Benchmarking non-first-come-first-served component allocation in an assemble-to-order system

Abstract

In multi-product, multi-component periodic review assemble-to-order systems, the component allocation policy has substantial impact on performance. In this paper, the impact of two classes of non-first-come-first-served allocation rules (non-FCFS) are evaluated; i.e., the last-come-first-served-within-one-period, and the product-based-priority-within-time-windows approach. The demand fulfillment rates for these rules are expressed analytically, which are then used to formulate base stock level optimization problems as chance-constrained programs. Moreover, three mathematical programs are used as benchmarks to test the non-FCFS rules. They maximize both the average cycle service level and the aggregate fill-rate, and separately minimize the operational cost per period under the FCFS rule. We show that these simple non-FCFS rules can significantly improve service measures and often outperform the benchmarks, though operational costs increase moderately. Importantly, the value of these rules is greatest when the customer service is used as a point of differentiation.

Appendix

Proof of Theorem 1

We need only to prove the first half of the claim, since the second half follows. Using the $E_{i,t}$ notation, the net inventory at the end of period $t$ is $(S_i-D_i[t-L_i-1,t-1])+D_{i,t-L_i-1}-E_{i,t}=S_i-D_i[t-L_i,t-1]-E_{i,t}$ under the LCFP rule. We need to show that $(S_i-D_i[t-L_i,t-1]-E_{i,t})^+=(S_i-D_i[t-L_i,t])^+$, which comes from the following four cases:

Case 1: $S_i \ge D_i[t-L_i-1,t-1]$ and $(S_i-D_i[t-L_i-1,t-1])^++D_{i,t-L_i-1} \ge D_{i,t}$. Then $E_{i,t}=D_{i,t}$, and $(S_i-D_i[t-L_i,t-1]-E_{i,t})^+=(S_i-D_i[t-L_i,t])^+$.

Case 2: $S_i \ge D_i[t-L_i-1,t-1]$ and $(S_i-D_i[t-L_i-1,t-1])^++D_{i,t-L_i-1} < D_{i,t}$. Then $E_{i,t}=S_i-D_i[t-L_i,t-1]$, and $(S_i-D_i[t-L_i,t-1]-E_{i,t})^+=0$. On the other hand, $S_i-D_i[t-L_i,t]=S_i-D_i[t-L_i-1,t-1]+D_{i,t-L_i-1}-D_{i,t} < 0$. So $(S_i-D_i[t-L_i,t])^+=0$.

Case 3: $S_i < D_i[t-L_i-1,t-1]$ and $D_{i,t-L_i-1} \ge D_{i,t}$. Then $E_{i,t}=D_{i,t}$, and $(S_i-D_i[t-L_i,t-1]-E_{i,t})^+=(S_i-D_i[t-L_i,t])^+$.

Case 4: $S_i < D_i[t-L_i-1,t-1]$ and $D_{i,t-L_i-1} < D_{i,t}$. Then $E_{i,t} = D_{i,t-L_i-1}$, and $(S_i-D_i[t-L_i,t-1]-E_{i,t})^+=(S_i-D_i[t-L_i-1,t-1])^+ = 0$. On the other hand, $(S_i-D_i[t-L_i,t])^{+} = 0$.$\square$

Proof of Theorem 2

Note $E_{i,t}=(S_i-D_i[t-L_i-1,t-1])^{+} + D_{i,t-L_i-1}< D_{i,t}$ if and only if the demand $D_{i,t}$ is not completely met in period $t$. Similarly, in period $t+s$, the amount of available on-hand inventory used to satisfy the demand in the same period (i.e., $D_{i,t+s}$) is $E_{i,t+s}$. If there is no LCFP, then in period $t+v(v \ge 1)$, the net inventory after satisfying the demands prior to period $t+1$ is $S_i-D_i[t-L_i+v,t]$. However, we must also consider the amount of inventory committed under the LCFP rule, i.e., $\sum _{s=1}^{v}E_{i,t+s}$. So the net inventory after meeting the demands prior to period $t+1$ and the LCFP commitment will be $S_i-D_i[t-L_i+v,t] - \sum _{s=1}^{v}E_{i,t+s}$. Therefore, if $E_{i,t}<D_{i,t}$, the demand $D_{i,t}$ will be satisfied before or in period $t+v$ if and only if $S_i-D_i[t-L_i+v,t] - \sum _{s=1}^{v}E_{i,t+s} \ge 0$.

$\square$

Proof of Theorem 3

We need only to prove the first half of the claim, since the second half follows. Under the FCFS rule, the net inventory for demands in period $t$ in period $t+w_j$ is $S_i-D_i[t-L_i+w_j,t-1]$. The PTW rule will lead to some addition and subtraction to this quantity. Consider the net inventory just before satisfying the product $j$ demand $a_{ij}P_{j,t}$. For any product $k$ with higher priority, i.e., $k<j$, and a demand $a_{ik}P_{k,s}$ that arrives in a period $s$ at or after period $t(s \ge t)$, if $s+w_k \le s+w_j$, then we should satisfy the demand $a_{ik}P_{k,s}$ before $a_{ij}P_{j,t}$. Therefore, we subtract this value $a_{ik}P_{k,t}$ from $S_i-D_i[t-L_i+w_j,t-1]$. On the other hand, consider any product $k$ with lower priority i.e., $k>j$, and a demand $a_{ik}P_{k,s}$ that arrives before period $t(s < t)$. If $s+w_k \ge s+w_j$, then we should satisfy the demand $a_{ik}P_{k,s}$ after $a_{ij}P_{j,t}$. Therefore, we add this value to $S_i-D_i[t-L_i+w_j,t-1]$. Counting all the additions and subtractions, we will obtain the net inventory just before satisfying the demand $a_{ij}P_{j,t}$ in period $t+w_j$ as in the theorem.$\square$

Proof of Theorem 4

Note that both the value $\sum _{k: k < j}\sum _{s: s\ge t, s+w_k \le t+w_j}a_{ik}P_{k,s}$ and the value $\sum _{k: k>j}\sum _{s: s<t, s+w_k \ge t+w_j}a_{ik}P_{k,s}$ represent the inventory commitment quantities violating the FCFS rule. Under the FCFS rule, the demand $\sum _{k: k < j}\sum _{s: s\ge t, s+w_k \le t+w_j}a_{ik}P_{k,s}$ should be met no earlier than $a_{ij}P_{j,t}$, which is violated under the PTW rule; the demand $\sum _{k: k>j}\sum _{s: s<t, s+w_k \ge t+w_j}a_{ik}P_{k,s}$ should be met earlier than $a_{ij}P_{j,t}$, which is also violated under the PTW rule. When we consider period $t+w_j+\delta _j$, these amounts do not change with respect to $\delta _j$. Therefore, in formulation (2), they are still the same as in Theorem 3. However, there are new order arrivals since period $t+w_j$, therefore $S_i-D_i[t-L_i+w_j,t-1]$ changes to $S_i-D_i[t-L_i+w_j+\delta _j,t-1]$. Combining these, formulation (2) is obtained.$\square$

Proof of Theorem 5

The random variables $X^j_{it}$ are the linear combinations of random variables with a multivariate normal distribution. Therefore, for a given $j$, the $X^j_{it}$’s also follow a multivariate normal distribution (see Theorem 1 in Zhang (1997)). Note the multivariate normal distribution has a log-concave density function (see Theorem 4.2.1 in Tong (1988)). According to the Theorem 2 in Prkopa (1980), the constraint for a single $j$ in the above formulation defines a convex set of $S$. Since the intersection of convex sets is still convex, the conclusion follows.$\square$

Proof of Theorem 6

We must show that any feasible component allocation decision that follows the FCFS rule must satisfy all the constraints in (6), and vice versa.

Firstly, note that $O_i^{L_i+1}=D_{i,t}$, which implies that the demands in period $t$ for component $i$ can be satisfied no later than period $t+L_i+1$. So we can claim that the demands in period $t$ will be met no later than period $t+L+1$. This is the equation $\sum _{k=0}^{L+1}x_{jk}=P_{j,t}$. Secondly, the assembly in any period $t+k$ can not use more on-hand inventory than available, i.e., $\sum _{\mu =0}^{k}\sum _{j=1}^{n}a_{ij}x_{j\mu } \le O^k_i$. According to FCFS, an earlier arrival of component supply will always be used to satisfy an earlier arrival of product demand. Therefore, whenever $O^k_i=D_{i,t}$, we must have $\sum _{\mu =0}^{k}\sum _{j=1}^{n}a_{ij}x_{j\mu } = D_{i,t}$. Thirdly, we assume that all the demands are integer numbers, and the bill-of-materials matrix only contains integer entries, so we have $x_{jk} \in \mathbb {Z}_{+}$ for all $x_{jk}$.

On the other hand, if $x \in X$, the definition of $O^k_i$’s will guarantee that the component allocation for the demands in period $t$ will be made after the demands of previous periods are met. The constraints $\sum _{\mu =0}^{k^*}\sum _{j=1}^{n}a_{ij}x_{j\mu } = D_{i,t}$ and the definition of $k^*$ will guarantee that the demands in period $t$ will be satisfied at the earliest possible time, so that the demands in period $t$ will always be met before any demand afterwards can be satisfied. Therefore, any $x \in X$ defines a feasible component allocation decision that follows FCFS.$\square$

Proof of Theorem 7

When $S \rightarrow \infty$, according to the definition of $O^k_i,\,O^0_i \rightarrow D_{i,t}$ almost surely. When $O^0_i = D_{i,t}$, we have $k^*=0$, and according to the first and third constraints in (6), we have $x_{jk} \equiv 0$ for all $k>0$ and $x_{j0} = P_{j,t}$.$\square$

Proof of Corollary 2

Indeed, according to Theorem 7, when $S \rightarrow \infty$, we have $X \rightarrow \{x^0\}$ almost surely, which implies no stock-out in period $t$ (for classical type I service level, the time window equals 0). Therefore the conclusion follows.$\square$

Proof of Corollary 4

According to Theorem 7, when $S \rightarrow \infty$, we have $X \rightarrow \{x^0\}$ almost surely. This implies that the second and third terms (the remnant stock holding cost and the backlogging cost) in $f_3(S,\xi (\omega ),x)$ will become zero due to the definition of $x^0$. Note that the first term (the classical inventory holding cost) is irrelevant to the component allocation decisions. Therefore the conclusion follows.$\square$

1.1 Numerical experiment setting

In the component allocation problem, the base stock levels are input data. We generate the base stock levels by using a uniform safety factor for all the components. Specifically, the base stock level of component $i$ is set to $S_i=\sum _{j=1}^n a_{ij}\mu _j (L_i+1)+ k^*\sqrt{ \sum _{j=1}^n a_{ij}(L_i+1)\sigma _j^2}$, where $k^*$ is the uniform safety factor, and $\mu _j$ and $\sigma _j$ are the expectation and standard deviation of the single period demand of product $j$ separately. We increase the safety factor from 0 to 4, and set the step size as 0.5. So for each component allocation rule, we need to collect data for 9 safety factors.

The performance measures of the component allocation rules in Table 2 are obtained by simulation. We use a warmup period length of 1000. For each simulation, 30 batches are used, and in each batch, the simulation is run for 5000 consecutive periods. In the figures, each data point is the sample mean of the batch data. Note that the warmup period length is tested by a Welch graph method, and the batch independence is tested by both the Von Neumann test and the runs test. All the tests show that the simulation parameters are appropriate (cf. Law and Kelton 2002).

To obtain the benchmark data, we solve the mathematical programs (7), (8), (9) proposed in Sect. 5 for 5000 realizations. For each mathematical program, we collect the sample mean and sample standard deviation of the optimal objectives. In the figures, we use the sample means as the data points. The sample deviation data show that the realization number 5000 can guarantee a negligible statistical variance in the sample mean data.

Throughout the experiment, we use the ANSI C programming language, by incorporating the CPLEX12.4 Callable Library optimization package to solve (7), (8), and (9). In all the computations, we use a Linux workstation with dual 2.4GHz Intel Xeon processors and 2GB RAM. The running times of the simulations and the mathematical programs are collected. The original data are available upon request.