project management - wkupeople.wku.edu/bradley.fink680/class/cit492/mod3.pdf · project management...

PROJECT

MANAGEMENT

Brad Fink

28 February 2013

Trying to manage a project without project

management is like trying to play a football

game without a game plan

K. Tate (Past Board Member, PMI).

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Executive Summary

David Carhart runs a consulting company; his new project has several activities that need to be

completed in order to finish a new project. David needs to know how long this project will take

to complete as well as identifying which activities are critical. The route chosen is to draw an

Activity-on-Node Network Diagram.

The District Manager needs to know what the critical path is on a new project as well as the

length of the critical path. Having been tasked to provide this information, the best possible

solution is to draw an activity-on-arrow network diagram.

Robert Klassen, owner of an Ontario factory has provided data for the activity time estimates on

one of his production lines. Robert wants a product so he can visually see each activity with the

critical paths during the process and what the expected time of completion may be. He would

also like some type of work process chart to post on the company bulletin board for his

employees to view at as well.

Andrea McGee of McGee Carpet and Trim installs carpet in commercial offices. She has a

concern with the amount of time that it is taking to complete the projects as of late due to some

of her employees being unreliable. A list of activities along with the optimistic, most likely and

pessimistic times has been provided. Andrea wants to know what the completion time will be as

well as the variance for each activity, the total project completion time with the critical path, and

what the probability of finishing the project in Forty days or less.

Bill Fennema, president of Fennema Construction wants an activity-on-node network diagram

illustrating each task, also the duration and predecessor relationships of the activities. What Bill

needs to distinguish is the expected time for activity C and its variance. Bill would also like to

see the critical path with its estimated time, the activity variance along the critical path and the

probability of completing the project before week Thirty Six.

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Contents

David Carhart’s Consulting Company ........................................................................................3

AOA Network Diagram .................................................................................................................7

Robert Klassen’s Factory ..............................................................................................................9

McGee Carpet and Trim .............................................................................................................13

Fennema Construction ................................................................................................................19

Summary .......................................................................................................................................23

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ActivityImmediate

Predecessor(s)

Time

(In Days)

A N/A 3

B A 4

C A 6

D B 6

E B 4

F C 4

G D 6

H E,F 8

Carhart's AON Network Diagram

David Carhart’s Consulting Company

David Carhart Consulting is about to start work on a new project, in order for David to manage

his time and his team, he needs to know two things; how long will it take his team to complete

the project, and what are the critical activities?

To help identify David’s needs the following data in Table 1 has been provided in order to help

draw an Activity-on Node Network Diagram shown in Figure 1.

Following the activity chart of Table 1, the

activity-on-node network diagram can now be

constructed. In the activity column, the activities

are identified by (A – H) respectively. The

middle column is the immediate predecessor; this

is the activity that must happen before the next

activity can be started. The last column is the

time, in this case the time is in days.

In Table 1, notice activity B has a predecessor of

A; this means that activity A must be completed

before activity B can start. Activity H however is slightly different; both activities (E and F)

must be completed before activity H can start. Corresponding to each activity is the time activity

A will be completed in three days, in three days activity B can start. All this can be better

viewed in an activity-on-node network diagram shown in Figure 1 below.

Table 1 –Activity-on-Node Chart

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Figure 1 shows each activity and in the what order during the process, this will help all

communicate more efficiently, if one team

member is responsible for activity F, he will

be able to speak directly to the individual in

charge of activity C and so forth.

David also needs to know how many days this

project is going to take, to find this answer

David needs to do some basic math. Figure 2

will provide the abbreviations to make the process easier.

Starting with activity A, take the activity time which is three days, add that to the ES which is

zero, since activity A has no predecessor, this will give activity A an earliest end time of three

days. Continuing left to right the next activity’s earliest start time will be the previous activity’s

earliest end time. Referring to Figure 1, notice that H has two different arrows pointing towards

it, one from activity E and activity F, activity H will have an ES of whichever EF is greater, in

this case it will be activity F. Moving left to right and following the arrows, simply continue to

fill in the blanks until the last activity is reached. Since activity H is the last activity the LF and

LS need to be figured.

Figure 1 –Activity-on-Node Network

Diagram

ES: Earliest start time an activity can start

EF: Earliest an activity can finish

LS: Latest time an activity can start

LF: Latest time an activity can finish

Figure 2 -Abbreviations

Start A StartE

H

F

GD

C

B

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To find the LF and LS we need to work backwards, activity H will have an equal EF and LF,

take the LF and subtract the activity time, in this case it will be 21-8 which gives the LS equal to

21, and this will be the predecessors LF. All of the start times and end times can be better

viewed in Table 2 below as well as how to find out which activities are on a critical path.

Notice that activity H has an LF time of 21, this is actual work days, this does not take in account

of weekends or holidays, the actual calendar view will essentially be 28 days. Also, on the

furthest right column all the critical paths are given, for Davis all activities with the exception of

activity E are critical activities.

For a better birds-eye view, Figure 3 can be a much easier way for managers to map and

calculate all activities in a project.

ActivityTime in

DaysPred. ES EF LS LF SLACK

Critical

Path

A 3 N/A 0 3 0 3 0 Yes

B 4 A 3 7 3 7 0 Yes

C 6 A 3 9 3 9 0 Yes

D 6 B 7 13 7 13 0 Yes

E 4 B 7 11 9 13 2 No

F 4 C 9 13 9 13 0 Yes

G 6 D 13 19 13 19 0 Yes

H 8 E,F 13 21 13 21 0 Yes

Acivity-on-Node Time Computations Chart

Table 2 –Activity Time Computation Chart

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0 3 3

0 0 3A

3 4 7

3 0 7

B

3 6 9

3 0 9

C

9 4 13

9 0 13

F

7 6 13

7 0 13

D13 6 19

13 0 19

G

7 4 11

9 2 13

E

13 8 21

13 0 21

H

STARTFINISH

Early Start Duration Early Finish

Late Start Slack Late Finish

Task Name

LEGEND

Figure 3 –Activity-on-Node (Diagram option 2, Red Arrows indicate Critical Path)

Figure 3 is the same activity-on-node as shown in Figure 1 except this one uses boxes with each

activities (ES, EF, LS and LF) just another way to help managers see the big picture. The red

arrows also indicate which activity is part of the critical path as well.

Some might ask, why is activity E not part of the critical path, the reason for this is it has a slack

time of two days, so activity E can be delayed for two days without disrupting the actual

completion time of the entire project.

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Activity-on-Arrow Network Diagram

A work diagram is needed for the employees to view. This will enable them to know when

their portion of the job will start, and how long it is supposed to take them to complete the

activity before handing it over to the next level.

There are many ways to give them a visual look, but the activity-on-arrow network diagram

is being used for the ease of understandability of all employees which can viewed in Figure 4.

A

5

A

5

B

B2

2

C

4 F

5

D

H

I

EG

Start

5

52

3

5


Finish

Figure 4 –Activity-on-Arrow Network Diagram

Besides just the diagram, some other information is needed for management. Information such

as; which activity or activities are on a critical path and what is the length of the critical path.

The critical path is vital to a project, if there is any delay during one of the activities on a critical

path, this could end up being a show stopper.

The best way to find the answers is to use a time computation chart, this will show all the

earliest start time (ES), the latest start time (LS), the earliest finish time (EF), and the latest finish

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time (LF). Although these time are somewhat irrelevant to some, they are key to finding what is

being asked, and that again is what are the critical paths and how long is the critical path? Below

in Table 3, shows the activity-on-arrow time computation chart showing in the far right column

marked “Critical Path”.

Looking back at Figure 4 the critical path is displayed with red arrows as well as seen in Table 3.

Notice that the slack time is zero when there is a critical path. This means in laymen terms; there

is no room for error during that particular activity. To find the length of the critical path, add up

activities (B, D, and H), these three activities are on a direct path to the finish activity. The length

of the critical path is 3 days.

ActivityTime in


Critical

Path

A 5 N/A 0 5 0 5 0 Yes

B 2 A 5 7 6 8 1 No

C 4 A 5 9 5 9 0 Yes

D 5 B 7 12 8 13 1 No

E 5 B 7 12 9 14 2 No

F 5 C 9 14 9 14 0 Yes

G 2 E,F 14 16 14 16 0 Yes

H 3 D 1 15 13 16 1 No

I 5 G,H 16 21 16 21 0 Yes

Acivity-on-Arrow Time Computations Chart

Table 3 –Activity-on-Arrow Time Computation Chart

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Robert Klassen’s Factory

Task time estimates for a production line project at Robert Klassen’s Ontario factory are as

follows, shown in Table 4. Based off this information, David Klassen wants an activity-on-node

network diagram configured, below in Figure 5. Looking at Table 4, the first two activities (A

and B) have no predecessors; these will be the start

point for all other activities to follow. Activity C

has a predecessor of A, and no contact from activity

B. Before activity D can begin, activities (B and C)

must be completed. Activity F has a predecessor of

activity F, while activity G has both activities (E and

F) proceeding it. This can all be better viewed in a

network diagram. When looking at Figure 5, notice

that the nodes now are represented by using

rectangles.

Another noticeable difference is that the times are in

hours, which differs from most projects which usually deal with days or even weeks. Figure 5

shows the nodes in a way better adapted for finding all answers any manager to find with little

effort.

ActivityTime (In

Hours)

Immediate

Predecessors

A 6 N/A

B 7.2 N/A

C 5 A

D 6 B,C

E 4.5 B,C

F 7.7 D

G 4 E,F

Table 4 –Activity Estimates

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0 6 Hours 6

0 0 6

Activity A

0 7.2 7.2

3.8 3.8 11

Activity B

6 5 11

6 0 11

Activity C

11 6 17

11 0 17

Activity D

11 4.5 15.5

20.2 9.2 24.7

Activity E

17 7.7 24.7

17 0 24.7

Activity F

24.7 4 28.7

24.7 0 28.7

Activity G



LEGEND

Figure 5 –Activity-on-Node Activity Diagram

This is only the first part of what David needs, and by using this diagram, it will make finding

the critical path easier, which is the next part. Based off Figure 5, the times needed to find the

critical path has already been identified, all that is needed is some simple math; in Table 5 the

spreadsheet will provide all the relevant data showing the critical path(s).

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Looking at Table 5 there is more than one thing that will answer most of David’s questions. At

the bottom of the column indicated by (LF), the number 28.7 is the total number of hours to

complete all activities. Secondly, the critical paths are easily identified under the (Slack)

column, all slack times with zero are considered to be on a critical path, in this case activities (A,

C, F and G) are the critical paths.

Another way to look at how each activity works hand in hand with the other activities is seen

through a (Gantt) chart; named after Henry Gantt. This chart is very common due to the fact it

usually follows along or in some instances placed on a calendar. The thought here is, if you can

look and understand a calendar, you will be able to look and understand a Gantt chart. Figure 6

shows exactly what a Gantt chart looks like pertaining to Robert’s needs.

ActivityTime in


Critical

Path

A 6.0 N/A 0 6.0 0.0 6 0 Yes

B 7.2 N/A 0 7.2 3.8 11 3.8 No

C 5.0 A 6 11.0 6.0 11.0 0 Yes

D 6.0 B,C 11 17.0 11.0 17 0 Yes

E 4.5 B,C 11 15.5 20.2 24.7 9.2 No

F 7.7 D 17 24.7 17.0 24.7 0 Yes

G 4.0 E,F 24.7 28.7 24.7 28.7 0 Yes

Acivity-on-Arrow Time Computations ChartFigure 2.2 -Table Lamp Assembly Instructions

Table 5 –Klassen’s Factory Time Computation

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Again, looking at the Gantt chart in Figure 6, the activities (A and B) can easily be seen starting

at the same time with activity C starting when they are finished. Activities (D and E) start as

soon as activity C is complete. Since activity D is longer, activity F will start as soon as activity

D is complete. Lastly, activity G will begin as soon as activity F is complete.

In the Gantt chart, on each activity bar, the number of hours is projected; showing how long each

task is supposed to take. At the very top of the chart just above activity A is the task summary

timeline bar which shows how long all tasks summed up will take, again this shows that all tasks

will take 28.7 hours.

\

Figure 6 –Gantt chart

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McGee Carpet and Trim

To help Andrea with her situation, she would like to know what the expected time it will take to

complete a project. Within the project she has eleven activities to accomplish and would also

like to know the variance for each task.

To start this off there are some issues that need to be addresses, these concern the three different

times that will determine the expected time of completion and help derive to the variance of each

activity.

In Table 7, the three time required to find the information are provided, they are times (A, B and

M). Using a little math, all the data needed is simple enough, but first we need to understand

what the times given represent.

Time A is the optimistic time an activity will take if everything goes as planned, in

estimating this value, there should be only 1/100 chance that the time will actually be less

than time A.

Time B is the pessimistic time an activity will take assuming very unfavorable conditions

exists. While estimating this value, there should be only a small probability of again

1/100 chance the activity time will be greater than time B.

Time M is the most likely time, or the most reliable estimate of the time required to

complete an activity.

To figure out what each activity’s expected completion times will be, as stated earlier, an easy

mathematical calculation must be performed. The expected completion time equals (Time A

plus 4 times time M plus time B) divided by 6. Since the most likely time M is the most

realistic, its value is four times greater than the other two times, hence multiplying it by four.

Looking at activity A, the formula will look like 3+4*6+8/6 which equals 5.83; this is the

expected completion time of activity A. Table 6 will provide the rest of the expected times for

each activity.

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ActivityA M B

Immediate

Predecessor(s)

A 3 6 8 N/A

B 2 4 4 N/A

C 1 2 3 N/A

D 6 7 8 C

E 2 4 6 B,D

F 6 10 14 A,E

G 1 2 4 A,E

H 3 6 9 F

I 10 11 12 G

J 14 16 20 C

K 2 8 10 H,I

(A): Optimistic, (M): Most Likely, (B):

Pessimistic Times

A M B

A N/A 3 6 8 5.83

B N/A 2 4 4 3.67

C N/A 1 2 3 2.00

D C 6 7 8 7.00

E B,D 2 4 6 4.00

F A,E 6 10 14 10.00

G A,E 1 2 4 2.17

H F 3 6 9 6.00

I G 10 11 12 11.00

J C 14 16 20 16.33

K H,I 2 8 10 7.33

Expected Completion Times Expected

Completion

TimeActivity Predecessor

Variance Times


.

Table 7 – (A, B, and M)

times per Activity

Table 7 –

Expected

Activity

Completion

Times

Below in Table 7 are the

completed calculated expected

completion times for each

activity.

Table 6 –Optimistic, Most

Likely and Pessimistic Time

Chart

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Now the expected completion times have been calculated, the variance for each activity can be

determined. To find the variance of each activity, again Table 7 has provided all the data

required to calculate the mathematical formula. The variance is found by time B minus time A

divided by six, and the answer squared, this will be the variance, taking the data from time A it

will look as so; ((8-3)/6))², which gives time A, a variance of 0.69. Table 8 will complete the

rest of the activities variances.

With the expected completion times and the variance complete, the next step to Andrea needs to

do is find the (ES, LS, ES and EF) times, the critical path(s), the total project completion time

and the slack times. This may sound overwhelming, however, with the right tool all this can be

accomplished in one easy process, and that tool is the activity-on-node network diagram shown

in Figure 7 with all corresponding data in Table 9.

A M B

A N/A 3 6 8 0.69

B N/A 2 4 4 0.11

C N/A 1 2 3 0.11

D C 6 7 8 0.11

E B,D 2 4 6 0.44

F A,E 6 10 14 1.78

G A,E 1 2 4 0.25

H F 3 6 9 1.00

I G 10 11 12 0.11

J C 14 16 20 1.00

K H,I 2 8 10 1.78

Activity Variance Variance

Activity Predecessor

Variance Times

Table 8 –Activity

Variance

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0 5.83 5.83

7.17 7.17 13

A

0 3.67 3.67

5.33 5.33 9

B

0 2 2

0 0 2

C

9 4 13

9 0 13

E

13 10 23

13 0 23

F

2 7 9

2 0 9

D

13 2.17 15.17

15.83 2.83 18

G

23 6 29

23 0 29

H

15.17 11 26.17

18 2.83 29

I

2 16.33 18.33

2 0 18.33

J

29 7.33 36.33

29 0 36.33

K



LEGEND

McGee Carpet and Trim Arow-on-Node Network Diagram

Figure 7 –McGee Arrow-on-Node Network Diagram

Figure 7 is only a tool to help get the real information needed, all the data in each node is easily

entered into a spreadsheet to find all the information Andrea needs which is below in Table 9.

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In Table 9, all the times need for Andrea are easily viewed, however the real important times are

the total project completion times which is highlighted in green under the LF column. Next to

the LF column is the slack times, notice that all slack times of zero are also a (Yes) in the critical

path column, so activities (C,D,E,F,H,J and K) are all part of the critical path, this can also be

seen by looking at the red arrows back in Figure 7.

Now that all times have been computed, the critical paths have been identified; Andrea’s last

question can be answered. What is the probability that McGee Carpet and Trim will finish the

project in 40 days or less? Again a little math is needed, to find the answer Andrea must first

add up all the variances on the critical paths which are; (0.11, 0.11, 0.44, 1.78, 1.0, 1.0 and 1.78)

which sums up to 6.22. Next is to find the project standard deviation, which is taking the sum of

all critical path variances and getting the square root, √6.22 with a result of 2.49 days. So now

ActivityTime in

DaysPredecessors ES EF LS LF SLACK

Critical

Path

A 28.3 N/A 0 5.83 7.17 13 7.17 No

B 18.7 N/A 0 3.67 5.33 9 5.33 No

C 9.5 N/A 0 2 0 2 0 Yes

D 35.3 C 2 9 2 9 0 Yes

E 19.0 B,D 9 13 9 13 0 Yes

F 48.3 A,E 13 23 13 23 0 Yes

G 9.7 A,E 13 15.17 15.83 18 2.83 No

H 28.5 F 23 29 23 29 0 Yes

I 56 G 15.17 26.17 18 29 2.83 No

J 81.33 C 2 18.33 2 18.33 0 Yes

K 35.67 H,I 29 36.33 29 36.33 0 Yes

Table 9 –Computation Times

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Z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

-1.6 0.05480 0.05370 0.05262 0.05155 0.05050 0.04947 0.04846 0.04746 0.04648

-1.5 0.06681 0.06552 0.06426 0.06301 0.06178 0.06057 0.05938 0.05821 0.05705

-1.4 0.08076 0.07927 0.07780 0.07636 0.07493 0.07353 0.07215 0.07078 0.06944

-1.3 0.09680 0.09510 0.09342 0.09176 0.09012 0.08851 0.08691 0.08534 0.08379

-1.2 0.11507 0.11314 0.11123 0.10935 0.10749 0.10565 0.10383 0.10204 0.10027

-1.1 0.13567 0.13350 0.13136 0.12924 0.12714 0.12507 0.12302 0.12100 0.11900

-1 0.15866 0.15625 0.15386 0.15151 0.14917 0.14686 0.14457 0.14231 0.14007


Andrea knows that the deviation is 2.49 now she can continue to find out the probability of the

project being completed in forty or less days. Now Andrea has to subtract 40 days from

36.33days giving her -3.67, now divide that by the project deviation of 2.49 giving her -1.47.

Andrea needs to look at a Z Table, Table 10 and find out what the percentage is.

Since Andrea has a value of 1.47 she needs to look at the Z Table and scroll down until she

reaches 1.4, then scroll right until she reaches the 0.07 column, circled in red, the number that

links the 1.4 row and the 0.07 column is 0.07078; multiply that by 100 and Andrea has a 7.08%

chance of finishing within Forty days or less.

Table 10 –Z Table

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Fennema Construction

Bill Fennema has spent an exhausting amount of time developing tasks, the tasks duration times

and which tasks are to be completed in a sequenced order. Bill has provided this data to have an

activity-on-node network (AON) diagram be drawn. With the AON diagram he has also asked

for the expected time of completion for activity C as well as its variance. Figure 8 with start this

process with the AON Diagram.

Notice in Figure 8 that instead of each node containing letters representing each activity, they are

now represented by numbers; 1 represents activity A and 11 representing activity K. This style

of activity numbering will help Bill recognize the tasks he is used to using. The critical path is

symbolized by the nodes in red. The critical paths in Figure 8 are (A, C, F, H, J and K)

The piece of information Bill needs, is to find the expected time of completion for activity C, this

can be done with little effort utilizing an Excel spreadsheet, and Table 11 will show all expected

completion times, however only activity C will be highlighted.

2 5

7

8

9

10 11

6

1

4

3

Figure 8 – Fennema Construction Activity-on-Node Diagram

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Table 12 –

Variance

Values


Bill now has the expected time of completion not only for activity C, but all activities involved

with the project. All of the expected completion times were computed using the optimistic time

(A), most likely time (M) and pessimistic time (B) which Bill provided in his data. The next bit

of information needed is the variance of activity C, again the spreadsheet will easily configure

this shown in Table 12 below.

A M B

A N/A 4 8 10 7.67

B A 2 8 24 9.67

C A 8 12 16 12.00

D A 4 6 10 6.33

E B 1 2 3 2.00

F E,C 6 8 20 9.67

G E,C 2 3 4 3.00

H F 2 2 2 2.00

I F 6 6 6 6.00

J D,G,H 4 6 12 6.67

K I,J 2 2 3 2.17

Times

Fennema Contruction Expected Completion Times

Activity Predeccessor

Expected

Completion

Time

A M B

A N/A 4 8 10 7.67 1.00

B A 2 8 24 9.67 13.44

C A 8 12 16 12.00 1.78

D A 4 6 10 6.33 1.00

E B 1 2 3 2.00 0.11

F E,C 6 8 20 9.67 5.44

G E,C 2 3 4 3.00 0.11

H F 2 2 2 2.00 0.00

I F 6 6 6 6.00 0.00

J D,G,H 4 6 12 6.67 1.78

K I,J 2 2 3 2.17 0.03

Variance Times

Fennema Contruction Variance (Activity C)


Expected

Completion

Time

Table 11 –Expected

Completion Times

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Now that Bill knows what the activity-on-node looks like, what the critical path is, the expected

completion time for activity C as well as its variance, the estimated time of the critical path and

activity variance along the critical path is needed. To complete this, the Gantt chart will become

extremely useful; Figure 9 will show the critical activities along with the expected completion

times.

Looking at Figure 9, there are only six activities; these particular activities are on the critical path

of the project. The estimated time of completion can be seen on the top summary bar; in this

case the estimated completion time is 40.18 (Questionable) weeks. The reason for the question

mark is that this is in fact an estimated time. Since the current subject is “critical path”, the next

relevant topic is the variance of the critical path. Looking back to Table 12, when summing up

all of the variances on the critical path the total will end up being 10.03, refer to Table 13.

Figure 9 –Gantt chart (Critical Path)

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-1.5 0.06681 0.06552 0.06426 0.06301 0.06178

-1.4 0.08076 0.07927 0.07780 0.07636 0.07493

-1.3 0.09680 0.09510 0.09342 0.09176 0.09012

-1.2 0.11507 0.11314 0.11123 0.10935 0.10749

-1.1 0.13567 0.13350 0.13136 0.12924 0.12714

-1 0.15866 0.15625 0.15386 0.15151 0.14917

-0.9 0.18406 0.18141 0.17879 0.17619 0.17361

-0.8 0.21186 0.20897 0.20611 0.20327 0.20045

Z 0 0.01 0.02 0.03


Table 13 –Critical Path Variance

Another piece of information Bill needs to know is the possibility of completing the project

before week 36, looking at Table 13 above, take the 10.03 variance and calculate the square root,

which is the standard deviation of 3.17.

Using a calculator, subtract the expected completion time of 40.18 weeks from the 36 weeks Bill

is striving for, the result is -4.18, this is the Z value, divide the Z value by the standard deviation

and the final calculation is -1.32. Looking at the Z Table, in Figure 14 go down the Z column

until -1.3 is reached, move to the right under the column heading of 0.02, the result is 0.09342,

multiply that by 100 and the probability is 9.34% chance of completing the project before week

36.

A M B

A N/A 4 8 10 1.00

C A 8 12 16 1.78

F E,C 6 8 20 5.44

H F 2 2 2 0.00

J D,G,H 4 6 12 1.78

K I,J 2 2 3 0.03

10.03

Variance

Total Variance:

Times

Fennema Contruction Variance (Activity C)


Figure 14 –Z

Table

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Summary

As managers, it is necessary to realize all the tools available on the market that will assist them

in producing vivid and explosive reports. More importantly is knowing when and how to use

these applications. In the case of David Carhart and Andrea McGee, the tools of choice were an

Excel spreadsheet and Visio, the importance of knowing not only how to use the basic functions

of an application, but how to master them can be the difference between a poor document and an

eye opening report.

In the case of Robert Klassen and Fennema Construction, not only were Visio and a spreadsheet

were used, but also an application called Project was utilized, as most can understand, knowing

more than one application and using more than just one can have a much greater impact on the

upper management and the decision they face during that crucial time of an approval or

disapproval.

The old saying “If you don’t use it, you lose it” is as true today as when it was first said. If

current managers do not use the applications designed to direct them into better managers, there

are always others behind them ready to get their chance at managing and climbing up the

corporate ladder. The managers who choose not to utilize the tools or refuse to train themselves

will lose all the vital skills necessary to become professionally proficient. For those managers

who do take the time and conduct their own professional development, not only will their

employees look at them with confidence, but maybe, just maybe so will the corporate

supervisors.

project management - wkupeople.wku.edu/bradley.fink680/class/cit492/mod3.pdf · project management...

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