Question

Information on Activity-Time duration of a project is provided below. The expected project duration in weeks, is ____________. (in integer)

Hint:

Critical path = sequence of activities with the maximum expected duration. PERT expected time = (optimistic + 4×most likely + pessimistic)/6.

Answer 1

Correct Answer: 39

Step 1: Compute expected time of each activity using PERT formula.
\[ t_e = \frac{a + 4m + b}{6} \] \[ \begin{aligned} t_A &= \frac{4 + 4(15) + 20}{6} = 14.0
t_B &= \frac{4 + 4(8) + 12}{6} = 8.0
t_C &= \frac{6 + 4(11) + 16}{6} = 11.0
t_D &= \frac{12 + 4(13) + 20}{6} = 14.0
t_E &= \frac{3 + 4(8) + 13}{6} = 8.0
t_F &= \frac{25 + 4(35) + 45}{6} = 35.0 \end{aligned} \] Step 2: Determine all possible project paths. Using predecessor–successor relations: - Path 1: \(1 \rightarrow 2 \rightarrow 3 \rightarrow 4\) Activities: A → C → D \[ T_1 = 14 + 11 + 14 = 39\ \text{weeks} \] - Path 2: \(1 \rightarrow 3 \rightarrow 4\) Activities: B → D \[ T_2 = 8 + 14 = 22\ \text{weeks} \] - Path 3: \(1 \rightarrow 4\) Activities: F \[ T_3 = 35\ \text{weeks} \] - Path 4: \(1 \rightarrow 2 \rightarrow 4\) Activities: A → E \[ T_4 = 14 + 8 = 22\ \text{weeks} \] Step 3: Critical path is the longest-duration path. \[ T_{\max} = \max(39,22,35,22) = 39\ \text{weeks} \] \[ \boxed{39} \]