A new three-machine shop scheduling: complexity and approximation algorithm

Abstract

The paper investigates a new three-machine shop scheduling problem that arises from many production systems, such as the garment assembly line, etc. In such scenarios, each job consists of three operations, each of which has to be non-preemptively processed by one specific machine. In contrast with the classical three-machine shop scheduling, the processing order of the operations of each job is partially restricted. In particular, the first two operations are ordered and all the same for all jobs, while the third operation is not restricted. The objective is to minimize the makespan. We show the problem is NP-hard in the ordinary sense and present a polynomial time approximation algorithm with a worst case performance ratio of $\frac{3}{2}$.

Appendix

Proof of Theorem 3.1

We prove it by a reduction from the classical Partition problem: given n+1 positive integers a ₁,…,a _n ,b with $\sum_{i=1}^{n}a_{i}=2b$, it is asked whether there are two disjoint subsets N ₁ and N ₂ of N such that $\sum_{i\in N_{1}}a_{i}=\sum_{i\in N_{2}}a_{i}=b$?

For any partition instance, we construct the following scheduling instance of $F_{2}O||C_{\mathit{max}}$: 3n+1 jobs {J ₀,J ₁,…,J _3n } with p ₁₀=p ₂₀=p ₃₀=b, p _1i =p _2,i+n =p _3,i+2n =a _i and p _2i =p _3i =p _1,i+n =p _3,i+n =p _1,i+2n =p _2,i+2n =0, for 1≤i≤n. The threshold is y=3b.

We will prove that a schedule with makespan no more than y is equivalent to the YES answer to the partition instance. In fact, if the answer to the partition instance is YES, then a feasible schedule for $F_{2}O||C_{\mathit{max}}$ can be obtained in the following way (see in Fig. 6). Job J ₀ is started at time 0,b and 2b on M ₁, M ₂ and M ₃, respectively. For M ₁, we further schedule the jobs {J _i |i∈N} from time b in an arbitrary order. For M ₂, we process the jobs {J _i+n |i∈N ₁} and the jobs {J _i+n |i∈N ₂} from time 0 to b and time 2b to 3b, respectively. And for M ₃, we schedule the jobs {J _i+2n |i∈N} in an arbitrary order from time 0. Clearly, this very schedule must generate a makespan of y=3b exactly.

Fig. 6

A schedule for the instance of our problem

On the other hand, if there exists a schedule whose makespan is no larger than y=3b, then the three operations of job J ₀ must be started at time 0, b and 2b. W.l.g.o., let s ₁₀=0, s ₂₀=b and s ₃₀=2b. Since $\sum_{i=1}^{3n+1}p_{2,i-1}=b+\sum_{i=1}^{n}a_{i}=3b$, there will be no idling before time 3b on M ₂. Hence, both the jobs scheduled before and after J ₀ must have a total processing time of b on this machine. In other words, there are two disjoint subsets N ₁ and N ₂ of N, such that $\sum_{i\in N_{1}}p_{i+n}=\sum_{i\in N_{2}}p_{i+n}=b$, i.e., $\sum_{i\in N_{1}}a_{i}=\sum_{i\in N_{2}}a_{i}=b$. This indicates that the partition instance must answer YES. □