# 14.7 Scenarios and Exercises

## 14.7.1 Capacity Determination

The following exercise was developed on the basis of a communication from Barry Firth, to whom we extend many thanks.

A plant runs 10 * 8 hour shifts per normal week. A work center in the plant has 5 identical machines, each requiring one operator to run it. This is a machine-paced work center (a machine capacity). Operators get a total of 1 hour for breaks, and they usually take their breaks at the same time. Each machine requires one episode of planned maintenance per week of 3 hours, scheduled by the planner. During the last 6 weeks, the performance data in Figure 14.7.1.1 were recorded:

Fig. 14.7.1.1       Capacity performance data.

Questions:

a.    What is the theoretical capacity in machine hours per normal week (5 days)?

b.    Taking into account scheduled nonproduction events, what is the availability (as a percentage) of machine time per normal week, without considering operator constraints?

c.    What is the availability (as a percentage) of machine time per normal shift, taking into account the normal working conditions for operators?

d.    If the tactical utilization is targeted to be 90%, what value should be used for the utilization factor of machine time for capacity rating purposes?

e.    What is the demonstrated capacity per normal week of this work center? (Adjust the data for week 2 to correct for the short week.)

f.    What was the actual utilization (as a percentage) through the 6 weeks in review?

g.    What was the work center efficiency through the 6 weeks in review?

h.    If planned efficiency is targeted to be 85%, and taking into account your answer to question (d), what was the rated capacity per normal week?

i.     Compare your answers to questions (a), (e), and (h). What should we do now?

a.    Theoretical capacity = 400 hours per normal week
= (5 machines) * (10 shifts) * (8 hours per shift and machine).

b.    Downtime due to maintenance is 15 hours per week. Therefore, the availability factor is (400 – 15) / 400 = 96.25%.

c.    Downtime due to operator breaks is 1 hour per shift of 8 hours. Therefore, the availability factor is 7 / 8 = 87.5%.

d.    Assuming that maintenance cannot be effected during operator breaks, the utilization factor is 87.5% * 96.25% * 90% ≈ 75.80%.

e.    Demonstrated capacity is expressed as standard hours produced (row 4 in the table above). The adjusted output for week 2 is 160 * 5 / 4 = 200 hours. Over 6 weeks, the mean is (1340 + 40) / 6 = 230 standard hours per week.

f.    During the 6 weeks in review, production has run for 1600 machi­ne hours (row 2 in the table above) out of a possible 2320 hours (= 5*400 + 320). Therefore, actual utilization = 1600 / 2320 ≈ 69.0%.

g.    Actual efficiency = standard hours produced divided by actual hours worked
= 1340 / 1600 = 83.75%.

h.    Rated capacity = 400 hours * 75.8% * 85% ≈ 258 (standard) hours.

i.     Demonstrated capacity (230 hours) is too low compared to rated capacity (258 hours). However, in week 4, the output (280 hours) exceeded 258 hours. Check whether the measurements are still requi­red. If so, check for exceptional events, calculating actual utilization and efficiency for each week. Decide whether to make adjustments to planned utilization or efficiency.

## 14.7.2 Algorithms for Load Profile Calculation

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.

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.

## 14.7.3 Rough-Cut Capacity Planning

Figure 14.7.3.1 shows the network plan for a production order.

Fig. 14.7.3.1       Rough-cut network plan with two rough-cut work centers.

a.    Complete the network plan: Calculate the earliest start date and the latest start date for each operation. What is the lead-time margin (the slack time), and what is the critical path? Determine the slack of all operations not on the critical path.

b.    Following the technique introduced in Section 14.4.1, determine the resource profiles for rough-cut work centers 1 and 2, as well as the resource profile for the combination of rough-cut work centers 1 and 2.

c.    Figure 14.7.3.2 shows the preload of rough-cut work center 2. Load the resource profile for rough-cut work center 2 with infinite loading. Determine the earliest completion date for the operations of rough-cut work center 2. Further, determine the load and the deferred earliest completion date for the operations of rough-cut work center 2 without overloading the capacities.

Fig. 14.7.3.2       Preload of rough-cut work center 2.

Solutions:

a.    Lead-time margin is 1. Operations 1, 3, 6, 8, 10, 11, and 12 make up the critical path. Operations 2, 4, 7, and 9 could be deferred by 4 time periods, operation 5 by 7 periods.

b.    The figure that follows shows the results for rough-cut work center 2 as well as for the combination of both rough-cut work centers 1 and 2. The length of the arrow indicates the number of time units for a possible deferring of the start date of operations not on the critical path.

c.    The earliest possible completion date for the operations of rough-cut work center 2 using infinite loading is — as the above figure shows — at the end of period 14. For finite loading, the next figure shows the result: an earliest possible completion date at the end of period 15. Note: Because operation 9 is not on the critical path, parts of its load can be deferred to later periods in order to prevent overload.