A note: flowshop scheduling with linear deterioration and job-rejection

Abstract

We study a scheduling problem on an $m$-machine flowshop with linear deterioration of job processing times and job rejection. The objectives are minimum makespan and minimum total load, subject to an upper bound on the total permitted rejection cost. The problems are NP-hard (since the single machine makespan minimization version was shown to be hard), and we introduce pseudo-polynomial dynamic programming algorithms, thus proving that both problems are NP-hard in the ordinary sense.

Appendix: Numerical examples

Numerical example 1 (minimizing makespan)

Consider the following input: $m=3,n=6$, $U=48$;

Deterioration rates: ${\alpha }_{j}=\left(0.910, 0.800, 0.680, 0.880, 0.740, 0.440\right)$;

(Generated uniformly in the interval ($\text{0,1}$).)

Basic processing times: ${\beta }_{j}=\left(1, 1, 4, 8, 9, 6\right)$;

(Integers generated uniformly in the interval $\left[\text{1,10}\right]$.)

It follows that: $\frac{{\alpha }_{j}}{{\beta }_{j}}=\left(0.910, 0.800, 0.170, 0.110, 0.082, 0.073\right)$.

Note that jobs are sorted (and numbered) in non-increasing order of $\frac{{\alpha }_{j}}{{\beta }_{j}}$.

Job rejection costs are: ${r}_{j}=\left(38, 41, 16, 21, 26, 43\right)$;

(Integers generated uniformly in the interval [$\text{1,50}]$.)

Applying DP1, we obtain the following optimal solution:

The set of rejected jobs: $R=\left\{{J}_{4}, {J}_{5}\right\}$, implying that the total rejection cost is $\sum _{ j\in R}{r}_{j}=47\le 48=U$.

The sequence of the processed jobs: $A=\left({J}_{1}, {J}_{2},{J}_{3}, {J}_{6}\right)$.

The actual processing times on the machines:

Machine 1: ${p}_{1j}=\left(1.000, 1.800, 5.904, 9.830\right)$,

Machine 2: ${p}_{2j}=\left(1.910, 3.438, 9.919, 14.283\right)$,

Machine 3: ${p}_{3j}=\left(3.648, 6.567, 16.663, 23.995\right)$.

The optimal makespan is ${C}_{max}=56.812$.$\hfill\square$

Numerical example 2 (minimizing total load)

We solve again a 6-job 3-machine flowshop problem.

Deterioration rates: ${\alpha }_{j}=\left(0.74, 0.99, 0.62, 0.90, 0.64, 0.05\right)$;

(Generated uniformly in the interval ($\text{0,1}$).)

Basic processing times: ${\beta }_{j}=\left(1, 2, 4, 7, 6, 8\right)$;

(Integers generated uniformly in the interval $\left[\text{1,10}\right]$.)

It follows that: $\frac{{\alpha }_{j}}{{\beta }_{j}}=\left(0.740, 0.495, 0.155, 0.129, 0.107, 0.006\right)$.

Jobs are sorted in non-decreasing order of $\frac{{\alpha }_{j}}{{\beta }_{j}}$.

Job rejection costs are: ${r}_{j}=\left(21, 9, 13, 5, 7, 18\right)$;

(Integers generated uniformly in the interval [$\text{1,30}]$.)

The upper bound on the total permitted rejection cost is: $U=22$.

The optimal solution obtained by DP2 consists of the following:

The set of rejected jobs: $R=\left\{{J}_{2}, {J}_{4}, {J}_{5}\right\}$.

The total rejection cost is: $\sum _{ j\in R}{r}_{j}=21\le 22=U$.

The sequence of the processed jobs: $A=\left({J}_{1}, {J}_{3}, {J}_{6}\right)$.

The actual processing times on the machines, and the total load:

Machine 1: ${p}_{1j}=\left(1.000, 4.620, 8.381\right)$. ${C}_{max}^{\left(1\right)}=13.901.$

Machine 2: ${p}_{2j}=\left(1.740, 7.484, 8.695\right)$. ${C}_{max}^{\left(2\right)}=22.596.$

Machine 3: ${p}_{3j}=\left(3.028, 12, 125, 11. 895\right)$. ${C}_{max}^{\left(3\right)}=34.491.$

The optimal total load is $TL=70.988$.$\hfill\square$