Computational complexity and algorithms for two scheduling problems under linear constraints

Abstract

This paper considers two different types of scheduling problems under linear constraints. The first is the single-machine scheduling problem with minimizing total completion time, while the second is the no-wait two-machine flow shop scheduling problem with minimizing makespan. For these two problems, a set of jobs is required to be scheduled to one or two machines. In contrast to the classic scheduling problems, the processing times of jobs are not fixed constants and are not predetermined. The decision-maker only knows that they should satisfy a system of given linear constraints. For both problems, the goal is to determine the processing time for each job and find the schedule that minimizes a particular criterion, namely, the total completion time or the makespan. First, we study the computational complexity and show that both the problems under linear constraints are NP-hard. These hardness results significantly differ from their traditional scheduling counterparts, as both of those are solvable in polynomial time. Then we propose polynomial time exact or approximation algorithms for various special cases. By utilizing the existing scheduling algorithms and the properties of linear programming, we demonstrate that both problems are polynomially solvable when the total number of linear constraints is a fixed constant.

Inapproximability of the SSLC problem

In this appendix, we demonstrate that based on some complexity assumptions, we are able to show that the problem (3) might not be approximated within some constant factor under some assumption. First, we state some known properties of the increasing ordered weighted average function (3a) in problem (3).

Theorem A.1

(Ogryczak and Śliwiński 2003, Ogryczak and Śliwiński 2010, Theorem 3) The objective function (3a) is a piecewise linear concave function of $\varvec{x}$.

By Theorem A.1, the problem (3) is a concave minimization problem under linear constraints. It is well-known that such problem always has a globally optimal solution at its extreme point. By the assumption of the constraints $A\varvec{x}\ge \varvec{b}$, $\varvec{x}\ge 0$, and the objective function (3a) is bounded from below over $\varvec{x}\ge 0$, we have the following statement.

Theorem A.2

(Benson 1995, Section 4) The concave minimization problem (3) always has an optimal solution which is an extreme point of the polyhedron defined by the constraints (3b) and (3c).

Now we present a gap-preserving reduction from the unweighted vertex cover to the problem (3). The hardness result is described as Theorem A.3.

Theorem A.3

Suppose that the (unweighted) vertex cover problem cannot be approximated within a constant factor of $\rho$ ($1< \rho < 2$), then problem (3) does not admit any approximation algorithms with a factor of $\rho ^2/2 -\epsilon$ for any $\epsilon >0$.

Proof

Given an instance of the vertex cover problem with $G = (V,E)$, $|V| = n$ and $|E| = m$, an integer K. We construct an instance of problem (3) with n jobs and m linear constraints:

$$\begin{aligned}&x_u + x_v \ge 1,&~~\forall (u, v) \in E,\end{aligned}$$

(A.1a)

$$\begin{aligned}&\varvec{x} \ge 0, \end{aligned}$$

(A.1b)

where $x_u$ corresponds to the vertex $u\in V$. We show that (1) if the vertex cover instance has a vertex cover with size at most K, then there is a feasible solution to (A.1) with objective value at most $\frac{K(K+1)}{2}$; (2) if all the vertex covers of the instance have size at least $\rho K$, then all the feasible solutions to (A.1) must have objective value at least $\frac{\rho K(\rho K+1)}{4}$

For the first direction, let $V'$ be the vertex cover, we set $x_u=1$ for each $u\in V'$ and 0 otherwise. Since the vertex cover has size at most K, the objective value is at most $K+(K-1)+\cdots +1 \le \frac{K(K+1)}{2}$.

Next, we prove the second direction by contradiction. Suppose that there exists a feasible solution to (A.1) with an objective value less than $\frac{\rho K(\rho K+1)}{4}$. By Theorem A.2, there is an extreme point of (A.1) which has objective value less than $\frac{\rho K(\rho K+1)}{4}$. Note that any extreme point for the polyhedron defined by vertex cover constraints (A.1) is half-integral, that is, must have value 0, 1/2, or 1 for each variable (see, e.g., Vazirani 2001, Section 14.3). Thus, we can find a half-integral solution to (A.2) which has an objective value less than $\frac{\rho K(\rho K+1)}{4}$. Let $K'$ be the total number of variables with positive values, and $\varTheta _1(\varvec{x}), \ldots , \varTheta _{K'}(\varvec{x})$ be the values of those variables where $\varTheta _1(\varvec{x}) \le \cdots \le \varTheta _{K'}(\varvec{x})$. Then the objective value of this solution is $K'\varTheta _{1}(\varvec{x})+(K'-1)\varTheta _{2}(\varvec{x})+\cdots \varTheta _{K'}(\varvec{x}) < \frac{\rho K(\rho K+1)}{4}$. Now we select all these $K'$ vertices corresponding to the variables with positive values. These vertices constitute a vertex cover with size exactly $K'$, since the constraints (A.1a) are satisfied. The half-intergality of the solution implies that $2\varTheta _{i}(\varvec{x})) \ge 1$ for each $i\in \{1, \ldots , K'\}$. Thus we have $\frac{K'(K'+1)}{2} = K' + (K'-1) + \cdots + 1 \le 2(K'\varTheta _1(\varvec{x})+(K'-1)\varTheta _{2}(\varvec{x})+\cdots \varTheta _{K'}(\varvec{x})) < \frac{\rho K(\rho K+1)}{2}$. Note that $K'$ is a nonnegative integer, it follows that $K' < \rho K$ as otherwise $\frac{K'(K'+1)}{2} \ge \frac{\rho K(\rho K+1)}{2}$. Therefore, we can find a vertex cover with size less than $\rho K$, which contradicts to the previous assumption that all the vertex covers of the instance have a size of at least $\rho K$.

Assume that problem (3) admits a $(\rho ^2/2 -\epsilon )$-approximation algorithm for some $\epsilon >0$. If $K+1 \le \frac{\rho ^2-\rho }{2\epsilon }$, then we can simply apply a brute-force enumeration in polynomial time (since $\rho$ and $\epsilon$ are fixed constants) and distinguish whether the instance has a vertex cover with size at most K or all the vertex covers must have size at least $\rho K$. Now we assume that $K +1 > \frac{\rho ^2-\rho }{2\epsilon }$. We can use the $(\rho ^2/2 -\epsilon )$-approximation algorithm distinguish the instance of the SSLC problem with objective value at most $\frac{K(K+1)}{2}$ or at least $\frac{K(K+1)}{2}\left( \frac{\rho ^2}{2} -\epsilon \right)$. Note that $\frac{K(K+1)}{2}\left( \frac{\rho ^2}{2} -\epsilon \right) <\frac{K(K+1)}{2}\left( \frac{\rho ^2}{2} -\frac{\rho ^2-\rho }{2(K+1)}\right) = \frac{\rho K(\rho K+1)}{4}$, and thus we can use this approximation algorithm to distinguish a vertex cover problem instance with size at most K or at least $\rho K$ by the reduction above, which leads to a contradiction. Therefore, we can conclude that problem (3) does not have a $(\rho ^2/2 -\epsilon )$-approximation algorithm for any $\epsilon >0$, provided that the (unweighted) vertex cover problem does not have a $\rho$-approximation algorithm. $\Box$

It is well-known that based on the unique games conjecture, the vertex cover is hard to approximate within $2-\epsilon$ for any $\epsilon$ (Khot and Regev 2008). As a corollary, problem (3) also has the inapproximability result based on the unique game conjecture.

Corollary A.1

Under the unique 2-prover-1-round game conjecture, problem (3) is hard to be approximated with any factor better than 2.

We remark that under the assumption $P \ne NP$, the currently best inapproximability result for vertex cover is $\rho = 1.3606$ for sufficiently large vertex degree (Dinur and Safra 2005). However, $1.3606^2/2\approx 0.926<1$ and hence we cannot obtain any inapproximability result better than Theorem 1 from the reduction in Theorem A.3 based on $P \ne NP$.