Intended learning outcomes: Explain and practice the use of the lead-time-stretching factor as well as probable scheduling.
The following exercise will allow you to practice the use of the lead-time-stretching factor as well as probable scheduling. It uses the same network example as in Figure 13.7.4.1.
Solve the two probable scheduling problems shown in Figure 13.7.5.1. Hint: First, calculate a new lead-time-stretching factor using the formula in the lower part of Figure 13.3.6.3, based on an appropriate solution of one of the four problems in the previous exercise (13.7.4) as an initial solution.
Fig. 13.7.5.1 Two probable scheduling problems.
Some common problems in the calculation process that can lead to errors are
- Not understanding the goal and principles of probable scheduling
- Not understanding the formula for recalculation of the lead-time-stretching factor in probable scheduling
- Not choosing the most appropriate last calculation as initial solution for recalculation of the
lead-time-stretching factor
Solutions:
(Again, ESD stands for earliest start date, ECD for earliest completion date, LSD for latest start date, LCD for latest completion date, STREFAC for lead-time-stretching factor.)
a. Use problem (c) in the previous exercise (13.7.4) as an initial solution. STREFAC(new) = (6 – 4) / (7 – 4) * 0.5 = 2/3 * 0.5 = 1/3. =>
ESD(op10) = 1, ECD(op10) = 1.5; ESD(op20) = 2.5, ECD(op20) = 3; ESD(op30) = 1, ECD(op30) = 1.5; ESD(op40) = 4.7, ECD(op40) =5.7; ESD(order) = 0, ECD(order) = 6.
Note that the upper path is critical. The lead-time margin of the lower path is 2/3 = 0.667.
b. Use problem (a) in the previous exercise (13.7.4) as an initial solution.
STREFAC(new) = (16 – 4) / (12 – 4) * 1 = 12/8 * 1 = 1.5. =>
ESD(op10) = 4.5, ECD(op10) =5; ESD(op20) = 9.5, ECD(op20) = 10; ESD(op30) = 4.5, ECD(op30) =5; ESD(op40)=14.5, ECD(op40)=15.5; ESD(order) = 0, ECD(order) = 17 (!).
Note that the lower path is critical. The lead-time margin of the upper path is 4.
Because the desired ECD(order) of 16 has not been met (can you say why this is the case?), an additional iteration is necessary: recalculation with
STREFAC(new) = (16 – 4) / (17 – 4) * 1.5 = 12/13 * 1.5 ≈ 1.4
will yield the desired solution.
Course section 13.7: Subsections and their intended learning outcomes
13.7.1 Exercise: Queues as an Effect of Random Load Fluctuations (1)
Intended learning outcomes: Answer a number of questions using the relevant formulas in queuing theory.
13.7.2 Scenario: Queues as an Effect of Random Load Fluctuations (2)
Intended learning outcomes: Experience average wait time as a function of capacity utilization in a job shop environment with random arrivals, execution of operations in order of arrival as well as operation times subject to a determinate distribution with mean and coefficient of variation.
13.7.3 Exercise: Network Planning
Intended learning outcomes: Calculate a scheduled network with incomplete data for six operations and a start operation (administration time. Determine the critical path.
13.7.4 Exercise: Backward Scheduling and Forward Scheduling
Intended learning outcomes: Explain and solve the forward and backward scheduling problems that is calculation of start and completion dates for the order and each operation, as well as the critical path and lead-time margin.
13.7.5 Exercise: The Lead-Time-Stretching Factor and Probable Scheduling
Intended learning outcomes: Explain and practice the use of the lead-time-stretching factor as well as probable scheduling.
13.7 Scenarios and Exercises
Intended learning outcomes: Assess queues as an effect of random load fluctuations. Calculate examples for network planning, backward scheduling, forward scheduling, the lead-time stretching factor, and probable scheduling.
