C.40:5.6 - Develop a scheduling rule rather than one schedule

A constructed workshop model has one machine, all jobs available at time zero, no interruptions or setup times, and known processing times p and due times d. For an order, completion time is the cumulative processing time; total tardiness is the sum of max(0, completion − d). Lower total tardiness is the receiving criterion. These are stipulated teaching conditions, not an operations recommendation or measured workplace effect.

The material under development is a rule that constructs an order from job data. Two executable incumbents are S, shortest processing time first, and E, earliest due time first; ties use job number. A third candidate H uses S if any job has p > d, otherwise E. This condition can be evaluated on new inputs, unlike retaining a literal order for one job set. The outer development proposes a rule; each inner application sorts the supplied jobs, computes completion times and returns an order plus total tardiness.

On A, S completes jobs at times 1, 3 and 7; only job 2 is late, by 3. E completes them at 4, 6 and 7, with none late. On C, E’s individual tardiness values are 4, 3 and 3; S’s are 0, 0 and 6. Across equally weighted A–C, the mean totals are 11/3 for S, 10/3 for E and 2 for H. Thus H is the best of these three on this selection set. No search over an unspecified larger space or evidence about future workloads follows.

Now apply the selected H to a previously unused set D: (4,3), (1,4), (2,7). Its p > d condition selects S, giving order 2,3,1, completion times 1,3,7 and tardiness 4. E gives order 1,2,3, completion times 4,5,7 and tardiness 2. The earlier improvement did not transfer. Once D is used to repair H, it is part of development, not an untouched final examination.

Inspect the failed condition: knowing that one job must already be late does not establish which ordering minimizes total tardiness. A feasible repair Q constructs both S and E orders, computes their total tardiness with the supplied model, and returns the better order, choosing E on a tie. Q returns totals 0, 0, 6 and 2 on A–D. It costs two order constructions and comparisons per use. Its result is guaranteed to be no worse than either supplied order under this model, because it compares those same two exact values. It is not a globally optimal scheduling rule for all inputs. A complete enumeration of the six orders is another affordable alternative for three jobs and can settle each of these cases directly; a larger population search is unnecessary here.

Suppose search is accelerated by judging only jobs completed by time 3. On A, S has completed two jobs and E none, but the final tardiness comparison favors E. That cheap progress signal cannot replace the receiving criterion. It may select trials only with a justified relation to final outcomes and a return when ordering reverses. In this small model the full calculation is cheaper than developing such a predictor. In an expensive simulator, the same distinction can justify a calibrated predictor and occasional complete trials instead.

If actual execution has already completed job 1, Q cannot select an order that places another job before it. A continuation comparison must hold that performed prefix fixed and order only the remaining jobs. It answers a different question from starting the rule at time zero. If processing times become uncertain or depend on order, the exact calculation above no longer supplies the receiving result; obtain an adequate scheduling model and trial basis before selecting on its predictions. Preserve the deterministic result within its original conditions.