--- title: "PERT mean and SD of two projects" subject: "Project Engineering" chapter: 3 key: "semester_7/project_engineering:pert-mean-sd-two-projects" source: https://www.bctengineer.com/q/semester_7/project_engineering/pert-mean-sd-two-projects site: BCT Engineering --- # PERT mean and SD of two projects Project Engineering · Chapter 3: Project Planning and Scheduling - Asked in: 2068 Baisakh (old course), 8 marks - Video: [Project Scheduling 2 -Calculating variance and probabilities -PERT/CPM](https://www.youtube.com/watch?v=2124ZV4wbQ8) — Joshua Emmanuel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/project_engineering/pert-mean-sd-two-projects - Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3 ## Question Define PERT and discuss its uses. Project A and B have to, tm and tp as 5, 8, 12 and 6, 8, 11 respectively. Find the mean and standard deviation. Which project is better and more certain? ## Short answer Using PERT formulas, mean = (t_o+4t_m+t_p)/6 and SD = (t_p–t_o)/6. For Project A: mean 8.17, SD 1.17; for Project B: mean 8.17, SD 0.83. Both have the same expected time, but Project B is more certain because of its smaller standard deviation. ## Answer #### 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** ($t_o=6,\ t_m=8,\ t_p=11$): $$ \begin{aligned} t_e &= \frac{6 + 4(8) + 11}{6} = \frac{49}{6} = 8.17 \\ \sigma &= \frac{11 - 6}{6} = 0.83, \quad \text{Variance} = 0.69 \end{aligned} $$ | Project | Mean $t_e$ | SD $\sigma$ | Variance | |---|---|---|---| | A | 8.17 | 1.17 | 1.36 | | B | 8.17 | 0.83 | 0.69 | 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. PERT is most suitable for which kind of project? A. Repetitive work with accurately known durations B. Projects whose cost is the only concern C. New or research-type projects with uncertain durations D. Very small jobs with a single activity Answer: C. PERT is a probabilistic technique used when activity times are uncertain, as in new, R and D or first-time projects. (Key point: Where PERT is used) 2. Project A has t_o = 5, t_m = 8 and t_p = 12. What is its mean time t_e? A. 8.17 B. 8.00 C. 8.50 D. 8.33 Answer: A. t_e = (5 + 4x8 + 12)/6 = 49/6 = 8.17. (Key point: Expected time of project A) 3. Project B has t_o = 6 and t_p = 11. What is its standard deviation? A. 1.17 B. 0.83 C. 1.36 D. 0.69 Answer: B. sigma = (t_p - t_o)/6 = (11 - 6)/6 = 0.83; 0.69 is its variance. (Key point: Standard deviation of project B) 4. Both projects have mean 8.17, with SD 1.17 (A) and 0.83 (B). Which is better and why? A. A, because its larger SD gives more flexibility B. Both are equal because the means are equal C. A, because its pessimistic time is larger D. B, because its completion time is less scattered about the mean Answer: D. With equal means, the smaller SD and variance of B mean lower risk of delay, so B is more certain. 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