Activity	Duration (in weeks)
	t₀	tₘ	tₚ
1–2	2	4	12
1–3	2	4	6
1–4	3	4	11
2–5	3	6	6
3–4	2	5	14
3–5	3	3	3
4–5	3	6	15
5–6	2	5	8

$\text{The expected project length is \_\_\_\_\_\_ weeks (in integers).}$

The correct answer is 22.

To determine the expected project length, we use the Project Evaluation and Review Technique (PERT). First, we calculate the expected duration ($t_e$) for each activity using the standard PERT formula:
$t_e=\frac{t_0+4t_m+t_p}{6}$
where:
$t_0$ is the optimistic time,
$t_m$ is the most likely time, and
$t_p$ is the pessimistic time.

Let's calculate the expected duration for each activity based on the values in the table:

1. Activity 1–2:
$t_e=\frac{2+4\left(4\right)+12}{6}=\frac{2+16+12}{6}=\frac{30}{6}=5$ weeks

2. Activity 1–3:
$t_e=\frac{2+4\left(4\right)+6}{6}=\frac{2+16+6}{6}=\frac{24}{6}=4$ weeks

3. Activity 1–4:
$t_e=\frac{3+4\left(4\right)+11}{6}=\frac{3+16+11}{6}=\frac{30}{6}=5$ weeks

4. Activity 2–5:
$t_e=\frac{3+4\left(6\right)+6}{6}=\frac{3+24+6}{6}=\frac{33}{6}=5.5$ weeks

5. Activity 3–4:
$t_e=\frac{2+4\left(5\right)+14}{6}=\frac{2+20+14}{6}=\frac{36}{6}=6$ weeks

6. Activity 3–5:
$t_e=\frac{3+4\left(3\right)+3}{6}=\frac{3+12+3}{6}=\frac{18}{6}=3$ weeks

7. Activity 4–5:
$t_e=\frac{3+4\left(6\right)+15}{6}=\frac{3+24+15}{6}=\frac{42}{6}=7$ weeks

8. Activity 5–6:
$t_e=\frac{2+4\left(5\right)+8}{6}=\frac{2+20+8}{6}=\frac{30}{6}=5$ weeks

Next, we identify all possible paths from the start node (1) to the end node (6) and calculate their total expected durations:

Path A: 1–2–5–6
Duration = $t_e(1–2)+t_e(2–5)+t_e(5–6)=5+5.5+5=15.5$ weeks

Path B: 1–3–5–6
Duration = $t_e(1–3)+t_e(3–5)+t_e(5–6)=4+3+5=12$ weeks

Path C: 1–3–4–5–6
Duration = $t_e(1–3)+t_e(3–4)+t_e(4–5)+t_e(5–6)=4+6+7+5=22$ weeks

Path D: 1–4–5–6
Duration = $t_e(1–4)+t_e(4–5)+t_e(5–6)=5+7+5=17$ weeks

The critical path is the path with the longest expected duration, which is Path C (1–3–4–5–6) with a total expected duration of 22 weeks. Thus, the expected project length is 22 weeks.