• Lecture on PERT
• Table 2-8 Activities, events, duration estimates, expected durations, variances, and standard deviations for Pert example problem

Total float & free float can be incorporated

Lag time avoids the calculation of LS and LF times.

FF = Lag time = ES of an activity – EF preceding activity

FF = 0 FF = 2

TF = 2 TF = 2

TF = LF - EF

FF = ES func {left ( stack {following # activity} right )}- EF func {left ( stack {this # activity} right )}

TF = LS - ES

TF of 30 = 2

FF of 30 = 0 lag

FF of 50 = 2

Lecture on PERT

Given:

Example:T_E = 243 days

S_T = 22.4

What is the probability of finishing a construction job in 277 days or less?

Solution:We know that there is a 50% probability of finishing in 243 days or less and 50% probability of finishing in 243 days or more.

Since T_E = 243 days

D = 277 days

D > T_E

Go to Table 2-9, the area under the right-hand portion of the distribution is approxi mately = 0.44; combining this area with the left half of the curve yields a probability of:

0.50 + 0.44 = 0.94 or a 95% probability of finishing within 277 days.

Remember that the calculated probability of finishing in exactly 277 days = 0, which means quite different from finishing "within" 277 days (which means 277 days or less).

;P = Pessimistic

;O = Optimistic

Table 2-8 Activities, events, duration estimates, expected durations, variances, and standard deviations for Pert example problem

Activity			Duration, days
Code	From	To	Optimistic	Most Probable	Pessimistic	Expected Duration Days	Variance	Standard Deviation
A	1	2	3	6	15	7.0	4.00	2.0
B	1	6	2	5	14	6.0	4.00	2.0
C	2	3	6	12	30	14.0	16.00	4.0
D	2	4	2	5	8	5.0	1.00	1.0
E	3	5	5	11	17	11.0	4.00	2.0
F	4	5	3	6	15	7.0	4.00	2.0
G	5	8	1	4	7	4.0	1.00	1.0
H	6	7	3	8	31	11	21.78	4.7
I	7	8	3	17	37	18	32.11	5.7

variances along the critical path In our example,

S_T² =4.00 + 16.00 + 4.00 +1.00 = 25.00

and the overall project standard deviation (S_T) is [from Eq. (2.6)]

S_T = (25.00)^1/2 = 5.0 days

Once the overall expected project duration (T_e) and standard deviation (S_e) are known, the probabilities associated with completing the project within a certain length of time may be calculated

with confidence. Turning again to statistics, we see that, although the shape of individual activity frequency distribution may be quite skewed and different from each other, the overall project frequency distribution can be closely approximated by the well-known normal distribution ( the familiar bell-shaped curve also known as the Gaussian distribution). The inference of this is known as the central limit theorem and is covered in detail in texts dealing with probability and statistics. The shape of the normal distribution curve is shown in Figure 2-15. It is symmetrical about the average and can be completely described by only two values: its average and its standard deviation. In other words, two normal distributions are identical if their respective averages and standard deviations are the same. Knowing these two values, the probability (area) under any portion of the curve can be calculated. For example, using the overall expected project duration and standard deviation, we see that

T_e +/- 1.0S_T will account for 68.2 percent of the area under the curve

T_e +/- 2.0S_T will account for 95.4 percent of the area under the curve

T_e +/- 3.0S_T will account for 99.9+ percent of the area under the curve

Table 2-9 contains the areas for portions of the normal distribution

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