Consider a job scheduling problem with 4 jobs J1, J2, J3, J4 and with…

In job sequencing with unit processing time on a single processor, a schedule is a permutation (ordering) of the jobs; the job placed at position k (1-indexed, counting from the start) finishes at time k. A schedule is feasible only when every job's completion time is less than or equal to its own deadline — a single job finishing even one time unit late makes the whole schedule infeasible.

1. J4 is scheduled first, so it finishes at time 1; its deadline is 2, so 1 ≤ 2 holds.

2. J1 is scheduled second, so it finishes at time 2; its deadline is 4, so 2 ≤ 4 holds.

3. J2 is scheduled third, so it finishes at time 3; its deadline is 2, so 3 ≤ 2 fails — J2 misses its deadline by one time unit.

4. Because J2 already violates its deadline, the schedule (J4, J1, J2, J3) is infeasible regardless of when J3 finishes.

Cross-check against the other three orders: in each of them, the two deadline-2 jobs (J2 and J4) occupy positions 1 and 2 in some order, so both finish by time 2; the two deadline-4 jobs (J1 and J3) then occupy positions 3 and 4, comfortably finishing by time 4. Only the order that pushes a deadline-2 job to position 3 — as this schedule does with J2 — breaks feasibility, confirming that (J4, J1, J2, J3) is the one infeasible schedule among the four.