A project consists of four activities with information as shown in the table.What is the overall variance of the project's duration? $\begin{array}{|l|l|l|l|}
\hline&&&\text { Standard }\
\text { Activity } & \text { Predecessor } & \text { Length } & \begin{array}{l}
\text { Deviation }
\end{array} \
\hline \text { A } & - & 10 & 2 \
\hline \text { B } & \text { A } & 12 & 3 \
\hline \text { C } & \text { A } & 16 & 2 \
\hline \text { D } & \text {B,C} & 14 & 3\
\hline
\end{array}$

A)13
B)17
C)26
D)35

Question

A project consists of four activities with information as shown in the table.What is the overall variance of the project's duration? $\begin{array}{|l|l|l|l|}
\hline&&&\text { Standard }\
 \text { Activity } & \text { Predecessor } & \text { Length } & \begin{array}{l}
\text { Deviation }
\end{array} \
\hline \text { A } & - & 10 & 2 \
\hline \text { B } & \text { A } & 12 & 3 \
\hline \text { C } & \text { A } & 16 & 2 \
\hline \text { D } & \text {B,C} & 14 & 3\
\hline 
\end{array}$

A)13
B)17
C)26
D)35

Quizplus · Accepted Answer

The overall variance of the project's duration is calculated by summing the variances of activities on the critical path. The critical path is the longest path through the project, which in this case is A-C-D (10+16+14=40) or A-B-D (10+12+14=36). Since A-C-D is longer, it's the critical path. The variances of activities A, C, and D are 4 (2^2), 4 (2^2), and 9 (3^2) respectively. Summing these gives 4+4+9=17.

A Project Consists of Four Activities with Information as Shown