# 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.

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.