Single-machine serial-batch delivery scheduling with two competing agents and due date assignment

Abstract

We consider a set of single-machine batch delivery scheduling problems involving two competing agents under two due date assignment models. Belonging to one of the two agents, each job is processed and delivered in a batch to its agent, where the jobs in each batch come from the same agent. The jobs in a batch are processed sequentially and the processing time of a batch is equal to the sum of the processing times of the jobs in it. A setup time is required at the start of each batch. The dispatch date of a job equals the delivery date of the batch it is in, i.e., the completion time of the last job in the batch. There is no capacity limit on each delivery batch, and the cost per batch delivery is fixed and independent of the number of jobs in the batch. The due date of each job is a decision variable, which is to be assigned by the decision maker using one of two due date models, namely the common and unrestricted due date models. Given the due date assignment model, the overall objective is to minimize one agent’s scheduling criterion, while keeping the other agent’s criterion value from exceeding a threshold given in advance. Two kinds of scheduling criteria are involved: (i) the total cost comprising the earliness, tardiness, job holding, due date assignment, and batch delivery costs; and (ii) the total cost comprising the earliness, weighted number of tardy jobs, job holding, due date assignment, and batch delivery costs. For each of the problems considered, we show that it is $\mathcal {NP}$-hard in the ordinary sense and admits a fully polynomial-time approximation scheme.

Appendix: Proof of Lemma 5.4

Proof of Lemma 5.4

We prove the lemma by induction on j and k, respectively. It is clear that the lemma holds for (1, 0) or (0, 1). As the induction hypothesis, for any given k, we first assume that the lemma holds for any $j=l-1$, i.e., for any eliminated label $(x,y,e_A,e_B,a_A,a_B,c_A,c_B,f^A,f^B)\in \mathcal {S}(j-1,k)$, there exists a label $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A, \widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ such that $\widetilde{a}_A\ge a_A,\widetilde{a}_B\ge a_B,\widetilde{c}_A\le c_A,\widetilde{c}_B\le c_B$, $\widetilde{f}^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B\le f^B$. We show that the lemma holds for $j=l$.

Consider an arbitrary label $(x,y,e_A,e_B,a_A,a_B,c_A,c_B,f^A,f^B)\in \mathcal {S}(j,k)$, extended from a label in $\mathcal {S}(j-1,k)$. Then $f^B\le U^B$ must hold by the constraint. First, we assume that job $J_j^A$ is scheduled as early in the corresponding partial schedule. While implementing Algorithm $ConTAA_{\varepsilon }$, two cases need to be considered.

Case 1 $x=1$. In this case, the label $(x,y,e_A,e_B,a_A,a_B,c_A,c_B,f^A,f^B)$ is constructed from $(0,y,0,e_B,a_A,a_B,c_A,c_B,f^A-n_A\gamma ^Ap_j^A-s^A-\varphi ^A,f^B)\in \mathcal {S}(j-1,k)$ if $e_A=1$ and $e_B=0$, and from $(1,y,e_A-1,e_B,a_A,a_B,c_A,c_B,f^A-(n_A\gamma ^A+e_A\theta ^A)p_j^A,f^B)\in \mathcal {S}(j-1,k)$ if $e_A>1$. When $e_A=1$ and $e_B=0$, according to the induction hypothesis, there exists a label $(0,y,0,e_B,\widehat{a}_A,\widehat{a}_B,\widehat{c}_A,\widehat{c}_B,\widehat{f}^A,\widehat{f}^B)\in \mathcal {S}(j-1,k)$ such that $\widehat{a}_A\ge a_A,\widehat{a}_B\ge a_B,\widehat{c}_A\le c_A,\widehat{c}_B\le c_B$, $\widehat{f}^A\le \Delta ^{j+k-1}(f^A-n_A\gamma ^Ap_j^A-s^A-\varphi ^A)$, and $\widehat{f}^B\le f^B\le U^B$. Then the label $(x,y,e_A,e_B,\widehat{a}_A,\widehat{a}_B,\widehat{c}_A,\widehat{c}_B,\widehat{f}^A+n_A\gamma ^Ap_j^A+s^A+\varphi ^A,\widehat{f}^B)$ will be generated during the label generation process. However, this label may be eliminated by another label $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A,\widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ through the elimination rules in Algorithm $ConTAA_{\varepsilon }$. If it is eliminated by $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A,\widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ through elimination rule (1), we have $\widetilde{a}_A\ge \widehat{a}_A\ge a_A,\widetilde{a}_B\ge \widehat{a}_B\ge a_B,\widetilde{c}_A\le \widehat{c}_A\le c_A,\widetilde{c}_B\le \widehat{c}_B\le c_B$, $\widetilde{f}^A\le \Delta (\widehat{f}^A+n_A\gamma ^Ap_j^A+s^A+\varphi ^A)\le \Delta (\Delta ^{j+k-1}(f^A-n_A\gamma ^Ap_j^A-s^A-\varphi ^A)+n_A\gamma ^Ap_j^A+s^A+\varphi ^A)\le \Delta \Delta ^{j+k-1}f^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B\le \widehat{f}^B\le f^B\le U^B$. Otherwise, we have $\widetilde{a}_A\ge \widehat{a}_A\ge a_A,\widetilde{a}_B\ge \widehat{a}_B\ge a_B,\widetilde{c}_A\le \widehat{c}_A\le c_A,\widetilde{c}_B\le \widehat{c}_B\le c_B$, $\widetilde{f}^A\le \widehat{f}^A\le \Delta ^{j+k-1}f^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B=\widehat{f}^B\le f^B\le U^B$. Similarly, the result also holds for the case where $e_A>1$.

Case 2 $x=2$. In this case, the label $(x,y,e_A,e_B,a_A,a_B,c_A,c_B,f^A,f^B)$ is constructed from $(2,y,0,e_B,a_A,a_B,c_A,c_B,f^A-n_A\gamma ^Ap_j^A-s^A-\varphi ^A,f^B-n_B\gamma ^Bp_j^A)\in \mathcal {S}(j-1,k)$ if $e_A=1$, and from $(2,y,e_A-1,e_B,a_A,a_B,c_A,c_B,f^A-(n_A\gamma ^A+e_A\theta ^A)p_j^A,f^B-n_B\gamma ^Bp_j^A)\in \mathcal {S}(j-1,k)$ if $e_A>1$. Analogous to proof of Case 1, there exists a label $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A,\widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ such that $\widetilde{a}_A\ge a_A,\widetilde{a}_B\ge a_B,\widetilde{c}_A\le c_A,\widetilde{c}_B\le c_B$, $\widetilde{f}^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B\le f^B\le U^B$.

It follows that the induction hypothesis holds for $j=l$ when job $J_j^A$ is scheduled as early in the corresponding partial schedule.

Now, we turn to the case where job $J_j^A$ is scheduled as tardy in the corresponding partial schedule. While implementing Algorithm $ConTAA_{\varepsilon }$, two cases need to be considered.

Case 3 $a_A=1$. In this case, the label $(x,y,e_A,e_B,a_A,a_B,c_A,c_B,f^A,f^B)$ is constructed from $(x,0,e_A,e_B,c,a_B,c,c_B,f^A-\beta ^A(c_A+l-1-e_A)p_j^A-s^A-\varphi ^A,f^B-\beta ^B(c_B+k-e_B-a_B)p_j^A)\in \mathcal {S}(j-1,k)$ if $e_A=l-1$ and $e_B=k$, and from $(x,y,e_A,e_B,c,a_B,c,c_B,f^A-\beta ^A(c_A+l-1-e_A)p_j^A-s^A-\varphi ^A,f^B-\beta ^B(c_B+k-e_B-a_B)p_j^A)\in \mathcal {S}(j-1,k)$ if $e_A<l-1$ or $e_B<k$ for some $1\le c\le c_A$. When $e_A=l-1$ and $e_B=k$, according to the induction hypothesis, there exists a label $(x,0,e_A,e_B,\widehat{a}_A,\widehat{a}_B,\widehat{c}_A,\widehat{c}_B,\widehat{f}^A,\widehat{f}^B)\in \mathcal {S}(j-1,k)$ such that $\widehat{a}_A\ge c,\widehat{a}_B\ge a_B,\widehat{c}_A\le c,\widehat{c}_B\le c_B$, $\widehat{f}^A\le \Delta ^{j+k-1}(f^A-\beta ^A(c_A+l-1-e_A)p_j^A-s^A-\varphi ^A)$, and $\widehat{f}^B\le f^B-\beta ^B(c_B+k-e_B-a_B)p_j^A)$, implying that $\widehat{a}_A=\widehat{c}_A=c$ since $\widehat{a}_A\le \widehat{c}_A$. Since $\widehat{f}^B+\beta ^B(c_B+k-e_B-a_B)p_j^A\le f^B\le U^B$, the label $(x,y,e_A,e_B,1,\widehat{a}_B,c_A,\widehat{c}_B,\widehat{f}^A+\beta ^A(c_A+l-1-e_A)p_j^A+s^A+\varphi ^A,\widehat{f}^B+\beta ^B(c_B+k-e_B-a_B)p_j^A)$ will be generated during the label generation process since $c\le c_A$. However, this label may be eliminated by another label $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A,\widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ through the elimination rules in Algorithm $ConTAA_{\varepsilon }$. If it is eliminated by $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A,\widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ through elimination rule (1), we have $\widetilde{a}_A\ge 1= a_A,\widetilde{a}_B\ge \widehat{a}_B\ge a_B,\widetilde{c}_A\le \widehat{c}_A=c\le c_A,\widetilde{c}_B\le \widehat{c}_B\le c_B$, $\widetilde{f}^A\le \Delta (\widehat{f}^A+\beta ^A(c_A+l-1-e_A)p_j^A+s^A+\varphi ^A)\le \Delta (\Delta ^{j+k-1}(f^A-\beta ^A(c_A+l-1-e_A)p_j^A-s^A-\varphi ^A)+\beta ^A(c_A+l-1-e_A)p_j^A+s^A+\varphi ^A)\le \Delta \Delta ^{j+k-1}f^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B\le \widehat{f}^B+\beta ^B(c_B+k-e_B-a_B)p_j^A\le f^B\le U^B$. Otherwise, we have $\widetilde{a}_A\ge 1= a_A,\widetilde{a}_B\ge \widehat{a}_B\ge a_B,\widetilde{c}_A\le c_A,\widetilde{c}_B\le \widehat{c}_B\le c_B$, $\widetilde{f}^A\le \widehat{f}^A\le \Delta ^{j+k-1}f^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B=\widehat{f}^B\le f^B\le U^B$. Similarly, the result also holds for the case where $e_A<l-1$ or $e_B<k$.

Case 4 $a_A>1$. In this case, the label $(x,y,e_A,e_B,a_A,a_B,c_A,c_B,f^A,f^B)$ is constructed from $(x,y,e_A,e_B,a_A-1,a_B,c_A,c_B,f^A-\beta ^A(c_A+l-e_A)p_j^A-a_A\theta ^Ap_j^A,f^B-\beta ^B(c_B+k-e_B-a_B)p_j^A)\in \mathcal {S}(j-1,k)$ if $y=1$, and from $(x,y,e_A,e_B,a_A-1,a_B,c_A,c_B,f^A-\beta ^A(c_A+l-e_A)p_j^A-a_A\theta ^Ap_j^A,f^B-\beta ^B(k-e_B-a_B)p_j^A)\in \mathcal {S}(j-1,k)$ if $y=2$. Analogous to proof of Case 3, there exists a label $(x,y,e_A,e_B,\widetilde{a}_A,\widetilde{a}_B,\widetilde{c}_A,\widetilde{c}_B,\widetilde{f}^A,\widetilde{f}^B)$ such that $\widetilde{a}_A\ge a_A,\widetilde{a}_B\ge a_B,\widetilde{c}_A\le c_A,\widetilde{c}_B\le c_B$, $\widetilde{f}^A\le \Delta ^{j+k}f^A$, and $\widetilde{f}^B\le f^B\le U^B$.

It follows that the induction hypothesis holds for $j=l$ when job $J_j^A$ is scheduled as tardy in the corresponding partial schedule.

In a similar way, we can show that for any given j, the result holds for $k=l$ when it is valid for any $k=l-1$.

Thus, the induction hypothesis holds in each case and the result follows. $\square$