 7 ) time algorithm for the MRULS model. In this paper, we show that the MRULS model with uncertain production cost function, defined by f and p, and uncertain demand d, can be solved in O(n 6 ) time. We improve this running time to O(n 5 ) when only the demand is uncertain and to O(n 4 ) when only the production cost function is uncertain. 2. Formulation and general results We assume that f, p, and d are subject to interval uncertainty, i.e., f U f = {f : f i [ f, f i i ], i = 1,..., n}, and similarly for p and d. Let u = (f, p, d) U := U f U p U d denote a realization of the uncertain data. We denote by y a first-stage setup decision, i.e., before knowing the actual realization u, and ŷ an optimal setup decision once u has been revealed. Then, for a given firststage decision vector y and a realization of uncertainty u, the regret is defined as follows, ( f i y i p i x i h i I i ) ( f i ŷ i p i ˆx i h i Îi ). Hence, the MRULS model is formulated as min max [ ( f i y i p i x i h i I i ) y {0,1} n u U Preprint submitted to Operations Research Letters October 31, 2013

2 ( f i ŷ i p i ˆx i h i Îi )], which is equivalent to min max max y {0,1} n ŷ {0,1} n [ f i (y i ŷ i ) u U h i (I i Î i )]. p i (x i ˆx i ) We introduce some notations that will be used throughout the paper. We define d i j := j k=i d k, h i j := j k=i h k, and α i j := p i h i, j1. In addition, we define α i j = p h i, j1 i and α i j = p i h i, j1. We introduce the period n 1 with d n1 = 0 and f n1 = p n1 = h n1 = 0. Without loss of optimality, we assume that y n1 = 1 = ŷ n1. For a given first-stage decision vector y, we define m i := max{k : y k = 1, 1 k i}, i.e., the production period that satisfies the demand in period i. We also define l i := min{k : y k = 1, i k n 1}, where if l i = n 1, then there is no production in [i, n]. Similarly, we define ˆm i and ˆl i for ŷ. Note that (1) α i j α l j α ir α lr for r j, (2) while in any optimal solution of the second-stage ULS model, α ˆm j, j α ˆmi, j for j > i. (3) For given y, ŷ, f, and p, m i and ˆm i are known and we can rewrite the MRULS model as follows: max [ d U d f i (y i ŷ i ) (α mi, i α ˆmi, i)d i ]. (4) Using (4), Zhang [4] shows that, without loss of optimality, we only need to consider the bounds of the uncertainty interval of d if f and p are given. We state the result here without a proof. Lemma 2.1. [4] For a given first-stage decision vector y, realizations of f and p, and a second-stage decision vector ŷ, there exists an optimal scenario d to (1), satisfying: di = d i if α mi,i < α ˆmi,i d i if α mi,i > α ˆmi,i d i or d i if α mi,i = α ˆmi,i. We derive a similar result for the production cost function. For given y, ŷ, and d, the MRULS model is equivalent to max max [ f i (y i ŷ i ) p i (x i ˆx i )], (5) f U f p U p in which the total inventory holding costs are eliminated. Lemma 2.2. For a given first-stage decision vector y, a realization d, and a second-stage decision vector ŷ, there exist optimal scenarios f and p to (1), satisfying: ( f i, p i ) = ( f i, p i ) if x i < ˆx i ( f i, p i ) if x i > ˆx i ( f, p ) or ( f i i i, p i ) if x i = ˆx i. Proof. If x i < ˆx i, it is easy to see that the result holds from (5) since (x i ˆx i ) < 0 and (y i ŷ i ) 0. Similarly for the two other cases. 3. The algorithm In this section we propose a dynamic programming (DP) algorithm that decomposes the MRULS problem into subplans, defined by the regeneration intervals at the first level and the incoming and outgoing inventory at the second level The dynamic programming algorithm We construct a network with nodes (k, ˆm k, ˆl k ) in which the DP moves from node (k, ˆm k, ˆl k ) to (t 1, ˆm t1, ˆl t1 ), defining the subplan [(k, ˆm k, ˆl k ), (t 1, ˆm t1, ˆl t1 )], for 1 k t n, 1 ˆm k k ˆl k ˆl t1, and ˆm k ˆm t1 t 1 ˆl t1 n 1. In this subplan, [k, t] is a regeneration interval for the first-stage ULS model, while in the second stage, d k,ˆl k 1 units enter the subplan and d t1,ˆl t1 1 leave. This shows the correctness of the decomposition at the flow level. The incoming inventory is produced at period ˆm k. If ˆm k is not a period in [k, t], i.e., ˆm k < k, then x ˆmk d ˆmk,k1 d ˆmk,ˆl k 1 = ˆx ˆm k, and by Lemma 2.2 α ˆm k,i = α ˆm k,i for i = ˆm k,..., ˆl k 1. (6) This shows the validity of the decomposition at the uncertainty level. We denote by φ(k, ˆm k, ˆl k, t 1, ˆm t1, ˆl t1 ) the minimax regret for periods [k, t] within the subplan [(k, ˆm k, ˆl k ), (t 1, ˆm t1, ˆl t1 )]. Let G(t1, ˆm t1, ˆl t1 ) be the minimax regret over the first t periods, where ˆm t1 is the last second-stage production period in [1, t 1], and H(k, t 1, ˆm t1, ˆl t1 ) be the minimax regret over the first t periods, where k is the last first-stage production period in [1, t 1] and ˆm t1 the last second-stage production period. We can derive a 2

3 recursion for G(t 1, ˆm t1, ˆl t1 ) and H(k, t 1, ˆm t1, ˆl t1 ): G(1, 1, 1) =0 G(t 1, ˆm t1, ˆl t1 ) = min 1 k t H(k, t 1, ˆm t1, ˆl t1 ) H(k, t 1, ˆm t1, ˆl t1 ) = max {G(k, ˆm k, ˆl k ) 1 ˆm k k ˆl k ˆl t1 ˆm k ˆm t1 for 1 ˆm t1 t 1 ˆl t1 n 1. φ(k, ˆm k, ˆl k, t 1, ˆm t1, ˆl t1 )}, If all values of φ(k, ˆm k, ˆl k, t 1, ˆm t1, ˆl t1 ) are known, we can show that the outer DP runs in O(n 6 ) time. Indeed, for given t, ˆm t1, and ˆl t1, calculating G(t 1, ˆm t1, ˆl t1 ) can be done in O(n) time due to the minimization over k. For each H(k, t 1, ˆm t1, ˆl t1 ), we need to consider the maximization over ˆm k and ˆl k, which takes O(n 2 ) time. Therefore, all values of H(k, t 1, ˆm t1, ˆl t1 ) and G(t 1, ˆm t1, ˆl t1 ) can be calculated in O(n 6 ) time. Next, we will show that all costs φ(k, ˆm k, ˆl k, t 1, ˆm t1, ˆl t1 ) can also be obtained in O(n 6 ) time. Therefore, the DP solves the MRULS model in O(n 6 ) time Subplan properties Below, we present two useful properties of the subplans. In Proposition 3.3, we derive the structure of an optimal d, and its proof uses two lemmas in Zhang [4], while Proposition 3.4 enables us to handle f, p, and d sequentially. Lemma 3.1. [4] For a given first-stage regeneration interval [k, t], a realization of p, and a second-stage decision vector ŷ, if α kk > α ˆmk,k, there exists an optimal scenario d such that d i = d i for i = k,..., t. Lemma 3.2. [4] For a given first-stage regeneration interval [k, t], a realization of p, and a second-stage decision vector ŷ, if α kk α ˆmk,k and ŷ i = 0 for i = k 1,..., t, there exists an optimal scenario d such that di = d i for i = k,..., t. Proposition 3.3. Given the subplan [(k, ˆm k, ˆl k ), (t 1, ˆm t1, ˆl t1 )] with ˆm t1 t, a realization of p, and a second-stage decision vector ŷ, there exists an optimal d such that: If ˆm k < ˆm t1, (d k,..., d t ) = b(i) : (d k,..., d t ) if α kk > α ˆmk,k b(ii) : (d k,..., d t ) if α kk α ˆmk,k and α k, ˆmt1 α ˆmt1, ˆm t1 b(iii) : (d k,..., d γ1, d γ,..., d t ) if α kk α ˆmk,k and α k, ˆmt1 α ˆmt1, ˆm t1, in which γ = min{i : ˆl k i ˆm t1, ŷ i = 1, α ki > α ii }. and if ˆm k = ˆm t1, { (d a(i) : k,..., (dk,..., d d t ) = t ) if α kk > α ˆmk,k a(ii) : (d k,..., d t ) if α kk α ˆmk,k; Proof. Note that, by construction, m i = k for i = k,..., t. Cases a(i) and b(i): The result follows directly from Lemma 3.1. Please note that the condition ˆm t1 t is not required. Case a(ii): Since ˆm k = ˆm t1, ŷ i = 0 for i = k1,..., t, and the result follows from Lemma 3.2. Please note that the condition ˆm t1 t is not required. Case b(ii): From Lemma 2.1, it will be enough to show that α ki α ˆmi,i for i = k,..., t. From the statement of the proposition, the inequality holds for i = k. For i = ˆm t1,..., t, the inequality holds by observing that α ki α ˆmt1,i from (2) and the fact that ˆm i = ˆm t1. It remains to analyze i = k 1,..., ˆm t1 1. By (3), α ˆmt1, ˆm t1 α ˆmi, ˆm t1. Combining this inequality with α k, ˆmt1 α ˆmt1, ˆm t1, we obtain α k, ˆmt1 α ˆmi, ˆm t1 and the desired inequality follows again using (2). Case b(iii): Using a similar argument as in Case b(ii), the result holds for periods i = k and i = ˆm t1,..., t. For k 1,..., ˆm t1 1, we need to consider whether there exists a period i, k 1 i ˆm t1 1, such that ŷ i = 1 and α ki > α ii. If such a period does not exists, then α ki α ˆmi,i for i = k 1,..., ˆm t1 1, and the result follows by letting γ = ˆm t1. Otherwise, from the definition of γ, α k, ˆmi α ˆmi, ˆm i, and using (2), α ki α ˆmi,i for i = k 1,..., γ 1. For i = γ,..., ˆm t1 1, we have α kγ > α γγ, which implies α ki > α γi by (2), while α γi > α ˆmi,i by (3). Combining the two inequalities, the result follows by Lemma 2.1. Proposition 3.4. Given the subplan [(k, ˆm k, ˆl k ), (t 1, ˆm t1, ˆl t1 )], a realization of d, and a second-stage decision vector ŷ, there exists an optimal p such that: (p k,..., p t ) = where θ = k I : (p k,..., p t ) if ˆl k = ˆl t1 II : (p k,..., pˆl k 1,,..., p ) pˆl k t if k < ˆl k < ˆl t1 III : (p k,..., p θ1, p θ,..., p t ) min{i : k < i t, ŷ i = 1} if k = ˆl k < ˆl t1, if ŷ i = 0 for i = k 1,..., t, otherwise. 3

4 Proof. We consider each case separately. Case I: In this case ˆl k t 1 and therefore ˆx i = 0 x i for i = k,..., t. We can assume without loss of optimality that p i = p i for i = k,..., t by Lemma 2.2. Case II: In this case ˆl k t. Since k < ˆl k, we can use the same argument as in Case I to show that the result holds for i = k,..., ˆl k 1. For i = ˆl k,..., t, the result follows by using x i = 0 ˆx i. Case III: We need to consider whether ŷ has a production period in [k 1, t]. If such a period does not exist, x k = d kt d k,ˆl t1 = ˆx k and x i = 0 = ˆx i for i = k 1,..., t. So, we can assume p i = p for i = k,..., t, and the desired i result follows by choosing θ = k. Now suppose that such a production period exists and θ = min{i : k < i t, ŷ i = 1}. By construction we have x k = d kt d k,θ1 = ˆx k and x i = 0 = ˆx i for i = k 1,..., θ 1, which implies that p i = p i for i = k,..., θ 1. For i = θ,..., t, x i = 0 ˆx i implies that p i = p i and the desired result follows Minimax regret of a subplan In this section we calculate φ(k, ˆm k, ˆl k, t 1, ˆm t1, ˆl t1 ). We start with diagonal subplans, i.e., ˆm t1 t. For convenience, we let L j i be the optimal objective value of the ULS model defined between periods i and j with all uncertain parameters at lower bounds. We define L j i in a similar fashion, but where the demands are at their upper bounds Case I: ˆl k = ˆl t1, ˆm t1 t Using Proposition 3.4, (p k,..., p t ) = (p k,..., p t ), while α ˆm k, i = α ˆmk, i, i = k,..., t, by (6). In addition, according to Proposition 3.3, we need to consider demand scenarios in Cases a(i)-a(ii). Therefore, φ = f k t α kid i t α ˆm k, i d i if α kk > α ˆmk,k f k t α kid i t α ˆm k, i d i if α kk α ˆmk,k. (7) Case II(b)(i): α kk > α ˆmk,k In this case, (d k,..., d t ) = (d k,..., d t ). In the second stage, there is no production in [k, ˆl k 1] or [ ˆm t1 1, t]. Therefore, φ = f k ˆl k 1 α ˆmk, i d i L ˆm t11 ˆl k f ˆmt1 i= ˆm t1 α ˆmt1, i d i. Case II(b)(ii): α kk α ˆmk,k and α k, ˆm t1 α ˆmt1, ˆm t1 (8) In this case, φ is identical to (8) with (d k,..., d t ) = (d k,..., d t ), i.e., φ = f k ˆl k 1 α ˆmk, i d i L ˆm t11 ˆl k f ˆmt1 i= ˆm t1 α ˆmt1, i d i. Case II(b)(iii): α kk α ˆmk,k and α k, ˆm t1 > α ˆmt1, ˆm t1 We need to go through all candidates to γ, i.e., ˆl k γ ˆm t1 where α kγ > α γγ, and force ŷ γ = 1 and (d k,..., d t ) = (d k,..., d γ1, d γ,..., d t ). By optimizing over all these candidates we get φ = f k L γ1 ˆl k γ1 [ α kγ >α γγ max ˆl k γ ˆm t1 L ˆm t11 γ ˆl k 1 α ˆmk,i d i ] f ˆmt Case III: k = ˆl k < ˆl t1, ˆm t1 t i= ˆm t1 α ˆmt1, i d i. Since ˆm k = k, we cannot have demand scenarios a(i) or b(i). If ˆm t1 = k, the demands in both stages are satisfied at period k and φ = 0. Otherwise, k < ˆm t1 t and α k, ˆm t1 = α k, ˆmt1 while α ˆm t1, ˆm t1 = α ˆmt1, ˆm t1. Case III(a)(ii): ˆm k = ˆm t Case II: k < ˆl k < ˆl t1, ˆm t1 t We have (p k,..., p t ) = (p k,..., pˆl k 1, pˆl k,..., p t ) by Proposition 3.4, while again α ˆm k, i = α ˆm k, i, i = k,..., ˆl k 1. In addition, according to Proposition 3.3, we need to consider demand scenarios in Cases (b)(i)-(b)(iii). As argued above, φ = 0. Case III(b)(ii): ˆm k < ˆm t1 and α k, ˆmt1 α ˆmt1, ˆm t1 In this case (d k,..., d t ) = (d k,..., d t ). In the second stage, we need to go through all candidates for θ, k < θ 4

5 ˆm t1. We force ŷ θ = 1, ŷ i = 0 for i = k,..., θ 1, and (p k,..., p t ) = (p k,..., p θ1, p θ,..., p t ), and optimise over all these candidates. Therefore, φ = max [ k<θ ˆm t1 i=θ L ˆm t11 θ ] f ˆmt1 i= ˆm t1 α ˆmt1, i d i. Case III(b)(iii): ˆm k < ˆm t1 and α k, ˆmt1 > α ˆmt1, ˆm t1 Besides θ, we need to optimize over all the candidates for γ as in Case II(b)(iii): φ = max k<θ γ θ γ ˆm t1 α kγ >α γγ f ˆmt1 γ1 [ i=θ i= ˆm t1 α ˆmt1, i d i Non-diagonal subplans L γ1 θ L ˆm t11 γ ] In this section, we focus on non-diagonal subplans, i.e., ˆm t1 = ˆl t1 = t 1. Note that if ˆl k = ˆl t1, then (7) still holds. Therefore, we will assume that ˆl k t, i.e., there is at least one second-stage production period in [k, t]. For given k, ˆm k and ˆl k, let us define Φ(t) = φ(k, ˆm k, ˆl k, t, t, t). If ˆm k < k, we have ˆl k 1 Φ(ˆl k ) = f k α ki di ˆl k 1 α ˆmk, i d i Φ(t 1) = max{ max [φ(k, ˆm k, ˆl k, t, ˆm t, t 1) ˆl k ˆm t <t (α kt α ˆmt,t )d t ], Φ(t) f t (α kt α tt )d t }, for t = ˆl k,..., n, where the direction of bounds of α follows from k < ˆm t t, and di is determined using Lemma 2.1. If ˆm k = k, the first period to be considered by the recursion is ˆl k 1, and the initial condition should be replaced by Φ(ˆl k 1) = The running time In this section we show that all values of φ can be calculated in O(n 6 ) time. Note that L j i, L j i and j s=i α rsd s can be calculated in O(n 3 ) time for all values of r i j in a preprocessing procedure. For diagonal subplans, φ can be calculated in constant time, except for Cases II(b)(iii) and III(b)(ii), in which it takes O(n) time, and for Case III(b)(iii), in which it takes O(n 2 ) time. The desired running time follows by observing that there are O(n 4 ) Case III subplans, while for Case II(b)(iii) there are only O(n 5 ) relevant subplans since φ(k, ˆm k, ˆl k, t 1, ˆm t1, a 1) = φ(k, ˆm k, ˆl k, t 1, ˆm t1, a), (9) with a > t 1 Once the values of φ for all diagonal subplans are available, it is straightforward to see from Section that all non-diagonal subplans can be handled in O(n 5 ) time. 4. Single type uncertainty In this section, an O(n 5 ) time algorithm is proposed for demand uncertainty only (the MRULS d model), and an O(n 4 ) time one for production cost function uncertainty only (the MRULS p model) Demand uncertainty When only the demand is uncertain, the nodes of the network proposed in Section 3 can be simplified to (k, ˆm k ) and the number of subplans is reduced to O(n 4 ). Let φ d (k, ˆm k, t 1, ˆm t1 ) denote the minimax regret for periods [k, t] within the subplan [(k, ˆm k ), (t 1, ˆm t1 )]. By including the O(n 3 ) time preprocessing procedure in Section 3, we show below that any diagonal subplan can be computed in at most O(n 2 ) time. Noting that φ d (k, ˆm k, t 1, b) = φ d (k, ˆm k, t, b) (α kt α bt )d t, for k b t, all diagonal subplans can be handled in O(n 5 ) time, and the desired running time for the MRULS d model follows. To calculate φ d (k, ˆm k, t1, ˆm t1 ), ˆm t1 t, we use Proposition Case a: ˆm k = ˆm t1 Please note that ˆm k = ˆm t1 implies ˆm t1 t. We have φ d = f k t α kid i t α ˆm k, id i if α kk > α ˆmk,k f k t α kid i t α if α kk α ˆmk,k ˆm k, id i and ˆm k < k if α 0 kk α ˆmk,k and ˆm k = k Case b: ˆm k < ˆm t1 t Case b(i): α kk > α ˆmk,k In this case, ˆm k < k. If η is the first second-stage pro- 5

6 duction period in [k 1, ˆm t1 ], then φ d = f k f ˆmt1 η1 min [ k<η ˆm t1 i= ˆm t1 α ˆmt1, id i. α ˆmk, id i L ˆm t11 η ] Case b(ii): α kk α ˆmk,k and α k, ˆmt1 α ˆmt1, ˆm t1 In this case, φ d is similar to that in Case b(i), where (d k,..., d t ) = (d k,..., d t ). Thus, φ d = f k f ˆmt1 if ˆm k < k, and φ d = f k if ˆm k = k. η1 min [ k<η ˆm t1 i= ˆm t1 α ˆmt1, id i L ˆm t11 k f ˆmt1 α ˆmk, id i L ˆm t11 ] η i= ˆm t1 α ˆmt1, id i Case b(iii): α kk α ˆmk,k and α k, ˆmt1 > α ˆmt1, ˆm t1 In addition to η, and similarly to Cases IIb(iii) and IIIb(iii) in Section 3.3, we need to optimize over candidates for γ. Therefore, φ d = f k L γ1 η if ˆm k < k, and max k<η γ η γ ˆm t1 α kγ >α γγ γ1 [ L ˆm t11 γ γ1 φ d = f k max [ α ki d k γ ˆm i t1 α kγ >α γγ f ˆmt1 α ˆmt1, id i i= ˆm t1 if ˆm k = k. η1 α ˆmk,id i ] f ˆmt1 i= ˆm t1 α ˆmt1, id i L γ1 k 4.2. Production cost function uncertainty L ˆm t11 γ ] The nodes of the network for the MRULS p model can be simplified to (k, ˆl k ) and again the number of subplans is reduced to O(n 4 ). Let φ p (k, ˆl k, t 1, ˆl t1 ) denote the minimax regret for periods [k, t] within the subplan [(k, ˆl k ), (t 1, ˆl t1 )], and L p (i, t, j), i t j, denote the optimal objective value of a ULS model from periods i to t with demands (d i,..., d t1, d t, j1 ) and setup and unit production costs at their lower bounds. We have φ p = f k p k d kt if ˆl k = ˆl t1 f k p k d kt L p (ˆl k, t, ˆl t1 ) if k < ˆl k < ˆl t1 max{p k d t1,ˆl t1 1, max k<θ t[p k d θt L p (θ, t, ˆl t1 )]} if k = ˆl k < ˆl t1. For a given value of j, L p (i, t, j) can be calculated in O(n 3 ) time for all values of i and t. Hence we can calculate all L p (i, t, j) in O(n 4 ) time in a preprocessing procedure. It is then straightforward to see that φ p (k, ˆl k, t 1, ˆl t1 ) can be calculated in constant time for ˆl k = ˆl t1 and k < ˆl k < ˆl t1. For k = ˆl k < ˆl t1, the running time is O(n), but there are only O(n 3 ) of these subplans. Therefore, the overall running time remains at O(n 4 ), and the desired running time for the MRULS d model follows. 5. Conclusions In this paper, we have formulated the minimax regret ULS model with both production cost function and demand uncertainty. We gave a theoretical analysis to reduce the number of optimal uncertainty scenarios and proposed an algorithm that solves the problem in O(n 6 ) time. Lower running times were found for the case in which only one single type of uncertainty is present. In particular, an O(n 4 ) time algorithm is proposed for production cost function uncertainty only, and an O(n 5 ) time one for demand uncertainty only. The latter reduces the running time of the O(n 7 ) time algorithm proposed in [4]. Bibliography [1] H. M. Wagner, T. Whitin, A dynamic version of the economic lot size model, Management Science 5 (1958) [2] D. Bertsimas, D. Brown, C. Caramanis, Theory and applications of robust optimisation, SIAM Review 53 (2011) [3] P. Kouvelis, G. Yu, Robust Discrete Optimization and its Applications, Kluwer Academic, Norwell, MA, [4] M. Zhang, Two-stage minimax regret robust uncapacitated lot-sizing problems with demand uncertainty, Operations Research Letters 39 (5) (2011)