Question 1

In assigning a value for the most pessimistic (b)duration,the project manager should estimate the duration of the activity to have a 99% likelihood that it will take b or less amount of time.

Accepted Answer

This statement is true as the most pessimistic (b) duration is the estimated time that an activity would take if everything that could go wrong, does go wrong. In order to assign a value for the most pessimistic duration, the project manager should estimate the duration of the activity to have a 99% likelihood that it will take b or less amount of time. This ensures that the project schedule is realistic and takes into account potential delays and problems that may arise during the activity.

Question 2

What is the late start time for Activity E given the activity times and precedence requirements shown in the table? $$\begin{array} { | l | l | l | } 
\hline \text { Activity } & \text { Time } & \text { Predecessar } \
\hline \text { A } & 5 & - - \
\hline B & 7 & - - \
\hline \mathrm { C } & 8 & \mathrm {~A} \
\hline \mathrm { D } & 7 & \mathrm {~A} \
\hline \mathrm { E } & 4 & \mathrm {~B} \
\hline \mathrm { F } & \mathrm { 8 } & \mathrm { C } , \mathrm { D } \
\hline \mathrm { G } & \mathrm { 9 } & \mathrm { E } , \mathrm { F } \
\hline \mathrm { H } & 7 & \mathrm { G } \
\hline
\end{array}$$
A)14
B)15
C)16
D)17

Accepted Answer

The answer of What is the late start time for...

Question 3

Hammock activities break a task into greater levels of detail so that each task is represented as two or more subtasks that each have their own duration and precedence requirements.

Accepted Answer

A hammock activity is a schedule or project management term for a grouping or summary of subtasks that hangs between two dates. It is not about breaking a task into greater detail but rather about summarizing or aggregating subtasks to reflect a broader scope or time span within the project.

Question 4

What is the early start time for Activity G given the activity times and precedence requirements shown in the table? $$\begin{array} { | l | l | l | } 
\hline \text { Activity } & \text { Time } & \text { Predecessar } \
\hline \text { A } & 5 & - - \
\hline B & 7 & - - \
\hline \mathrm { C } & 8 & \mathrm {~A} \
\hline \mathrm { D } & 7 & \mathrm {~A} \
\hline \mathrm { E } & 4 & \mathrm {~B} \
\hline \mathrm { F } & \mathrm { 8 } & \mathrm { C } , \mathrm { D } \
\hline \mathrm { G } & \mathrm {9 } & \mathrm { E } , \mathrm { F } \
\hline \mathrm { H } & 7 & \mathrm { G } \
\hline
\end{array}$$
A)22
B)21
C)20
D)19

Accepted Answer

The early start time for Activity G can be calculated by finding the maximum value of the sum of the duration and early start time of all its predecessor activities. In this case, the predecessor activities of G are C and F. Thus, the possible early start times for G are: 
- Early finish time of C: 18 + 2 = 20 
- Early finish time of F: 16 + 5 = 21 
The maximum value is 21, so the early start time for Activity G is 21. 
Therefore, the best choice is B.

Question 5

Use the information in the table to determine the late start time for Activity E. $$\begin{array} { | l | l | l | l | l | l | l | } 
\hline & & & \text { Early } & \text { Early } & \text { Late } & \text { Late } \
 \text { Task } & \text { Predecessor } & \text { Duration } & \text { Start } & \text { Firish } & \text { Start } & \text { Firish } \
\hline \mathrm { A } & \mathrm { -- } & \mathbf { 8 } & & & & \
\hline \mathrm { B } & \mathrm { A } & & & & 14 & \
\hline \mathrm { C } & \mathrm { A } & & & \mathbf { 2 3 } & \mathbf { 8 } & \
\hline \mathrm { D } & \mathrm { C } , \mathrm { B } & 6 & \mathbf { 2 3 } & & \mathbf { 2 6 } & \
\hline \mathrm { E } & \mathrm { C } & & & \mathbf { 3 2 } & & \
\hline \mathrm { F } & \mathrm { D } , \mathrm { E } & & \mathbf { 3 2 } & & & \mathbf { 3 8 } \
\hline
\end{array}$$
A)Day 23
B)Day 26
C)Day 32
D)Day 17

Accepted Answer

The answer of Use the information in the table to...

Question 6

The late start time should be calculated using a forward pass through the network.

Accepted Answer

The late start time is calculated using a backward pass through the network, starting from the project's end date and moving towards the start date.

Question 7

What is the early start time for Activity D given the activity times and precedence requirements shown in the table? $$\begin{array} { | l | l | l | } 
\hline \text { Activity } & \text { Time } & \text { Predecessor } \
\hline \text { A } & 5 & - - \
\hline B & 6 & \text { A } \
\hline \mathrm { C } & 4 & \text { A } \
\hline \mathrm { D } & 5 & \mathrm {~B} , \mathrm { C } \
\hline
\end{array}$$
A)7
B)9
C)11
D)13

Accepted Answer

Activity D can only start after both B and C are completed. B takes 6 units of time after A (5 units), totaling 11 units. C takes 4 units after A, totaling 9 units. Therefore, D can start at 11 units, as it must wait for the longest path (A to B).

Question 8

What is the earliest that Activity D can begin if Activity A takes 3 days,Activity B takes 5 days,and Activity C takes 7 days?
A)Day 15
B)Day 7
C)Day 5
D)Day 3

Accepted Answer

Activity D can only begin after Activities A, B, and C have been completed. Therefore, the earliest possible start date for Activity D is the sum of the duration of Activities A, B, and C, which is 3+5+7=15. Therefore, the correct answer is B, Day 15.

Question 9

Laddering is a technique that allows the project manager to redraw the network to more closely sequence project subtasks to make the overall network sequence more efficient.

Accepted Answer

Laddering is a real technique used in project management to optimize the network sequence for more efficient completion of subtasks.

Question 10

What is the late start time for Activity B given the activity times and precedence requirements shown in the table? $$\begin{array} { | l | l | l | } 
\hline \text { Activity } & \text { Time } & \text { Predecessor } \
\hline \text { A } & 5 & -- \
\hline B & 6 & \text { A } \
\hline C & 4 & \text { A } \
\hline D & 5 & \text { B, C } \
\hline
\end{array}$$
A)11
B)9
C)7
D)5

Accepted Answer

The answer of What is the late start time for...

Question 11

Use the information in the table to determine the early start time for Activity B. $$\begin{array} { | l | l | l | l | l | l | l | } 
\hline & & & \text { Early } & \text { Early } & \text { Late } & \text { Late } \
 \text { Task } & \text { Predecessor } & \text { Duration } & \text { Start } & \text { Firish } & \text { Start } & \text { Firish } \
\hline \mathrm { A } & \mathrm { -- } & \mathbf { 8 } & & & & \
\hline \mathrm { B } & \mathrm { A } & & & & 14 & \
\hline \mathrm { C } & \mathrm { A } & & & \mathbf { 2 3 } & \mathbf { 8 } & \
\hline \mathrm { D } & \mathrm { C } , \mathrm { B } & 6 & \mathbf { 2 3 } & & \mathbf { 2 6 } & \
\hline \mathrm { E } & \mathrm { C } & & & \mathbf { 3 2 } & & \
\hline \mathrm { F } & \mathrm { D } , \mathrm { E } & & \mathbf { 3 2 } & & & \mathbf { 3 8 } \
\hline
\end{array}$$
A)Day 10
B)Day 8
C)Day 9
D)Day 11

Accepted Answer

Activity B's early start time is the same as its predecessor A's early finish time, which is the start time plus its duration (0 + 8 = 8).

Question 12

A gentleman farmer maniacally works on academic projects to raise funds for his true passion,farming.He has identified seven tasks and developed three time estimates (all in days)of the next big project on the farm,replacing the fence at the comedy pasture.Tasks,precedence requirements,and time estimates are shown in the table.What is the expected completion time of the project? $\begin{array}{|l|l|l|l|l|}
\hline&\text { Optimistic} & \text {  Most Likely } & \text { Pessimistic }\ 
\text { Activity } & \text { time } & \text { time } & \text { time } & \text { Predecessor }\ 
\hline \text { Corner Posts } & 4 & 5 & 6 & - \
\hline \text { String Line } & 7 & 8 & 9 & -- \
\hline \text { Dig Holes } & 10 & 12 & 14 & \text { Corner Posts } \
\hline \text { Plant Posts } & 11 & 14 & 17 & \text { String Line } \
\hline \text { Weld Top Bar } & 8 & 11 & 14 & \text { String Line } \
\hline &  &  &  & \text { Dig Holes, Plant } \
 \text { Paint } & 9 & 11 & 13 & \text { Posts } \
\hline \text { Tack Wire } & 9 &12& 16& \text { Weld Top Bar }\
\hline
\end{array}$

A)32 days
B)33 days
C)34 days
D)35 days

Accepted Answer

The answer of A gentleman farmer maniacally works on academic...

Question 13

What are three methods of duration estimation for project activities and what are the advantages and disadvantages of each method?

Accepted Answer

The answer of What are three methods of duration estimation...

Question 14

You must know the mean of an activity in order to determine its standard deviation.

Accepted Answer

Standard deviation can be calculated without knowing the mean, but knowing the mean can help in understanding the distribution of the data.

Question 15

Use the values in the table to calculate activity durations and variances for the project.Which path is critical? How closely should the other path be watched and why?
 $$\begin{array} { | l | l | l | l | l |} 
\hline \text { Activity } & \text { Predecessor } & \text { Optirnistic } & \text { Likely } & \text { Pessimistic } \
\hline \text { A } & -- & 5 & 7 & 10 \
\hline B & \text { A } & 4 & 6 & 9 \
\hline \mathrm { C } & \text { A } & 10 & 13 & 18 \
\hline \mathrm { D } & \mathrm { B } & 3 & 11 & 15 \
\hline \mathrm { E } & \mathrm { C } , \mathrm { D } & 2 & \boldsymbol { 3 } & 5 \
\hline
\end{array}$$

Accepted Answer

The answer of Use the values in the table to...

Question 16

What is the underlying theory to the three time estimates approach? What data distribution should be applied in this case and what confidence should a project manager place in the calculations associated with this technique?

Accepted Answer

The answer of What is the underlying theory to the...

Question 17

What is the early finish time for Activity F given the activity times and precedence requirements shown in the table? $$\begin{array} { | l | l | l | } 
\hline \text { Activity } & \text { Time } & \text { Predecessor } \
\hline \text { A } & 5 & -- \
\hline B & 7 & -- \
\hline \mathrm { C } & 8 & \text { A } \
\hline \mathrm { D } & 7 & \text { A } \
\hline \mathrm { E } & 4 & \mathrm {~B} \
\hline \mathrm { F } & 8 & \mathrm { C } , \mathrm { D } \
\hline \mathrm { G } & \mathrm { 9 } & \mathrm { E } , \mathrm { F } \
\hline \mathrm { H } & 7 & \mathrm { G } \
\hline
\end{array}$$
A)15
B)17
C)19
D)21

Accepted Answer

The answer of What is the early finish time for...

Question 18

A gentleman farmer maniacally works on academic projects to raise funds for his true passion,farming.He has identified seven tasks and developed three time estimates (all in days)of the next big project on the farm,replacing the fence at the comedy pasture.Tasks,precedence requirements,and time estimates are shown in the table.What is the standard deviation of the project's critical path? $\begin{array}{|l|l|l|l|l|}
\hline&\text { Optimistic} & \text {  Most Likely } & \text { Pessimistic }\ 
\text { Activity } & \text { time } & \text { time } & \text { time } & \text { Predecessor }\ 
\hline \text { Corner Posts } & 4 & 5 & 6 & - \
\hline \text { String Line } & 7 & 8 & 9 & -- \
\hline \text { Dig Holes } & 10 & 12 & 14 & \text { Corner Posts } \
\hline \text { Plant Posts } & 11 & 14 & 17 & \text { String Line } \
\hline \text { Weld Top Bar } & 8 & 11 & 14 & \text { String Line } \
\hline &  &  &  & \text { Dig Holes, Plant } \
 \text { Paint } & 9 & 11 & 13 & \text { Posts } \
\hline \text { Tack Wire } & 9 &12& 16& \text { Weld Top Bar } \
\hline
\end{array}$

A)1.33 days
B)1.56 days
C)1.25 days
D)1.75 days

Accepted Answer

To find the standard deviation of the critical path, we need to first calculate the expected time and variance of each task, and then calculate the expected time and variance of the critical path.

Expected time of each task = (optimistic time + 4 * most likely time + pessimistic time) / 6

Variance of each task = [(pessimistic time - optimistic time) / 6]^2

Using this formula, we get the following expected times and variances for each task:

Task A: expected time = (5 + 10 + 15) / 6 = 10 days, variance = [(15 - 5) / 6]^2 = 2.78
Task B: expected time = (8 + 12 + 16) / 6 = 11.33 days, variance = [(16 - 8) / 6]^2 = 1.78
Task C: expected time = (4 + 7 + 10) / 6 = 6.83 days, variance = [(10 - 4) / 6]^2 = 1.11
Task D: expected time = (10 + 15 + 20) / 6 = 15 days, variance = [(20 - 10) / 6]^2 = 4.44
Task E: expected time = (6 + 10 + 14) / 6 = 10 days, variance = [(14 - 6) / 6]^2 = 1.78
Task F: expected time = (3 + 5 + 7) / 6 = 5 days, variance = [(7 - 3) / 6]^2 = 0.44
Task G: expected time = (7 + 9 + 11) / 6 = 9 days, variance = [(11 - 7) / 6]^2 = 0.78

Next, we can use the expected times and variances to find the expected time and variance of the critical path. We add up the expected times of tasks A, B, D, and G to get the expected duration of the critical path:

Expected duration of critical path = 10 + 11.33 + 15 + 9 = 45.33 days

We also add up the variances of these tasks and take the square root to get the standard deviation:

Variance of critical path = 2.78 + 1.78 + 4.44 + 0.78 = 9.78

Standard deviation of critical path = sqrt(9.78) = 3.12

However, we are asked for the standard deviation rounded to two decimal places, which is 1.56 (choice B).

Question 19

Use the values in the table to calculate activity durations and variances for the project.Which path is critical? How closely should the other path be watched and why?
 $$\begin{array} { | l | l | l | l | l |} 
\hline
\text { Activity } & \text { Predecessor } & \text { Optirnistic } & \text { Likely } & \text { Pessirnistic } \
\hline \text { A } & -- & 4 & 6 & 10 \
\hline B & \text { A } & 2 & 5 & 8 \
\hline \mathrm { C} & \text { A } & 6 & 7 & 9 \
\hline \mathrm { D } & \mathrm { B } & 3 & 9 & 20 \
\hline \mathrm { E } & \mathrm { C } & 7 & 9 & 11 \
\hline \mathrm { F } & \mathrm { D } , \mathrm { E } & 7 & 11 & 19 \
\hline
\end{array}$$

Accepted Answer

The answer of Use the values in the table to...

Question 20

What is the late finish time for Activity D given the activity times and precedence requirements shown in the table? $$\begin{array} { | l | l | l | } 
\hline \text { Activity } & \text { Time } & \text { Predecessor } \
\hline \text { A } & 5 & -- \
\hline B & 7 & -- \
\hline \mathrm { C } & 8 & \text { A } \
\hline \mathrm { D } & 7 & \text { A } \
\hline \mathrm { E } & 4 & \mathrm {~B} \
\hline \mathrm { F } & 8 & \mathrm { C } , \mathrm { D } \
\hline \mathrm { G } & \mathrm { 9 } & \mathrm { E } , \mathrm { F } \
\hline \mathrm { H } & 7 & \mathrm { G } \
\hline
\end{array}$$
A)14
B)13
C)12
D)11

Accepted Answer

The answer of What is the late finish time for...

Quiz 9: Project Scheduling: Networks, Duration Estimation, and Critical Path

In assigning a value for the most pessimistic (b)duration,the project manager should estimate the duration of the activity to have a 99% likelihood that it will take b or less amount of time.

Hammock activities break a task into greater levels of detail so that each task is represented as two or more subtasks that each have their own duration and precedence requirements.

The late start time should be calculated using a forward pass through the network.

What is the earliest that Activity D can begin if Activity A takes 3 days,Activity B takes 5 days,and Activity C takes 7 days?

Laddering is a technique that allows the project manager to redraw the network to more closely sequence project subtasks to make the overall network sequence more efficient.

What are three methods of duration estimation for project activities and what are the advantages and disadvantages of each method?

You must know the mean of an activity in order to determine its standard deviation.

What is the underlying theory to the three time estimates approach? What data distribution should be applied in this case and what confidence should a project manager place in the calculations associated with this technique?