Chapter 7, Problem 32P

a

Summary Introduction

Interpretation:

Expected completion time of project.

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

Calculate the estimated task time:

Task	Immediate predecessors	Optimistic time	Most likely time	Pessimistic time	Estimated task time	Variance
A	None	1	2	3	2	0.11
B	A	2	4	6	4	0.44
C	None	2	4	6	4	0.44
D	B and C	1	2	9	3	1.78

Calculate the estimated task time for task A:

It is calculated by dividing the value attained by adding optimistic time, pessimistic time and four times of most likely time with the value 6.

  $\begin{array}{c}\text{Estimated}\text{task}\text{time}\text{of}\text{Task}\text{A}\text{=}\frac{\text{Optimistic}\text{+}\text{(4×}\text{Most}\text{likely)}\text{+Pessimistic}}{\text{6}}\\ \text{=}\frac{1\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\text{+}\text{(4×}2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\text{)}\text{+}\text{3}\text{hour}}{\text{6}}\\ \text{=}\frac{12\text{hours}}{\text{6}}\\ \text{=}2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\end{array}$

Hence, the estimated task time for task A is 2 hours.

Calculate the estimated task time for task B:

It is calculated by dividing the value attained by adding optimistic time, pessimistic time and four times of most likely time with the value 6.

  $\begin{array}{c}\text{Estimated}\text{task}\text{time}\text{of}\text{Task}\mathrm{B}\text{=}\frac{\text{Optimistic}\text{+}\text{(4×}\text{Most}\text{likely)}\text{+Pessimistic}}{\text{6}}\\ \text{=}\frac{2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\text{+}\text{(4×}4\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\text{)}\text{+}6\text{hour}}{\text{6}}\\ \text{=}\frac{24\text{hours}}{\text{6}}\\ \text{=}4\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\end{array}$

Hence, the estimated task time for task B is 4 hours.

Calculation of estimated task time for Task C:

It is calculated by dividing the value attained by adding optimistic time, pessimistic time and four times of most likely time with the value 6.

  $\begin{array}{c}\text{Estimated}\text{task}\text{time}\text{of}\text{Task}\mathrm{C}\text{=}\frac{\text{Optimistic}\text{+}\text{(4×}\text{Most}\text{likely)}\text{+Pessimistic}}{\text{6}}\\ \text{=}\frac{2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\text{+}\text{(4×}4\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\text{)}\text{+}6\text{hour}}{\text{6}}\\ \text{=}\frac{24\text{hours}}{\text{6}}\\ =\text{4 hours}\end{array}$

Hence, the estimated task time for Task C is 4 hours.

Calculate the estimated task time for Task D:

It is calculated by dividing the value attained by adding optimistic time, pessimistic time and four times of most likely time with the value 6.

  $\begin{array}{c}\text{Estimated}\text{task}\text{time}\text{of}\text{Task}\mathrm{D}\text{=}\frac{\text{Optimistic}\text{+}\text{(4×}\text{Most}\text{likely)}\text{+Pessimistic}}{\text{6}}\\ \text{=}\frac{1\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\text{+}\text{(4×}2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\text{)}\text{+}9\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{\text{6}}\\ \text{=}\frac{18\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{6}\\ \text{=}3\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\end{array}$

Hence, the estimated task time for task D is 3 hours.

Calculate the variance for Task A:

It is calculated by dividing the value attained subtracting the optimistic time from the pessimistic time with the value of 6. Then, square the attained value.

  $\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{A}={\left(\frac{\mathrm{P}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{c}-\mathrm{O}\mathrm{p}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{c}}{6}\right)}^2$

  $\begin{array}{l}\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{A}={\left(\frac{3\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}-1\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ ={\left(\frac{2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ =0.11\end{array}$

Calculate the variance for Task B:

It is calculated by dividing the value attained subtracting the optimistic time from the pessimistic time with the value of 6. Then, square the attained value.

  $\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{B}={\left(\frac{\mathrm{P}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{c}-\mathrm{O}\mathrm{p}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{c}}{6}\right)}^2$

  $\begin{array}{l}\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{D}={\left(\frac{6\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}-2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ ={\left(\frac{4\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ =0.44\end{array}$

Hence, the variance for task B is 0.44

Calculate the variance for Task C:

It is calculated by dividing the value attained subtracting the optimistic time from the pessimistic time with the value of 6. Then, square the attained value.

  $\begin{array}{l}\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{D}={\left(\frac{6\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}-2\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ ={\left(\frac{4\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ =0.44\end{array}$

Hence, the variance for Task C is 0.44

Calculate the variance for Task D:

It is calculated by dividing the value attained subtracting the optimistic time from the pessimistic time with the value of 6. Then, square the attained value.

  $\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{D}={\left(\frac{\mathrm{P}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{c}-\mathrm{O}\mathrm{p}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{c}}{6}\right)}^2$

  $\begin{array}{l}\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{T}\mathrm{a}\mathrm{s}\mathrm{k}\mathrm{D}={\left(\frac{9\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}-1\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ ={\left(\frac{8\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{6}\right)}^2\\ =1.78\end{array}$

Hence, the variance for Task D is 1.78

Project network diagram for given information:

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

  

A

The critical path in the project network:

The paths in the given project network are as follows:

A-B-D

C-D

There are two paths in the project network.

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

Hence, the critical path is A-B-D and the project completion time is 9 days.

b

Summary Introduction

Interpretation:

Time duration in the beginning of reconfiguration.

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

The late start time for activity C:

According to the project network, the late start time for activity C is 2 day.

c

Summary Introduction

Interpretation:

Probability that warehouse will finish 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

The probability to finish the operation, which started at 1:00 a.m. and finished at 11:00 a.m. on the same day:

The value of Z should be calculated to find the probability to find the operation.

It is calculated by dividing the value attained by subtracting the completion time from the time (time gap from 1:00 a.m. to 11:00 a.m.) with the square root of the variance of critical path (sum of the variance of critical path).

  $\text{Z}\text{=}\frac{\text{Time}\text{-}\text{Completion}\text{time}}{\sqrt{\text{Variance}\text{of}\text{critical}\text{path}}}$

  $\begin{array}{l}\text{Z}\text{=}\frac{10\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\text{-}9\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\sqrt{0.11+0.44+1.78}}\\ =\frac{1\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\sqrt{2.33}}\\ =0.6551\end{array}$

Hence, the value of Z is 0.6551

For Z= 0.6551, the normal distribution table value is 0.7422. Thus, the probability is 74%.