A tight linear time $\frac{13}{12}$-approximation algorithm for the $P2 || C_{\max }$ problem

Abstract

We consider problem $P2 || C_{\max }$ where the goal is to schedule n jobs on two identical parallel machines to minimize the makespan. We focus on constant factor approximation algorithms with complexity independent from the required accuracy. We exploit the famous Longest Processing Time (LPT) rule that requires to sort jobs in non-ascending order of processing times and then to assign one job at a time to the machine whose load is smallest so far. We propose an approximation algorithm that applies LPT to a subset of 2k jobs, then considers a single step of local search on the obtained subschedule and finally applies list scheduling to the remaining jobs. This algorithm, when applied for $k=5$, reaches a tight $\frac{13}{12}$-approximation ratio improving the ratio of $\frac{12}{11}$ proposed by He et al. (Nav Res Logist 47:593–601, 2000). We use Mathematical Programming to analyze the approximation ratio of our approach. As a byproduct, we also show how to improve the FPTAS of Sahni for any $n > 1/\epsilon$ so as to reach an approximation ratio $(1 + \epsilon )$ with time complexity $O(n + \frac{1}{\epsilon ^3})$.

Appendix: Extended linear formulation of constraint (10)

A linear formulation of constraint (10) can be expressed by introducing for each pair of jobs i, j the binary variables $v^\prime _{ij}$, $v^{\prime \prime }_{ij}$ and $v^{\prime \prime \prime }_{ij}$. Variable $v^\prime _{ij}$ is equal to 1 iff $p_i-p_j \le \frac{\delta }{2}$, variable $v^{\prime \prime }_{ij}$ is equal to 1 iff $\frac{\delta }{2} < p_i-p_j \le \delta$ and variable $v^{\prime \prime \prime }_{ij}$ is equal to 1 iff $\delta < p_i-p_j$. Correspondingly, $\forall i < j \; | \; w_{1,i}= w_{2,j} = 1$, the following set of big-M constraints (for a reasonable large value of M, e.g. $M=1000$) are introduced.

$$\begin{aligned} v^{\prime }_{ij}+v^{\prime \prime }_{ij}+v^{\prime \prime \prime }_{ij} =1; \end{aligned}$$

(15)

$$\begin{aligned} \frac{\delta }{2} - p_i + p_j \le M v^\prime _{ij} \end{aligned}$$

(16)

$$\begin{aligned} -\frac{\delta }{2} + p_i - p_j \le M(1 - v^\prime _{ij}) \end{aligned}$$

(17)

$$\begin{aligned} \delta - p_i + p_j \le M (v^\prime _{ij} + v^{\prime \prime }_{ij}) \end{aligned}$$

(18)

$$\begin{aligned} -\delta + p_i - p_j \le M v^{\prime \prime \prime }_{ij} \end{aligned}$$

(19)

$$\begin{aligned} \hat{\delta } \ge p_i - p_j - M(1 - v^\prime _{ij}) \end{aligned}$$

(20)

$$\begin{aligned} \hat{\delta } \ge \delta - p_i + p_j - M(1 - v^{\prime \prime }_{ij}) \end{aligned}$$

(21)

Indeed, constraint (15) indicates that either $v^\prime _{ij}=1$, or $v^{\prime \prime }_{ij}=1$ or $v^{\prime \prime \prime }_{ij}=1$.

Then, if $v^\prime _{ij} =1$, constraint (17) implies that $p_i-p_j\le \frac{\delta }{2}$. Correspondingly, constraints (16, 18, 21) are inactive, while constraint (19) that implies that $p_i-p_j\le \delta$ is dominated by constraint (17). Hence, the swap induces the makespan reduction $\hat{\delta } = p_i - p_j$ through constraint (20) in combination with the objective function (1).

Else, if $v^{\prime \prime }_{ij} = 1$, constraint (16) implies that $p_i-p_j\ge \frac{\delta }{2}$, while constraint (19) implies that $p_i-p_j\le \delta$. Also, constraints (17, 18, 20) are inactive. Hence, the binding constraint is in this case constraint (21) that is satisfied as an equality, that is $\hat{\delta } = \delta - p_i + p_j$. Correspondingly, due to constraint (9), the new makespan will be on machine $M_2$ and its value in the objective function (1) will be $C_{max}^{M_2} + p_i - p_j$.

Else, $v^{\prime \prime \prime }_{ij} = 1$. In this case, constraint (16) induces $p_i-p_j \ge \frac{\delta }{2}$, and constraint (18) induces $p_i-p_j \ge \delta$, that is swap will not occur as it can only worsen the objective function value. Besides, constraints (17, 18, 20, 21) are inactive. Correspondingly, as $\hat{\delta }$ must be positive or null, it has negative coefficient in the objective function and is not further constrained, we have $\hat{\delta }=0$.