---
title: "Expected time of contractors (PERT)"
subject: "Project Engineering"
chapter: 3
key: "semester_7/project_engineering:pert-contractor-expected-time"
source: https://www.bctengineer.com/q/semester_7/project_engineering/pert-contractor-expected-time
site: BCT Engineering
---

# Expected time of contractors (PERT)

Project Engineering · Chapter 3: Project Planning and Scheduling

- Asked in: 2072 Kartik
- 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-contractor-expected-time
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-3

## Question

Find out the expected time of each contractor to complete a given project having the following details. Also find out which contractor you prefer for operation and why?

| Contractor | to | tm | tp |
|---|---|---|---|
| A | 5 | 7 | 13 |
| B | 6 | 11 | 12 |
| C | 3 | 5 | 7 |

## Short answer

Using te = (t_o+4t_m+t_p)/6 and σ = (t_p–t_o)/6. Contractor A: te = 7.67, σ = 1.33; B: te = 10.33, σ = 1.00; C: te = 5.00, σ = 0.67. Contractor C is preferred because it has the shortest expected time and smallest standard deviation, making it the fastest and most certain.

## Answer

Expected time $t_e = \dfrac{t_o + 4t_m + t_p}{6}$; standard deviation $\sigma = \dfrac{t_p - t_o}{6}$ (the unit of time is as given).

**Contractor A** ($5, 7, 13$): $t_e = \dfrac{5 + 28 + 13}{6} = \dfrac{46}{6} = 7.67$, $\sigma = \dfrac{13 - 5}{6} = 1.33$, variance $= 1.78$

**Contractor B** ($6, 11, 12$): $t_e = \dfrac{6 + 44 + 12}{6} = \dfrac{62}{6} = 10.33$, $\sigma = \dfrac{12 - 6}{6} = 1.00$, variance $= 1.00$

**Contractor C** ($3, 5, 7$): $t_e = \dfrac{3 + 20 + 7}{6} = \dfrac{30}{6} = 5.00$, $\sigma = \dfrac{7 - 3}{6} = 0.67$, variance $= 0.44$

| Contractor | $t_o$ | $t_m$ | $t_p$ | $t_e$ | $\sigma$ | Variance |
|---|---|---|---|---|---|---|
| A | 5 | 7 | 13 | 7.67 | 1.33 | 1.78 |
| B | 6 | 11 | 12 | 10.33 | 1.00 | 1.00 |
| C | 3 | 5 | 7 | 5.00 | 0.67 | 0.44 |

#### Choice of contractor

Contractor **C** is preferred because:
- it has the shortest expected time (5.00 against 7.67 and 10.33);
- it also has the smallest standard deviation (0.67) and variance (0.44), so its completion time is the most certain, and the risk of delay is the least.

(Cost and quality are assumed equal; if they differ, they must be considered along with time.)

**Answer: $t_e$ = 7.67 (A), 10.33 (B), 5.00 (C); choose contractor C.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Contractor A has t_o = 5, t_m = 7 and t_p = 13. What is t_e?
   A. 8.00
   B. 7.67
   C. 9.00
   D. 7.00
   Answer: B. t_e = (5 + 28 + 13)/6 = 46/6 = 7.67. (Key point: Expected time of contractor A)

2. Contractor B has t_o = 6, t_m = 11 and t_p = 12. What is t_e?
   A. 11.00
   B. 9.67
   C. 10.33
   D. 9.00
   Answer: C. t_e = (6 + 44 + 12)/6 = 62/6 = 10.33. (Key point: Expected time of contractor B)

3. What is the standard deviation of contractor C (3, 5, 7)?
   A. 0.67
   B. 0.44
   C. 1.00
   D. 0.33
   Answer: A. sigma = (7 - 3)/6 = 0.67; 0.44 is the variance. (Key point: Standard deviation formula)

4. Why is contractor C preferred?
   A. It has the largest pessimistic time
   B. It has the shortest expected time and the smallest SD
   C. It has the shortest expected time but the largest SD
   D. It has the largest variance, so most flexibility
   Answer: B. C has te = 5.00 (shortest) and SD 0.67 (smallest), so it is both quickest and most certain. (Key point: Choosing the contractor)

## Related in Project Engineering

- [Network errors; expected time of engineers](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Anetwork-errors-expected-time-engineers) — Project Engineering, chapter 3
- [Bar chart; expected time and variance](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Abar-chart-draw-adv-te-variance) — Project Engineering, chapter 3
- [PERT mean and SD of two projects](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Apert-mean-sd-two-projects) — Project Engineering, chapter 3
- [CPM vs PERT; te and variance for x, y](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-pert-te-variance-x-y) — Project Engineering, chapter 3
- [CPM network, building project (A to K)](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acpm-network-eleven-act-s73) — Project Engineering, chapter 3
- [Project environment and dynamic operation](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aproject-environment-dynamic-example) — Project Engineering, chapter 1

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