# 5.7.3 Scenario: Finite Forward Scheduling

### Intended learning outcomes: Perform finite forward scheduling for eight products manufactured on three machines by using a Gantt-type chart.

Your company owns one lathe (M1), one milling machine (M2), and one drilling machine (M3). A working day lasts eight hours. As Figure 15.7.3.1 shows, eight products (P1, P2, P3,  . . . , P8) are manufactured on these machines. Each product loads these machines in a different sequence. For simplicity, assume that there is no inter­operation time.

Fig. 15.7.3.1       Eight products manufactured on three machines.

Perform finite forward scheduling for the next three days. The normal working time of 8 hours per day has to be respected, as do the sequence of the operations for each order given by Figure 15.7.3.1 and the following three priority rules:

1.    No idle time on the machine

2.    Operation with the shortest processing time

3.    Longest remaining lead time for the order

The Gantt-type chart planning board in Figure 15.7.3.2 will help you to perform the task. Note the first orders on each machine. The order for product P1 has been chosen for machine M1 because of the third priority rule.

Fig. 15.7.3.2       Gantt-type chart for finite forward scheduling.

Discuss whether other priority rules would result in a better solution with regard to work in process.

Solution:

The total load is 21 hours on machine 1, 20 hours on machine 2, and 24 hours on machine 3. Thus, machine 3 is fully loaded, and priority rule 1 makes full sense. There are solutions for this problem that schedule the other two machines without idle time, respecting the sequence of operations for all eight orders. One of these solutions can be found by simply following the priority rules.

Replacing the second and the third priority rule by the rule shortest remaining lead time would result in considerably less work in process. However, strict application of this rule not only results in idle time on machine 3, but also creates delays for order 3 and order 6: They cannot be finished at the end of the third day. Both effects cannot be tolerated because these orders are started too late. As a consequence, there must be some rule giving them priority at some time, thereby augmenting work in order.

## Course section 15.7: Subsections and their intended learning outcomes

• ##### 15.7.1 Exercise: Load-Oriented Order Release (Loor)

Intended learning outcomes: Explain the Loor algorithm with given data for a scheduling problem, considering anticipation, loading percentage, and conversion factor.

• ##### 15.7.2 Exercise: Corma — Capacity-Oriented Materials Management

Intended learning outcomes: Describe results of applying the capacity-oriented materials management (Corma) principle in order release.

• ##### 5.7.3 Scenario: Finite Forward Scheduling

Intended learning outcomes: Perform finite forward scheduling for eight products manufactured on three machines by using a Gantt-type chart.

• ##### 15.7.4 Scenario: Order Picking

Intended learning outcomes: Differentiate between the main characteristics of several picking strategies, by listing the advantages and disadvantages of each, and deriving possible fields of application.

• ##### 15.7 Scenarios and Exercises

Intended learning outcomes: Calculate examples for load-oriented order release (Loor) and for finite forward scheduling. Assess characteristics of capacity-oriented materials management (Corma) and of order Picking.