Market Segmentation and Product Technology Selection for Remanufacturable Products: Appendix

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1 Market Segmentation and Product Technology Selection for Remanufacturable Products: Appendix Laurens G. Debo Graduate School of Industrial Administration Carnegie-Mellon University Pittsburgh, PA 523, USA L. Beril Toktay Technology and Operations Management INSEAD Fontainebleau, France Luk N. Van Wassenhove Technology and Operations Management INSEAD Fontainebleau, France Some properties of the revenue function Lemma 4 i The Hessian H of the revenue function R is of the form H a + b a a a. ii If F F k and n > 0, then η θ + κ η θ > 0 and η θ + κ η θ > 0 θ [0, imply a < 0 and b < 0. Proof. i Recall that p N and p R denote the prices of new and remanufactured products, respectively, and that the net utility that a consumer of type θ derives from buying a new product, a remanufactured product, and no product, is θ p N, ηθ p R, and 0, respectively. In a given period, consumers choose which product to buy based on the utility that they derive in that period from this purchase. Without loss of generality, we only consider cases where p R ηp N ; if p R

2 were larger than ηp N, no remanufactured products would be sold, and the price p R could be reduced to the level ηp N without affecting the demand for either product. Let p denote the vector p N, p R. Then Ω N p. {θ [0, ] : θ p N θ p R } is the set of consumer types who purchase a new product. Ω R p is defined analogously as the set of consumer types who purchase a remanufactured product. Define the marginal consumers θ l p and θ h p such that θ l is indifferent between buying no product and buying a remanufactured product, and θ h is indifferent between buying a remanufactured product and a new product. Since ηθ is a strictly increasing function, Ω R p [θ l p, θ h p and Ω N p [θ h p, ], where θ l p and θ h p satisfy ηθ l p R and θ h ηθ h p N p R. A- Taking the derivative of these two equalities with respect to p N and p R gives θ l p R η θ l, θ l p N 0, θ h p R η θ h and θ h p N η θ h. Let n and r denote the volume of consumers who purchase new and remanufactured products, respectively, and define ν. n, r. Then n Ω N p df θ θ h f θdθ Fθ h and r Ω R p df θ θ h θ l f θdθ Fθ h Fθ l. Taking the derivative of these two equalities with respect to n and r gives η θ h f θ h p N + η θ h f θ h p R 0 η θ h f θ h p N + η θ h f θ h p R 0 η θ h f θ h p N η θ h f θ h + η θ l f θ l η θ h f θ h p N η θ h f θ h + η θ l f θ l pr pr The simultaneous solution of these four equations yields p N θ l η fθ l η θ h fθ h p R η θ l fθ l. and p N p R Since Rν np N ν + rp R ν, we obtain, using the chain rule, that [ [ [ R, R ] p N + n p N + r p R, p R + n p N + r p R ] p N η θ h fθ h + η θ l fθ l Fθ h η θ l fθ l Fθ h Fθ l, p R η θ l fθ l Fθ l ]. 2

3 Define G N θ. θ Fθ fθ [ R, R and G R θ. ηθ η θ Fθ fθ. Using A-, we find ] [ G N θ h G R θ h + G R θ l, G R θ l ]. A-2 Taking the derivative of Rν and Rν obtain the elements of the Hessian H: with respect to r and n, and following similar steps, we Since θ l 2 R 2 G N θ h G R θ h θ h + G R θ l θ l and 2 R 2 2 R G R θ l θ l. θ l fθ l, we see that H G N θ h G R θ h θ h. a + b a a a A-3 with a G R θ l fθ l and b ii We now specialize this matrix to the case F θ θ κ, or fθ κ θ κ. For this distribution, G N θ θ θ κ, G R θ ηθ η θ θ κ A-4 so G N θ κ + and G R θ η θ κ + η θ θ κ. Since n Fθ h, we have θ h n κ. Therefore, θ h κ θ h κ. Substituting, we find a η θ l κ + η θ l θ l κ κ θ l κ b κ + η θ h + η θ h θ h κ κ θ h κ. n > 0 θ h <. In this case, the last term in both expressions is positive since θ l θ h. When the conditions η θ + κ η θ > 0 and η θ + κ η θ > 0 hold θ [0,, this is sufficient for the first term in both expressions to be positive. We conclude that a < 0 and b < 0. In particular, if ηθ θ with 0,, then η θ and η θ 0, so these conditions are satisfied. In the remainder of the paper, we will assume ηθ is such that a < 0 and b < 0 for the family of distributions F F k when n > 0. In this case, H has the following properties: 2 R H > 0, 2 R < 0 R is strictly concave. Property 2 < 0 New and remanufactured products are imperfect substitutes. Property 2 3

4 2 R < 0 Property R < 2 R 2 Property 4 2 R 2 2 > 2 R Property 5 2 Assumptions A useful condition for characterizing the optimal path is that the solution to 3 is interior to the feasible region. In addition, to characterize q using first order conditions, we need q <. To this end, we introduce Assumption and Assumption 2, and we provide conditions on model parameters that assure that Assumption holds if F F κ Lemma 5. We will assume throughout the derivations that in addition to those assumptions already introduced in the text, the following assumptions hold: The maximizer n, r of 3 satisfies n + r <. Assumption c n 0 <, c r q < q and q < : c q < v q for q < q. Assumption 2 c q > 0 and c nq + qc r q > 0. Assumption 3 c n q and c r q are finite for all q [0, ]. Assumption 4 F F κ. Assumption 5 η θ + κ η θ > 0, η θ + κ η θ > 0, η θ < θ [0,. Assumption 6 Lemma 5 For a consumer profile with F F κ, there exists κ 0 > 0 such that Assumption is satisfied for all κ 0, κ 0. Proof. The Lagrangian has the form L πn, r + vi + qn r + µ n n + µ r r + λ n r + τ I r. 4

5 r, n, λ, µ n, µ r, τ satisfy the K-T conditions: π + qv I + qn r + µ n λ 0 π v I + qn r + µ r λ τ 0 λ n r 0, µ n n 0, µ r r 0, τ I r 0 λ 0, µ n 0, µ r 0, ν 0 Assume that r + n and µ n µ r 0Then π n, n + qv I + qn r λ 0, or λ G N θ h G R θ h +G R 0 c n q+qv I+qn r. Define yθ h. G N θ h G R θ h +G R 0 θ h θ h κ ηθ h +η θ h θ h κ +η0 η 0 κ. Let θ be the maximizer of y. If y θ c n 0+v 0 < 0, then λ < 0 for all θ h [0, ], then λ < 0, which is inconsistent with the K-T conditions. This condition can be rewritten as κ < a. η θ < with Assumption 6, a exists. θ η θ+η θ [v 0+ θ+η0 η θ c n0] +. Since Assumption 2 holds and Assume that r + n and µ n 0, but µ r > 0. Then n, r 0, and τ 0, and π, 0 + qv I + q λ 0, or λ κ c nq + qv I + q. If κ < â. [v 0 c n0] +, then λ < 0, which is inconsistent with the K-T conditions. With Assumption 2, â exists. Assume that r + n and µ r 0, but µ n > 0. i Assume I. Then n 0, r, and τ 0. In this case, π 0, v I λ 0, or λ η0 η 0 κ c r q v I. If κ < ã. η 0 η0, then λ < 0, which is inconsistent with the K-T conditions. ii Assume I <. Then r contradicts the K-T condition r I. Assume that r + n and µ r > 0, µ n > 0. Then n r 0, which contradicts the assumption r + n. We conclude that if κ < κ 0. mina,â,ã, then r + n <. 3 Proofs Proof of Lemma. Smith and McCardle 2002 consider a Markov decision process of the form v k x k sup ak A k {r k a k, x k + k E[v k x k a k, x k ]} for k > 0, v 0 x 0 0. Here, a k and x k Θ are the decision variable and the state variable, respectively, in period k. The authors 5

6 establish conditions under which the properties of the reward functions r k are inherited by the value function vk. In particular, they prove the following: Proposition 5 Smith and McCardle Let U be the set of functions on X satisfying a C3 closed convex cone property P and let P be a joint extension of P on AxΘ. If, for all k, a the reward functions r k a k, x k satisfy P and b the transitions x k a k, x k satisfy P U, then each vk satisfies P and lim k vk, if it exists, also satisfies P. The corresponding variables in our problem are the following: a ν, A D {r I}, x I, Θ [0,,, and x I + qn r. We would like to show that the value function is concave. Concavity is a C3 property. The authors show that for convex action and state spaces A and Θ, joint concavity on AxΘ is a joint extension of concavity on Θ. In our problem, the reward function r k ν k, I k πν k and is independent of I k. Since π is concave, condition a is satisfied. Since our recursion is deterministic and the transition function is linear, it trivially satisfies condition b. In addition, since we have a discounted-cost formulation with a bounded reward function, V lim k v k exists. We conclude that V I; q is a concave function of I. Note that if the path {n t, r t, t 0} is feasible for I 0 I, then it is feasible for any I 0 > I. Thus, V I; q is non-decreasing in I. This concludes the proof of part i. The Hessian H of fν, I with respect to ν satisfies H < 0 and H > 0 so f is strictly concave in ν. A unique maximizer of f on A therefore exists for all t and the optimal path {ν t n t,r t, t 0} is unique. Proof of Lemma 2. Part i: We want to show that if c q < v q, then there exists a feasible path with a positive discounted profit. We proceed by constructing such a path. Pick ε > 0 and consider the path P ε {ε, 0,ε, qε, ε, qε,...} I0. Using the Taylor series expansion of Rν around the point 0,0, we can write R ε, 0 R 0, 0+ε R0,0 +o ε ε R0,0 +o ε and R ε, qε R 0, 0 + ε R0,0 + qε R0,0 + o ε ε R0,0 + q R0,0 + o ε, where we used the 6

7 fact that R0, 0 0. Define Ṽε; q. {ν t P ε,t 0} t πν t, q. Then, Ṽ ε; q R ε, 0 c n q ε + R 0, 0 ε R 0, 0 ε ε ε t0 t R ε, qε c n q ε qc r q ε t + o ε c n q ε + + q c n q + qc r q + t R 0, 0 t R 0, 0 + ε + q R 0, 0 ε t R 0, 0 t t t c n q + qc r q t ε v q c q + o ε R 0, 0 To show that Ṽε; q > 0 for some ε > 0, consider Ṽ ε + q + q + oε t c n q + qc r q + oε t0 R 0, 0 + o ε c n q ε qc r q ε R 0, 0 Ṽ lim ε;q ε 0+ ε v q c q + oε lim ε 0+ ε v q c q. Since the latter expression is strictly positive if cq < v q we conclude that there exists an ε > 0 such that Ṽ ε; q > 0 if cq < v q. Part ii: We now show that if cq v q, then V q 0 and it is optimal not to sell anything. To do this, we first show that there exists a feasible path with zero discounted profit. Next, we find a non-positive upper bound on the discounted profit on any path under the condition cq v q. The discounted profit on the path {0, 0, 0, 0,0, 0,...} I0 is 0. Therefore V q 0. Take any P I0. By the definition of I0, T t0 qn t r t+ 0 for any T. We will now show by induction that T t0 t qn t r t+ 0. We will first establish that T t0 t qn t r t+ T T t0 qn t r t+ by induction on T. For T 0, this condition holds with equality. For any T, assume that T t0 t qn t r t+ 7

8 T T t0 qn t r t+ induction step. Then T T t qn t r t+ t qn t r t+ + T qn t r t+ t0 t0 T T qn t r t+ + T qn t r t+ t0 T T qn t r t+ + T qn t r t+ t0 T T qn t r t+, t0 where the inequality follows from the induction hypothesis and the next step from the fact that < multiplies a positive term. This proves the induction hypothesis. Since T t0 qn t r t+ 0, we conclude that T t0 t qn t r t+ 0, or q T t0 t n t T t t r t. Multiplying the inequality by and taking the limit for T, we obtain t r t q t n t. t t0 A-6 We can use this property to derive an upper bound on V,P, the discounted profits for path P. The profits are: V,P R ν 0 c n qn 0 + t R ν t c n q n t c r q r t t R 0, 0 n 0 c n qn 0 R 0, 0 + t R 0, 0 n t + r t c n qn t c r qr t t R 0, 0 c n q n 0 R 0, 0 R 0, 0 + t c n q n t + c r q r t t R 0, 0 R 0, 0 c n q t n t + c r q q t n t t0 t0 R 0, 0 R 0, 0 c n q + q c r q t n t t0 v q c q t n t. t0 where the first inequality holds because of the concavity of R ν and the second inequality holds by A-6. We conclude that cq v q implies V q 0, which is achieved on the path 8

9 {0, 0, 0, 0,0, 0,...}. Proof of Lemma 3. Consider the path P q {n s, 0,ñ, r,ñ, r,ñ, r...} I0 and let V,P q denote corresponding discounted profit on this path. We have that Rñ, r Rn s,0 c n q. Hence, Rns,0 Rñ, r. By Property and Property 2, Rn,r c n q and is a strictly decreasing function of r and n. Therefore, either ñ n s and r 0 or ñ < n s and r > 0. Then, I qn s qñ r. As qñ r, we have that I t+ I t + qñ r I t t. Since I r, it follows that I t r t and P I0. By the definition of n s and ñ, r, the elements of policy P achieve the optimal profit over the feasible region in each period starting with the initial condition I 0 0. Therefore, P achieves the optimal profit: V,P q V q. We will show that the derivative of V,P q with respect to q is negative. V,P q t0 t πν t,q. πn s,0,q π π s + π π c nqn s, where we used πns,0 n 0. π νt,q π ñ + π r + π c nqñ c rq r t where we used π, π 0, 0 by the definition of ñ, r. Summing, we find that V,P q c n q n s + ñ c r q r ; Assumption 4 assures that the sum converges. Since n s ñ, V P, q c n q ñ c r q r. By assumption, c n q > 0. If, in addition, c r q 0, then V P, q < 0. If c r q < 0, then V P, q c n q qc r qñ since r qñ. Invoking Assumption 3, we again obtain V P, q < 0. Lemma 6 Assume c q < v q. Recall that the policy function g I I + qn I r I and consider the interval X [0, g 0]. Then, gi > 0 for I X and either i g I < 0, g I < and r I I for I X, or ii there exists I X such that g I < 0, g I < and r I I for I [0, I], and g I 0, r I < I for I I, g 0]. In addition, if qñ < r, gi < I on I, g0]. Proof In this lemma, we prove that the policy function g has one of the two forms shown in Figure 6 decreasing or U-shaped on [0, g0], which is the relevant interval, as we shall show. For notational simplicity, we suppress the dependence of πν, q and V I, q on q in this proof. We make the following conjecture concerning n I, r I: There exists an interval [0, Ĩ] on which n I > 0 and r I I, i.e. there exists an interval over which the constraint r I in 3 is binding. Assuming that the conjecture is correct, we characterize n I and g I in Step 9

10 gi gi I g0 I I I g0 I Case i Case ii Figure 6: Policy function for Lemma 6 a. We validate the conjecture in Step b, and determine whether it holds for the entire interval [0, g 0]. If so, then part i is proven with Ĩ g0; otherwise, we determine the exact interval [0, I] for which the conjecture holds. We make a second conjecture concerning n I,r I: On I, g 0], n I > 0 and 0 < r I < I. Steps 2a and 2b characterize g under these assumptions and validate this conjecture, respectively, proving part ii. Step a: Characterization of n I and g I under the conjecture r I I and n I > 0 on [0, Ĩ]. Consider the maximization problem in 3. By Lemma Assumption, r I + n I <. In addition, we conjectured that n I > 0, so n I is an interior solution. Finally, we conjectured a boundary solution r I I. Therefore, n I and r I jointly satisfy π n I, I π n I,I + q V qn I I V qn I I 0. A-7 > 0. A-8 0

11 We now characterize gi using A-7: Taking the derivative of A-7 with respect to I and using the chain rule, we can calculate n. dn I di : We find n 2 π n I, I 2 n + 2 π n I, I + q 2 V qn I n 0. 2 πn I,I 2 πn I,I 2 +q 2 V qn I < 0 where the inequality follows by Property, Property 2 and the concavity of V. Since gi qn I under the conjecture, g I qn < 0 and 2 πn,i 2 πn,i 2 πn,i 2 q 2 V qn,i 2 πn,i n < the former step because V qn 0 and the latter 2 step by Property 4. Thus g I <. To summarize, we have shown that g I < 0, g I < and n I strictly decreases if we assume that r I I and n I > 0. Step b: Validation of the conjecture We now need to show that there exists a range [0, Ĩ] on which n I > 0 and A-8 is satisfied. Let us take the derivative of the two terms in A-8 with respect to I. We find πn I,I I 2 π 2 π n + 2 π 2 V r qn < 0 and 2 π I V qn I qv qn I n > 0. A-8 holds at 2 r H +q 2 +qv qn I 0. As the first term in A-8 strictly decreases in I and the second term strictly increases in I, one of the following two cases is true on [0, g0]: either i A-8 holds I [0, g0] or ii there exists I 0, g0] for which A-8 is satisfied at equality. Expressing this more precisely, we have one of the following two cases: i πn g0,g0 V qn g0 > 0. Then g I < 0, g I < and n I strictly decreases on X [0, g0]. Since g I <, we have g g0 > 0, that is, n g0 > 0 since g qn on this range which, together with the fact that n I strictly decreases on this range, validates the conjecture that n I > 0 over [0, g 0]. ii There exists I. sup I : πn I,I V qn I > 0 < g0 The conjecture that r I I on [0, Ĩ] is validated with Ĩ I since A-8 holds on [0, I, with equality holding only at I I. Then g I < 0, g I < and n I strictly decreases on [0, I]. Step 2a: Characterization of g I under the conjecture 0 < r I < I and n I > 0 on I, g 0] We now complete case ii by characterizing n I, r I over I, g0] We conjecture that 0 < r I < I and n I > 0 for I I, g0]. In this case, n I and r I satisfy the first order

12 conditions of the right hand side of 3: π n I, r I π n I,r I + q V gi I V gi I 0. A-9 0. A-0 We also know that V, n, r jointly satisfy V I π n I,r I + V g I with gi I + qn I r I. Taking the derivative of this expression with respect to I gives V I π n + π r + V gig I with r. dr I di. Taking the derivative of gi and evaluating it at n I, r I gives g I + qn r. Substituting g I in the previous expression, collecting terms, and simplifying using A-9 and A-0, we find V I V g I. A- V I V g Ig I. A-2 By Lemma, V I 0. First consider the case V I 0. By A-, if V I 0, we have V g I 0. Then, by A-9 and A-0 and the definition of ν, n I, r I ñ, r. Thus, g I I + qñ r, and g I > 0. Note that if qñ < r, then gi < I. Next consider the case V I > 0. By A-, if V I > 0, then, as <, we must have that V g I > V I, and by the concavity of V, g I < I. We conclude that if qñ < r, then gi < I on I, g0]. To characterize g I when V I > 0, consider the following three subcases: a V g I < 0 and V I < 0. Since 0 < <, and A-2 must hold, g I > 0. b V g I < 0 and V I 0. Since > 0 and A-2 must hold, g I 0. c V g I 0. Taking the derivative of A-9 and A-0 with respect to I and using V g I 0 gives a system of two equations in the two unknowns n I and r I whose only solution is n I 0 and r I 0. In this case, g I > 0. Thus, we conclude that g I 0 on I, g0]. Step 2b: Validation of the conjecture 2

13 We will first determine the sign of n and r. Taking the derivative of A-9 and A-0 with respect to I, we obtain 2 π n + 2 π I 2 + q 2 2 V 2 2 π q 2 V n + I 2 I 2 r q 2 V I 2 q 2 V 2 π + 2 V r 2 V 2 I 2 I 2 A-3 from which we solve for n, r : n r 2 V I 2 H + 2 π 2 + q 2 2 π 2 + 2q 2 π 2 V I 2 q 2 π q 2 π + 2 π 2 2 π 2. The signs of each term can easily be determined: 2 V I 2 0, H + 2 π + q 2 2 π π 2 2 q 2 V I 2 0, q 2 π 2 2 π > 0 and q 2 π + 2 π 2 < 0 by Lemma, Property 2, Property 3 and Property 4. Therefore, n 0, r 0 and n n. Showing that n < is equivalent to showing that With Property 5 we obtain q q 2 π 2 + 2q 2 π 2 π 2 V 2 I 2 < H. and with Property 4, we obtain further that q q 2 π 2 + 2q 2 π > 2 π, q q 2 π 2 + 2q 2 π > 2 π 2. A-4 Together with 2 V I 2 < 0 and H > 0, we have that A-4 is satisfied. We conclude that n <. Since r I I > 0 and we have proven that r 0 on I, g0], we conclude that r I > 0 on I, g0]. However, since we have proven that n < 0, it is not as immediate that n I > 0 I I, g0]. On the other hand, as n < both on I, g0] and on [0, I] by part i, we have that n g0 > 0. Because n < 0 on [0,g0], we conclude that n I > 0 on I, g0] is also verified. Lemma 7 Define ν qn r. If c q < v q and qñ < r, then,. n, r D simultaneously satisfying Rν + q Rν c q and i There exists a unique I [0, g0] such that I solves g I I. In addition, g I < 0 and g I <. Moreover, r I. ii The region [gi, I] is a capture region, that is, I t [gi, I] implies that I t+ gi t [gi, I]. 3

14 iii Starting with I 0 0, there exists a T q 0 such that r t I t t T q. Moreover, lim t I t I. Proof Part i: We start by conjecturing that there exists I such that g I I, g I < 0 and g I <. Then, I gi I + qn I r I, implying that qn I r I. A-5 By Lemma 6, we know that g I 0 and g I < if and only if r I I, so under the above conjecture, I satisfies A-7: π n I, I + q V qn I I 0. A-6 Also recall from Lemma 6 that V I π n I, I + V qn I for any I such that r I I. Using the chain rule, V π I I + π + qv qn I I I π + qv qn I + π π where the last equality follows by substituting A-7. In particular, under the above conjecture, this equality holds for I I, yielding V I πn I, I A-7 In addition, since I r I qn I, V Substituting this equation in A-6, we find: qn I πn I, qn I. A-8 π n I, qn I + q π n I, qn I 0. A-9 Consider N n. πn,qn +q πn,qn. As N n 2 πn,qn 2 +q 2 πn,qn +q 2 πn,qn + q 2 πn,qn < 2 0 since all terms are negative, the solution to A-9, if it exists, is unique. Now, we show that such a solution exists. N 0 v q cq by definition. Since v q > cq we have N0 > 0. Note that n [0, +q ], as n + qn. Thus, if we show that N +q < 0, then, we will have proven that N n 0 has a unique solution n such that n > 0 and n+q n <. The proof of Lemma 5 develops conditions on κ that satisfy πn, n conclude that π +q, q +q + q π +q, q +q + q πn, n < 0. Noting that at n +q, r n, we < 0, which can be rewritten as N 4 +q < 0. We further

15 conclude that n I solving A-9 is unique and that n I, qn I intd. In addition, since dn I di < 0 in this range as shown in the proof of part i in Lemma 6, I is unique. Recall that ν. n, r satisfying πν + q πν 0, qn r and ν D simultaneously. Comparing A-5 and A-9 with the definition of ν, and noting that n I, qn I intd, we conclude that n n I and r r I under the conjecture. In addition, since I r I under the conjecture, we have that r I. We now need to show that the conjecture is true. By Lemma 6, I satisfies the conjecture if πn I,I V qn I > 0, or, using A-5 and substituting A-8, if πn I,qn I > 0, or, since > 0, π n I,qn I Let the functions r n and r n be defined by πn,rn > 0. A-20 0 and πn,rn 0. Consider Figure 7. We will show that if qñ < r, then the solution to A-9 lies on the segment determined by r qn, r < r n, and r > rn marked in bold in Figure 7, and that on this segment, πn,qn this case, I satisfies the conjecture. > 0. In r rn * n,r * qn r, r r n, r r n qn r 0 n su n rn Figure 7: r < r n, and r > rn for Lemma 7 5

16 Since R ν is concave, ñ, r is the unique intersection point of r n and r n. Note that by differentiating πn,rn 2 πn,rn + 2 πn,rn n 2 rñ rñ r, we find ñ r ñ < r ñ 0 and πn,rn 0, we obtain 2 πn,rn πn,rn n 0 and 0. Evaluating these expressions at n ñ and using the fact that 2 πñ, r 2 πñ, r 2 2 πñ, r < 2 πñ, r 2 and ñ 2 πñ, r 2 2 πñ, r 2 πñ, r 2 2 πñ, r. It follows that 0 < 2 π 2 π π 2 π H which is true by Property. Therefore, r n crosses r n from above, validating Figure 7. Thus, for n > ñ, we have r n < r n. As r > qñ, it follows that the line r qn lies below the point ñ, r, further validating Figure 7. Furthermore, as 2 π πn,r < 0 we have that < 0 for r > r n and πn,r > 0 for r < r n. Therefore, on the segment determined by r qn, r < r n, and r > rn, marked in bold in Figure 7, we have that πn,r < 0 and πn,r > 0. A-9 and A-20 being simultaneously satisfied means that the first term of A-9 must be negative and the second term must be positive. Therefore, the solution n, qn to A-9 lies on the segment determined by r qn, r < r n, and r > rn. Thus, from A-20, it follows that I satisfies the conjecture. We have shown that there exists a unique solution to gi I on [0, I]. To complete the proof, we need to show that gi I has a unique solution on [0, g0]. When qñ < r, Lemma 6 shows that gi < I for I I, g0]. In addition, g I < 0 for I [I, I] [0, I]. So gi < I for I I, g0] under the condition qñ < r and the equality gi I admits only one solution on [0, g0]. Part ii: Let {I t } be the sequence obtained by starting with I 0 0 and applying g successively, i.e., I t gi t. Define L. [gi, I] [0, g0]. L [gi, I {I } I, I]. Pick I t L. If gi I t < I, I t+ gi t ggi < gi + I gi I where the first inequality follows because g is strictly decreasing in this region, and the second inequality follows because, in addition, g < in this region. If I t I, then I t+ I by the definition of I. If I < I t I, then gi I t+ gi t < I t where the first inequality follows because gi is the minimum value that the function g attains on [0, g0], and the second inequality follows since gi < I in this region. Putting it all together, we conclude that gi I t+ < I. In other words, I t [gi, I] implies that I t+ [gi, I. Therefore, if there exists a finite time T q such that I Tq L, then I t L t T q. 6

17 Part iii: We will prove this result separately for cases i and ii in Lemma 6. For case i, consider gg0. Because g <, gg0 < g0. If 0 I t g0, then 0 < gg0 I t+ gi t g0, where the first inequality follows because g > 0 and the last two inequalities follow because g is strictly decreasing. The interval [0, g0] is therefore a capture region and I t [0, g0] t starting with I 0 0. By the properties of ri in case i, r t I t t 0 and T q 0. In case ii, gi 0 g0 > I. We need to prove that there exists a T q such that I t > I if t T q, and I Tq [gi, I]. To prove this, let us start by supposing that no such T q exists. Then I < I t g0 t. In this region, I t+ gi t < I t, so {I t } is a strictly decreasing sequence. Because this sequence is in the bounded interval I < I t g0, it must converge to I, that is, lim t I t I. However, I cannot be a limit point of this sequence since gi < I. By contradiction, it cannot be true that I < I t g0 t, and there exists a finite T q such that I Tq I. In addition, since g increasing on I, g0], gi gi I I, g0], so I Tq gi Tq gi. We therefore conclude that there exists a finite T q such that I Tq [gi, I]. By ii, I t [gi, I] t T q and rt I t by the properties of r I on [0, I] in case ii of Lemma 6. Finally, since I ming0, I is the unique point such that gi I, lim t I t I. Proof of Proposition : The proof of this proposition draws on Lemma 6 which characterizes the policy function g when cq < vq and on Lemma 7 that characterizes the optimal path when cq < vq and qñ < r. Derivation of 4. From Lemmas 2 and 3, it follows that we can restrict our attention to cases where q Q and from Lemma 7, it follows that we can split problem in two parts when qñ < r. In particular, for t T q, we can focus on solutions of the form r t I t and I t+ gi t I t + qn t r t qn t. For simplicity, the feasible region is suppressed in the maximization problems below. where V q. T q max t π n t, r t, q + Tq V ITq, q, t0 A-2 V ITq, q. max τ π n Tq+τ, I Tq+τ, q τ0 max π n Tq, I Tq, q + τ+ π n Tq+τ+, qn Tq+τ, q τ0 A-22 7

18 and T q I Tq qn 0 + qn t r t t A-23 We now establish 4 in two steps, first for the case where t T q and then for the case 0 t < T q. In the case that T q 0, the part 0 t < T q can be omitted. Taking the derivative of V I, q with respect to I and evaluating it at I I Tq, we obtain: I V ITq, q π n T q, I Tq, q A-24 Taking the derivative of the sum in the right-hand side of A-22 with respect to n Tq+τ and using the chain rule for τ 0, we obtain first order conditions that are satisfied by the optimal sequence νt, t T q : π n τ T, q+τ r T q+τ or, equivalently, R n T+τ, r T+τ π n + τ+ T, q+τ+ r T q+τ+ q 0, for τ 0,,... A-25 + q R n T+τ+, r T+τ+ c q, for τ 0,,... A-26 We have thus established 4 for t T q. Let us now turn to t < T q. Taking the derivative of the sum in A-2 with respect to n t, t 0,,..T q, we obtain t πn t,r t t πn t,r t + q Tq V I Tq,q I, t 0,,..T q, where the equality follows because I Tq t definition of I Tq in A Tq V I Tq,q I Tq I t q by the Similarly, taking the derivative of the sum in A-2 with respect to r t, t, 2,..T q, we obtain t πn t,r t + Tq V I Tq,q I Tq I t t πn t,r t Tq V I Tq,q I, t, 2,..T q. The optimal sequence {ν t, t 0,,...,T q } satisfies the first order conditions obtained by equating these expressions to 0. Using r0 0, A-24 and dividing through by t, we can write: π n Tq t Tq,r Tq,q πn t,r t,q πn t,r t,q + q π Tq t 0, t 0,,..., T q n Tq,r Tq,q 0, t, 2,..., T q Substituting the second set of equalities into the first set, we find π n t,r t, q + q π n t,rt, q, t 0,, 2,...,T q. A-27 8

19 From the second set of equalities, we find that n T q, I Tq, q Tq t π Substituting in A-27, we obtain π n t,r t, q π n t,r t, q π n t+, r t+, q, t,..., T q A-28 + q π n t+, r t+, q, t 0,, 2,...,T q Thus, we have R n t,r t + q R n t+, r t+ c q for t 0,,..., T q, A-29 Together with A-26, we obtain 4. Derivation of 5. We again proceed in two parts. For t T q, we focus on solutions of the form r t I t and I t+ gi t I t + qn t r t qn t. It is convenient to define the following recursive relationship for t T q : V I t, q max [π n t, I t, q + V qn t, q]. Let n I t 0 n t+i t argmax [π n t, I t, q + V qn t, q]. By Assumption, we are assured that n is an interior solution 0 n t+i t and satisfies the first-order condition [πnt,it,q+v qn t,q] 0. We can further write V I t, q π n t,i t, q + V qn t,q t T q. Taking the derivative of this expression with respect to q for each t T q, and simplifying using the first order conditions satisfied by n t, we obtain τ V ITq+τ, q τ+ V ITq+τ+, q π n τ T, I q+τ T q+τ, q + τ+ V ITq+τ+, q I for τ 0. Adding these equalities over τ 0 and taking the limit as τ, we obtain V ITq, q τ π n T, I q+τ T q+τ, q + V ITq+τ+, q n T I q+τ. τ0 n T q+τ By definition, πn T q+τ, I T q+τ, q Rn T q+τ, I T q+τ c n qn t+τ c r qr t+τ, so πn T+τ,I T+τ,q c nqn t+τ c rqr t+τ. In addition, πn T+τ,I T+τ,q I V ITq+τ+, q π Rn T+τ,I T+τ n T q+τ+, I T q+τ+, q c r q. Using A-24,. A-30 9

20 Putting these two together, we find V ITq, q R n τ Tq+τ+, r Tq+τ+, q c r q c n q n T q+τ c r q rt q+τ τ0 A-3 We now turn to t < T q. Let us first rewrite A-2: V q T q t0 t π n t,rt, q+ Tq V ITq, q Tq t0 t π n t,rt, q + Tq V qn 0 + T q t qn t rt,q. We find V q T q t0 where we used d I 0 + Tq t0 qn t r t,q dq obtain V q τ0 t π n t,r t, q + Tq V I Tq, q I T q t0 n t + Tq V I Tq, q T q t0 n t. Collecting terms, and using A-24 and A-3, we T q t Tq t π n T q, I Tq, q c n q n t c r q r t t0 π n + Tq τ T q+τ+, I T q+τ+, q c n q n T q+τ c r q rt q+τ or with A-28 we obtain: T q V q t π n t+, r t+, q c n q n t c r q rt t0 π n + Tq τ T q+τ+, I T q+τ+, q c n q n T q+τ c r q rt q+τ τ0 Finally, we can rewrite the previous expression using the definition π n, r, q R n, r c n q n c r q r: V q T q t R n t+, r t+ c r q c n q n t c r q rt t0 R n + Tq τ T q+τ+, I T q+τ+ c r q c n q n T q+τ c r q rt q+τ. τ0 Note that the terms in the latter expression have the same structure for 0 t < T q as well as for T q t. Thus, we have obtained 5: V q t R n t+, r t+ c r q c n q n t c r q rt. t0 20

21 Proof of Proposition 2: The function V q was defined exclusive of the initial fixed investment cost kq; the discounted profit at time 0 equals V q kq. A sufficient condition for the existence of a q > 0 is therefore V 0 k 0 > 0. Let us evaluate V 0. As q 0, and I 0 0, we have that r t 0 t 0. Evaluating 4 at q 0 gives Rν t c 0 t 0. By the definition of n su, ν t n su, 0 t 0. Evaluating 5 at q 0 and ν t n su, 0 t 0, we find V 0 R nsu, 0 c r 0 c n 0 n su. For a linear consumer profile with η θ θ, it can easily be shown that Rn,0 see A-2 and A-4. As by definition Rnsu,0 c n 0, we obtain 7. Rn,0 Proof of Proposition 3. We first derive dnsu dκ. If { c n 0 c r 0} > c n 0, the sign of d dκ is identical to the sign of dnsu dκ. By definition, n su is determined by Rnsu,0 c n 0. For F θ θ κ F κ, the marginal consumer, θ su, is defined by n su θ su κ. From A-2, we know that Rnsu,0 G N θ su. Using A-4, we obtain θ su θsu κ c n 0 or θ su +cn0κ +κ. Plugging θ su into n su θ su κ, we obtain n su cn0κ κ. +κ Finally, dnsu dκ κ cn0κ +κ ln cn0κ +κ + +κ. Since the first term is non-negative, we obtain that dnsu dκ < 0 if and only if c n 0 > +κ κ e /+κ. As 0 > +κ κ e /+κ and c n 0 0, the latter condition is always satisfied. This completes our proof that dnsu dκ < 0, or, that d dκ < 0 if { c n 0 c r 0} > c n 0. We now consider d dc n0. d dc n0 n su c n 0 + { c n 0 c r 0} c n 0 dnsu dc n0. The first term is positive. dn su dc n0 κ cn0κ +κ κ κ +κ < 0. We conclude that depending on the magnitude of dn su dc n0, be either positive or negative. d dc n0 could Proof of Proposition 4. This proof is done in two steps. First, we characterize an approximation of q for. Second, we show how this approximation depends on κ. In previous results developed for a given remanufacturability level q, νt was determined for q, but this dependence was suppressed in the notation. From now on, we work with q, so νt is determined for q. We 2

22 again suppress this dependence in the notation. Step a. Approximate characterization of q : Since r0 0, we can rewrite 5, evaluated at q, as V q t R n t+, r t+ c r q c n q n t c r q rt+. t0 Multiplying this equation by, and separating it into two parts, we obtain V q T q t R n t+, r t+ c r q c n q n t c r q rt+ t0 + R n t+, q n t c r q c n q n t q c r q n t. tt q t tt q t We find that lim V q lim R n t+, q n t c r q c n q n t q c r q n t since the first term above is finite. q satisfies the first-order condition V q V q k q. Then lim V q 0. From the previous expression, we have lim tt q t k q, or, R n t+, q n t c r q c n q n t q c r q n t 0, or, lim We approximate Rn t+,q n t around ν n, q n. ν I q using Taylor series expansion: R n t+, q n t R tt q t R n t+, q n t c q n t 0. A R tt q t+ n t+ n + q 2 R 2 n t n + o ν t+ ν, A-33. where R Rν. Substituting into A-32 and collecting terms, we find 0 lim R c q t n t tt q 2 R +lim n t+ n + q 2 R 2 n t n + o νt+ ν n t. 22

23 From Lemma 6, we have that at I, A-7 is satisfied and A-8 is satisfied with equality, therefore πn I,I +q πn I,I 0. Remember from Lemma 7 that πn I,I +q πn I,I 0 with I qn I A-9. Therefore, as, I I. Thus for t T q, n t+, q n t n, q n as and the second term in parentheses converges to 0 in the last equality. Since n t > 0 t, expressing the dependence of q on the parameter with the expression q, we conclude that the following must hold for the first term to also be equal to 0: R n, q lim n c q 0. Therefore, for, R n,q n c q. Let q be such that R n, q n c q. A-34 q approximates the optimal remanufacturability level q. With A-9 evaluated at q, we obtain R n, q n + q R n, q n c q, which, with A-34 gives R n, q n c q q c q. A-35 In conclusion, A-34 and A-35 approximately determine q, n. This completes our approximate characterization of q. We will now specialize this characterization to F F κ. The marginal consumers θ, l are defined by n θ, κ h and r θ, κ l θ, h A-34 and A-35 as a function of θ, l, θ, h, q : G R θ, l c q and G N θ, h, θ, h κ for F F κ. We can rewrite + GR θ, l c q q c q. Furthermore, we can write A-5 as a function of θ, l, θ, h, q θ :, l κ θ, + q. Defining h κ c l q. c q and c h q. c q + q c q, we obtain that the triple θ, l, θ, h, q satisfies the following conditions: G N θ l c l q and G N θ h c h q A-36 θ l θ h + q κ. A-37 23

24 With Lemma 4, we have that G N θ θ θ κ. Substituting in A-36, solving for θ l, θ h and substituting these expressions into A-37, we find that the following condition that must be satisfied by q : c lq c hq + q κ. A-38 This completes our approximate characterization of q for F F κ and. Step 2. Characterizing q as a function of κ. If > 0, then q 0, q. It can easily established that c l q strictly increases in q. This follows from c l q c q and from c q > 0 Assumption 3. Similarly, it can easily be established that c h q strictly decreases in q. This follows from c h q + q c q and from c q > 0. As c l q strictly increases and c h q strictly decreases in q, the left hand side of A-38 strictly decreases in q. The right hand side strictly increases with q. Therefore, q solving A-38 is unique. Let us denote the dependence of q on κ explicitly with the notation q κ. Note that increasing κ decreases the right-hand side of A-38 but does not affect its left-hand side. Therefore, q κ increases in κ. Proof of Proposition 5. In this proof, we use the approximate characterization of q developed in the proof of Proposition 4 for. In particular, we use that q solves A-38. Remember from A- that we can write p N and p R as a function of θ l and θ h : p R θ l and p N θ l + θ h. The profit margin on the remanufactured product at the solution n, q n is thus Mr. θ, l c r q. By A-36, θ, l G c q N. For F F κ, we have G N c κ +c > c for c <. As from A-38, it follows that c q κ + <, we obtain that c q G N > c q, which yields θl c r q > c nq + q c r q > 0, where the last inequality follows from Assumption 3. θ, h The profit margin on the new product at the solution n, q n is Mn. θ, l + c n q. With A-36 and for F F κ, we obtain θ, l G N and c q c q + κ + κ 24

25 θ, h G N cq +q c q cq +q c q + κ + κ. We can rewrite M n as M n c qκ + c qκ + q κ c qκ + κ + κ κq κ c n q κ + q κc r q κ c n q κ + κ c n q κ From this equation, we see that the margin on the new product becomes negative for large enough values of κ, and that an increase in the cost of the new product works in the same direction as an increase in κ. Proof of Proposition 6. In this proof, we use the approximate characterization of q developed in the proof of Proposition 4 for. Let c r q c r0 + c r q, with c r 0 0. We calculate dq step, dn and dr step 2. Step : Calculation of dq Let us first introduce some notation: c q. c n q + qc r q, c l q. c q and c h q. c q + q c q. Recall c q. c n q+qc r q, c l q. c q and c h q. c q + q c q. Some algebraic manipulation yields c q c q + qc r0, c l q c r0 + c l q and c h q c r0 + c h q. Plugging in these expressions for c l q and c h q in condition A-38, we obtain that q satisfies c r0+c l q c r0+c h q Differentiation of both sides of A-39 with respect to c r0 gives: +c l q dq c r0+c h q + +c h q dq c r0+c h q From the latter equation, we can solve for dq : dq κ + q κ c r0+c h q + q κ. A-39 c r0+c l q 2 κ + q κ dq c r0+c h q 2 c + l q + c r0+c l q c r0+c h q c h q c r0+c l q 25

26 or, making use of A-39: dq κ + + q κ c r0 +c l q +q + c l q c h q + q κ. A-40 Since c q is convex by Assumption 3, c q is convex. Taking the derivative of c l q and c h q with respect to q we obtain c l c q q > 0 and c h c q + q q < 0. A-4 With A-39 and A-4, we observe that the sign of each of the terms in the previous expression is positive. We conclude that dq < 0. Step 2: Calculation of dn and dr. Define the marginal consumers θ l, θ h such that n θ h κ, r θ l κ θ h κ and r q n. Then, we need to calculate dn κ θh κ dθh and dr κ θh κ dθh κ θl κ dθl κ θh κ dθh + q κ dθ κ l where the last equality follows from A- 38. Using A-36, we can solve for θ l, θ h : Note that θ h and θ l < 0 and c h q dq c r0 + c hq + κ + κ c r0 + c lq + κ + κ dθ h dθ l > 0, therefore, the sign of + c h q dq + κ + c l q dq + κ A-42 A-43 dθ h is indeterminate. As the sign of dn is the same as the sign of dθ h, we find that dn 0. One can easily find examples of both cases. Using A-42 and A-43 we find dθh + q κ κ dθ l + c h q dq + q κ κ + κ + q κ κ c l q dq. The sign of this expression determines the sign of dr. We can substitute A-40 in the previous expression and obtain dθh + q κ κ dθl + + q κ c r0 +c l q κ + κ +q + q κ c h q + c l q. c r0 +c l q +q c h q + q c κ κ + l q κ 26

27 Recall that c l q. c q c h q +c l q c q c q we obtain dθh + q κ κ dθ l and c h q. c q + q c q. Substituting the following equality + κ + q c q + c q q c q in the latter expression, + + q κ c r0 +c l q κ +q + q 2 κc q c r0 +c l q +q c h q + q κ κ + c l q κ < 0. Noting that c h q < 0, c l q > 0 by A-4, we conclude by inspection of all terms that dr < 0. Proof of Proposition 7. This proof is structured as follows: First, we fix p U I and calculate the dynamic Cournot competition between the manufacturer and remanufacturers by determining r e pu.,i I i N and n e p U. I. As the remanufacturers are symmetric, we have that re p U.,i I re p U. I i N step. Second, we determine the market clearing price, p e U I, such that the obtained Cournot equilibrium satisfies Nrp e e U. I I for pe U. > 0, or Nre p e U. I < I for pe U. 0 step 2. In step 3, we take the derivative of the equilibrium value function of the manufacturer with respect to q, at q 0. In this way, we obtain e and note that it is exactly the same as. Step : Cournot competition between the manufacturer and N remanufacturers. Fixing p U I and r i I i N, the manufacturer s problem can be written as the following DP: V pu.,n I max n 0 n +V pu.,n p N n, N r i I + qp U I + qn i I + qn N r i I i N r i I c n q i A-44 i.e. the manufacturer chooses for every I a new product quantity of n. Fixing p U I, n I and r j I j N,j i, the remanufacturer s problem can be written as the following DP: V pu.,r,i I max r i r i +V pu.,r,i p R n I, r i + N j,j i I + qn I r i r j I p U I c r q N j,j i r j I A-45 27

28 i.e. each remanufacturer chooses for every I a remanufacturing quantity r i. For a fixed p U I, let n e p U. I be the maximizer of A-44, with r i I i N r e pu.,i I and let be i N re p U.,i I the maximizer of A-45, for n I n e p U. I and r j I j N,j i r e pu.,j I. Then, j N,j i n e pu. r I, e pu.,i I determine the dynamic Cournot equilibrium for a fixed p U I. n e p U. I i N satisfies the FOC of A-44 with respect to n: 0 p N n, N rp e U.,i I c n q + qp U I + qn i +n p N +q d di V p U.,N n, N i re p U.,i I I + qn + q 2 p U N rp e U.,i I and rp e U.,i I satisfies the FOC of A-45 with respect to r: 0 p R n e p U. I, r + i N j,j i I + qn N rp e U.,i I i N i rp e U.,j I c r q p U I p R n e p U. I,r + N j,j i re p U.,j I +r d di V p U.,R I + qn e p U. I r N j,j i rp e U.,i I rp e U.,j I Using the symmetry of A-45, let us suppress the index i and use rp e U. I instead. The previous equations reduce then to: 0 p N n e p U. I,Nre p U. I c n q + qp U I + qn e p U. I Nre p U. I +n p N n e pu. I, Nre pu. I + q 2 p U I + qn e pu. I Nre pu. I +q d di V p U.,N I + qn e pu. I Nre pu. I A-46 and 0 p R n e p U. I, Nre p U.,j I c r q p U I p R n e p U. I, Nre p U.,j I +r d di V p U.,R I + qn e pu. I Nre pu.,j I A-47 28

29 Let Vp e U.,N I and V e p U.,R,i I denote the value functions in A-44 and A-45 evaluated at n e pu. r I, e pu.,i I. Taking the derivative of these functions with respect to I, we find i N d di V e p U.,N I ne p U. I p N and +qn e p U. Ip U + d di V p U.,N n e p U. I, N i re p U.,i I N i d di re p U.,i I N I + qn e p U. I rp e U.,i I i N I + qn e p U.,i I rp e U.,i I i N i N i n e pu. I, re pu.,i I + j i re pu.,j I d di re p U.,i I d di re p U.,i I d p R di V p e U.,R,i I re p U.,i I { d di ne p U. I p R n e p U. I,re p U.,i I + j i re p U.,j I d + di re p U.,j I p U I} j i + d di V p U.,R,i I + qn e p U. I re p U.,i I rp e U.,j I j i ; + q d di ne p U. I j i d di re p U.,j I Again, with symmetry with respect to i N, we obtain: n e pu. I,Nre pu. I d di V e p U.,N I ne p U. I p N +{qn e p U. Ip U + d di V p U.,N Nrp e I I + qn e pu. I Nre pu. I I + qn e pu. I Nre pu. I N d di re pu. I }A-48 n e pu. I,Nre pu. I d p R di V p e U.,R I re p U. I { p R n e p U. I,Nre p U. I + + d di V p U.,R d di ne p U. I N d di re p U. I p U I} I + qn e pu. I Nre pu. I + q d di ne pu. I N d di re pu. A-49 I We thus have four equations A-44, A-45, A-48 and A-49 determining the four unknowns n e p U. I, re p U. I, d di V e N I and d di V e R I, for a given p U I. 29

30 Step 2: Equilibrium p e U I. As mentioned above, the market clearing price, p e U I is such that the obtained Cournot equilibrium satisfies Nr e p e U. I I for pe U. > 0, or Nre p e U. I < I for pe U. 0. Let ne I, r e I denote n e p U. I,re p U. I and V e R I,V e N I denote d di V p e e U.,R I, d di V p. e e U.,N I Consider p e U I > 0 and Nre I I. Substituting these in A-44, A-45, A-48 and A-49, we obtain: p N n e I, I c n q + qp e U qne I + n e I pn n e I,I + q 2 p e U qne I + qvn e qne I 0 V e R I I N p R n e I, I c r q p e U I + I N pr n e I,I V e N I ne I p Nn e I,I p R n e I,I V e R qne I 0 n e I + p Rn e I,I N N pe U I + VR e qne I N + qne I A-50 The solution to this set of four equations determines the four unknowns n e I, p e U I, V e N I and VR e I. In the next step, we study the case of I 0 and q 0. Under these conditions, we have that N i re i 0 0. We will use A-50, and validate that pe U 0 > 0. Step 3: Derivative of VN e I, q with respect to q evaluated at q 0. In this step, we reintroduce the dependence of all previous expressions with respect to q. Taking the partial derivative of VN e I, q with respect to q yields V e N I, q n e I c n q + p e U qn e I + qn e Ip e U qn e I +n e I V e N I + qne I Nr e I,q I + V N e I + qne I Nr e I,q. Evaluated for I 0 and q 0, can write the previous expression as V N e 0, 0 n e 0 p e U 0 c n q + n e 0 V N e 0, 0. A-5 I Substituting I 0 and q 0 in the first equation of A-50, we obtain p N n e 0, 0 c n 0 + n e 0 p Nn e 0,0 0, which can be rewritten as Rne 0,0 c n 0, and is solved by n su see definition of n su. Substituting I 0 and q 0 in the fourth equation of A-50, we obtain VR e 0 V e R 0 N, from which it follows that V e R 0 0. Plugging the latter result in the second equation of A-50, we obtain p R n su, 0 c r 0 p e U 0. 30

31 Using the proof of Lemma 4, we see that p R n su, 0 p N n su, 0 and that p N n su, 0 > Rn su,0. Therefore, p e U 0 p N n su, 0 c r 0 > Rnsu,0 c r 0 c n 0 c r 0. Thus, if c n 0 > c r 0, then p e U 0 > 0, which validates our assumption in step 2. Finally, from the third equation of A-50, we obtain VN e 0 n su p Nn su,0. These relationships can be substituted back in A-5: V N e 0, 0 p N n su, 0 n su p R n su, 0 c r 0 + n su c n 0 or V e N 0, 0 R n nsu, 0 su c r 0 c n 0 Thus, taking the fixed investment costs into account, we obtain e. V e N 0, 0 and we observe that e. k 0 R n nsu, 0 su c r 0 c n 0 k 0 4 Reference Smith, J. E., K. F. McCardle Structural Properties of Stochastic Dynamic Programs. Operations Research