Single machine scheduling with two competing agents, arbitrary release dates and unit processing times

Abstract

We study various single machine scheduling problems with two competing agents, unit processing times and arbitrary integer release dates. The problems differ by the scheduling criterion used by each of the two agents, and by the variant of the bicriteria problem that has to be solved. We prove that when the scheduling criterion of either one of the two agents is of a max-type, then all considered variants of the bicriteria problem are solvable in polynomial time. However, when the two agents have a sum-type of scheduling criterion, several variants of the bicriteria problem become $\mathcal {NP}$-hard.

Appendices

Appendix 1: A numerical example that illustrates the implementation of Algorithm 1

Consider an instance with five jobs in set $\mathcal {J}^{(2)}$, where the job’s parameters are given in Table 4 below.

Table 4 Agent’s 2 job’s parameters

Below, we implement Algorithm 1 on job set $\mathcal {J}^{(2)}$ to construct our premier schedule.

[Initialization] $D=\emptyset$

• Step 1: The jobs in $\mathcal {J}^{(2)}$ are already numbered according to the ERD rule.

• Step 2: The deadlines are given in Table 4, and at the end of this step we have that $D=(D_{[1]},D_{[2]},D_{[3]},D_{[4]})=(2,3,9,10)$. Moreover, we set $D_{[0]}=0$.

• Step 3: We set $\mathcal {I}_{1}=\{1,2,3\}$; $\mathcal {I}_{2}=\{1,2\}$; $\mathcal {I}_{3}=\mathcal {I}_{5}=\{6,7,8,9,10\}$; and $\mathcal {I}_{4}=\{5,6,7,8,9\}$. Thus, $\mathcal {I}=\{0,1,2,3,5,6,7,8,9,10\}$ and $\mathcal {L}(k)=\emptyset$ for any $k\in \mathcal {I}$.

• Step 4: At the end of this step we have $\mathcal {L}(2)=\{J_{2}^{(2)}\}$, $\mathcal {L}(3)=\{J_{1}^{(2)}\}$, $\mathcal {L}(9)=\{J_{4}^{(2)}\}$ and $\mathcal {L}(10)=\{J_{5}^{(2)},J_{3}^{(2)}\}$.

• Step 5: We set $k=D_{[4]}=10$ and $j=4$.

• Step 6:

  • Iteration 1: Job $J_{5}^{(2)}$ is at the head of $\mathcal {L}(k=10)$. Thus, $h=5$. Since $k-1=9\ge r_{5}^{(2)}=8$, we set $S_{5}^{(2)}=10$ and $\mathcal {L}(10)=\mathcal {L}(10)\diagdown J_{5}^{(2)}=\{J_{3}^{(2)}\}$. Since $\mathcal {L}(10)\ne \emptyset$ and $\mathcal {L}(9)\ne \emptyset$, we set $j=3$. Then, we merge $\mathcal {L}(10)$ into $\mathcal {L}(9)$ such that $\mathcal {L}(9)=\{J_{4}^{(2)},J_{3}^{(2)}\}$. Since $\mathcal {L}(9)\ne \emptyset$, we set $k=9$.

  • Iteration 2: Job $J_{4}^{(2)}$ is at the head of $\mathcal {L}(k=9)$. Thus, $h=4$. Since $k-1=8\ge r_{4}^{(2)}=8$, we set $S_{4}^{(2)}=9$ and $\mathcal {L}(9)=\mathcal {L}(9)\diagdown J_{4}^{(2)}=\{J_{3}^{(2)}\}$. Since $\mathcal {L}(9)\ne \emptyset$ and $\mathcal {L(}8)=\emptyset$, we do not update the value of j. Then, we merge $\mathcal {L}(9)$ into $\mathcal {L}(8)$ such that $\mathcal {L}(8)=\{J_{3}^{(2)}\}$. Since $\mathcal {L}(8)\ne \emptyset$, we set $k=8$.

  • Iteration 3: Job $J_{3}^{(2)}$ is at the head of $\mathcal {L}(k=8)$. Thus, $h=3$. Since $k-1=7\ge r_{3}^{(2)}=6$, we set $S_{3}^{(2)}=8$ and $\mathcal {L}(8)=\mathcal {L}(8)\diagdown J_{3}^{(2)}=\emptyset$. Since $\mathcal {L}(8)=\emptyset$, we set $j=2$. Then, we merge $\mathcal {L}(8)$ into $\mathcal {L}(7)$ such that $\mathcal {L}(7)=\emptyset$. Since $\mathcal {L}(7)=\emptyset$, we set $k=D_{[2]}=3$.

  • Iteration 4: Job $J_{1}^{(2)}$ is at the head of $\mathcal {L}(k=3)$. Thus, $h=1$. Since $k-1=2\ge r_{1}^{(2)}=0$, we set $S_{1}^{(2)}=3$ and $\mathcal {L}(3)=\mathcal {L}(3)\diagdown J_{1}^{(2)}=\emptyset$. Since $\mathcal {L}(3)=\emptyset$, we set $j=1$. Then, we merge $\mathcal {L}(3)$ into $\mathcal {L}(2)$ such that $\mathcal {L}(2)=\{J_{2}^{(2)}\}$. Since $\mathcal {L}(2)\ne \emptyset$, we set $k=2$.

  • Iteration 5: Job $J_{2}^{(2)}$ is at the head of $\mathcal {L}(k=2)$. Thus, $h=2$. Since $k-1=1\ge r_{2}^{(2)}=1$, we set $S_{2}^{(2)}=2$ and $\mathcal {L}(2)=\mathcal {L}(2)\diagdown J_{2}^{(2)}=\emptyset$. Since $\mathcal {L}(2)=\emptyset$, we set $j=0$. Then, we merge $\mathcal {L}(2)$ into $\mathcal {L}(1)$ such that $\mathcal {L}(1)=\emptyset$. Since $\mathcal {L}(1)=\emptyset$, we set $k=D_{[0]}=0$.

•  [Output] $S^{(2)}=(S_{1}^{(2)},S_{2}^{(2)},\ldots ,S_{n_{2}}^{(2)})=(3,2,8,9,10)$, where $S_{j}^{(2)}$ is the time slot in which $J_{j}^{(2)}$ is scheduled in the premier solution. Figure 1 below illustrates the resulting premier solution (schedule).

Fig. 1

The premier schedule

Appendix 2: A numerical example that illustrates the implementation of Algorithm 2

In addition to the instance provided in Appendix  2 for the jobs in set $\mathcal {J}^{(2)}$, we include also three jobs that belongs to agent 1, where the job’s parameters are given in Table 5 below.

Table 5 Agent’s 1 job parameters

In order to construct an optimal schedule for the corresponding $1\left| {\textit{CO}},r_{j}^{(i)},p_{j}^{(i)}=1\right| \epsilon (f_{\Sigma }^{(1)}/f_{\max }^{(2)})$ problem, we next implement Algorithm 2 on the entire job set $\mathcal {J}^{(1)}\cup \mathcal {J}^{(2)}$.

[Input]: $S^{(2)}=(S_{1}^{(2)},S_{2}^{(2)},S_{3}^{(2)},S_{4}^{(2)},S_{5}^{(2)})=(3,2,8,9,10)$ {the output of Phase 1}.

[Initialization] Set $k=1$.

Step 1: The jobs in $\mathcal {J}^{(1)}$ are already sorted according to the ERD rule. We also sort the jobs of the second agent in a non-decreasing order of $S_{j}^{(2)}$ such that $\mathcal {J}^{(2)}=\{J_{2}^{(2)},J_{1}^{(2)},J_{3}^{(2)},J_{4}^{(2)},J_{5}^{(2)}\}$.

Step 2:

• Iteration 1: Jobs $J_{1}^{(1)}$ and $J_{2}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=0\ge \min \{r_{1}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=0$ we do not reset the value of k. Since $S_{h}^{(2)}=S_{2}^{(2)}=2\ne k=1$ and that $k-1=0<r_{h}^{(1)}=r_{1}^{(1)}=1$, Case 3 holds. Since $J_{1}^{(2)}$ is the first job in $\mathcal {J}^{(2)}$ with a release date not later than $k-1=0$, we set $S_{1}^{(2)}=1$, $\mathcal {J}^{(2)}=\{J_{2}^{(2)},J_{3}^{(2)},J_{4}^{(2)},J_{5}^{(2)}\}$ and $k=2$.

• Iteration 2: Jobs $J_{1}^{(1)}$ and $J_{2}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=1\ge \min \{r_{1}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=1$, we do not reset the value of k. Since $S_{h}^{(2)}=S_{2}^{(2)}=2=k$, Case 1 holds. Thus, we set $\mathcal {J}^{(2)}=\mathcal {J}^{(2)}\diagdown J_{h}^{(2)}=\{J_{3}^{(2)},J_{4}^{(2)},J_{5}^{(2)}\}$ and $k=3$.

• Iteration 3: Jobs $J_{1}^{(1)}$ and $J_{3}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=2\ge \min \{r_{1}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=1$, we do not reset the value of k. Since $S_{h}^{(2)}=S_{3}^{(2)}=8\ne k=3$ and $k-1=2>r_{h}^{(1)}=r_{1}^{(1)}=1$, Case 2 holds. Thus, we set $S_{1}^{(1)}=3$, $\mathcal {J}^{(1)}=\mathcal {J}^{(1)}\diagdown J_{h}^{(1)}=\{J_{2}^{(1)},J_{3}^{(1)}\}$ and $k=4$.

• Iteration 4: Jobs $J_{2}^{(1)}$ and $J_{3}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=3\ge \min \{r_{2}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=3$, we do not reset the value of k. Since $S_{h}^{(2)}=S_{3}^{(2)}=8\ne k=4$ and $k-1=3=r_{h}^{(1)}=r_{2}^{(1)}=3$, Case 2 holds. Thus, we set $S_{2}^{(1)}=4$, $\mathcal {J}^{(1)}=\mathcal {J}^{(1)}\diagdown J_{h}^{(1)}=\{J_{3}^{(1)}\}$ and $k=5$.

• Iteration 5: Jobs $J_{3}^{(1)}$ and $J_{3}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=4<\min \{r_{3}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=6$, we set $k=7$. Since $S_{3}^{(2)}=8\ne k=7$ and $k-1=6<r_{3}^{(1)}=10$, Case 3 holds. Since $J_{3}^{(2)}$ is the first job in $\mathcal {J}^{(2)}$ with a release date not later than $k-1=6$, we set $S_{3}^{(2)}=7$, $\mathcal {J}^{(2)}=\{J_{4}^{(2)},J_{5}^{(2)}\}$ and $k=8$.

• Iteration 6: Jobs $J_{3}^{(1)}$ and $J_{4}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=7<\min \{r_{3}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=8$, we set $k=9$. Since $S_{h}^{(2)}=S_{4}^{(2)}=9=k$, Case 1 holds. Thus, we set $\mathcal {J}^{(2)}=\mathcal {J}^{(2)}\diagdown J_{h}^{(2)}=\{J_{5}^{(2)}\}$ and $k=10$.

• Iteration 7: Jobs $J_{3}^{(1)}$ and $J_{5}^{(2)}$ are at the head of $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$, respectively. Since $k-1=9>\min \{r_{3}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{r_{j}^{(2)}\}\}=8$, we do not reset the value of k. Since $S_{h}^{(2)}=S_{4}^{(2)}=10=k$, Case 1 holds. Thus, we set $\mathcal {J}^{(2)}=\mathcal {J}^{(2)}\diagdown J_{h}^{(2)}=\emptyset$ and $k=11$.

• Iteration 8: Job $J_{3}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$. Since $k-1=10=\min \{r_{3}^{(1)},{\min _{J_{j}^{(2)}\in \mathcal {J}^{(2)}}}\{ r_{j}^{(2)}\} \}=10$ we do not reset the value of k. Since $\mathcal {J}^{(2)}=\emptyset$ and $k-1=10=r_{3}^{(1)}=10$, Case 2 holds. Thus, we set $S_{3}^{(1)}=11$, $\mathcal {J}^{(1)}=\emptyset$ and $k=12$. Since $\mathcal {J}^{(1)}\cup \mathcal {J}^{(2)}=\emptyset$, the entire set of jobs is scheduled.

Fig. 2

The optimal schedule for the numerical example

[Output] $S^{(1)}=(S_{1}^{(1)},S_{2}^{(1)},S_{3}^{(1)})=(3,4,11)$ and $S^{(2)}=(S_{1}^{(2)},S_{2}^{(2)},S_{3}^{(2)},S_{4}^{(2)},S_{5}^{(2)})=(1,2,7,9,10)$. Figure 2 below illustrates the resulting schedule.

Appendix 3: A numerical example that illustrates the implementation of Algorithm 4

Consider an instance of the $1\left| {\textit{CO}},r_{j}^{(i)},p_{j}^{(i)}=1\right| \#(\Sigma C_{j}^{(1)},\Sigma C_{j}^{(2)})$ problem, where $n_{1}=5$ and $n_{2}=3$. The release dates of the jobs in $\mathcal {J}^{(1)}$ and $\mathcal {J}^{(2)}$ are given in Tables 6 and 7 below.

Table 6 Agent’s 1 job parameters

Table 7 Agent’s 2 job parameters

Below, we implement Algorithm 4 to find the entire set of Pareto-optimal solutions.

Step 1: We call Procedure “obtaining Pareto-optimal solution $\pi _{1}$” (Algorithm 3) and below we present the implementation of this procedure on the above numerical example.

[Initialization] The jobs in each $\mathcal {J}^{(i)}$ are already numbered according to the ERD rule. We set $\pi _{1}=\emptyset$, $F^{(1)}(\pi _{1})=0$, $F^{(2)}(\pi _{1})=0$ and $r=k_{1}=1$.

[Core]

• Iteration 1: Job $J_{1}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{1}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{1}-1=0\ge \min \{r_{1}^{(1)},r_{1}^{(2)}\}=0$, we do not reset the value of $k_{1}$. Since $k_{1}-1=0\ge r_{1}^{(1)}=0$, we set $\pi _{1}=\{J_{1}^{(1)}\}$, $F^{(1)}(\pi _{1})=1$, $\mathcal {J}^{(1)}=\{J_{2}^{(1)},J_{3}^{(1)},J_{4}^{(1)},J_{5}^{(1)}\}$ and $r=k_{2}=2$.

• Iteration 2: Job $J_{2}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{1}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{2}-1=1\ge \min \{r_{2}^{(1)},r_{1}^{(2)}\}=0$, we do not reset the value of $k_{2}$. Since $k_{2}-1=1\ge r_{2}^{(1)}=0$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)}\}$, $F^{(1)}(\pi _{1})=1+2=3$, $\mathcal {J}^{(1)}=\{J_{3}^{(1)},J_{4}^{(1)},J_{5}^{(1)}\}$ and $r=k_{3}=3$.

• Iteration 3: Job $J_{3}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{1}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{3}-1=2\ge \min \{r_{3}^{(1)},r_{1}^{(2)}\}=1$, we do not reset the value of $k_{3}$. Since $k_{3}-1=2\ge r_{3}^{(1)}=1$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)}\}$, $F^{(1)}(\pi _{1})=3+3=6$, $\mathcal {J}^{(1)}=\{J_{4}^{(1)},J_{5}^{(1)}\}$ and $r=k_{4}=4$.

• Iteration 4: Job $J_{4}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{1}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{4}-1=3\ge \min \{r_{4}^{(1)},r_{1}^{(2)}\}=1$, we do not reset the value of $k_{4}$. Since $k_{4}-1=3<r_{4}^{(1)}=4$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)},J_{1}^{(2)}\}$, $F^{(2)}(\pi _{1})=4$, $\mathcal {J}^{(2)}=\{J_{2}^{(2)},J_{3}^{(2)}\}$ and $r=k_{5}=5$.

• Iteration 5: Job $J_{4}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{2}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{5}-1=4\ge \min \{r_{4}^{(1)},r_{2}^{(2)}\}=1$, we do not reset the value of $k_{5}$. Since $k_{5}-1=4\ge r_{4}^{(1)}=4$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)},J_{1}^{(2)},J_{4}^{(1)}\}$, $F^{(1)}(\pi _{1})=6+5=11$, $\mathcal {J}^{(1)}=\{J_{5}^{(1)}\}$ and $r=k_{6}=6$.

• Iteration 6: Job $J_{5}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{2}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{6}-1=5\ge \min \{r_{5}^{(1)},r_{2}^{(2)}\}=1$, we do not reset the value of $k_{6}$. Since $k_{6}-1=4<r_{5}^{(1)}=9$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)},J_{1}^{(2)},J_{4}^{(1)},J_{2}^{(2)}\}$, $F^{(2)}(\pi _{1})=4+6=10$, $\mathcal {J}^{(2)}=\{J_{3}^{(2)}\}$ and $r=k_{7}=7$.

• Iteration 7: Job $J_{5}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and job $J_{3}^{(2)}$ is at the head of $\mathcal {J}^{(2)}$. Since $k_{7}-1=6<\min \{r_{5}^{(1)},r_{3}^{(2)}\}=8$, we set $k_{7}=9$. Since $k_{7}-1=8<r_{5}^{(1)}=9$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)},J_{1}^{(2)},J_{4}^{(1)},J_{2}^{(2)},J_{3}^{(2)}\}$, $F^{(2)}(\pi _{1})=10+9=19$, $\mathcal {J}^{(2)}=\emptyset$, $r=8$ and $k_{8}=10$.

• Iteration 8: Job $J_{5}^{(1)}$ is at the head of $\mathcal {J}^{(1)}$ and there are no jobs in $\mathcal {J}^{(2)}$. Since $k_{8}-1=9\ge \min \{r_{5}^{(1)}\}=9$, we do not reset the value of $k_{8}$. Since $k_{8}-1=9\ge r_{5}^{(1)}=9$, we set $\pi _{1}=\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)},J_{1}^{(2)},J_{4}^{(1)},J_{2}^{(2)},J_{3}^{(2)},J_{5}^{(1)}\}$, $F^{(1)}(\pi _{1})=11+10=21$, $\mathcal {J}^{(1)}=\emptyset$ and $k_{9}=11$.

[Output] The triplet $(\pi _{1},F^{(1)}(\pi _{1}),F^{(2)}(\pi _{1}))=(\{J_{1}^{(1)},J_{2}^{(1)},J_{3}^{(1)},J_{1}^{(2)},J_{4}^{(1)},J_{2}^{(2)},J_{3}^{(2)},J_{5}^{(1)}\},21,19)$.

Figure 3 below illustrates the schedule represented by $\pi _{1}$ for the above numerical example.

Fig. 3

Schedule $\pi _{1}$

We set $\Pi _{e}=\{(\pi _{1},F^{(1)}(\pi _{1}),F^{(2)}(\pi _{1}))\}$, and $k=1$.

Step 2:

• Iteration 1 ($i_{2}=1$): Since $J_{3}^{(1)}$ is the latest job of agent 1 that precedes job $J_{1}^{(2)}$ in permutation $\pi _{1}$, we set $l=3$. Since $\min \{C_{1}^{(2)}(\pi _{1})-1-C_{0}^{(2)}(\pi _{1}),C_{1}^{(2)}(\pi _{1})-1-r_{1}^{(2)}\}=\min \{3,2\}=2$, we can perform only two pairwise interchanges with job $J_{1}^{(2)}$. The first with job $J_{l+1-1}^{(1)} =J_{3}^{(1)}$, which results in $\pi _{2}=\{J_{1}^{(1)},J_{2}^{(1)},J_{1} ^{(2)},J_{3}^{(1)},J_{4}^{(1)},J_{2}^{(2)},J_{3}^{(2)},J_{5}^{(1)}\}$ with $F^{(1)}(\pi _{2})=F^{(1)}(\pi _{1})+1=22$ and $F^{(2)}(\pi _{2})=F^{(2)}(\pi _{1})-1=18$ (we include the triplet $(\pi _{2},F^{(1)}(\pi _{2}),F^{(2)}(\pi _{2}))$ in $\Pi _{e}$) and the second with job $J_{l+1-2}^{(1)}=J_{2}^{(1)}$, which results in $\pi _{3}=\{J_{1}^{(1)},J_{1}^{(2)},J_{2}^{(1)},J_{3} ^{(1)},J_{4}^{(1)},J_{2}^{(2)},J_{3}^{(2)},J_{5}^{(1)}\}$ with $F^{(1)} (\pi _{3})=F^{(1)}(\pi _{2})+1=23$ and $F^{(2)}(\pi _{3})=F^{(2)}(\pi _{2})-1=17$ (we include the triplet $(\pi _{3},F^{(1)}(\pi _{3}),F^{(2)}(\pi _{3}))$ in $\Pi _{e}$). We set $k=k+2=3$.

• Iteration 2 ($i_{2}=2$): Since $J_{4}^{(1)}$ is the last job of agent 1 before $J_{2}^{(2)}$ in $\pi _{3}$, we have that $l=4$. Since $\min \{C_{2}^{(2)}(\pi _{3})-1-C_{1}^{(2)}(\pi _{3}),C_{2}^{(2)}(\pi _{3} )-1-r_{2}^{(2)}\}=\min \{3,4\}=3$, we can perform only three pairwise interchanges with job $J_{2}^{(2)}$. The first is with job $J_{l+1-1} ^{(1)}=J_{4}^{(1)}$, which results in $\pi _{4}=\{J_{1}^{(1)},J_{1}^{(2)} ,J_{2}^{(1)},J_{3}^{(1)},J_{2}^{(2)},J_{4}^{(1)},J_{3}^{(2)},J_{5}^{(1)}\}$ with $F^{(1)}(\pi _{4})=F^{(1)}(\pi _{3})+1=24$ and $F^{(2)}(\pi _{4} )=F^{(2)}(\pi _{3})-1=16$ (we include the triplet $(\pi _{4},F^{(1)}(\pi _{4}),F^{(2)}(\pi _{4}))$ in $\Pi _{e}$); the second is with job $J_{l+1-2} ^{(1)}=J_{3}^{(1)}$, which results in $\pi _{5}=\{J_{1}^{(1)},J_{1}^{(2)} ,J_{2}^{(1)},J_{2}^{(2)},J_{3}^{(1)},J_{4}^{(1)},J_{3}^{(2)},J_{5}^{(1)}\}$ with $F^{(1)}(\pi _{5})=F^{(1)}(\pi _{4})+1=25$ and $F^{(2)}(\pi _{5} )=F^{(2)}(\pi _{4})-1=15$ (we include the triplet $(\pi _{5},F^{(1)}(\pi _{5}),F^{(2)}(\pi _{5}))$ in $\Pi _{e}$); and the last is with job $J_{l+1-3}^{(1)}=J_{2}^{(1)}$, which results in $\pi _{6}=\{J_{1}^{(1)} ,J_{1}^{(2)},J_{2}^{(2)},J_{2}^{(1)},J_{3}^{(1)},J_{4}^{(1)},J_{3}^{(2)} ,J_{5}^{(1)}\}$ with $F^{(1)}(\pi _{6})=F^{(1)}(\pi _{5})+1=26$ and $F^{(2)} (\pi _{6})=F^{(2)}(\pi _{5})-1=14$ (we include the triplet $(\pi _{6},F^{(1)} (\pi _{6}),F^{(2)}(\pi _{6}))$ in $\Pi _{e}$). We set $k=k+3=6$.

• Iteration 3 ($i_{2}=3$): Since $J_{4}^{(1)}$ is the last job of agent 1 before $J_{3}^{(2)}$ in $\pi _{6}$, we have that $l=4$. Since $\min \{C_{3}^{(2)}(\pi _{6})-1-C_{2}^{(2)}(\pi _{6}),C_{3}^{(2)}(\pi _{6})-1-r_{3}^{(2)}\}=\min \{5,0\}=0$, we cannot perform any pairwise interchange with job $J_{3}^{(2)}$. We set $k=k+0=6$.

[Output] Set $\Pi _{e}=\{(\pi _{1},21,19),(\pi _{2},22,18),(\pi _{3},23,17),(\pi _{4},24,16),(\pi _{5},25,15),(\pi _{6},26,14)\}$ (Tables 1, 2, 3).