Two-agent preemptive Pareto-scheduling to minimize the number of tardy jobs and total late work

Abstract

We consider the single-machine preemptive Pareto-scheduling problem with two competing agents A and B, where agent A wants to minimize the number of its jobs (the A-jobs) that is tardy, while agent B wants to minimize the total late work of its jobs (the B-jobs). We provide an \(O(nn_{A}\log n_{A}+n_B\log n_B)\)-time algorithm that generates all the Pareto-optimal points, where \(n_A\) is the number of the A-jobs, \(n_B\) is the number of the B-jobs, and \(n=n_A+n_B\).

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Acknowledgements

We would like to thank the Associate Editor and two anonymous referees for their many helpful comments and suggestions. This research was supported in part by the NSFC under grant numbers 12071442 and 11771406.

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Correspondence to Jinjiang Yuan.

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Appendix

Appendix

Now we give an instance to demonstrate the execution of Algorithm 5.1. Let \({\mathcal {J}}=\{J^A_{1}, J^A_{2}, \ldots ,J^A_{6}, J^B_{1}, \ldots , J^B_{4}\}\) be the instance displayed in Table 1.

Table 1 The job instance \({\mathcal {J}}\)

Note that \(P^*=1+\max \{d^A_{n_A}, d^B_{n_B}\}=1+\max \{25,26\}=27\). Let \(\Omega =\Omega ({\mathcal {J}}^A, {\mathcal {J}}^B)\). The key steps in applying Algorithm 5.1 to solve the instance are as follows:

(i) Generate the schedule \(\sigma _0^B=(J_1^B, J_2^B, J_3^B, J_4^B)\) and calculate the Y-value \(Y^{(0)}=T_{\max }(\sigma _0^B)=1\). Then generate the schedule \(\sigma ^{B(Y^{(0)})}\), and the forbidden intervals \({\mathcal {I}}^{B(Y^{(0)})}=\{h_1,h_2,h_3,h_4,h_5\}\) is determined, where \(h_1=[0,4]\), \(h_2=[5,10]\), \(h_3=[12,18]\), \(h_4=[22,25]\), and \(h_5=[27,28]\). Then, for each \(y\in (Y^{(0)},P_B]=(1, 19]\), \(\sigma ^{B(y)}\) and \({\mathcal {I}}^{B(y)}\) can be easily generated. The forbidden intervals of \({\mathcal {I}}^{B(Y^{(0)})}={\mathcal {I}}^{B(1)}\) are displayed in Fig. 1.

Fig. 1
figure1

The forbidden intervals of \({\mathcal {I}}^{B(1)}\)

(ii) Generate the schedule \(\sigma ^{A(Y^{(0)})}=\sigma ^{A(1)}\) and calculate the U-value \(U^{(0)}=\sum U_j^A(\sigma ^{A(1)})=4\). Then \(\sigma ^{(1)}=(\sigma ^{A(1)}, \sigma ^{B(1)})\) is a Pareto-optimal schedule corresponding to \((4,1)\in \Omega \), as displayed in Fig. 2.

Fig. 2
figure2

Schedule \(\sigma ^{(1)}\) corresponding to \((4,1)\in \Omega \)

(iii) Calculate \(Y^{(1)}= Y^{(0)}+|h_1|=5\), generate \(\sigma ^{A(5)}\), and calculate the U-value \(U^{(1)}=U(5)=3\). The schedule \(\sigma ^{A(5)}\) is displayed in Fig. 3.

Fig. 3
figure3

Schedule \(\sigma ^{A(5)}\) corresponding to \(U^{(1)}=3\)

(iv) Now \(U^{(0)}=4>3=U^{(1)}\). Calculate \(Y(3)=4\) and generate \(\sigma ^{A(4)}\). Then \(\sigma ^{(4)}=(\sigma ^{A(4)}, \sigma ^{B(4)})\) is a Pareto-optimal schedule corresponding to \((3,4)\in \Omega \), as displayed in Fig. 4.

Fig. 4
figure4

Schedule \(\sigma ^{(4)}\) corresponding to \((3,4)\in \Omega \)

(v) Calculate \(Y^{(2)}= Y^{(1)}+|h_2|=10\), generate \(\sigma ^{A(10)}\), and calculate the U-value \(U^{(2)}=U(10)=2\). The schedule \(\sigma ^{A(10)}\) is displayed in Fig. 5.

Fig. 5
figure5

Schedule \(\sigma ^{A(10)}\) corresponding to \(U^{(2)}=2\)

(vi) Now \(U^{(1)}=3>2=U^{(2)}\). Calculate \(Y(2)=7\) and generate \(\sigma ^{A(7)}\). Then \(\sigma ^{(7)}=(\sigma ^{A(7)}, \sigma ^{B(7)})\) is a Pareto-optimal schedule corresponding to \((2,7)\in \Omega \), as displayed in Fig. 6.

Fig. 6
figure6

Schedule \(\sigma ^{(7)}\) corresponding to \((2,7)\in \Omega \)

(vii) Calculate \(Y^{(3)}= Y^{(2)}+|h_3|=16\), generate \(\sigma ^{A(16)}\), and calculate the U-value \(U^{(3)}=U(16)=0\). The schedule \(\sigma ^{A(16)}\) is displayed in Fig. 7.

Fig. 7
figure7

Schedule \(\sigma ^{A(16)}\) corresponding to \(U^{(3)}=0\)

(viii) Now \(U^{(2)}=2>0=U^{(3)}\). For \(u=U^{(2)}-1=1\), we calculate \(Y(u)=Y(1)=11\), and generate \(\sigma ^{A(11)}\). Then \(\sigma ^{(11)}=(\sigma ^{A(11)}, \sigma ^{B(11)})\) is a Pareto-optimal schedule corresponding to \((1,11)\in \Omega \), as displayed in Fig. 8.

Fig. 8
figure8

Schedule \(\sigma ^{(11)}\) corresponding to \((1,11)\in \Omega \)

(ix) For \(u=U^{(3)}=0\), calculate \(Y(u)=Y(0)=16\) and generate \(\sigma ^{A(16)}\). Then \(\sigma ^{(16)}=(\sigma ^{A(16)}, \sigma ^{B(16)})\) is a Pareto-optimal schedule corresponding to \((0,16)\in \Omega \), as displayed in Fig. 9.

Fig. 9
figure9

Schedule \(\sigma ^{(16)}\) corresponding to \((0,16)\in \Omega \)

(x) Finally, we conclude that \(\Omega =\{(4,1), (3,4), (2,7), (1,11),(0,16)\}\) and \(\sigma ^{(1)}, \sigma ^{(4)},\sigma ^{(7)},\sigma ^{(11)}\), and \(\sigma ^{(16)}\) are the corresponding Pareto-optimal schedules.

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He, R., Yuan, J., Ng, C.T. et al. Two-agent preemptive Pareto-scheduling to minimize the number of tardy jobs and total late work. J Comb Optim 41, 504–525 (2021). https://doi.org/10.1007/s10878-021-00697-2

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Keywords

  • Scheduling
  • Two agents
  • Pareto-scheduling
  • Number of tardy jobs
  • Total late work