The Effect of Remanufacturability of End-of-Use Returns on the Optimal Intertemporal Production Decisions of a Newsvendor

Abstract

In this paper we study a joint manufacturing–remanufacturing problem of a manufacturer under demand uncertainty. Supply of remanufacturable units is constrained by the availability of used and returned cores, which depends on previous supply of new units. Potential cost savings due to remanufacturing in later periods may induce the manufacturer to reduce its short-term profits by artificially increasing its supply in earlier periods. For dealing with this trade-off we formulate an intertemporal optimization problem in a classical two-period newsvendor setting. Exploiting first period information when taking second period supply decisions we provide analytical insights into the optimal strategy and compare this optimal strategy with a previously proposed heuristic, where both first and second period decisions are committed to prior to period 1. Through a comprehensive numerical study we evaluate the associated profit implications.

Appendix

Proof

Proof of Proposition 1: Let λ _s , λ _r and λ ₂ correspond to the shadow prices of constraints (4), (5) and (6), respectively. Then, for given s ₁ > 0, the KKT conditions associated with the problem

$$\displaystyle{ \max _{q_{2},q_{2}^{r}}\pi _{2} }$$

(21)

subject to constraints (4)–(6) are given by

$$\displaystyle\begin{array}{rcl} F_{2}(q_{2} + q_{2}^{r})& =& \frac{p_{2} - (c_{2}-\delta ) +\lambda _{r} -\lambda _{s}} {p_{2}} {}\end{array}$$

(22)

$$\displaystyle\begin{array}{rcl} F_{2}(q_{2} + q_{2}^{r})& =& \frac{p_{2} - c_{2} +\lambda _{2}} {p_{2}} {}\end{array}$$

(23)

$$\displaystyle\begin{array}{rcl} q_{2}& \geq & 0{}\end{array}$$

(24)

$$\displaystyle\begin{array}{rcl} q_{2}^{r}& \geq & 0{}\end{array}$$

(25)

$$\displaystyle\begin{array}{rcl} \gamma s_{1} - q_{2}^{r}& \geq & 0{}\end{array}$$

(26)

$$\displaystyle\begin{array}{rcl} \lambda _{s}& \geq & 0{}\end{array}$$

(27)

$$\displaystyle\begin{array}{rcl} \lambda _{r}& \geq & 0{}\end{array}$$

(28)

$$\displaystyle\begin{array}{rcl} \lambda _{2}& \geq & 0{}\end{array}$$

(29)

$$\displaystyle\begin{array}{rcl} \lambda _{s}(\gamma s_{1} - q_{2}^{r})& =& 0{}\end{array}$$

(30)

$$\displaystyle\begin{array}{rcl} \lambda _{r}q_{2}^{r}& =& 0{}\end{array}$$

(31)

$$\displaystyle\begin{array}{rcl} \lambda _{2}q_{2}& =& 0{}\end{array}$$

(32)

Let us first prove that q ₂ ^r > 0 is always true. We will show this by contradiction. Let q ₂ ^r = 0. Given γ > 0 and s ₁ > 0 this implies that γ s ₁ − q ₂ ^r > 0 and consequently λ _s  = 0. From (22) and (23) we get $F_{2}(q_{2}) = \frac{p_{2}-(c_{2}-\delta )+\lambda _{r}} {p_{2}} = \frac{p_{2}-c_{2}+\lambda _{2}} {p_{2}}$. When q ₂ = 0 the resulting $F_{2}(0) = \frac{p_{2}-c_{2}+\lambda _{2}} {p_{2}}> 0$ contradicts our assumption F ₂(0) = 0 since p ₂ > c ₂. Alternatively, q ₂ > 0 implies λ ₂ = 0, which yields $F_{2}(q_{2}) = \frac{p_{2}-(c_{2}-\delta )+\lambda _{r}} {p_{2}} = \frac{p_{2}-c_{2}} {p_{2}}$, which holds when $\delta +\lambda _{r} = 0$. Since δ > 0 and λ _r  ≥ 0 this can never be true. Thus, q ₂ ^r = 0 always leads to a contradiction and it follows that q ₂ ^r > 0 and consequently λ _r  = 0.

Now assume that q ₂ > 0. This implies that λ ₂ = 0 and from (23) we get $F_{2}(q_{2} + q_{2}^{r})\ =\ \frac{p_{2}-c_{2}} {p_{2}}$. Together with (22) this yields $\delta -\lambda _{s} = 0$. It follows that λ _s  = δ > 0 and consequently $\gamma s_{1} - q_{2}^{r} = 0$. Since γ > 0 and s ₁ > 0 we get q ₂ ^r = γ s ₁ > 0. Using $F_{2}(q_{min})\ =\ \frac{p_{2}-c_{2}} {p_{2}}$ it follows that $q_{2} + q_{2}^{r} = q_{min}$. Thus, $q_{2}^{r} =\gamma s_{1} = q_{min} - q_{2} <q_{min}$. This concludes the proof of Case 3.

Now consider q ₂ = 0, i.e. λ ₂ ≥ 0. From (22) and (23) we get $F_{2}(q_{2}^{r}) = \frac{p_{2}-(c_{2}-\delta )-\lambda _{s}} {p_{2}} = \frac{p_{2}-c_{2}+\lambda _{2}} {p_{2}}$. We continue by distinguishing two scenarios. First, let q ₂ ^r < γ s ₁. It follows that λ _s  = 0. This implies that $F_{2}(q_{2}^{r}) = \frac{p_{2}-(c_{2}-\delta )} {p_{2}}$, i.e. q ₂ ^r = q _max . As a result, q _max  < γ s ₁. This concludes the proof of Case 1. Second, let q ₂ ^r = γ s ₁. Now λ _s  ≥ 0 gives $F_{2}(q_{2}^{r}) = \frac{p_{2}-(c_{2}-\delta )-\lambda _{s}} {p_{2}} = \frac{p_{2}-c_{2}+\lambda _{2}} {p_{2}}$. Thus, F ₂(q _max ) ≥ F ₂(q ₂ ^r) ≥ F ₂(q _min ) and consequently q _max  ≥ q ₂ ^r = γ s ₁ ≥ q _min , which concludes the proof of Case 2.

Proof

Proof of Proposition 2

Let us first consider the case where γ q ₁ > q _max . In that case, there are the following four possible first generation outcomes leading to different optimal second generation strategies.

• 1. d ₁ > q ₁

  $\pi = -c_{1}\:q_{1} + p_{1}\:q_{1} +\beta \pi _{2,1}$

  Here first generation sales are given by supply q ₁ and consequently returns are sufficient to cover q _max in generation 2 yielding expected profit π _2, 1.

• 2. d ₁ ≤ q ₁ & γ ⋅ d ₁ > q _max : 

  $\pi = -c_{1}\:q_{1} + p_{1}\:d_{1} +\beta \pi _{2,1}$ In this setting, demand is limiting first generation sales. Yet as above, the associated returns are sufficient to cover q _max in generation 2 yielding expected profit π _2, 1.

• 3. q _max  ≥ γ ⋅ d ₁ ≥ q _min : $\pi = -c_{1}\:q_{1} + p_{1}\:d_{1} +\beta g(\gamma \:d_{1})$ Again demand is limiting first generation sales but associated returns are now insufficient to cover q _max in generation 2. Yet, the associated returns are sufficient to cover q _min in generation 2. Thus, we are in Case 2 described in Proposition 1. However, now second generation profits depend on the actual level of d ₁ as shown by the term g(γ d ₁).

• 4. q _min  > γ ⋅ d ₁: $\pi = -c_{1}\:q_{1} + p_{1}\:d_{1} +\beta k(\gamma \:d_{1})$ In this final setting, demand is limiting first generation sales and the associated returns are insufficient to even cover q _min in generation 2. Thus, dual sourcing is optimal and we are in Case 3 described in Proposition 1. Similarly as above, second generation profits depend on the actual level of d ₁ as shown by the term k(γ d ₁).

We can now turn to the second case where q _max  ≥ γ q ₁ ≥ q _min . Now, first generation sales are no longer sufficient to cover q _max in generation 2 and we are left with three possible first generation outcomes leading to different optimal second generation strategies.

• 1. d ₁ > q ₁

  $\pi = -c_{1}\:q_{1} + p_{1}\:q_{1} +\beta g(\gamma \:q_{1})$

  Here first generation sales are given by supply q ₁. The associated returns are sufficient to cover q _min in generation 2. Thus, we are in Case 2 described in Proposition 1. However, now second generation profits depend on the actual level of q ₁ as shown by the term g(γ q ₁).

• 2. d ₁ ≤ q ₁ & γ ⋅ d ₁ ≥ q _min : 

  $\pi = -c_{1}\:q_{1} + p_{1}\:d_{1} +\beta g(\gamma \:d_{1})$

  Here first generation sales are given by demand d ₁. The associated returns are sufficient to cover q _min in generation 2. Thus, we are still in Case 2 described in Proposition 1. However, now second generation profits depend on the actual level of d ₁ as shown by the term g(γ d ₁).

• 3. q _min  > γ ⋅ d ₁: 

  $\pi = -c_{1}\:q_{1} + p_{1}\:d_{1} +\beta k(\gamma \:d_{1})$

  In this final setting, demand is limiting first generation sales and the associated returns are insufficient to even cover q _min in generation 2. Thus, dual sourcing is optimal and we are in Case 3 described in Proposition 1. Similarly as above, second generation profits depend on the actual level of d ₁ as shown by the term k(γ d ₁).

The third case arises when q _min  > γ q ₁, i.e. first generation supply and consequently sales are insufficient to even cover the minimum supply of generation 2 units. In this case, dual sourcing is always necessary and optimal and we are left with two possible first generation outcomes leading to different optimal second generation strategies.

• 1. d ₁ > q ₁

  $\pi = -c_{1}\:q_{1} + p_{1}\:q_{1} +\beta k(\gamma \:q_{1})$

  Here first generation sales are given by supply q ₁. Since the associated returns are insufficient to even cover q _min in generation 2 we are in Case 3 described in Proposition 1. Second generation profits depend on the actual level of q ₁ as shown by the term k(γ q ₁).

• 2. d ₁ ≤ q ₁

  $\pi = -c_{1}\:q_{1} + p_{1}\:d_{1} +\beta k(\gamma \:d_{1})$

  Now demand is limiting first generation sales and second generation profits depend on the actual level of d ₁ as shown by the term k(γ d ₁).

Integrating over the possible values of d ₁ again yields the proposed expected profit function in all three cases.

Proof

Proof of Proposition 3 The optimality conditions for q _1, 1, q _1, 2 and q _1, 3 are readily obtained from setting the first derivative of the associated expected profit functions given by Proposition 2 equal to zero.

Proof

Proof of Proposition 4 Let us first observe that at most one of the candidate solutions q _1, 1, q _1, 2 and q _1, 3 can fall within its admissible range, since q _1, 3 ≥ q _1, 2 ≥ q _1, 1 while at the same time their boundaries prescribe the opposite order. Furthermore, we know that at the boundaries between two cases their respective profits are identical. Thus, from concavity it follows that whenever one of the candidate solutions falls within its admissible range, it is the optimal solution, since its associated profit exceeds the respective boundary profits of the other two cases.

Using this logic, we can begin with computing an arbitrary candidate solution q _1, 1, q _1, 2 or q _1, 3. The flowchart in Fig. 1 shows this approach starting with q _1, 1. When this solution falls within its admissible range, the optimum has been found and the algorithm stops. Otherwise, the next case needs to be considered. To compute as little solutions as possible the algorithm exploits the information on q _1, 1 as well as the fact that q _1, 3 ≥ q _1, 2 ≥ q _1, 1 when deciding which case to consider next. Specifically, whenever γ q _1, 1 > q _min Case 3 can never be optimal and needs not be considered, since γ q _1, 3 > q _min violates its admissible range.