The complexity of CO-agent scheduling to minimize the total completion time and total number of tardy jobs

Abstract

In this paper, we revisit a two-agent scheduling problem on a single machine. In this problem, we have two competing agents A and B, which means that the job set of agent A and the job set of agent B are disjoint. The objective is to minimize the total completion time of agent A, under the constraint that the total number of tardy jobs of agent B is no larger than a given bound. The complexity of this problem was posed as open in Agnetis et al. (Oper Res 52:229–242, 2004). Leung et al. (Oper Res 58:458–469, 2010a, b. https://doi.org/10.1287/opre.1090.0744ec) showed that the problem is binary NP-hard. However, their NP-hardness proof has a flaw. Here, we present a new NP-hardness proof for this problem. Our research shows that the problem is still NP-hard even if the jobs of agent A have a common processing time.

Appendix: Proof of Observation 2.2

Proof

The proof of Lemma 1 in Leung et al. (2010b) is based on the assumption that there exists a solution for the given instance of the even–odd partition, that is, there is a partition $(I_1, I_2)$ of the index set $\{1, 2, \ldots , 2n\}$ with $|I_1\cap \{2i-1, 2i\}|= 1$ and $|I_2\cap \{2i-1, 2i\}|= 1$ for each $i=1, 2, \ldots , n$ such that $\sum _{j\in I_1} a_j= \sum _{j\in I_2} a_j= H$. From Observation 2.1, the condition $\sum _{j\in I_1} a_j= \sum _{j\in I_2} a_j= H$ is equivalent to condition (C1). The aim of their Lemma 1 is to show that there is a schedule such that $\sum C_j^a\le TC$ and $\sum U_j^b\le n$. This is achieved by constructing a schedule $\pi$.

To guarantee that their constructed schedule $\pi$ satisfies $\sum U_j^b(\pi )\le n$, the R-job must be on time in the schedule. Then we have

$$\begin{aligned}&\mathop \sum \limits _{i:2i-1\in I_1} p_{2i-1}+\mathop \sum \limits _{i:2i\in I_1} p_{2i}+ \mathop \sum \limits _{i=1}^{n} x_{i}+ L\\&\quad = \mathop \sum \limits _{i:2i-1\in I_1}a_{2i-1}+\mathop \sum \limits _{i:2i\in I_1}(a_{2i}+(l_{i}-1)\sigma _i)\\&\qquad + \,\mathop \sum \limits _{i=1}^{n} x_{i}+ L \\&\quad = \mathop \sum \limits _{i=1}^{n}x_{i}+ H+ L+\mathop \sum \limits _{i:2i\in I_1} (l_{i}-1)\sigma _i \\&\quad = \mathop \sum \limits _{i=1}^{n}x_{i}+ H+ L+\mathop \sum \limits _{i:2i\in I_1} l_{i}\sigma _i\\&\qquad -\,\mathop \sum \limits _{i:2i\in I_1}\sigma _i+\mathop \sum \limits _{i:2i-1\in I_1} l_{i}\sigma _i - \mathop \sum \limits _{i:2i-1\in I_1} l_{i}\sigma _i \\&\quad = \mathop \sum \limits _{i=1}^{n} x_{i}+ H+ L+\mathop \sum \limits _{i=1}^{n}l_{i}\sigma _i\\&\qquad -\,\mathop \sum \limits _{i:2i-1\in I_1}l_{i}\sigma _i -\mathop \sum \limits _{i:2i\in I_1}\sigma _i \\&\quad = \mathop \sum \limits _{i=1}^{n} x_{i}+ H+ L+\mathop \sum \limits _{i=1}^{n}l_{i}\sigma _i\\&\qquad -\,\mathop \sum \limits _{i:2i-1\in I_1}l_{i}\sigma _i -\frac{1}{2}\mathop \sum \limits _{i=1}^{n} \sigma _i\\&\quad \le d_{R}= \mathop \sum \limits _{i=1}^{n} x_{i}+ \left[ H+ \frac{1}{2}\mathop \sum \limits _{i=1}^{n} (l_{i}-1)\sigma _i\right] + L. \end{aligned}$$

Thus, we have $\sum _{i:2i-1\in I_1}l_{i}\sigma _i\ge \frac{1}{2}\sum _{i=1}^{n}l_{i}\sigma _i$.

On the other hand, to guarantee that the total completion time of agent A is no more than TC, we should have

$$\begin{aligned}&\mathop \sum \limits _{i:2i-1\in I_1}(n-i+1)p_{2i-1}+\mathop \sum \limits _{i:2i\in I_1} (n-i)p_{2i}\\&\qquad +\,\mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i}\\&\quad = \mathop \sum \limits _{i:2i-1\in I_1}(n-i+1)a_{2i-1}+\mathop \sum \limits _{i:2i\in I_1}(n-i)(a_{2i}\\&\qquad +\,(l_{i}-1)\sigma _i)+\mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i}\\&\qquad +\,\mathop \sum \limits _{i:2i-1\in I_1}(n-i)(a_{2i}+(l_{i}-1)\sigma _i)\\&\qquad -\,\mathop \sum \limits _{i:2i-1\in I_1} (n-i)(a_{2i}+(l_{i}-1)\sigma _i)\\&\quad = \mathop \sum \limits _{i=1}^{n}(n-i)(a_{2i}+(l_{i}-1)\sigma _i)\\&\qquad +\,\mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i}+\mathop \sum \limits _{i:2i-1\in I_1} a_{2i-1}\\&\qquad -\,\mathop \sum \limits _{i:2i-1\in I_1} (n-i)(a_{2i}- a_{2i-1}+(l_{i}-1)\sigma _i)\\&\quad = \mathop \sum \limits _{i=1}^{n}(n-i)(a_{2i}+(l_{i}-1)\sigma _i)+ \mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i} \\&\qquad +\,\mathop \sum \limits _{i:2i-1\in I_1} (a_{2i-1}-(n-i)l_{i}\sigma _i)\\&\quad = \mathop \sum \limits _{i=1}^{n}(n-i)(a_{2i}+(l_{i}-1)\sigma _i)\\&\qquad + \,\mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i}+\mathop \sum \limits _{i:2i-1\in I_1} l_{i}\sigma _{i}\\&\quad \le TC=\mathop \sum \limits ^{n}_{i=1}(n-i)(a_{2i}+(l_i-1)\sigma _i)\\&\qquad +\mathop \sum \limits ^{n}_{i=1}(n-i+1)x_i+\frac{1}{2}\mathop \sum \limits ^{n}_{i=1}l_i\sigma _i. \end{aligned}$$

Thus, we have $\sum _{i:2i-1\in I_1}l_{i}\sigma _i\le \frac{1}{2}\sum _{i=1}^{n}l_{i}\sigma _i$.

From the above discussion, we can see that the correctness of their Lemma 1 needs the correctness of condition (C2), i.e., $\sum _{i:2i-1\in I_1}l_{i}\sigma _i = \sum _{i:2i-1\in I_2}l_{i}\sigma _i= \frac{1}{2}\sum _{i=1}^{n}l_{i}\sigma _i$. Consequently, the correctness of their Lemma 1 requires that

The proof of Lemma 10 in Leung et al. (2010b) is based on the assumption that there exists a solution for problem $1||\sum C_j^a\le TC:\sum U_j^b\le n$ on the constructed scheduling instance. The aim is to show that there exists a solution for the instance of the even–odd partition.

Their proof also uses the results in their Lemmas 2–9. We do not need to state all of the steps, but the R-job must be on time, and the total completion time for agent A is no more than TC. Then, from Lemmas 2–9, we have

$$\begin{aligned}&\mathop \sum \limits _{i=1}^{n}p_{2i}+\mathop \sum \limits _{i=1}^{n}x_i+L-\mathop \sum \limits _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\\&\quad = \mathop \sum \limits _{i=1}^{n}(a_{2i}+(l_{i}-1)\sigma _i)+ \mathop \sum \limits _{i=1}^{n} x_{i}+ L\\&\qquad -\,\mathop \sum \limits _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\\&\quad = H+\frac{1}{2}\mathop \sum \limits _{i=1}^{n}\sigma _i+\mathop \sum \limits _{i=1}^{n}(l_{i}-1)\sigma _i+\mathop \sum \limits _{i=1}^{n}x_{i}+ L\\&\qquad -\,\mathop \sum \limits _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\\&\quad = \mathop \sum \limits _{i=1}^{n}x_{i}+ H+ L+\mathop \sum \limits _{i=1}^{n}l_{i}\sigma _i-\frac{1}{2}\mathop \sum \limits _{i=1}^{n}\sigma _i\\&\qquad -\,\mathop \sum \limits _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i \\&\quad \le d_{R}= \mathop \sum \limits _{i=1}^{n} x_{i}+\left[ H+ \frac{1}{2}\mathop \sum \limits _{i=1}^{n} (l_{i}-1)\sigma _i\right] + L, \end{aligned}$$

and

$$\begin{aligned}&\mathop \sum \limits _{i=1}^{n}(n-i)p_{2i}+\mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i}\\&\qquad +\,\mathop \sum \limits _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\\&\quad = \mathop \sum \limits _{i=1}^{n}(n-i)(a_{2i}+(l_{i}-1)\sigma _i)\\&\qquad +\,\mathop \sum \limits _{i=1}^{n}(n-i+1)x_{i} +\mathop \sum \limits _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\\&\quad \le TC=\mathop \sum \limits ^{n}_{i=1}(n-i)(a_{2i}+(l_i-1)\sigma _i)\\&\qquad +\,\mathop \sum \limits ^{n}_{i=1}(n-i+1)x_i+\frac{1}{2}\mathop \sum \limits ^{n}_{i=1}l_i\sigma _i. \end{aligned}$$

Hence, we have that

$$\begin{aligned} \sum _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\ge \frac{1}{2}\sum _{i=1}^{n}l_{i}\sigma _i \text{ and } \sum _{i:P_{2i-1} \text{ is } \text{ early }}l_i\sigma _i\le \frac{1}{2}\sum _{i=1}^{n}l_{i}\sigma _i. \end{aligned}$$

That is,

Note that condition ($\hbox {C2}'$) is potentially implied in the proof of their Lemma 10. After this, the authors of Leung et al. (2010b) directly conclude that the instance of the even–odd partition has a solution. In the normal deduction, we should add the following procedure.

Let $I_1= \{2i-1: P_{2i-1} \text{ is } \text{ early }\}\cup \{2i: P_{2i} \text{ is } \text{ early }\}$ and $I_2= \{1, 2, \ldots , 2n\}\setminus I_1$. The results in their Lemmas 2-9 imply that $|I_1\cap \{2i-1, 2i\}|= 1$ and $|I_2\cap \{2i-1, 2i\}|= 1$ for each $i=1, 2, \ldots , n$. Now condition ($\hbox {C2}'$) can be rewritten as condition (C2), i.e., $\sum _{i:2i-1\in I_1}l_i\sigma _i= \sum _{i:2i-1\in I_2}l_i\sigma _i= \frac{1}{2}\sum _{i=1}^{n}l_{i}\sigma _i$. Thus, from Observation 2.1, the correctness of their Lemma 10 requires that

From equations (A) and (B), we conclude that if their Lemmas 1 and 10 are correct, then for every partition $(I_1, I_2)$ of the index set $\{1,2, \ldots , 2n\}$ with $|I_1\cap \{2i-1, 2i\}|= 1$ and $|I_2\cap \{2i-1, 2i\}|= 1$ for each $i=1, 2, \ldots , n$, the two conditions (C1) and (C2) are equivalent. $\square$