3.1. Commitment lead time

The manufacturer uses the commitment lead time to improve the operational efficiency of the system. As stated above, the commitment lead time is the time difference between the customer order and the corresponding demand, i.e., the customer places her orderψ time units before her arrival, i.e., demand. According to Hariharan and Zipkin (Citation1995) and Ahmadiet al. (Citation2019) a single-location inventory system with commitment lead time is equivalent to a single-location inventory system where the commitment lead time is subtracted from the component replenishment lead time. The same is valid for the ATO system under consideration.

We define intervals${\Psi }_1=[0,l_2],{\Psi }_2=[l_2,l_1],$ and$\Psi ={\Psi }_1\cup {\Psi }_2=[0,l_1].$ The lengths of intervals${\Psi }_2$ and${\Psi }_1$ represent the replenishment lead times of components 1 and 2 in the equivalent serial system, respectively. By considering commitment lead time and subtracting it from the replenishment lead time of each component, we can find the updated replenishment lead timesL₁ andL₂ in the equivalent serial system as follows:$$L_1=\max (0,(l_1-l_2)-\max (0,\psi -l_2\left)\right),$$and$$L_2=\max (0,l_2-\psi ).$$

The ATO system and its corresponding equivalent serial system are shown inFigure 2. Notice that when$\psi \ge l_1\ge l_2,$ i.e.,$\psi \notin \Psi ,$ we have$L_1=L_2=0.$ This is an ideal service situation with zero holding and backordering costs. On the other hand, when$\psi \in \Psi ,$ there are two options for the pair (L₁,L₂) depending on whether$\psi \in {\Psi }_1$ or$\psi \in {\Psi }_2.$ We have:$$(L_1,L_2)=\{\begin{array}{cc}(l_1-l_2,l_2-\psi ),\hfill & \hspace{1em}\psi \in {\Psi }_1\hfill \\ (l_1-\psi ,0),\hfill & \hspace{1em}\psi \in {\Psi }_2\hfill \\ (0,0),\hfill & \hspace{1em}\psi \notin \Psi .\hfill \end{array}$$

From the first two options we observe that, depending on the length of commitment lead time$\psi \in \Psi ,$ one of the component replenishment lead times depends onψ. If$\psi \in {\Psi }_1,$L₂ depends onψ and if$\psi \in {\Psi }_2,$L₁ depends onψ. Accordingly, we divide our analysis into two cases; when$\psi \in {\Psi }_1$ and$\psi \in {\Psi }_2.$ In the former case we have a two-stage serial inventory system, whereas in the latter case we have a single-location inventory system, i.e.,$L_2=0.$ It is important to remember thatψ is a decision variable. Hence, the optimal case depends on the system parameters.

3.2. Net inventory levels

The manufacturer manages the replenishment processes of both components. Hence, the inventory system is managed centrally; the manufacturer has all the information on component net inventory levels and can coordinate the replenishment processes to avoid having excess inventory of one component and out-of-stock situation for the other component. The manufacturer would like to avoid such imbalances to the extent possible.

Zipkin (Citation2000) provides a detailed explanation on adjusting component inventories for avoiding imbalances. For an assembly system with two components, the component with the longer lead time uses an ordinary base-stock policy whereas the policy of the component with the shorter lead time should be adjusted to avoid the imbalances mentioned above. We define$D(t_1,t_2]$ as the cumulative customer demand from timet₁ tot₂. Accordingly, the net inventory of component 1 at any timet can be written as$I{N}_1(t)=s_1-D(t-(L_1+L_2),t].$ The net inventory of component 2 should be adjusted such that the net inventories of both components are equal at time$t+L_2,$ i.e.,$I{N}_1(t+L_2)=I{N}_2(t+L_2).$ This can be achieved by adjusting the inventory order position for component 2 as$\mathit{IO}{P}_2(t)=\min (s_2,I{N}_1(t)+I{O}_1^{+}(t)),$ where$I{O}_1^{+}(t)$ is the portion of the outstanding component 1 replenishment orders at timet that arrives before time$t+L_2.$

As defined above,$I{O}_1^{+}(t)$ is the portion of the outstanding component 1 replenishment orders at timet that arrives before time$t+L_2.$ Accordingly,$I{O}_1^{-}(t)$ is the portion of the outstanding component 1 replenishment orders at timet that has not arrived before time$t+L_2.$ Refer toFigure 3 for a graphical representation of these two portions.

We define$I{O}_1(t)$ as the outstanding component 1 orders at timet. Since$I{O}_1(t)=I{O}_1^{+}(t)+I{O}_1^{-}(t),\mathit{IO}{P}_2(t)$ can alternatively be written as$\mathit{IO}{P}_2(t)=\min (s_2,I{N}_1(t)+I{O}_1(t)-I{O}_1^{-}(t)).$ Both components are supplied by suppliers with infinite capacity. Therefore, every replenishment order is fulfilled by the suppliers immediately. This implies that the inventory order position, defined as$\mathit{IO}{P}_1(t)=I{N}_1(t)+I{O}_1(t),$ satisfies the equality$\mathit{IO}{P}_1(t)=s_1,\forall t.$ Hence, we have$\mathit{IO}{P}_2(t)=\min (s_2,s_1-I{O}_1^{-}(t)).$ In addition, since each outstanding order corresponds to a demand,$I{O}_1^{-}(t)$ is equal to the cumulative demand during$(t-(L_1+L_2),t-L_2].$ Hence,$\mathit{IO}{P}_2(t)=\min (s_2,s_1-D(t-(L_1+L_2),t-L_2]).$

Using the stationary and independent increments property of the Poisson process, we can rewrite the net inventory of component 1 at timet as$I{N}_1(t)=s_1-(D(t-(L_1+L_2),t-L_2]+D(t-L_2,t]).$ As a result, in steady-state the component net inventory levels are(1)$$I{N}_1=s_1-(X+Y),$$(1)and(2)$$I{N}_2=\min (s_2,s_1-X)-Y,$$(2)whereX andY are cumulative demand duringL₁ andL₂, respectively.

3.3. Remnant inventory levels

In an ATO system, when a customer order arrives, the ability of the manufacturer to satisfy the corresponding demand depends on the net inventory levels of the components. The manufacturer is able to meet the demand at a specific point of time if the minimum of the net inventories of both components is positive, i.e.,$\min (I{N}_1,I{N}_2)>0;$ otherwise, the demand is backordered.

By definition of the base-stock policy, a customer order for the end product implies orders for each individual component. If component inventory levels are positive, one unit of each component is dedicated to the specific customer order. We call the component inventory that is dedicated to specific customer ordersremnant inventory and use$\Delta I_j$ to represent the remnant inventory level of componentj. In our setting, we have the remnant inventory only when one component is available and another one is missing (de Kok,Citation2003). Hence, at any point in time, there can be at most one component type in remnant inventory. The steady-state remnant inventory level of each component can be calculated as$$\Delta I_1=\max (0,B_2-B_1)=\max (0,\max (0,-I{N}_2)-\max (0,-I{N}_1\left)\right),$$and$$\Delta I_2=\max (0,B_1-B_2)=\max (0,\max (0,-I{N}_1)-\max (0,-I{N}_2\left)\right).$$

Here,B₁ andB₂ are steady-state backorder levels for components 1 and 2, respectively. It is obvious that a delayed delivery of one component does not only result in backordering cost of the end product, but also can result in inventory holding cost for the other component if it is already in stock as remnant inventory. Next, we determine the expected remnant inventory levels for both components.

Proposition 1.

The expected remnant inventory levels for a centralized ATO system with two components and Poisson customer demand for the end product are$$\begin{array}{c}\mathbb{E}\{\Delta I_1\}=\sum _{x=0}^{s_1-s_2}(\sum _{y=s_2+1}^{s_1-x}(y-s_2)P_2(y)+(s_1-s_2-x)(1-F_2(s_1-x)))P_1(x),\\ \mathbb{E}\{\Delta I_2\}=0.\end{array}$$

Here,$F_j(x)$ and$P_j(x)$ are the cumulative distribution and probability mass functions of a Poisson random variable with mean${\mu }_j=\lambda L_j,$ respectively. According toProposition 1, when component inventories are controlled centrally by the manufacturer, the orders for component 2 can be synchronized with the orders for component 1 such that$\mathbb{E}\{\Delta I_2\}=0.$ Component 2 has a shorter replenishment lead time than component 1. The manufacturer needs one unit from both components to satisfy a unit of demand. The manufacturer can place orders with the supplier of component 2 such that the component 2 never needs to wait for the arrival of component 1. This is why there is no remnant inventory for component 2. In addition, if$s_1\le s_2,$ the expected remnant inventory level of component 1 is zero, i.e.,$\mathbb{E}\{\Delta I_1\}=0.$ When$s_1\le s_2,$ if there is a demand and if there is at least one unit of component 1, there is also at least one unit of component 2 in the inventory. This is why there is no remnant inventory for the components. Except this case with$s_1\le s_2,$ even under a centralized control, the expected remnant inventory level of component 1 is positive. Under a decentralized control scheme both expectations can be positive.

3.4. Long-run average cost

Having the expressions for the net inventory levels and the expected remnant inventory levels, we are ready to derive the expression for the long-run average cost. For that purpose, we defineI_j as on-hand inventory level of componentj andB as the number of backorders of the end product. We use$\mathit{s}$ to represent the base-stock level pair (s₁,s₂). Then, for an arbitrary base-stock level pair$\mathit{s}$ and commitment lead timeψ we can write the long-run average cost$C_a(\mathit{s},\psi )$ as follows:$$C_a(\mathit{s},\psi )=\mathbb{E}\left\{\sum _{j=1}^2{h}_j(I_j+\Delta I_j)+\mathit{pB}\right\}+c\lambda \psi .$$

Note that we append the subscripta to indicate that this is the conventional cost of an ATO system.$C_a(\mathit{s},\psi )$ is the sum of long-run average holding, backordering and commitment costs. Holding cost of each component is comprised of on-hand inventory holding cost, and remnant inventory holding cost. The manufacturer incurs backordering cost if a customer demand is satisfied more thanψ time units after the corresponding order is given. Finally, the manufacturer incurs commitment cost. It is important to note that the random variables$I_j,\Delta I_j,$ andB depend on$\mathit{s}$ andψ. Our purpose in the remainder of this section is to rewrite the expression for$C_a(\mathit{s},\psi )$ such that these dependencies are made explicit.

First, we rewrite$C_a(\mathit{s},\psi )$ by using the following equalities;$I_j={[I{N}_j]}^{+},$ forj = 1, 2,$\Delta I_1=\max (0,{[I{N}_2]}^{-}-{[I{N}_1]}^{-}),\Delta I_2=\max (0,{[I{N}_1]}^{-}-{[I{N}_2]}^{-})$ and$B=\max ({[I{N}_1]}^{-},{[I{N}_2]}^{-}),$ where${[x]}^{+}=\max (0,x)$ and${[x]}^{-}=\max (0,-x).$ We have(3)$$\begin{array}{l}{C}_a(\mathit{s},\psi )=h_1\left(\mathbb{E}\right\{{\left[I{N}_1\right]}^{+}\}+\mathbb{E}\{\max (0,{\left[I{N}_2\right]}^{-}-{\left[I{N}_1\right]}^{-})\left\}\right)\\ +h_2\left(\mathbb{E}\right\{{\left[I{N}_2\right]}^{+}\}+\mathbb{E}\{\max (0,{\left[I{N}_1\right]}^{-}-{\left[I{N}_2\right]}^{-})\left\}\right)\\ +p\mathbb{E}\left\{\max \right({\left[I{N}_1\right]}^{-},{\left[I{N}_2\right]}^{-}\left)\right\}+c\lambda \psi .\end{array}$$(3)

The first and second rows are expected holding costs for on-hand and remnant inventory levels of components 1 and 2, respectively. The third row is the sum of expected backordering and commitment costs. Next, we rely on simple algebra to obtain the following equivalent expression for$C_a(\mathit{s},\psi ).$ We refer the reader to theAppendix for the derivations:(4)$$\begin{array}{l}{C}_a(\mathit{s},\psi )=\sum _{j=1}^2{h}_j\mathbb{E}\left\{I{N}_j\right\}+(p+h_1+h_2)\mathbb{E}\left\{\max \right({\left[I{N}_1\right]}^{-},{\left[I{N}_2\right]}^{-}\left)\right\}+c\lambda \psi .\hfill \end{array}$$(4)

Rosling (Citation1989) and Chen and Zheng (Citation1994) show that a two-component, single end product ATO system can be reduced to an equivalent two-stage serial system with modified replenishment lead timesL_i (refer toFigure 2). In the equivalent serial system, the local holding costs for stages 1 and 2 areh₁ and$h_1+h_2,$ respectively. According to Shang and Song (Citation2003) the long-run average cost of the serial system can be written as$$\begin{array}{cc}{C}_s(\mathit{s},\psi )=\hfill & \sum _{j=1}^2{h}_j\mathbb{E}\left\{I{N}_j\right\}+(p+h_1+h_2)\mathbb{E}\left\{{\left[I{N}_2\right]}^{-}\right\}+c\lambda \psi .\hfill \end{array}$$

Note that we use subscripts to indicate the long-run average cost of the equivalent serial system. Under a centralized control scheme, the expression for$\mathbb{E}\left\{B\right\},$ as derived in theAppendix, is as follows:(5)$$\begin{array}{c}\mathbb{E}\left\{B\right\}=\mathbb{E}\left\{\max \right({\left[I{N}_1\right]}^{-},{\left[I{N}_2\right]}^{-}\left)\right\}=\mathbb{E}\left\{{\left[I{N}_2\right]}^{-}\right\}.\hfill \end{array}$$(5)

This result is consistent withProposition 1. According toProposition 1, we have$\mathbb{E}\{\Delta I_2\}=0,$ which means that component 2 never waits for the arrival of component 1 to satisfy a customer demand. Hence, if there is a backorder, it is due to a missing component 2. This is why the expected number of customer backorders equals the expected number of component 2 backorders.

Although all the terms of$C_s(\mathit{s},\psi )$ and$C_a(\mathit{s},\psi )$ look exactly the same, there is a small difference that results from the fact that in the ATO system there is no holding cost for in-transit inventory, whereas in the serial system the holding cost for items in-transit to stage 2 is included. Hence, the total cost of the serial system exceeds the total cost of the assembly system by$h_1\lambda L_2$ (Zhang,Citation2006). Then, for an arbitrary$\psi \in \Psi ,$ we have$C_a(\mathit{s},\psi )=C_s(\mathit{s},\psi )-h_1\lambda L_2.$ It is a well-known result that, for an arbitrary$\psi \in \Psi ,$ the total cost of the equivalent serial system$C_s(\mathit{s},\psi )$ can be derived using the recursive method proposed by Shang and Song (Citation2003). By subtracting$h_1\lambda L_2$ and applying some simple algebra we can rewrite the expression for$C_a(\mathit{s},\psi )$ in terms of system parameters and decision variables. We refer the reader to theAppendix for the details:(6)$$C_a(\mathit{s},\psi )={\mu }_1\left(p-\left((p+h_1)F_1(s_1-s_2-1)+(p+h_1+h_2){\sum }_{t=s_1-s_2}^{s_1-1}{F}_2(s_1-t-1)P_1(t)\right)\right)+s_1\left(-p+\left((p+h_1)F_1(s_1-s_2)+(p+h_1+h_2){\sum }_{t=s_1-s_2+1}^{s_1}{F}_2(s_1-t)P_1(t)\right)\right)+{\mu }_2\left(p-(p+h_1+h_2)\left(F_2(s_2-1)F_1(s_1-s_2-1)+{\sum }_{t=s_1-s_2}^{s_1-1}{F}_2(s_1-t-1)P_1(t)\right)\right)+s_2(p+h_1+h_2)F_1(s_1-s_2)\left(F_2(s_2)-\frac{p+h_1}{p+h_1+h_2}\right)+c\lambda \psi .$$(6)

When$\psi \in {\Psi }_j,F_j(x)$ and$P_j(x)$ are independent ofψ.

4.1. Optimization of the component base-stock levels

For an arbitrary$\psi \in \Psi ,$ determining the base-stock levels which minimize$C_a(\mathit{s},\psi )$ is a well-known optimization problem in the multi-echelon inventory theory literature. We have$$\begin{array}{c}{C}_a({\mathit{s}}^{\psi },\psi )=\underset{{}_{\mathit{s}}}{\min }{C}_a(\mathit{s},\psi ),\hfill \end{array}$$where$C_a({\mathit{s}}^{\psi },\psi )$ is a tight lower bound for$C_a(\mathit{s},\psi ).$${\mathit{s}}^{\psi }$ represents the optimal pair ($s_1^{\psi },s_2^{\psi }$) corresponding to a$\psi \in \Psi .$ To compute the optimal echelon base-stock levels, two nested convex functions should be minimized. More specifically, for a given$\psi \in \Psi ,$ we need to solve the recursive optimization equations which have been proposed by Shang and Song (Citation2003). Knowing that the optimal base-stock levels of a two-component ATO system are equal to the optimal echelon base-stock levels of the equivalent serial system, we can formulate the following theorem, which states the inequalities that need to be satisfied by the optimal base-stock levels.

Theorem 1.

In a two-component ATO system with a given commitment lead time$\psi \in \Psi ,$ the optimal base-stock levels of components 1 and 2 are two non-negative integers$s_1^{\psi }$ and$s_2^{\psi },$ which satisfy the inequalities:$$F_2(s_2^{\psi }-1)<\frac{p+h_1}{p+h_1+h_2}\le F_2\left(s_2^{\psi }\right),$$and$${\Pi }_1(s_1^{\psi }-1,s_2^{\psi })<\frac{p}{p+h_1+h_2}\le {\Pi }_1(s_1^{\psi },s_2^{\psi }),$$where$F_j(x)$and$P_j(x)$are the cumulative distribution and probability mass functions of a Poisson random variable with mean${\mu }_j=\lambda L_j$and$${\Pi }_1(x,y)=\left(\frac{p+h_1}{p+h_1+h_2}\right)F_1(x-y)+\sum _{n=x-y+1}^x{F}_2(x-n)P_1\left(n\right).$$

Theorem 1 states that the optimal component base-stock levels depend on the commitment lead time. For each value ofψ, obtaining the optimal base-stock levels needs two consecutive steps. First, the optimal base-stock level of component 2,$s_2^{\psi }$ should be calculated, and then by using this value in the second inequality, the optimal base-stock level of component 1,$s_1^{\psi }$ is calculated. The optimal base-stock levels of a two-component ATO system are equal to the optimal echelon base-stock levels of the equivalent serial system (Rosling,Citation1989; Shang and Song,Citation2003). Our ATO system and its equivalent serial system are depicted inFigure 2. It is known that the optimal base-stock policy for a two-location serial systems can be computed through minimizing two nested convex functions recursively, starting from the location closest to the customer, i.e., location where component 2 is stored (Shang and Song,Citation2003).Theorem 1 is consistent with this result and it provides the inequalities that need to be satisfied by the optimal component base-stock levels.

We would like to note that, when$\psi \in {\Psi }_2,$ we have${\mu }_2=0$ and$s_2^{\psi }=0.$ Hence, when$\psi \in {\Psi }_2$ the manufacturer relies on a BTO strategy for component 2. In this case,EquationEquation (6)(6)$$C_a(\mathit{s},\psi )={\mu }_1\left(p-\left((p+h_1)F_1(s_1-s_2-1)+(p+h_1+h_2){\sum }_{t=s_1-s_2}^{s_1-1}{F}_2(s_1-t-1)P_1(t)\right)\right)+s_1\left(-p+\left((p+h_1)F_1(s_1-s_2)+(p+h_1+h_2){\sum }_{t=s_1-s_2+1}^{s_1}{F}_2(s_1-t)P_1(t)\right)\right)+{\mu }_2\left(p-(p+h_1+h_2)\left(F_2(s_2-1)F_1(s_1-s_2-1)+{\sum }_{t=s_1-s_2}^{s_1-1}{F}_2(s_1-t-1)P_1(t)\right)\right)+s_2(p+h_1+h_2)F_1(s_1-s_2)\left(F_2(s_2)-\frac{p+h_1}{p+h_1+h_2}\right)+c\lambda \psi .$$(6) reduces to$$\begin{array}{c}{C}_a\left(\right(s_1,0),\psi )={\mu }_1(p-(p+h_1\left)F_1\right(s_1-1\left)\right)+s_1(-p+(p+h_1\left)F_1\right(s_1\left)\right)+c\lambda \psi .\hfill \end{array}$$

$C_a((s_1,0),\psi )$ represents the long-run average cost of a single-location inventory system. We refer to Ahmadiet al. (Citation2019) for detailed analysis of the single-location system with commitment lead time.

With the next result, we prove the behavior of the optimal base-stock levels with respect to the commitment lead time.

Theorem 2.

The optimal base-stock level of component j,$s_j^{\psi },$ is a piecewise-constant and non-increasing function of$\psi \in \Psi$ for j = 1, 2.

According toTheorem 2, the manufacturer can operate with less inventory if the commitment lead time gets longer. In fact, for a sufficiently long commitment lead time, the manufacturer can optimize his long-run average cost by setting both component base-stock levels to zero. This would imply using a BTO strategy. InFigure 4, we illustrate how the optimal component base-stock levels change with respect to the commitment lead time.

Next, we are interested in the behavior of the average cost function with respect to the commitment lead time. InFigure 5, for a specific parameter setting with$\lambda =0.4,h=[5,7],p=20,l=[8,5]$ andc = 6, we draw the average cost function for multiple base-stock level pairs over$\Psi ={\Psi }_1\cup {\Psi }_2,$ where${\Psi }_1=[0,5]$ and${\Psi }_2=[5,8].$ The vertical dotted line atψ = 5 shows the separation of the intervals${\Psi }_1$ and${\Psi }_2.$ We also indicate the tight lower bound of these cost functions. This tight lower bound is$C_a({\mathit{s}}^{\mathit{\psi }},\psi ).$ This is the function that needs to minimized for obtaining the optimal commitment lead time. Note that the function$C_a({\mathit{s}}^{\mathit{\psi }},\psi )$ does not have a well-defined behavior. This is why finding the optimal commitment lead time requires a different approach.

4.2. Unit commitment cost thresholds

In the previous section, we analyzed the properties of the average cost function and the optimal base-stock levels when the commitment lead time is fixed to a certain value. In this section, we define unit commitment cost thresholds that are used to identify the optimal commitment lead time.

First, we define a set, which helps us with providing a better explanation for the subsequent analysis. The set is related to the end points of the sub-domains of$\Psi ,$ which are$0,l_2,$ andl₁.

Definition 1.

Let${\Psi }_1=[0,l_2]$and${\Psi }_2=[l_2,l_1]$be the sub-domains of the commitment lead times in a two-component ATO system. Then, we define a terminal set$\mathcal{T}$as$\mathcal{T}=\{0,l_2,l_1\}.$

Using the data inFigure 4, we have$\mathcal{T}=\{0,5,8\}.$

Next, we define three unit commitment cost thresholds. Each definition uses two elements of the set$\mathcal{T}.$ We usec_ik,i <k, to represent the threshold that usesith andkth elements of the set$\mathcal{T}.$ We have$$\begin{array}{c}{c}_{12}=\frac{C_a({\mathit{s}}^0,0)-C_a({\mathit{s}}^{l_2},l_2)}{\lambda l_2}+c\\ c_{13}=\frac{C_a({\mathit{s}}^0,0)-C_a({\mathit{s}}^{l_1},l_1)}{\lambda l_1}+c\\ c_{23}=\frac{C_a({\mathit{s}}^{l_2},l_2)-C_a({\mathit{s}}^{l_1},l_1)}{\lambda (l_1-l_2)}+c.\end{array}$$

Although it seems like the unit commitment cost thresholds depend on unit commitment cost,c, –c comes out of the fractions in each expression and it cancels out$+c.$ Remember that${\mathit{s}}^{\psi }$ is the optimal base-stock level vector for commitment lead timeψ. Note that we have defined a unit commitment cost threshold for each pair of points in the terminal set. The unit commitment cost thresholds ensure the equality of the long-run average costs when the commitment lead time is set to the corresponding pair of points. For example, when$c=c_{12},$ we have$C_a({\mathit{s}}^0,0)=C_a({\mathit{s}}^{l_2},l_2).$ InFigure 6, we provide three numerical examples with the same parameter settings where the unit commitment costs are set toc₁₂,c₁₃, andc₂₃.

With the following lemma, we define multiple different relationships among unit commitment cost thresholds.

Lemma 1.

Consider a two-component ATO system and unit commitment cost thresholds c₁₂,c₁₃and c₂₃.The following results hold;

1. $(l_1-l_2)c_{23}=l_1{c}_{13}-l_2{c}_{12},$

2. either$c_{12}\le c_{13}\le c_{23}$or$c_{23}<c_{13}<c_{12}$holds.

Remark 1.

When the component replenishment lead times are the same, i.e.,l₁ =l₂,we have$c_{12}=c_{13}$and$c_{23}=\infty .$In this case, there is a unique unit commitment cost threshold$c_0=c_{12}=c_{13}$which is exactly the same as unit commitment cost threshold in the single-location problem (see Ahmadiet al. (Citation2019)for more details).

Next, using the unit commitment cost thresholds, we define lower bounds. These bounds are used for obtaining the optimal commitment lead time.

Definition 2.

For all$t_i,t_j\in \mathcal{T}$and t_i < t_j, define$C_{L{B}_{\mathit{ij}}}(\psi )$as$$C_{L{B}_{\mathit{ij}}}(\psi )=C_a({\mathit{s}}^{t_i},t_i)-\lambda (c_{\mathit{ij}}-c)(\psi -t_i),$$where$C_{L{B}_{\mathit{ij}}}(\psi )$is a linear function which connects the points$\left(t_i,C_a({\mathit{s}}^{t_i},t_i)\right)$and$\left(t_j,C_a({\mathit{s}}^{t_j},t_j)\right)$.We call this function a linear lower bound.

Based on this definition, in a two-component ATO system, there exist three linear lower bounds;$C_{L{B}_{12}}(\psi ),C_{L{B}_{23}}(\psi ),$ and$C_{L{B}_{13}}(\psi ).$ The slope of each linear lower bound depends on the corresponding unit commitment cost thresholds. In a two-component ATO system, the lower bounds form a triangle. FromLemma 1, we know that either$c_{12}\le c_{13}\le c_{23}$ or$c_{23}<c_{13}<c_{12}$ holds. When$c_{12}<c_{13}<c_{23}$ holds the triangle points up and when$c_{23}<c_{13}<c_{12}$ holds the triangle points down. Refer toFigure 7 for examples. Since in the former case there are two potential minimum points and in the latter case there are three potential minimum points we call them Case 2 and Case 3, respectively. Note that when$c_{12}=c_{13}=c_{23},$ we have a straight line with two potential minimum points.

Given an ATO system with a specific parameter setting, we can determine the Case number by calculating two of the three unit commitment cost thresholds only. Suppose thatc₁₂ andc₁₃ are calculated. Then we know that if$c_{12}\le c_{13}$ holds, we have Case 2; otherwise, we have Case 3.

4.3. Optimization of the commitment lead time

In this section we characterize the optimal preordering strategy. First, we formulate a conjecture which identifies a lower bound on$C_a({\mathit{s}}^{\psi },\psi ).$ If the lower bound is known, we can also indicate whichψ values are candidates for being optimal. Then, it is enough to evaluate the cost$C_a({\mathit{s}}^{\psi },\psi )$ only at these points.

Conjecture 1.

In a two-component ATO system with commitment lead time ψ and lower bounds$C_{L{B}_{\mathit{ij}}}(\psi )$that connect the points$(t_i,C_a({\mathit{s}}^{t_i},t_i))$and$(t_j,C_a({\mathit{s}}^{t_j},t_j))$, where$t_i,t_j\in \mathcal{T}$such that t_i < t_j, for all$\psi \in {\Psi }_1$we have

1. $C_{L{B}_{13}}(\psi )\le C_a({\mathit{s}}^{\psi },\psi )$if$c_{12}\le c_{13}\le c_{23}$or

2. $C_{L{B}_{12}}(\psi )\le C_a({\mathit{s}}^{\psi },\psi )$if$c_{23}<c_{13}<c_{12}.$

We present the above result as a conjecture since we do not have a proof for it. We test the correctness of the conjecture by generating random instances. Our belief on the correctness of the conjecture is confirmed in every single instance. We provide more detail on our numerical setup and results inSection 5.

Theorem 3.

In a two-component ATO system with commitment lead time ψ with$C_{L{B}_{13}}(\psi )\le C_a({\mathit{s}}^{\psi },\psi )$when$c_{12}\le c_{13}\le c_{23}$and$C_{L{B}_{12}}(\psi )\le C_a({\mathit{s}}^{\psi },\psi )$when$c_{23}<c_{13}<c_{12}$, the optimal commitment lead time${\psi }^{*}$is either zero or equal to the replenishment lead time of one of the components, i.e.,${\psi }^{*}\in \left\{0,l_2,l_1\right\}.$

Theorem 3 characterizes the optimal commitment lead time. When the optimal commitment lead time is equal to zero, the preorder strategy is not beneficial to the ATO system and the optimal component ordering strategy is a BTS strategy. When the optimal commitment lead time is equal tol₂, the optimal strategy of components 1 and 2 are BTS and BTO strategies, respectively. When the optimal commitment lead time is equal tol₁, the optimal strategy for both components is an BTO strategy.

Next, we use the unit commitment cost thresholds to provide a guideline on when to set the commitment lead time to$0,l_2$ orl₁. For an arbitrary parameter setting, the optimal commitment lead time can be determined using the results inTheorem 4.

Theorem 4.

In a two-component ATO system with commitment lead time ψ and unit commitment cost c, with$C_{L{B}_{13}}(\psi )\le C_a({\mathit{s}}^{\psi },\psi )$ when$c_{12}\le c_{13}\le c_{23}$ and$C_{L{B}_{12}}(\psi )\le C_a({\mathit{s}}^{\psi },\psi )$ when$c_{23}\le c_{13}\le c_{12},$ the optimal commitment lead time${\psi }^{*}$ is$${\psi }^{*}=\{\begin{array}{cc}0\hfill & \hspace{1em}\mathit{\text{if}}\hspace{1em}\mathit{\text{Case}}2,c\ge c_{13}\hspace{1em}\text{or}\hspace{1em}\mathit{\text{Case}}3,c\ge c_{12}\hfill \\ l_1\hfill & \hspace{1em}\mathit{\text{if}}\hspace{1em}\mathit{\text{Case}}2,c\le c_{13}\hspace{1em}\text{or}\hspace{1em}\mathit{\text{Case}}3,c\le c_{23}\hfill \\ l_2\hfill & \hspace{1em}\mathit{\text{if}}\hspace{1em}\mathit{\text{Case}}3,c_{23}\le c\le c_{12},\hfill \end{array}$$where, we have Case 2 if$c_{12}\le c_{13}\le c_{23}$and Case 3 if$c_{23}<c_{13}<c_{12}.$

Recall that for any parameter setting we end up with either Case 2 or Case 3. In Cases 2 and 3 there are two and three potential optimal solutions, respectively. In addition to indicating the optimal commitment lead time,Theorem 4 tells us when the preordering strategy is beneficial or not. The preordering strategy is not beneficial if${\psi }^{*}=0.$ Hence, if$c\ge c_{13}$ in Case 2, the preordering strategy is not beneficial. In Case 3, if$c\ge c_{12},$ the preordering strategy should not be preferred.

Theorem 4 specifies the optimal component ordering strategy for different values of the unit commitment cost. InFigure 8 we provide a visual representation of the results inTheorem 4. More specifically, we identify the values of unit commitment cost for which BTO or BTS strategy is optimal. In Case 2, for any value of unit commitment cost the manufacturer should use the same strategy for both components, i.e., the switch from BTO to BTS happen at the same threshold valuec₁₃. In Case 3, based on the value of unit commitment cost, the manufacturer may use the same or different strategies for the components. For example, when$c_{23}<c<c_{12},$ the optimal strategy is BTS and BTO for component 1 and 2, respectively.

For concision, define the following mathematical notations:${[x]}^{+}=$ max$(x,0),{[x]}^{-}=$ max$(-x,0),x\vee y=$ max(x,y) and$x\wedge y=$ min(x,y).