# 13.7 Scenarios and Exercises

## 13.7.1 Queues as an Effect of Random Load Fluctuations (1)

Answer the following questions using the relevant formulas in queuing theory (refer to Figure 13.2.2.4):

a.    How many parallel workstations are needed to have an expected wait time of less than 10 hours, if capacity utilization is 0.95, the mean of the operation time is 2 hours, and the coefficient of variation of the operation time is 1?

b.    The capacity is 10 hours. How much does the expected wait time increase if load rises from 4 to 8 hours?

c.    How is the expected wait time affected when the coefficient of variation increases from 1 to 2?

Solutions:

a.    s = 0.95 / (1 – 0.95)  *  (1 + (1 * 1)) / 2  *  2 / 10) = 3.8. Thus, with four workstations, the expected wait time will be 9.5 hours.

b.    Capacity utilization increases from 4/10 to 8/10. Therefore, the respective factor in the formula for the expected wait time increases from 0.4 / (1 – 0.4) = 2/3 to 0.8 / (1 – 0.8) = 4. The new factor is 4 / (2/3) = 6 times greater than the old factor. Thus, the expected wait time increases by a factor of 6.

c.    The respective factor in the formula for the expected wait time increases from (1 + (1 * 1)) / 2 = 1 to (1 + (2 * 2)) / 2 = 2.5. Thus, the expected wait time increases by the factor 2.5.

## 13.7.2 Queues as an Effect of Random Load Fluctuations (2)

Figure 13.2.2.3 shows the average wait time as a function of capacity utilization in a job shop environment with random arrivals, execution of operations in order of arrival (or according to random selection from the queue), as well as operation times (OT) subject to a determinate distribution with mean M(OT) and coefficient of variation CV(OT). We reproduced the effect shown in Figure 13.2.2.3 by means of a simulation, which you can view at this URL:

Start the simulation by clicking on the given arrival rate and execution (service) rate on the gray button to the far left at the bottom of the figure and watch the number of elements in the system. Stop the simulation by clicking on the middle of the three buttons (or empty the system by clicking the button to the far right). Now change the input rate to bring it closer and closer to the execution rate and observe the rising number of elements in the queue. You will see the exploding number of elements in the system as soon as, for an execution rate of 60 per unit of time, the arrival rate is 58 and higher.

## 13.7.3 Network Planning

Figure 13.7.3.1 shows a scheduled network with incomplete data for 6 operations and a start operation (administration time).

Fig. 13.7.3.1       Scheduled network (for you to complete).

a.    For each process, please fill in the earliest start date (ESD) and the latest start date (LSD) in the scheduled network. What is the critical path, that is, the path with the longest duration? What is its lead-time margin, that is, the slack time?

b.    The operation time for operation 6 has not yet been determined. What is the longest possible time for operation 6 (lead-time margin = zero)?

Solutions:

a.    For the time being, as long as the time for operation 6 is still open, the longest path is (start – op1 – op3 – op5 – end). Lead-time margin = 7.

b.    As soon as the time for operation 6 is greater than 4, the longest path is (start – op1 – op3 – op6 – end). The longest possible time for op6 is 11.

## 13.7.4 Backward Scheduling and Forward Scheduling

Here, you will practice some backward and forward scheduling. Figure 13.7.4.1 presents a simple network, including a legend showing the lead- time elements used.

Solve the forward and backward scheduling problems (calculation of start and completion dates for the order and each operation, as well as the critical path and lead-time margin) listed in Figure 13.7.4.2:

Fig. 13.7.4.1       Scheduled network.

a.    Common forward scheduling.
b.    Common backward scheduling.
c.    Forward scheduling with a different lead-time-stretching factor, that is, a different order urgency, to accelerate or slow down the order.
d.    Forward scheduling with lead-time-stretching factor = 0, which results in the lead time as the sum of operation times plus the technical inter­operation times.

Fig. 13.7.4.2       Various forward and backward scheduling problems.

Some common problems in the calculation process lead to the following potential errors:

• Calculating incorrect start date and due dates, not respecting inter­operation times multiplied by the stretching factor
• Multiplying technical waiting time by the stretching factor
• Incorrectly calculating the longest path in a network
• Not understanding the principle of forward or backward scheduling

Solutions:
(ESD stands for earliest start date, ECD for earliest completion date, LSD for latest start date, LCD for latest completion date.)

a.    ESD(op10) = 3, ECD(op10) = 3.5; ESD(op20) = 6.5, ECD(op20) = 7; ESD(op30) = 3, ECD(op30) = 3.5; ESD(op40) = 10, ECD(op40) = 11; ESD(order) = 0, ECD(order) = 12. Note the critical path in determi­ning the ESD(op40): The lower path is critical. The lead-time margin of the upper path is 2.

b.    LCD(op40) = 15, LSD(op40) = 14; LCD(op30) = 9.5, LSD(op30) = 9; LCD(op20) = 11, LSD(op20) = 10.5; LCD(op10) = 7.5, LSD(op10)=7; LCD(order) = 16, LSD(order) = 4. Note that — again — the lower path is critical. The lead-time margin of the upper path is again 2.

c.    ESD(op10) = 1.5, ECD(op10) = 2; ESD(op20) = 3.5, ECD(op20) = 4; ESD(op30) = 1.5, ECD(op30) = 2; ESD(op40 = 5.5, ECD(op40) =6.5; ESD(order) = 0, ECD(order) = 7. Note that both paths are critical.

d.    ESD(op10) = 0, ECD(op10) = 0.5; ESD(op20) = 0.5, ECD(op20) = 1; ESD(op30) = 0, ECD(op30) = 0.5; ESD(op40) = 3, ECD(op40) = 4; ESD(order) = 0, ECD(order) = 4. Note that the critical path has changed. The upper path is now critical. The lead-time margin of the lower path is 2.

## 13.7.5 The Lead-Time-Stretching Factor and 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.