Computing Expected Activity Times and Variances, and the Expected Duration and Standard Deviation of Each Path The network diagram for a project is shown in the accompanying figure, with three time estimates for each activity. Activity times are in weeks. Do the following: a. Compute the expected

Question

Computing Expected Activity Times and Variances, and the Expected Duration and Standard Deviation of Each Path
The network diagram for a project is shown in the accompanying figure, with three time estimates for each activity. Activity times are in weeks. Do the following:
a. Compute the expected time for each activity and the expected duration for each path.
b. Identify the critical path.
c. Compute the variance of each activity and the variance and standard deviation of each path.

Accepted Answer

a.

		TIMES
Path	Activity	[latex]t_o[/latex]	[latex]t_m[/latex]	[latex]t_p[/latex]	[latex]t_0=\frac{t_o +4t_m +t_p}{6}[/latex]	Path Total
a–b–c	a	1	3	4	[latex]\left. \begin{matrix} 2.83 \ 4.00 \ 3.17 \end{matrix} \right\}[/latex]	10.00
	b	2	4	6
	c	3	3	5
d–e–f	d	3	4	5	[latex]\left. \begin{matrix} 4.00 \ 5.00\ 7.00 \end{matrix} \right\}[/latex]	16.00
	e	3	5	7
	f	5	7	9
g–h–i	g	2	3	6	[latex]\left. \begin{matrix} 3.33\ 6.00 \ 4.17 \end{matrix} \right\}[/latex]	13.50
	h	4	6	8
	i	3	4	6

b. The path that has the longest expected duration is the critical path. Because path d–e–f has the largest path total, it is the critical path.

c.

		TIMES
Path	Activity	[latex]t_o[/latex]	[latex]t_m[/latex]	[latex]t_p[/latex]	[latex]\sigma ^2_{oct}=\frac{(t_p-t_o)^2}{36}[/latex]	[latex]σ^2_{Path}[/latex]	[latex]σ_{path}[/latex]
a–b–c	a	1	3	4	[latex] \left. \begin{matrix} (4 − 1)^2/36 = 9/36 \ (6 − 2)^2/36 = 16/36 \ (5 − 2)^2/36 = 9/36 \end{matrix}\right\} [/latex]	34/36 = 0.944	0.97
	b	2	4	6
	c	2	3	5
d–e–f	d	3	4	5	[latex]\left.\begin{matrix} (5 − 3)^2/36 = 4/36 \(7 − 3)^2/36 = 16/36 \(9 − 5)^2/36 = 16/36 \end{matrix}\right\} [/latex]	36/36 = 1.00	1.00
	e	3	5	7
	f	5	7	9
g–h–i	g	2	3	6	[latex]\left. \begin{matrix} (6 − 2)^2/36 = 16/36 \(8 − 4)^2/36 = 16/36\(6 − 3)^2/36 = 9/36\end{matrix} \right\} [/latex]	41/36 = 1.139	1.07
	h	4	6	8
	i	3	4	6

Question 17.5: Computing Expected Activity Times and Variances, and the Exp...

Related Answered Questions

Verified Answer:

Expected times in weeks, as well as variances, for the major activities of an R&D project are depicted in the following table. Determine the probability that project completion time will be: a. 50 weeks or less b. More than 50 weeks ...

Verified Answer:

Costs for a project are $12,000 per week for as long as the project lasts. The project manager has supplied the cost and time information shown. Use the information to: a. Determine an optimum crashing plan. b. Graph the total costs for the plan. ...

Verified Answer:

Using the computing algorithm, determine the slack times for the following AOA diagram. Identify the activities on the critical path. ...

Verified Answer:

A path in a network has three activities. Their standard deviations are 1.50, 0.80, and 1.30. Find the path standard deviation. ...

Verified Answer:

The following table contains information related to the major activities of a research project. Use the information to do the following: a. Draw a precedence diagram using AOA. b. Find the critical path. c. Determine the expected length of the project. ...

Verified Answer:

Optimal Project Crashing Using the following information, develop the optimal time–cost solution. Project costs are $1,000 per day. ...

Verified Answer:

Computing Slack Times Compute slack times for the preceding example. ...

Verified Answer:

Computing the Latest Finishing and Starting Times Compute the latest finishing and starting times for the precedence diagram developed in Example 2. ...

Verified Answer:

Computing Earliest Starting and Finishing Times Compute the earliest starting time and earliest finishing time for each activity in the diagram shown in Figure 17.5. ...

Verified Answer: