Finite-Capacity Scheduling and Release

Lesson 20.4
12 minOpsida Teknoloji · Last reviewed: October 30, 2026

In one sentence

A finite schedule places every job on the real resource calendar without overlap; a release rule prevents jobs that will wait in the queue from being sent to the shop floor without limit.

By the end of this lesson

  • Calculate the setup and processing load of 24 jobs for the same machine.
  • Build a two-day calendar while preserving predecessor and delivery priorities.
  • Verify the finish of the jobs by accounting for machine closures.
  • Reject the infinite-capacity calculation and choose a controlled release rule.

Why it matters

Finite-capacity scheduling applies, on the real calendar, the fact that the same machine can do only one job at a time. Working back the job time from the demand date does not produce an executable schedule if the resource is assigned to another job in that interval. The calculation includes each job's ready time, predecessors, setup time and delivery condition. Intervals closed to production must also be visible in the calendar. That monthly total capacity is sufficient does not show that daily jobs will not overlap. A detailed schedule is a separate piece of work that tests the feasibility of the aggregate plan at job and resource level.

A priority rule and physical feasibility are different things. A job with an early delivery may have priority; but if its component is not ready or its predecessor is not finished, it cannot be started. A sequence is chosen from among the feasible jobs. If the setup time depends on the sequence, sequencing also changes the total load. Running the same family back to back can reduce setups; but it can delay another job with an early delivery. A sequence that only raises machine utilization therefore does not automatically give the best delivery result. When the rule used and its exceptions are written out, the purpose for which the schedule was built becomes clear.

Release is the authorization of a planned job to be run on the shop floor. Sending a job to the floor too early may not bring its finish forward; the queue in front of a single machine grows. Material is moved, space fills up and it becomes harder to follow which job has priority. While the schedule can show all future jobs, release can open only the near and feasible ones. When these two decisions are separated, the visibility of the plan is preserved and work in process is limited. A job whose material is missing or whose predecessor is incomplete is prevented from being treated as ready merely because it appears in the plan.

The schedule is not calculated once and forgotten. Actual finishes, breakdowns and material delays affect later jobs. At each deviation, instead of deleting the past record and writing a new target, the part that has not been executed is re-evaluated. The safety, quality and setup consequences of stopping a job that has already started are examined separately. A delivery promise must rest on a date that is possible in the renewed finite schedule. This lesson does not claim that the sequencing rule gives the globally best result; the aim is to build a schedule that satisfies the given constraints and whose intermediate results can be checked, and an executable release decision. For the numerical scenarios in the lesson quizzes as well: This dataset was generated synthetically for training; it does not represent a real business, employee or customer study.

Formula

Job block (min/job) = Setup (min/job) + Processing (min/job)

Resource load (min) = Sum of job blocks

Feasible start = The first sufficient open interval in which the resource is free, the material is ready and the predecessors are complete

Lateness (min) = max(0, Actual or planned finish − Due time)

Work in process (jobs) = Number of jobs released but not completed

Five elements distributed over three stations with precedence relationships preservedBody18 s1Inner part27 s1Gasket16 s2Lid23 s2Check14 s3St. 1Body 18Inner part 2745 ≤ 50St. 2Gasket 16Lid 2339 ≤ 50St. 3Check 1414 ≤ 50Target 50 s
In a separate precedence-network example, station loads of 18+27 = 45 s, 16+23 = 39 s and 14 s meet the 50 s target. The arrows show which elements must finish first. Distributing 98 s of total work over three stations does not mean that the single machine in this lesson can do three jobs at the same time. This dataset was generated synthetically for training; it does not represent a real business, employee or customer study.

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