Intended learning outcomes: Explain load profile calculation extending across several load periods.
One of the problems associated with the use of simple algorithms is that an operation can extend across several load periods (see Figure 14.2.2.2). This exercise will examine how manual or computer algorithms establish capacity and load in a load profile.
Use Figure 14.7.2.1 to enter the capacity or load curve (continuous or rectangular distribution within a time period) for a work center, given the problem outlined below.
Fig. 14.7.2.1 Load profile calculation.
a. Determine the start date of each period and enter it into the figure above, given 2 weekly periods of 3.5 days each (½ calendar week): Sunday morning to Wednesday noon and Wednesday noon to Saturday evening. The load profile starts with Sunday morning, May 9 (as indicated in the figure). The load profile covers 6 periods (3 weeks).
b. Allocate theoretical capacity to each of the 6 time periods, respecting the following data: At the work center, the plant runs one 8-hour shift per normal workday (8 a.m. to 12 p.m., 1 p.m. to 5 p.m.). The work center has 5 identical machines. Saturdays and Sundays are off. Furthermore, May 13 and May 24 are public holidays (in practice, these dates would change each year). Note that “today,” or the moment of the inquiry, is 7 a.m. on Wednesday, May 12.
c. Assume no existing load on the work center. For the following operation, allocate its standard load to the work center: Operation start date is Friday morning, May 14. Standard load (including setup) is 81 hours. The operation can be split on 2 machines, maximum.
Solutions:
a. The second period starts at Wednesday noon, May 12. The third period starts on Sunday morning, May 16. The fourth period starts at Wednesday noon, May 19. The fifth period starts on Sunday morning, May 23. The sixth period starts at Wednesday noon, May 26. The load profile ends before Sunday morning, May 30.
b. Note that there is either a Saturday or Sunday in each period of ½ calendar week. Thus, theoretical capacity per period with normal working days is
(5 machines) * (8 hours per day and machine) * (2.5 working days) = 100 hours.
Note that, in the first period, only 20 hours of capacity are left, because it is already Wednesday morning, May 12. Furthermore, in the second and the fifth periods there is one less working day due to public holidays, which results in only 60 hours of capacity for each of these periods.
c. The load has to be distributed to different periods. Only one working day is left during the period in which May 14 falls (the second period). Because only 2 machines can be used, a maximum of only 16 standard hours (note: not 40) can be loaded. During the third period, 2.5 working days on two machines allow the load of 40 hours. The same would be possible for the fourth period. However, only 25 hours are left to be loaded.
Course section 14.7: Subsections and their intended learning outcomes
14.7.1 Exercise: Capacity Determination
Intended learning outcomes: Determine theoretical capacity, availability, rated capacity, demonstrated capacity, actual utilization, work center efficiency, etc.
14.7.2 Exercise: Algorithms for Load Profile Calculation
Intended learning outcomes: Explain load profile calculation extending across several load periods.
14.7.3 Exercise: Rough-Cut Capacity Planning
Intended learning outcomes: Calculate, in a rough-cut network plan, the earliest start date and the latest start date for each operation. Determine the resource profiles as well as the load and the deferred earliest completion date for the operations without overloading the capacities.
14.7 Scenarios and Exercises
Intended learning outcomes: Determine capacity. Execute algorithms for load profile calculation. Elaborate an example for rough-cut capacity planning.
