PERT mean and SD of two projects

PERT

PERT is a probabilistic network technique used when activity times are uncertain. Three estimates are made for each activity: optimistic $t_o$, most likely $t_m$, pessimistic $t_p$.

$$t_e = \frac{t_o + 4t_m + t_p}{6}, \qquad \sigma = \frac{t_p - t_o}{6}$$

Uses:

• planning and control of new, research and development or first-time projects with uncertain durations;
• finding the expected project duration and the probability of finishing by a due date;
• identifying critical activities with large variance for special attention;
• scheduling, coordination and review of large programmes (event-oriented control).

Calculation

Project A ($t_o=5,\ t_m=8,\ t_p=12$):

$$\begin{aligned} t_e &= \frac{5 + 4(8) + 12}{6} = \frac{49}{6} = 8.17 \\ \sigma &= \frac{12 - 5}{6} = 1.17, \quad \text{Variance} = 1.36 \end{aligned}$$

Project B ($t_o=6,\ t_m=8,\ t_p=11$):

$$\begin{aligned} t_e &= \frac{6 + 4(8) + 11}{6} = \frac{49}{6} = 8.17 \\ \sigma &= \frac{11 - 6}{6} = 0.83, \quad \text{Variance} = 0.69 \end{aligned}$$

Both have the same expected time (8.17), but B has the smaller standard deviation and variance. So project B is better and more certain, because its completion time is less scattered about the mean, so the risk of delay is lower.

Answer: mean 8.17 for both; SD 1.17 (A) and 0.83 (B); project B is preferred.