Online Appendices: Inventory Control in a Spare Parts Distribution System with Emergency Stocks and Pipeline Information

Transcription

1 Online Appendices: Inventory Control in a Spare Parts Distribution System with Emergency Stocks and Pipeline Information Christian Howard christian.howard@vti.se VTI - The Swedish National Road and Transport Research Institute, Traffic and Road Users Johan Marklund johan.marklund@iml.lth.se Lund University, Industrial Management and Logistics Tarkan Tan t.tan@tue.nl Eindhoven University of Technology, Industrial Engineering and Innovation Sciences Ingrid Reijnen ingrid@rajko.nl Gordian Logistic Experts

2 Online Appendices: Inventory Control in a Spare Parts Distribution System with Emergency Stocks and Pipeline Information Online Appendix A - Proofs Proof of Lemma 1 From Figure 3 and Figure 4 it is clear that Lemma 1 holds when T = L. To establish the equivalence in the case of T < L we consider all possible states for the on hand inventories and show that all state transitions from one state to the other are the same in the two models. Hence, the steady state distributions must be identical. Assume that both systems are in a state with k units of stock on hand, and consider how a customer arrival affects the value of k in all relevant cases. If k 1 in both systems, an arriving customer receives an item from stock on hand, rendering k-1 units in stock. At the same time a regular replenishment order is placed, which will arrive in L time units. Thus, when k 1 the two systems behave exactly the same. If k = 0 and there is unreserved stock in the second part of the pipeline, i.e. N 1 = M 1 < S, an arriving customer is backordered in both systems and k = 0 still holds. As a regular replenishment order is placed in both systems, they behave equivalently in this case as well. Finally, consider a customer demand arriving when k = 0 and there are no unreserved items in the second part of the pipeline, i.e. N 1 = M 1 = S. In the TBB model this demand is lost to the emergency supplier. The system will remain in its current state until the unreserved order closest to the local warehouse reaches the second part of the pipeline, resulting in a transition to the state with k = 0 and N 1 = S 1. Given the same situation in the DS model, the customer demand is backordered and an item is ordered from the second supplier, with lead time T. However, because M 1 = S implies that the unreserved item closest to the local warehouse is farther than T time units away, the item from the second supplier will arrive at the local warehouse before any unreserved items, and leave the system immediately with the waiting customer. Therefore, despite the order to the second supplier, the DS system remains in the state k = 0 and M 1 = S after the customer demand has arrived. A transition to the state k = 0 and M 1 = S 1 will occur at the exact same time as in the TBB model, i.e. when the closest unreserved item reaches the second part of the pipeline. Once this occurs, the two systems will operate in exactly the same way again. 1

3 Proof of Corollary 2 Lemma 1 implies that α = α DS. Furthermore, from Figure 4 it is clear that the first station in the DS queuing model also is identical to the Erlang loss system and, hence, ψ = ψ DS. Because all fractions must add up to one it then follows directly that β = β DS. Proof of Lemma 2 In the TBB model we can only have waiting in anticipation of reserved pipeline stock, while in the DS model some customers will also be waiting for orders from the second supplier. However, because β = β DS, ψ = ψ DS and the waiting time for an order from the second supplier is always T time units it holds that According to Little s law EW DS EW T. (A.1) EW Therefore, by combining (A.1) and (A.2), we obtain DS EIL DS. (A.2) ( ) ( ) EW EW DS DS EIL T T. Proof of Proposition 1 For ease of exposition we suppress the subscript 0 and also the reference to the parameter L 0 for the random variables. From the definitions above we know that V D L distributed random variable the conditional probabilities for the variable distributed, i.e. For the variables D and P D k x x k! k e x, for D we have the following relationship k 0,1, E V D V E D E V L V L V D. Because N is a Poisson D are also Poisson, (A.3) which provides us with the lower bound for D V. To derive the upper bound, let f (x) be the probability density function for, and let G (y) x f (x)dx. The expected value and the bounds y 0 on imply that G(0) 0, G(y) 0 for 0 y L, and G(y) L for y L parts yields. Thus, integration by which means that 2 L 2 x f (x) dx L G(L) E G(y) dy L, 0 L 0 2 2

4 Combining (A.3) and (A.4) completes the proof. E 2 E 2 2 L 2 V. (A.4) Online Appendix B - Heuristic for the multi-echelon model The heuristic for determining optimal or near-optimal S and T is based on enumerating over both fixed values of T 0, and fixed lost sales probabilities at the support warehouse. Let ψ 0 be the fraction of support warehouse demand that is lost (transferred to the central warehouse) and assume that ψ 0 is given. The basic idea is to decouple the local warehouses by approximating each local warehouse s lost sales cost as c j (1 0)c j 0p j, j {1,,J}. Given c j we first determine the optimal S j and T j - values at each retailer j, and then determine the S 0 -value that results in the given value for ψ 0 (or close to the ψ 0 -value, considering the issue of integrality). Let Δ j be the step sizes for the threshold times T j, j {0,,J}, and let ε be the step size for ψ 0. Furthermore, denote S * and T * as the best solution found so far by the heuristic, with resulting total cost C *. We then have the following algorithm: Step 0. Initialize by setting C * =, T 0 = 0. Step 1. Set S 0 = 0 and ψ 0 = 1. Step 1.1. Compute c j = (1 ψ 0 )c j + ψ 0 p j, for j {1,,J}. Step 1.2. For each j {1,,J} optimize S j and T j and determine the corresponding ψ j -values using the single-echelon optimization procedure based on c = c j and Δ = Δ j. J Step 1.3. Compute 0 j j and find the highest S 0 such that the resulting lost sales j 1 probability is larger or equal to ψ 0, by increasing S 0 one unit at a time. Step 1.4. Compute C(S,T) from the S j and T j (j {0,,J}) obtained in steps 1.2 and 1.3. Step 1.5. If C(S,T) < C * then set C * = C(S,T), S * = S and T * = T. Step 1.6. Set ψ 0 = ψ 0 ε and if ψ 0 > 0 return to Step 1.1. Step 2. Set T 0 = T 0 + Δ 0. If T 0 L 0 then stop, otherwise return to Step 1. Online Appendix C - Single-echelon model Parameter effects on the (S,T) policy In this section we analyze the effects of using the (S,T) policy in the single-echelon system. The main motivation behind these tests is that it is easier to isolate and present the effects from specific parameters in a single-echelon setting. Moreover, because a local warehouse is an intricate part of any emergency (lateral) transshipment model, the insights gained about using pipeline information should be valuable for these types of models as well. 3

5 To gain insights about the value of using the threshold time to incorporate pipeline information, we focus on comparing the OPT policy (where we obtain S OPT and T OPT ) to the AR policy (where we obtain S AR for T AR = 0) and the resulting penalty ΔP AR. The basic test series consists of 3840 problem scenarios obtained through all combinations of the following parameter values: L {2, 4, 6, 8, 10, 12, 14, 16} h {0.01, 0.25, 0.5, 0.75, 1} b {1} c {0.01, 0.1, 1, 2, 5, 8, 10, 12, 15, 20, 50, 100} λ {0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4}. We only consider a single value of b because different cost ratios are achieved by varying h and c. For all scenarios we calculated ΔP AR with the step size for optimizing T set to Δ = The results show that this penalty can be quite high in many cases. The average ΔP AR over all 3840 scenarios was 65%, with a maximum of 414% and a minimum of zero. Figure C1 depicts the average ΔP AR as a function of p, L, λ and h. Average ΔP Average ΔP 150% 100% 50% 0% (a) c % 100% 50% 0% (b) L Average ΔP Average ΔP 150% 100% 50% 0% (c) λ 150% 100% 50% 0% (d) Figure C1. Average ΔP AR -values as functions of the input parameters h Figure C1 (Graph a) shows that, on average, ΔP AR increases with the cost for emergency shipments, as expected. In Graph b we see that the average ΔP AR decreases in L. Moreover, from Graph c we can conclude that λ appears only to have a marginal effect on ΔP AR. This also seems to hold true for a wider range of λ-values, as additional tests (960 problem scenarios) with λ = 0.1 and λ = 8 resulted in close to the same average ΔP AR (65.6% when λ = 0.1, 67.1% when λ = 8, compared to between 63.7% and 66.4% for the other λ-values). Graph d indicates that there is a positive correlation between ΔP AR and h. An increase in the holding cost means that it becomes more attractive to either backorder demand, or use emergency shipments. In the cases where pipeline stock is not considered, T is fixed 4

6 to zero and we cannot achieve a combination of these two options. Therefore the penalty is likely to increase. Because ΔP AR is increasing in c and h one can also conclude that the penalty must be decreasing in b. That is, if it is expensive to wait for orders in the replenishment pipeline it is not so costly to always request an emergency shipment. However, it is important to emphasize that this refers to a relative increase in the customer waiting cost, compared to the other costs. An absolute increase, will also lead to an increase in the emergency shipment cost, the size of which is determined by the underlying emergency shipment lead time. Turning to the structure of the solutions, ΔP AR is depicted against the optimal ratio T * /L in Figure C2. In Figure C2 we see that the lowest ΔP AR -values are found when T * /L is close to zero, and the highest values of ΔP AR occur when T * /L is close to 1. It is, however, interesting to note that a high value for T * /L does not always result in a high value for ΔP AR. That is, just because the optimal threshold time is to consider replenishment orders far away from the local warehouse does not necessarily imply that there is a high penalty for ignoring these orders completely. To explain this, Figure C3 depicts the fraction of demand that is backordered (β) against T * /L. Perhaps somewhat counter-intuitive, Figure C3 shows that there is no obvious relationship between T * /L and β. However, one needs to recall that the values of β and ΔP AR are also influenced by S OPT. Clearly, the variable S will affect the proportion of demand that is satisfied from stock on-hand. 400% ΔP 300% 200% 100% 0% Figure C2 T*/L 0.6 β Figure C3 T*/L 5

7 The variable T, in turn, affects the ratio of demand fulfillment between the pipeline stock and the emergency shipments. However, as indicated, it is difficult to predict the properties of the optimal (S,T) policy for a given scenario. This highlights the need for methods such as the one presented in this work, when using pipeline information in practice. In Figure C4 we provide a few examples of how the cost function changes as we increase T. In each example T is increased from 0 to L, while the base-stock level remains fixed at the same value S OPT. We observe that the shape of the function may change dramatically depending on the input data. Moreover, the cost function appears to be fairly smooth around the optimum, at least as long as T * is not at the extreme points 0 or L. This implies that the choice of threshold time is insensitive to small errors, as long as the item in question requires a mix of emergency shipments and pipeline stock. In all cases studied, we have observed that the cost function appears to be unimodal in T. C(S*,T) (a) c = 1 c = 100 T/L C(S*,T) (b) L = 2 L = 16 T/L C(S*,T) (c) λ = 0.5 λ = 4 T/L C(S*,T) T/L 0 (d) Figure C4. Graph (e): L=8, h=0.5, b=1, λ=2. Graph (f): h=0.5, b=1, c=10, λ=2. Graph (g): L=8, h=0.5, b=1, c=2. Graph (h): L=2, b=1, c=15, λ=0.5 h = 0.25 h = 1 Comparison to the dual-index policy, (S 1,S 2 ) Our analysis of the (S,T) policy addresses the scarcity of multi-echelon emergency shipment policies that incorporate pipeline information. However, for single-echelon systems, the previous literature does provide alternative policies to the (S,T) policy. In this section we compare the performance of our (S,T) policy to one such policy, the dual-index policy with complete backordering previously analyzed by Song and Zipkin (2009) (and also Moinzadeh and Schmidt, 1991). The dual-index policy consists of two base-stock levels S 1 and S 2 (S 1 S 2 ), where S 1 triggers regular orders and S 2 triggers emergency orders (see Song and Zipkin, 2009, for details). It is natural to compare our results to this 6

8 policy as they are closely related (see Section of the paper). The purpose of the comparison is twofold: Firstly, the aim is to provide a sense of when one policy might outperform the other. Secondly, because the dual-index policy allows for proactive emergency ordering (i.e., the placing of emergency orders before stock is depleted), comparison to our non-proactive (S,T) policy provides some indicative information on the potential benefit of using proactive emergency ordering. In addition to functioning differently compared to the (S,T) policy, the dual-index policy is based on a different formulation of the emergency shipment cost and of the emergency shipment lead time. Therefore, we will adjust the multi-echelon data provided by Volvo Parts (presented in Section 5.2 of the paper and Online Appendix E) to facilitate a straightforward single-echelon comparison. To that end, assume that each local warehouse presented in Table E1-E3 in Online Appendix E is a single-echelon system with parameters L, λ, h, and b as given by the tables. This provides 840 problem scenarios in total (70 multi-echelon scenarios with 12 local warehouses each). Furthermore, the emergency shipment lead time is equal to τ = 1 (i.e., we assume that the support warehouse can always deliver on time). For the (S,T) policy, emergency shipments can only be placed when a stockout occurs. This means that the customer cost of waiting for the emergency order (b ) is directly included in the emergency shipment cost, that is, c = c + bτ, where c denotes the handling and shipment cost associated with every emergency order. For the dual-index policy, the emergency shipment cost only refers to the fixed cost c, as the cost of waiting for emergency shipments is included in the expected negative inventory level. Thus, for each problem scenario we compare the expected total cost from applying the optimal (S,T) policy with emergency shipment cost c = c + bτ, with the optimal dual-index policy with cost c and emergency replenishment lead time τ (Song and Zipkins method is used for cost evaluation and policy optimization). Table C1 presents the average, median, maximum and minimum relative decrease in expected total costs for the (S,T) policy compared to the dual-index policy over all 840 problem scenarios. Note that a negative cost decrease indicates that the dual-index policy performed better than the (S,T) policy. Table C1. The expected cost decrease for the (S,T) policy. Average Median Maximum Minimum Cost decrease, (S,T) policy 1.3% 0.5% 48% -67% We see in Table C1 that average performance was essentially equivalent. However, as indicated by the maximum and minimum values, in some instances the (S,T) policy considerably outperforms the dual-index policy and vice versa. In total, the (S,T) policy performed better than the dual-index policy in 662 (79%) of the scenarios. The main driver behind the relative performance of the (S,T) policy appears to be the emergency handling and shipment cost c. When c is higher, the (S,T) policy performs better, and when c is lower the dual-index policy performs better (the estimated coefficient 7

9 of correlation between the relative cost decrease and c was 0.66). We also note that the optimal S 2 base-stock level was equal to zero in 707 (84%) of the scenarios. This means that for the cases considered proactive shipments were rarely used when optimizing the base stock levels in the dual index policy. Online Appendix D - Details on the model validation For each problem we obtained S OPT and T OPT (or S AR when T AR = 0) from our analytical model with Δ = 0.5 and ε = (except for problem scenarios where Δ = 2). We then searched for the best solution according to the simulation model by performing a complete search over these variables. We used the ad-hoc approach of only considering S at most five units above the S OPT (or S AR ) given by the analytical model (this is reasonable because the ψ-values were equal to zero for such high S- values). Each combination of variables was simulated for 5000 arrivals, excluding a system warm-up period of 500 arrivals. The 10 best candidate solutions were then simulated for an additional arrivals, making it plausible that the final solution with the lowest total cost is the optimal solution. Table D1-D 5 provide the resulting S OPT, T OPT and S AR when applying the analytical model. The solution obtained from the simulation optimization is given in bold font when it differs from the analytical solution. The resulting relative total cost increase of using the solution provided by the analytical model, compared to the simulation optimization, is given by ΔOPT and ΔAR, respectively. The tables also provide the fraction of demand that is satisfied by emergency shipments (i.e., the fraction that overflows from the local warehouse) given the optimal solution. These are denoted by ψ OPT and ψ AR. In problem scenarios 1-64 we assume that all local warehouses are identical. The input data are given by N = 12 or 24, L 0 = 2 or 4, L j = 4 or 8, b j = 20 or 80, c j = 30 or 60 and λ j = 0.1 or 0.4 (j = 1,, J), where the 64 combinations of low and high values constitute our set of problem scenarios. The remaining input parameters are the same for all problems: h j = 1 (j = 0,, J) and p j = 70 (j = 1,, J). The results are presented in Table D1 (N = 12) and Table D2 (N = 24). In problem scenarios we no longer assume that all local warehouses are identical. We consider two different demand rates: λ 1 and λ 2. In each scenario, local warehouses 1 to J/2 are identical with demand rate λ 1, and local warehouses J/2 + 1 to J are identical with demand rate λ 2. The input data for the 32 scenarios are given by N = 12 or 24, (λ 1, λ 2 ) = (0.003, 0.02) or (0.01, 0.15), L j = 4 or 8, b j = 100 or 400 and c j = 200 or 400 (j = 1,, J). Remaining input parameters are the same for all problems: h j = 1 (j {0,,J)), p j = 600 (j = 1,, J) and the support warehouse lead time is equal to half of the retailer lead time, that is, L 0 = 0.5L j. The results are provided in Table D3. In problem scenarios we consider 3 non-identical retailers. The input data consist of different combinations of the elements in the vectors L j = (1, 1.5, 2), b j = (3, 6, 12), c j = (1, 2, 4), 8

10 p j = (4, 6, 8), λ j = (0.4, 0.8, 1.2), whereas L 0 = 0.5 and h j = 1 (j = 1,, J). The input data are provided in Table D4 and the results are provided in Table D5. 9

11 Table D1. Data for scenarios Input data Results # L S L j b j c j λ j S OPT T OPT ψ OPT (%) ΔOPT S AR ψ AR (%) ΔAR (2, 1) (2, 1.5) 20 (3, 1) (1, 4) (1.5, 1.5) 2 (2, 4) (0, 2) (0, 3.5) 0 (0, 2) (0, 4) (0, 3.5) 0 0.3% (0, 5) 2 (1, 3) (2, 1) (0.5, 0.5) % (3, 1) 29 (3, 1) (2, 4) (0.5, 0.5) 4 (2, 4) 6 1.0% (3,4) (0, 2) (0, 1) 3 (0, 2) (1, 4) (0, 0.5) 4 (0, 5) (1, 2) (2, 1.5) 11 (2, 2) (3, 5) (2, 1) 9 (2, 6) 6 1.5% (3, 6) (1, 2) (0.5, 2.5) 9 (1, 2) (1, 6) (0.5, 2) 2 (2, 6) 6 0.6% (1, 7) (2, 2) (0.5, 0.5) 14 (2, 2) (2, 6) (0.5, 0.5) 5 (2, 6) (1, 2) (0, 0.5) 14 (1, 2) (1, 6) (0, 0.5) 5 (2, 6) (3, 1) (1.5, 1.5) 20 (4, 1) (1, 4) (2, 1.5) 2 0.8% (3, 4) 6 (0, 4) (0, 4) (0, 2) (0, 3.5) 0 (0, 2) (0, 4) (0, 3.5) 0 (0, 5) (3, 1) (0.5, 0.5) 26 (4, 1) (2, 4) (0.5, 0.5) 4 (3, 4) (0, 2) (0, 1) 3 (0, 2) (1, 4) (0, 0.5) 4 (0, 5) (2, 2) (2, 1.5) 11 (2, 2) (4, 5) (1.5, 1) 9 (3, 6) (1, 2) (0.5, 2.5) 9 0.4% (1, 2) 15 (0, 2) (2.5, 3) (1, 6) (0.5, 2) 2 0.3% (1, 7) 3 (0, 6) (1.5, 3) (2, 2) (0.5, 0.5) 14 (2, 2) (3, 6) (0.5, 0.5) 5 (3, 6) (1, 2) (0, 0.5) 14 (1, 2) (2, 6) (0, 0.5) 5 (1, 7) 3 Note. N = 12, p j = 70 (j = 1,, J) and h j = 1 (j = 0,, J) in all scenarios. Bold font indicates the alternative solution given by the simulation optimization. 10

12 Table D2. Data for scenarios Input data Results # L S L j b j c j λ j S OPT T OPT ψ OPT (%) S AR ψ AR (%) ΔAR (3, 1) (2, 1.5) 20 (4, 1) (1, 4) (1.5, 1.5) 2 (4, 4) (0, 2) (0, 3.5) 0 (1, 2) (0, 4) (0, 3.5) 0 (1, 5) (4, 1) (0.5, 0.5) 26 (4, 1) (3, 4) (0.5, 0.5) 4 (4, 4) (0, 2) (0, 1) 3 (1, 2) (2, 4) (0, 0.5) 4 (1, 5) (2, 2) (2, 1.5) 11 (3, 2) (4, 5) (2, 1) 9 (3, 6) 6 1.4% (4, 6) (1, 2) (0.5, 2.5) 9 (2, 2) (1, 6) (0.5, 2) 2 (3, 6) (3, 2) (0.5, 0.5) 14 (3, 2) (3, 6) (0.5, 0.5) 5 (3, 6) 6 0.5% (7, 5) (2, 2) (0, 0.5) 14 (2, 2) (2, 6) (0, 0.5) 5 (3, 6) (4, 1) (1.5, 1.5) 20 (6, 1) (2, 4) (1.5, 1.5) 2 (5, 4) (0, 2) (0, 3.5) 0 (1, 2) (0, 4) (0, 3.5) 0 (1, 5) (6, 1) (0.5, 0.5) 26 (6, 1) (4, 4) (0.5, 0.5) 4 (5, 4) (0, 2) (0, 1) 3 (1, 2) (2, 4) (0, 0.5) 4 (1, 5) (3, 2) (1.5, 1.5) 11 (4, 2) (7, 5) (1.5, 1) 9 (5, 6) 6 0.5% (6, 6) (1, 2) (0.5, 2.5) 9 (2, 2) (1, 6) (0.5, 2) 2 (4, 6) (4, 2) (0.5, 0.5) 14 (4, 2) (5, 6) (0, 0.5) 5 (5, 6) (2, 2) (0, 0.5) 14 (2, 2) (3, 6) (0, 0.5) 5 (4, 6) 6 Note. N = 24, p j = 70 (j = 1,, J) and h j = 1 (j = 0,, J) in all scenarios. Bold font indicates the alternative solution given by the simulation optimization. 11

13 Table D3. Data for scenarios Input data Results # N L j b j c j 100*λ j S OPT T OPT ψ OPT (%) S AR ψ AR (%) ΔAR (0.3, 2) (1, 0, 1) (2, 0, 2) (100, 4) (1, 0, 1) (100, 7) (1, 15) (0, 1, 3) (0, 4, 4) (0,0,0) (1, 1, 3) (4, 2) (0.3, 2) (0, 1, 1) (0, 4, 4) (0,0,0) (1, 1, 1) (1, 7) (1, 15) (0, 1, 3) (0, 4, 4) (0,0,0) (0, 1, 4) (4, 0) (0.3, 2) (1, 0, 1) (0, 0, 0) (100, 7) (1, 0, 1) (100, 7) (1, 15) (1, 1, 3) (0, 0, 0) (4, 2) (1, 1, 3) (4, 2) (0.3, 2) (1, 1, 1) (0, 2, 0) (1, 7) (1, 1, 1) (1, 7) (1, 15) (0, 1, 4) (0, 2, 2) (2, 0) (0, 1, 4) (4, 0) (0.3, 2) (1, 0, 1) (4, 0, 2) (100, 11) (2, 0, 1) (100, 14) (1, 15) (1, 1, 4) (4, 2, 2) (6, 1) (2, 1, 4) (7, 3) (0.3, 2) (1, 1, 1) (2, 4, 4) (1, 7) (0, 1, 2) (2, 1) (1, 15) (0, 1, 4) (0, 6, 6) (2, 0) (1, 1, 5) (7, 1) (0.3, 2) (2, 0, 1) (2, 0, 0) (100, 14) (2, 0, 1) (100, 14) (1, 15) (2, 1, 4) (2, 0, 0) (7, 3) (2, 1, 4) (7, 3) (0.3, 2) (0, 1, 2) (0, 2, 2) (2, 1) (0, 1, 2) (2, 1) (1, 15) (1, 1, 5) (0, 0, 0) (7, 1) (1, 1, 5) (7, 1) (0.3, 2) (1, 0, 1) (2, 0, 2) (100, 4) (2, 0, 1) (100, 7) (1, 15) (0, 1, 3) (0, 4, 4) (0, 0) (2, 1, 3) (4, 2) (0.3, 2) (0, 1, 1) (0, 4, 4) (0, 0, 0) (1, 1, 1) (1, 7) (1, 15) (0, 1, 3) (0, 4, 4) (0, 0, 0) (1, 1, 4) (4, 0) (0.3, 2) (2, 0, 1) (2, 0, 0) (100, 7) (2, 0, 1) (100, 7) (1, 15) (2, 1, 3) (2, 0, 0) (4, 2) (2, 1, 3) (4, 2) (0.3, 2) (1, 1, 1) (0, 2, 0) (1, 7) (1, 1, 1) (1, 7) (1, 15) (0, 1, 4) (0, 2, 2) (2, 0) (1, 1, 4) (4, 0) (0.3, 2) (2, 0, 1) (4, 0, 2) (100, 11) (2, 0, 1) (100, 14) (1, 15) (1, 1, 4) (4, 2, 2) (6, 1) (2, 1, 4) (7, 3) 0.9% (3, 1, 4) (0.3, 2) (1, 1, 1) (2, 4, 4) (1, 7) (0, 1, 2) (2, 1) (1, 15) (1, 1, 4) (2, 4, 4) (4, 0) (1, 1, 5) (7, 1) (0.3, 2) (2, 0, 1) (0, 0, 0) (100, 14) (2, 0, 1) (100, 14) (1, 15) (2, 1, 4) (2, 0, 0) (7, 3) (2, 1, 4) (7, 3) (0.3, 2) (0, 1, 2) (0, 2, 2) (2, 1) (0, 1, 2) (2, 1) (1, 15) (1, 1, 5) (0, 0, 0) (7, 1) (1, 1, 5) (7, 1) Note. L S = 0.5L j, p j = 600 (j = 1,, J) and h j = 1 (j = 0,,J) in all scenarios. Bold font indicates the alternative solution given by the simulation optimization. 12

14 Table D4. Input data for scenarios # L j b j c j p j λ j 97 (1, 1.5, 2) (3, 6, 12) (1, 2, 4) (4, 6, 8) (0.4, 0.8, 1.2) 98 (1, 1.5, 2) (3, 6, 12) (1, 2, 4) (4, 6, 8) (1.2, 0.8, 0.4) 99 (1, 1.5, 2) (3, 6, 12) (1, 2, 4) (8, 6, 4) (0.4, 0.8, 1.2) 100 (1, 1.5, 2) (3, 6, 12) (4, 2, 1) (4, 6, 8) (0.4, 0.8, 1.2) 101 (1, 1.5, 2) (12, 6, 3) (1, 2, 4) (4, 6, 8) (0.4, 0.8, 1.2) 102 (2, 1.5, 1) (3, 6, 12) (1, 2, 4) (4, 6, 8) (0.4, 0.8, 1.2) 103 (1, 1.5, 2) (3, 6, 12) (1, 2, 4) (8, 6, 4) (1.2, 0.8, 0.4) 104 (1, 1.5, 2) (3, 6, 12) (4, 2, 1) (4, 6, 8) (1.2, 0.8, 0.4) 105 (1, 1.5, 2) (12, 6, 3) (1, 2, 4) (4, 6, 8) (1.2, 0.8, 0.4) 106 (2, 1.5, 1) (3, 6, 12) (1, 2, 4) (4, 6, 8) (1.2, 0.8, 0.4) 107 (1, 1.5, 2) (3, 6, 12) (4, 2, 1) (8, 6, 4) (0.4, 0.8, 1.2) 108 (1, 1.5, 2) (12, 6, 3) (1, 2, 4) (8, 6, 4) (0.4, 0.8, 1.2) 109 (2, 1.5, 1) (3, 6, 12) (1, 2, 4) (8, 6, 4) (0.4, 0.8, 1.2) 110 (2, 1.5, 1) (12, 6, 3) (1, 2, 4) (4, 6, 8) (0.4, 0.8, 1.2) 111 (2, 1.5, 1) (3, 6, 12) (4, 2, 1) (4, 6, 8) (0.4, 0.8, 1.2) 112 (2, 1.5, 1) (12, 6, 3) (1, 2, 4) (4, 6, 8) (0.4, 0.8, 1.2) Note. N = 3, L 0 = 0.5, h j = 1 (j = 0,, J) in all scenarios. Table D5. Results for scenarios # S OPT T OPT ψ OPT (%) ΔOPT S AR ψ AR (%) ΔAR 97 (1, 0, 1, 3) (0.5, 0, 0.5, 0) (100, 44, 27) (1, 0, 2, 4) (100, 25, 14) 98 (1, 1, 1, 1) (0.5, 0.5, 0.5, 0.5) (38, 44,38) 1.4% (2, 1, 1, 1) (55, 55, 44) (1, 1, 2, 1) 99 (0, 1, 2, 3) (0, 1, 1, 0) (0, 5, 27) (1, 1, 2, 3) (29, 25, 27) 100 (2, 1, 1, 1) (0.5, 1, 0.5, 0) (0, 44, 71) (1, 1, 2, 3) (29, 25, 27) 0.6% (2, 1, 2, 2) 101 (1, 0, 1, 3) (0.5, 0, 0.5, 1) (100, 44, 9) (1, 0, 2, 4) (100, 25, 14) 102 (1, 0, 1, 3) (0.5, 0, 0.5, 0.5) (100, 44, 2) (1, 0, 2, 3) (100, 25, 9) 103 (0, 2, 2, 1) (0, 1, 1, 0.5) (0, 5, 38) (1, 2, 2, 1) (25, 25, 44) 104 (1, 2, 1, 0) (0.5, 1, 0.5, 0) (0, 44, 100) (1, 2, 2, 1) (25, 25, 44) 105 (1, 1, 2, 1) (0, 0, 0.5, 1.5) (55, 15, 17) (2, 1, 1, 1) (55, 55, 44) 1.6% (1, 2, 2, 1) 106 (1, 1, 1, 1) (0.5, 0.5, 0.5, 0.5) (64, 44, 17) (1, 2, 2, 1) (46, 25, 29) 107 (1, 1, 2, 2) (0, 1, 0.5, 0) (0, 15, 46) (1, 1, 2, 2) (29, 25, 46) 108 (1, 0, 1, 3) (0.5, 0, 0.5, 1) (100, 44, 9) (1, 1, 2, 3) (29, 25, 27) 109 (1, 0, 1, 3) (0.5, 0, 0.5, 0.5) (100, 44, 2) (1, 1, 2, 2) (44, 25, 25) 110 (1, 0, 2, 2) (0, 0, 0.5, 1) (100, 15, 0) (1, 0, 2, 3) (100, 25, 9) 111 (2, 1, 1, 1) (0.5, 1, 0.5, 0) (29, 44, 55) (1, 1, 2, 2) (44, 25, 25) 112 (1, 0, 2, 2) (0, 0, 0.5, 1) (100, 15, 0) 1.5% (1, 0, 2, 3) (100, 25, 9) (0.5, 1, 0.5, 1) Note. Bold font indicates the alternative solution given by the simulation optimization. 13

15 Appendix E - Problem scenarios for the multi-echelon model The multi-echelon model was applied with the step size for ψ 0 set to ε = In Table E1-E3, the values of ΔP AR, ΔP NR, ΔP QO and ΔP CO were obtained by simulation. The standard deviations of the ΔP-values are always less than 1% of ΔP. 14

16 Table E.1. Problem scenarios 1-25: input data, solution for the OPT policy, and ΔP-values for the AR, NR, QO and CO policies. # b c p 100*λ S OPT T OPT ΔP AR ΔP NR ΔP QO ΔP CO (6,5,4,3,3,3,2,2,2,2,2,2) (0,2,1,1,1,1,1,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 106% 0% 91% 0% (12,10,6,6,4,4,3,3,3,3,2,2) (0,3,2,2,2,2,2,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 79% 0% 60% 0% (4,2,2,2,2,2,2,2,1,1,1,1) (0,1,1,1,1,1,1,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 39% 0% 33% 0% (22,19,11,11,10,8,8,8,8,7,7,7) (0,4,4,3,3,3,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 39% 0% 26% 0% (8,6,5,5,5,4,4,4,4,3,3,3) (0,3,2,2,2,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 32% 0% 15% 0% (6,5,5,4,4,4,4,4,4,4,3,3) (0,2,2,2,2,1,1,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 32% 0% 20% 0% (17,9,9,6,6,6,6,4,4,4,3,3) (0,3,2,2,2,2,2,2,2,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 30% 0% 22% 3% (18,11,8,6,6,5,4,4,4,4,3,3) (0,4,3,3,2,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 29% 0% 13% 1% (9,9,6,6,5,5,5,5,4,4,4,4) (0,3,3,2,2,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 29% 0% 13% 2% (26,8,7,6,5,5,5,5,5,5,4,4) (0,5,3,3,2,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 29% 0% 14% 2% (12,7,6,5,5,4,4,4,3,3,3,3) (0,3,2,2,2,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 28% 0% 13% 2% (19,15,9,8,7,7,4,4,3,3,2,2) (0,4,4,3,3,3,3,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 26% 0% 8% 1% (7,7,7,5,4,4,4,4,4,4,4,4) (0,3,3,3,2,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 26% 0% 10% 2% (13,9,9,5,4,4,4,4,3,3,3,3) (0,3,2,2,2,1,1,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 25% 0% 13% 1% (10,10,5,4,4,4,4,4,4,4,4,3) (0,3,3,2,2,2,2,2,2,2,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 25% 0% 12% 3% (4,4,4,3,3,3,3,2,2,2,2,2) (0,1,1,1,1,1,1,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 25% 0% 17% 4% (11,11,9,7,6,6,5,5,4,4,4,3) (0,3,3,3,3,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 23% 0% 9% 1% (32,9,8,7,6,5,4,4,4,4,4,3) (0,5,2,2,2,2,2,2,2,2,2,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 22% 0% 13% 2% (9,2,1,1,1,1,1,0.6,0.6,0.3,0.3,0.3) (1,2,1,0,0,0,0,0,0,0,0,0,0) (3,4,4,0,0,0,0,0,0,0,0,0,0) 22% 20% 18% 0% (18,8,7,6,6,6,5,4,4,4,3,3) (0,4,3,3,3,2,2,2,2,2,2,2,2) (0,6,6,6,6,6,6,6,6,6,6,6,6) 19% 0% 8% 1% (8,8,8,6,6,5,5,5,4,4,4,4) (0,2,2,2,2,2,2,2,2,2,2,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 18% 0% 9% 1% (10,6,5,4,3,2,2,2,1,1,1,1) (0,3,2,2,2,2,1,1,1,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 16% 0% 10% 2% (23,9,8,8,7,6,5,3,2,2,2,1) (0,5,3,3,3,2,2,2,2,1,1,1,1) (0,6,6,6,6,6,6,6,6,6,6,6,6) 14% 0% 7% 0% (16,15,13,12,11,9,8,8,7,7,6,6) (1,4,4,4,3,3,3,3,3,3,2,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 14% 1% 2% 0% (6,5,5,5,5,4,4,4,3,3,3,3) (1,1,1,1,1,1,1,1,1,1,1,1,1) (3,2,2,2,2,2,2,2,2,2,2,2,2) 14% 9% 5% 0% 15

17 Table E2. Problem scenarios 26-50: input data, solution for the OPT policy, and ΔP-values for the AR, NR, QO and CO policies. # b c p 100*λ S OPT T OPT ΔP AR ΔP NR ΔP QO ΔP CO (14,5,5,4,4,3,2,2,1,1,1,1) (1,4,2,2,2,2,1,1,1,1,1,1,1) (3,2,2,2,2,2,2,2,2,2,2,2,2) 13% 5% 3% 0% (4,4,4,2,2,2,2,2,2,2,1,1) (1,1,1,1,1,1,1,1,1,1,1,0,0) (3,3,3,3,3,3,3,3,3,3,3,0,0) 13% 5% 9% 0% (4,3,2,2,2,2,1,1,1,1,1,1) (2,1,0,0,0,0,0,0,0,0,0,0,0) (3,2,0,0,0,0,0,0,0,0,0,0,0) 13% 50% 12% 0% (5,4,2,2,2,1,1,1,1,1,1,0.6) (2,1,1,1,0,0,0,0,0,0,0,0,0) (3,2,2,2,0,0,0,0,0,0,0,0,0) 12% 22% 11% 0% (8,8,5,5,4,3,3,3,3,3,3,3) (1,2,2,2,2,2,1,1,1,1,1,1,1) (3,2,2,2,2,2,2,2,2,2,2,2,2) 12% 9% 4% 0% (34,22,20,19,14,12,10,9,8,8,8,6) (1,7,5,5,5,4,4,3,3,3,3,3,3) (3,1,1,1,1,1,1,1,1,1,1,1,2) 11% 0% 0% 0% (14,8,8,7,6,6,6,5,4,4,4,4) (1,4,3,3,2,2,2,2,2,2,2,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 11% 4% 2% 0% (21,13,12,9,9,9,7,7,7,6,6,6) (1,5,3,3,3,3,3,2,2,2,2,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 10% 0% 1% 0% (12,7,7,6,5,4,4,4,4,4,4,4) (1,2,2,2,2,1,1,1,1,1,1,1,1) (3,2,2,2,2,2,2,2,2,2,2,2,2) 10% 11% 3% 0% (29,12,12,11,11,10,8,8,8,7,6,5) (1,6,3,3,3,3,3,3,3,3,3,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 10% 3% 3% 0% (5,3,3,2,2,2,2,2,2,2,1,1) (1,2,1,1,1,1,1,1,1,1,1,1,1) (3,2,2,2,2,2,2,2,2,2,2,2,2) 10% 5% 5% 0% (10,8,6,5,5,5,5,5,4,4,4,4) (1,3,3,2,2,2,2,2,2,2,2,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 10% 3% 3% 0% (9,8,8,7,6,6,6,5,5,5,5,4) (1,3,3,3,3,2,2,2,2,2,2,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 9% 3% 2% 0% (18,17,14,8,8,7,7,7,7,7,6,6) (1,4,4,4,3,3,3,3,3,2,2,2,2) (3,1,1,1,1,1,1,1,1,1,1,1,1) 8% 0% 0% 0% (6,5,5,5,4,4,4,4,4,4,4,4) (1,2,2,2,2,2,2,2,2,2,2,2,2) (3,2,2,2,2,2,2,2,2,2,2,2,2) 8% 2% 1% 0% (4,4,4,4,4,4,2,2,2,2,2,0.3) (1,1,1,1,1,1,1,1,1,1,1,1,0) (3,2,2,2,2,2,2,2,2,2,2,2,0) 8% 7% 3% 0% (24,15,15,14,11,11,8,6,6,5,5,5) (1,5,4,4,4,3,3,3,2,2,2,2,2) (2,1,1,1,1,1,1,1,1,1,1,1,1) 7% 7% 0% 0% (19,17,12,9,8,8,7,7,7,7,6,6) (1,4,4,3,3,3,3,3,2,2,2,2,2) (2,1,1,1,1,1,1,1,1,1,1,1,1) 7% 3% 0% 0% (6,4,4,4,2,2,2,2,2,2,2,2) (1,2,1,1,1,1,1,1,1,1,1,1,1) (3,1,1,1,1,1,1,1,1,1,1,1,1) 7% 14% 1% 0% (26,11,11,10,9,9,9,7,7,6,6,6) (1,6,3,3,3,3,3,3,3,3,3,3,2) (3,1,1,1,1,1,1,1,1,1,1,1,1) 7% 2% 0% 0% (11,7,6,6,4,4,4,4,4,3,3,3) (1,3,2,2,2,2,2,2,2,1,1,1,1) (3,1,1,1,1,1,1,1,1,1,1,1,1) 7% 8% 0% 0% (48,26,7,7,3,3,2,1,1,1,1,1) (1,7,4,2,2,1,1,1,1,1,1,1,1) (3,1,1,1,1,1,1,1,2,2,2,2,2) 7% 5% 0% 0% (17,16,15,14,13,11,9,8,8,8,7,7) (1,4,4,4,4,4,3,3,3,3,3,2,2) (3,1,1,1,1,1,1,1,1,1,1,1,1) 6% 2% 0% 0% (37,22,17,17,16,14,14,13,12,10,10,10) (1,7,5,4,4,4,4,4,4,4,3,3,3) (2,1,1,1,1,1,1,1,1,1,1,1,1) 6% 5% 0% 0% (56,29,19,18,14,14,14,13,10,9,8,8) (1,9,6,5,4,4,4,4,4,3,3,3,3) (2,1,1,1,1,1,1,1,1,1,1,1,1) 6% 4% 0% 0% 16

18 Table E3. Problem scenarios 51-70: input data, solution for the OPT policy, and ΔP-values for the AR, NR, QO and CO policies. # b c p 100*λ S OPT T OPT ΔP AR ΔP NR ΔP QO ΔP CO (37,27,26,25,22,21,18,18,15,14,13,12) (1,7,6,6,6,5,5,5,5,4,4,4,4) (1,1,1,1,1,1,1,1,1,1,1,1,1) 6% 4% 0% 0% (7,6,5,5,5,4,4,4,3,3,3,3) (1,2,2,2,2,2,2,2,2,1,1,1,1) (2,2,2,2,2,2,2,2,2,2,2,2,2) 6% 7% 2% 0% (12,7,7,7,6,6,5,5,4,4,4,4) (1,3,2,2,2,2,2,2,2,2,2,2,2) (3,1,1,1,1,1,1,1,1,1,1,1,1) 6% 7% 0% 0% (12,9,9,7,7,6,6,6,6,5,5,5) (1,3,3,3,3,3,2,2,2,2,2,2,2) (3,1,1,1,1,1,1,1,1,1,1,1,1) 6% 4% 0% 0% (12,8,6,5,5,5,4,4,4,3,3,3) (1,3,3,2,2,2,2,2,2,2,2,2,2) (2,1,1,1,1,1,1,1,1,1,1,1,1) 5% 9% 0% 0% (35,26,15,12,10,10,9,8,6,5,4,4) (1,6,5,4,3,3,3,3,3,2,2,2,2) (1,1,1,1,1,1,1,1,1,1,1,1,1) 5% 10% 0% 0% (16,11,3,3,3,3,2,1,1,0.6,0.3,0.3) (1,4,3,1,1,1,1,1,1,1,1,0,0) (3,1,1,1,1,1,1,1,1,1,1,0,0) 5% 10% 1% 0% (8,6,6,5,5,4,4,4,3,3,3,3) (1,2,2,2,2,2,2,2,1,1,1,1,1) (3,1,1,1,1,1,1,1,1,1,1,1,1) 5% 8% 0% 0% (44,29,20,20,18,16,13,12,11,11,11,10) (1,8,6,5,5,5,4,4,4,3,3,3,3) (1,1,1,1,1,1,1,1,1,1,1,1,1) 4% 4% 0% 0% (27,26,24,22,18,13,13,13,12,12,12,11) (1,6,6,5,5,4,4,4,4,4,4,4,3) (1,1,1,1,1,1,1,1,1,1,1,1,1) 4% 4% 0% 0% (14,14,11,11,10,8,8,4,4,4,4,4) (1,4,4,3,3,3,3,3,2,2,2,2,2) (1,1,1,1,1,1,1,1,1,1,1,1,1) 4% 9% 0% 0% (34,33,26,25,20,16,16,14,12,9,9,8) (0,7,7,6,6,5,5,5,4,4,3,3,3) (0,1,1,1,1,1,1,1,1,1,1,1,1) 4% 7% 0% 0% (25,14,10,6,4,4,4,2,2,2,1,1) (1,4,3,2,2,2,1,1,1,1,1,1,1) (2,1,1,1,1,1,1,1,1,1,1,1,1) 4% 8% 0% 0% (21,16,14,12,11,10,7,7,7,6,6,6) (1,4,3,3,3,2,2,2,2,2,2,2,2) (1,1,1,1,1,1,1,1,1,1,1,1,1) 4% 13% 0% 0% (16,11,11,10,7,7,6,6,6,6,6,5) (1,4,3,3,3,3,3,3,3,3,2,2,2) (1,1,1,1,1,1,1,1,1,1,1,1,1) 4% 7% 0% 0% (31,2,1,1,1,1,1,0.6,0.6,0.6,0.6,0.3) (2,5,1,0,0,0,0,0,0,0,0,0,0) (2,1,1,0,0,0,0,0,0,0,0,0,0) 3% 25% 1% 0% (15,7,5,4,3,3,3,2,2,2,2,2) (1,3,2,1,1,1,1,1,1,1,1,1,1) (1,1,1,1,1,1,1,1,1,1,1,1,1) 3% 21% 0% 0% (15,13,8,7,7,6,4,4,3,0.6,0.3,0.3) (0,4,3,2,2,2,2,2,2,1,1,0,0) (0,1,1,1,1,1,1,1,1,1,1,0,0) 3% 16% 0% 0% (16,7,5,4,3,2,2,1,0.6,0.6,0.3,0.3) (1,3,2,1,1,1,1,1,1,0,0,0,0) (1,1,1,1,1,1,1,1,1,0,0,0,0) 2% 30% 0% 0% (10,7,4,3,3,3,3,3,2,2,2,2) (0,2,2,1,1,1,1,1,1,1,1,1,1) (0,1,1,1,1,1,1,1,1,1,1,1,1) 2% 34% 0% 0% 17