Chapter 7, Problem 31P

a

Summary Introduction

Interpretation:

Critical path of this network.

Concept Introduction: The critical path is the arrangement of project activities which gives an estimated time duration under which the project will be completed. The project activities may include float activities which can be delayed focusing on the shortest time duration.

Explanation of Solution

Project network diagram for given information:

The below image shows the project network with the early start, early finish, late start and late finish:

  

The critical path in the project network:

The paths in the given project network are as follows:

A-E-J

A-C-F

B-D-G-I-J

B-D-H-I-J

There are four paths in the project network.

Out of the above mentioned path, the path that has the equal early start and late start at the same time early finish and late finish are the critical path.

Here two paths are satisfying the above condition, which are as follows:

A-C-F

B-D-H-I-J

When the network has more than one critical path, the path, which has the maximum project completion time, should be selected.

Hence, the critical path is B-D-H-I-J.

b

Summary Introduction

Interpretation:

Completion time of the project.

Concept Introduction: Early start time: The rule for the early start time of a task is that it is equal to the largest early finish time of the task’s immediate predecessors.

Early finish time: The early finish time of a task is the addition of both task time and early start time of the task.

Explanation of Solution

When the network has more than one critical path, the path, which has the maximum project completion time, should be selected.

A-C-F

B-D-H-I-J

Hence, the critical path is B-D-H-I-J.

As the critical path is determined as B-D-H-I-J, the project completion time is 11 days.

c

Summary Introduction

Interpretation:

The slack of activity A.

Concept Introduction: The critical path is the arrangement of project activities which gives an estimated time duration under which the project will be completed. The project activities may include float activities which can be delayed focusing on the shortest time duration.

Explanation of Solution

The slack of the activity is calculated by subtracting the early start from the late finish.

  $\begin{array}{l}\text{Slack}\text{of}\text{the}\text{activity}\text{=}\text{Late}\text{finish}\text{-}\text{Early}\text{finish}\\ \text{=}\text{1-1}\\ \text{=}0\end{array}$

Hence, the slack of activity A is 0.

d

Summary Introduction

Interpretation:

Task to be crashed and its cost.

Concept Introduction: The critical path is the arrangement of project activities which gives an estimated time duration under which the project will be completed. The project activities may include float activities which can be delayed focusing on the shortest time duration.

Explanation of Solution

The cost of crashing.

It is given that the paths should be crashed and the total time of the path should not exceed 8 days. The cost to crash Activity B is $500 per crash, Activity H is $200 per crash, and all eligible activities are $100 per crash.

Use the following steps to crash the activities:

The paths and the total task time to complete the path within 8 days is shown below:

Path A-E-J:

Total task time is calculated by adding the duration of all the activities in the path.

  $\begin{array}{l}\text{Total}\text{task}\text{time}\text{=}\text{Duration}\text{of}\text{Activity}\text{A+}\text{Duration}\text{of}\text{Activity}\text{E+}\text{Duration}\text{of}\text{Activity}\text{J}\\ \text{=}\text{1}\text{day+}\text{7}\text{days+}\text{1}\text{day}\\ \text{=}\text{9}\text{days}\end{array}$

Hence, the total task time is 9 days. It exceeds 8 days. Thus, the path should be crashed. Activity A and Activity J cannot be crashed as the duration is 1. Activity E should be crashed from 7 days to 6 days.

  $\text{Modified}\text{total}\text{task}\text{time}\text{=}\left(\begin{array}{l}\text{Duration}\text{of}\text{activity}\text{A+}\text{Duration}\text{of}\text{activity}\text{E}\\ \text{+}\text{Duration}\text{of}\text{activity}\text{J}\end{array}\right)$

  $\begin{array}{l}\text{=1}\text{day}\text{+}\text{2}\text{days+}\text{1}\text{day}\\ \text{=}\text{4}\text{days}\end{array}$

Hence, the total task time is 4 days. It does not exceed 8 days. Thus, the path cannot be crashed.

Path B-D-G-I-J:

Total task time is calculated by adding the duration of all activities in the path.

  $\text{Total}\text{task}\text{time}\text{=}\left(\begin{array}{l}\text{Duration}\text{of}\text{activity}\text{B+}\text{Duration}\text{of}\text{activity}\text{D}\\ \text{+}\text{Duration}\text{of}\text{activity}\text{G+}\text{Duration}\text{of}\text{activity}\text{I}\\ \text{+}\text{Duration}\text{of}\text{activity}\text{J}\end{array}\right)$

  $\begin{array}{l}\text{=}\text{4}\text{days+}\text{2}\text{days+}\text{2}\text{days+}\text{1}\text{day+}\text{1}\text{day}\\ \text{=}\text{10}\text{days}\end{array}$

Hence, the total task time is 10 days. It exceeds 8 days. Thus, the path should be crashed. However, it costs $500 to crash one day of Activity B. Thus, Activity D and Activity G can be crashed from 2 days to 1 day.

  $\text{Modified}\text{Total}\text{task}\text{time}\text{=}\left(\begin{array}{l}\text{Duration}\text{of}\text{activity}\text{B+}\text{Duration}\text{of}\text{activity}\text{D}\\ \text{+}\text{Duration}\text{of}\text{activity}\text{G+}\text{Duration}\text{of}\text{activity}\text{I}\\ \text{+}\text{Duration}\text{of}\text{activity}\text{J}\end{array}\right)$

  $\begin{array}{l}\text{=}\text{4}\text{days+}\text{1}\text{day+}\text{1}\text{day+}\text{1}\text{day+}\text{1}\text{day}\\ \text{=}\text{8}\text{days}\end{array}$

The cost for crashing is $200 as the crashing was done for two days in Activity D and Activity G. It is given that the activities except for Activity B and H cost $100 per day.

Path B-D-H-I-J

Total task time is calculated by adding the duration of all the activities in the path.

  $\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{t}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}=\left(\begin{array}{l}\mathrm{D}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{B}+\mathrm{D}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{D}\\ +\mathrm{D}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{H}+\mathrm{D}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{I}\\ +\mathrm{D}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{J}\end{array}\right)$

  $\begin{array}{l}=4\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}+1\mathrm{d}\mathrm{a}\mathrm{y}+3\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}+1\mathrm{d}\mathrm{a}\mathrm{y}+1\mathrm{d}\mathrm{a}\mathrm{y}\\ =10\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}\end{array}$

Hence, the total task time is 10 days. It exceeds 8 days. Thus, the path should be crashed. Activity I, Activity D (crashed in the previous step), Activity J cannot be crashed as the duration is 1 day. Activity B can be crashed. However, it costs $500 to crash one day of Activity B. Thus, Activity H can be crashed from 3 days to 2 days.

  $\text{Modified}\text{total}\text{task}\text{time}\text{=}\left(\begin{array}{l}\text{Duration}\text{of}\text{activity}\text{B+}\text{Duration}\text{of}\text{activity}\text{D}\\ \text{+}\text{Duration}\text{of}\text{activity}\text{G+}\text{Duration}\text{of}\text{activity}\text{I+}\\ \text{Duration}\text{of}\text{activity}\text{J}\end{array}\right)$

  $\begin{array}{l}\text{=}\text{4}\text{days+}\text{1}\text{day+}\text{1}\text{day+}\text{1}\text{day+}\text{1}\text{day}\\ \text{=}\text{8}\text{days}\end{array}$

The cost for crashing is $400 as the crashing was done for two days in Activity H. It is given Activity H costs $100 per day.

Total crashing cost:

Total crashing cost is calculated by adding all the crashing cost done in the previous steps.

  $\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{c}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\cos \mathrm{t}=\left(\begin{array}{l}\mathrm{C}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{c}\cos \mathrm{t}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{A}-\mathrm{E}-\mathrm{J}+\\ \mathrm{C}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\cos \mathrm{t}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{B}-\mathrm{D}-\mathrm{G}-\mathrm{I}-\mathrm{J}\\ +\mathrm{C}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\cos \mathrm{t}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{B}-\mathrm{D}-\mathrm{H}-\mathrm{I}-\mathrm{J}\end{array}\right)$

  $\begin{array}{l}=\mathrm{\$}100+\mathrm{\$}200+\mathrm{\$}400\\ =\mathrm{\$}700\end{array}$