Chapter 1, Problem 1SE

To determine

The total project time and all the critical paths.

Answer to Problem 1SE

The total project time is $\underset{\_}{18}$ and the critical path is $\underset{\_}{\text{B-D-G-I}}$.

Explanation of Solution

The given network diagram does not have any arrows.

Thus, by default consider all the edges are directed from left to right.

The possible paths are,

$\begin{array}{l}\text{A-C-F-H, A-C-F-I, A-D-F-H, A-D-F-I, A-D-G-I, A-D-G-J, B-D-F-H, B-D-F-I, B-D-G-I,}\\ \text{B-D-G-J, B-E-G-I and B-E-G-J}\end{array}$

Obtain the total time taken for the paths $\text{A-C-F-H and A-C-F-I}$.

$\begin{array}{c}\text{A-C-F-H}=3+7+2+5\\ =17\\ \text{A-C-F-I}=3+7+2+5\\ =17\end{array}$

Obtain the total time taken for the paths $\text{A-D-F-H and A-D-F-I}$.

$\begin{array}{c}\text{A-D-F-H}=3+5+2+5\\ =15\\ \text{A-D-F-I}=3+5+2+5\\ =15\end{array}$

Obtain the total time taken for the paths $\text{A-D-G-I and A-D-G-J}$.

$\begin{array}{c}\text{A-D-G-I}=3+5+4+5\\ =17\\ \text{A-D-G-J}=3+5+4+3\\ =15\end{array}$

Obtain the total time taken for the paths $\text{B-D-F-H and B-D-F-I}$.

$\begin{array}{c}\text{B-D-F-H}=4+5+2+5\\ =16\\ \text{B-D-F-I}=4+5+2+5\\ =16\end{array}$

Obtain the total time taken for the paths $\text{B-D-G-I and B-D-G-J}$.

$\begin{array}{c}\text{B-D-G-I}=4+5+4+5\\ =18\\ \text{B-D-G-J}=4+5+4+3\\ =16\end{array}$

Obtain the total time taken for the paths $\text{B-E-G-I and B-E-G-J}$.

$\begin{array}{c}\text{B-E-G-I}=4+4+4+5\\ =17\\ \text{B-E-G-J}=4+4+4+3\\ =15\end{array}$

The total project time is the largest time taken among the all possible paths.

Thus, the total project time is, $\underset{\_}{18}$.

The path which takes more time to complete the task is called the critical path.

Therefore, the critical path is $\underset{\_}{\text{B-D-G-I}}$.