Project Management

Project Management

Chapter 2Copyright 2013 Pearson Education, Inc. publishing as

Prentice Hall 02 -0 1

What is a Project?

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Project An interrelated set of activities with a definite starting and ending point, which results in a unique outcome for a specific allocation of resources.

What is Project Management?

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Project Management

A systemized, phased approach to defining, organizing, planning, monitoring, and controlling projects.

Defining and Organizing Projects

Defining the Scope and Objectives

Selecting the Project Manager and Team

Recognizing Organizational Structure


Planning Projects

Defining the Work Breakdown Structure

Diagramming the network Developing the schedule Analyzing cost-time

trade-offs Assessing risks


Work Breakdown Structure

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Work Breakdown StructureA statement of all the work that has to be completed.

ActivityThe smallest unit of work effort consuming both time and resources that can be planned and controlled.

Copyright 2013 Pearson Education, Inc. Publishing as Prentice Hall.

Purchase and deliver equipment

Construct hospital

Develop information system

Install medical equipment

Train nurses and support staff

Work Breakdown Structure

Select administration staff

Site selection and survey

Select medical equipment

Prepare final construction plans

Bring utilities to site

Interview applicants for

nursing and support staff

Organizing and Site PreparationPhysical Facilities and InfrastructureLevel 1

Level 0

Level 2

Relocation of St. Johns Hospital

02 -07

Diagramming the Network Program Evaluation and Review

Technique (PERT)

Critical Path Method (CPM)

Activity-on-Node approach


Establishing Precedence Relationships


Precedence RelationshipsDetermining the sequence for undertaking activities.

Example 2.1


FinishStart

A

B

C

D

E

F

G

H

I

J

K

ABCADBEBFAGCHDIAJE,G,HKF,I,J

Immediate Predecessor

Example 2.1


The following information is known about a project

Draw the network diagram for this project

Activity Activity Time (days)Immediate

Predecessor(s)A 7 B 2 AC 4 AD 4 B, CE 4 DF 3 EG 5 E

Application 2.1


Finish

G5

F3

E4

D4

Activity Activity Time (days)Immediate

Predecessor(s)A 7

B 2 AC 4 AD 4 B, CE 4 DF 3 EG 5 E

B2

C4

Start A7

Application 2.1


Developing the Schedule PathThe sequence of activities

between a projects start and finish.

Critical Path The sequence of activities

between a start and finish that takes the longest time to complete. Copyright 2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 14

Developing the Schedule Earliest start time (ES) - the latest

earliest finish time of any immediately preceding activities

Earliest finish time (EF) - the earliest start time plus its estimated duration EF = ES + t Latest finish time (LF) the earliest of the latest start times of any of the immediately following activities.

Latest start time (LS) - the latest finish time minus its estimated duration LS = LF t


Activity Slack - the maximum length of time an activity can be delayed without delaying the entire project

LF-EF or LS-ES

Developing the Schedule


Latest finish time

Latest start time

Activity

Duration

Earliest start time Earliest finish time

0

2

12

14

A

12

FinishStart

A

B

C

D

E

F

G

H

I

J

K

Path Time (wks)

A-I-K 33A-F-K 28A-C-G-J-K 67B-D-H-J-K 69B-E-J-K 43

Paths are the sequence of activities between a projects start and finish.

Example 2.2


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02 - 18

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12F

100

Earliest start time

12

Earliest finish time

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 63 69

Example 2.2


Example 2.2

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12F

10

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 63 690 12

The Critical Path takes 69 weeks


K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12F

10

0 9

9 33

9 1919 59

22 5712 22

59 63

12 27

12 22 63 690 12

48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

63 69

Example 2.2Latest start time

Latest finish time


K

6

C

10

G

35

J

4

H

40

B

9D

10

E

24

I

15

FinishStart

A

12F

10

0 9

9 33

9 1919 59

22 5712 22

59 63

12 27

12 22 63 690 12

48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

63 69

Example 2.2

S = 0

S = 2

S = 26

S = 0

S = 36

S = 2

S = 2

S = 41

S = 0

S = 0

S = 0

Developing a Schedule


Gantt chart

Application 2.2Calculate the four times for each activity in order to determine the critical path and project duration for the diagram in Application Problem 1.

Activity

Duration

Earliest Start (ES)

Latest Start (LS)

Earliest Finish (EF)

Latest Finish (LF)

Slack (LS-ES)

On the Critical Path?

A 7 0 0 7 7 0-0=0 YesB 2

C 4

D 4

E 4

F 3

G 5

The critical path is ACDEG with a project duration of 24 days.


Calculate the four times for each activity in order to determine the critical path and project duration.

Activity

Duration

Earliest Start (ES)

Latest Start (LS)

Earliest Finish (EF)

Latest Finish (LF)

Slack (LS-ES)

On the Critical Path?

A 7 0 0 7 7 0-0=0 YesB 2

C 4

D 4

E 4

F 3

G 5

7 9 9 11 9-7=2 No7 7 11 11 7-7=0 Yes

19 21 22 24 21-19=2 No19 19 24 24 19-19=0 Yes

11 11 15 15 11-11=0 Yes15 15 19 19 15-15=0 Yes

Application 2.2


Application 2.2


The critical path is ACDEG with a project duration of 24 days.

Start FinishA7

B2

C4

D4

E4

F3

G5

Analyzing Cost-Time Trade-Offs


Project CrashingShortening (or expediting) some activities within a project to reduce overall project completion time.Project Costs Direct Costs Indirect Costs Penalty Costs

Analyzing Cost-Time Trade-Offs


Project Costs Normal time (NT) is the

time necessary to complete an activity under normal conditions.

Normal cost (NC) is the activity cost associated with the normal time.

Crash time (CT) is the shortest possible time to complete an activity.

Crash cost (CC) is the activity cost associated with the crash time.

Cost to crash per period =CC NC

NT CT

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Cost-Time Relationships

Linear cost assumption

8000

7000

6000

5000

4000

3000

0

Dir

ect

cost

(do

llars

)

| | | | | |5 6 7 8 9 10 11

Time (weeks)

Crash cost (CC)

Normal cost (NC)

(Crash time) (Normal time)

Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks

5200

Figure 2.602 - 28

Analyzing Cost-Time Trade-OffsDetermining the Minimum Cost Schedule:

1. Determine the projects critical path(s).2. Find the activity or activities on the critical

path(s) with the lowest cost of crashing per week.3. Reduce the time for this activity until

a. It cannot be further reduced orb.Until another path becomes critical, orc. The increase in direct costs exceeds the

savings that result from shortening the project (which lowers indirect costs).

4. Repeat this procedure until the increase in direct costs is larger than the savings generated by shortening the project.


Example 2.3 DIRECT COST AND TIME DATA FOR THE ST. JOHNS HOSPITAL PROJECT

Activity Normal Time (NT) (weeks)

Normal Cost (NC)($)

Crash Time (CT)(weeks)

Crash Cost (CC)($)

Maximum Time Reduction (week)

Cost of Crashing per Week ($)

A 12$12,000

11$13,000

1 1,000

B 950,000

7 64,000 2 7,000

C 10 4,000 5 7,000 5 600D 10

16,0008 20,000 2 2,000

E 24120,000

14200,000

10 8,000

F 1010,000

6 16,000 4 1,500

G 35500,000

25530,000

10 3,000

H 401,200,000

351,260,000

5 12,000

I 1540,000

10 52,500 5 2,500

J 410,000

1 13,000 3 1,000

K 630,000

5 34,000 1 4,000

Totals$1,992,000 $2,209,500


Example 2.3Determine the minimum-cost schedule for the St. Johns Hospital project.Project completion time =69 weeks. Project cost = $2,624,000

Direct = $1,992,000 Indirect = 69($8,000) = $552,000 Penalty = (69 65)($20,000) = $80,000

AIK 33 weeksAFK 28 weeksACGJK 67 weeksBDHJK 69 weeksBEJK 43 weeks


Example 2.3STAGE 1Step 1. The critical path is BDHJK.Step 2. The cheapest activity to crash per week is J at $1,000, which is much less than the savings in indirect and penalty costs of $28,000 per week.Step 3. Crash activity J by its limit of three weeks because the critical path remains unchanged. The new expected path times are

ACGJK: 64 weeks BDHJK: 66 weeks

The net savings are 3($28,000) 3($1,000) = $81,000. The total project costs are now $2,624,000 - $81,000 = $2,543,000.Copyright 2013 Pearson Education, Inc. publishing as

Prentice Hall 02 - 32

Example 2.3


Finish

K6

I15

F10

C10

D10

H40

J1

A12

B9

StartG

35

E24

STAGE 1

Example 2.3STAGE 2Step 1. The critical path is still BDHJK.Step 2. The cheapest activity to crash per week is now D at $2,000.Step 3. Crash D by two weeks. The first week of reduction in activity D saves

$28,000 because it eliminates a week of penalty costs, as well as indirect costs.

Crashing D by a second week saves only $8,000 in indirect costs because, after week 65, no more penalty costs are incurred.

Updated path times areACGJK: 64 weeks and BDHJK: 64 weeks

The net savings are $28,000 + $8,000 2($2,000) = $32,000.

Total project costs are now $2,543,000 $32,000 = $2,511,000.


Example 2.3

Finish

K6

I15

F10

C10

D8

H40

J 1

A12

B9

Start G35

E24


STAGE 2

Example 2.3STAGE 3Step 1. The critical paths are BDHJK and A-C-G-J-K

Step 2. Activities eligible to be crashed:(A, B); (A, H); (C, B); (C, H); (G, B); (G, H) or

to crash Activity K We consider only those alternatives for which

the costs of crashing are less than the potential savings of $8,000 per week.

We choose activity K to crash 1 week at $4,000 per week.


Step 3. Updated path times are: ACGJK: 63 weeks

and BDHJK: 63 weeks Net savings are $8,000 - $4,000 = $4,000. Total project costs are $2,511,000 $4,000 =

$2,507,000.

Example 2.3

Finish

K5

I15

F10

C10

D8

H40

J1

A12

B9

StartG

35

E24


STAGE 3

Example 2.3STAGE 4Step 1. The critical paths are still BDHJK and ACGJK.

Step 2. Activities eligible to be crashed: (B,C) @ $7,600 per week.

Step 3. Crash activities B and C by two weeks. Updated path times are ACGJK: 61 weeks and BDHJK: 61 weeks The net savings are 2($8,000) 2($7,600) =

$800. Total project costs are now $2,507,000 $800 = $2,506,200.


Example 2.3

Finish

K5

I15

F10

C8

D8

H40

J1

A12

B7

StartG

35

E24


STAGE 4

Example 2.3Stage Crash

ActivityTime Reduction (weeks)

Resulting Critical Path(s)

Project Duration (weeks)

Project Direct Costs, Last Trial ($000)

Crash Cost Added ($000)

Total Indirect Costs ($000)

Total Penalty Costs ($000)

Total Project Costs ($000)

0 B-D-H-J-K

69 1,992.0

552.080.0

2,624.0

1 J 3 B-D-H-J-K

66 1,992.03.0

528.020.0

2,543.0

2 D 2 B-D-H-J-KA-C-G-J-K

64 1,995.04.0

512.00.0

2,511.0

3 K 1 B-D-H-J-KA-C-G-J-K

63 1,999.04.0

504.00.0

2,507.0

4 B, C 2 B-D-H-J-KA-C-G-J-K

61 2,003.015.2

488.00.0

2,506.2


Indirect project costs = $250 per day Penalty cost = $100 per day past day 14.

Project Activity and Cost Data

Activity

Normal Time (days)

Normal Cost ($)

Crash Time (days)

Crash Cost ($)

Immediate Predecessor(s)

A 51,000

41,200

B 5 800 32,000

C 2 600 1900

A, B

D 31,500

22,000

B

E 5 900 31,200

C, D

F 21,300

11,400

E

G 3 900 3900

E

H 5 500 3900

G

Application 2.3



1319

F2

1521

1313

G3

1616

55

D3

88

88

E5

1313

56

C2

78

00

B5

55

01

A5

56

Start

ESLS

IDDUR

EFLF

1616

H5

2121

Finish

Application 2.3

Project Activity and Cost DataActivit

y Crash Cost/DayMaximum Crash Time

(days)A 200 1B 600 2C 300 1D 500 1E 150 2F 100 1G 0 0H 200 2Normal Total Costs =

Total Indirect Costs = Penalty Cost = Total Project Costs =

$7,500$250 per day 21 days = $5,250

$100 per day 7 days = $700$13,450

Application 2.3


Step 1: The critical path is , and the project duration is

BDEGH21 days.

Step 2: Activity E on the critical path has the lowest cost of crashing ($150 per day). Note that activity G cannot be crashed.

Step 3: Reduce the time (crashing 2 days will reduce the project duration to 19 days) and re-calculate costs:

Normal Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost =

$7,500$150 2 days = $300$250 per day 19 days = $4,750$100 per day 5 days = $500$13,050

Application 2.3


Step 4: Repeat until direct costs greater than savings

(step 2) Activity H on the critical path has the next lowest cost of crashing ($200 per day).

(step 3) Reduce the time (crashing 2 days will reduce the project duration to 17 days) and re-calculate costs:

Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost =

$7,500 + $300 (the added crash costs) = $7,800$200 2 days = $400

$250 per day 17 days = $4,250$100 per day 3 days = $300

$12,750


Application 2.3

(step 4) Repeat

(step 2) Activity D on the critical path has the next lowest crashing cost ($500 per day).

(step 3) Reduce the time (crashing 1 day will reduce the project duration to 16 days) and re-calculate costs:

Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Cost =

$7,800 + $400 (the added crash costs) = $8,200$500 1 day = $500

$250 per day 16 days = $4,000$100 per day 2 days = $200

$12,900 which is greater than the last trial. Hence we stop the crashing process.

Application 2.3


The recommended completion date is day 17.

Trial

Crash Activi

ty

Resulting

Critical Paths

Reduction

(days)

Project

Duration

(days)

Costs

Last Trial

Crash

Cost Adde

d

Total Indire

ct Costs

Total Penalty

Costs

Total Projec

t Costs

0 B-D-E-G-H 21$7,500

$5,250 $700

$13,450

1 E B-D-E-G-H 2 19$7,500 $300

$4,750 $500

$13,050

2 H B-D-E-G-H 2 17$7,800 $400

$4,250 $300

$12,750Further reductions will cost more than the savings

in indirect costs and penalties.The critical path is B D E G H and the duration is 17 days.

Application 2.3


Assessing Risks Risk-management Plans

Strategic Fit

Service/Product Attributes

Project Team Capability

Operations


Assessing Risks

Statistical Analysis

Optimistic time (a)

Most likely time (m)

Pessimistic time (b)


Statistical Analysis

a m bMeanTime

Beta distribution

a m bMeanTime

3 3

Area under curve between a and b is 99.74%

Normal distributionCopyright 2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 50

Statistical Analysis The mean of the beta distribution

can be estimated byte =

a + 4m + b6

The variance of the beta distribution for each activity is

2 =b a

6

2


Example 2.4Suppose that the project team has arrived at the following time estimates for activity B (site selection and survey) of the St. Johns Hospital project:a = 7 weeks, m = 8 weeks, and b = 15

weeksa. Calculate the expected time and

variance for activity B.b. Calculate the expected time and

variance for the other activities in the project.


Example 2.4a. The expected time for activity B is

The variance for activity B is

te = = = 9 weeks7 + 4(8) + 15

6546

2 = = = 1.7815 7

6

286

2


Example 2.4b.The following table shows the expected

activity times and variances for this project. Time Estimates (week) Activity Statistics

Activity Optimistic (a)Most Likely

(m)Pessimistic

(b)Expected Time

(te)Variance

(2)A 11 12 13 12 0.11B 7 8 15 9 1.78C 5 10 15 10 2.78D 8 9 16 10 1.78E 14 25 30 24 7.11F 6 9 18 10 4.00G 25 36 41 35 7.11H 35 40 45 40 2.78I 10 13 28 15 9.00J 1 2 15 4 5.44K 5 6 7 6 0.11Copyright 2013 Pearson Education, Inc. publishing as

Prentice Hall 02 - 54

Application 2.4


The director of continuing education at Bluebird University just approved the planning for a sales training seminar. Her administrative assistant identified the various activities that must be done and their relationships to each other:

Application 2.4


Start

A

B

C

D

FinishE

F

G J

H

I

The Network Diagram is:

ActivityImmediate

Predecessor(s)Optimistic

(a)

Most Likely

(m)Pessimistic

(b)Expected Time (t)

Variance ()

A 5 7 8

B 6 8 12

C 3 4 5

D A 11 17 25

E B 8 10 12

F C, E 3 4 5

G D 4 8 9

H F 5 7 9

I G, H 8 11 17

J G 4 4 4

For the Bluebird University sales training seminar activities, calculate the means and variances for each activity.

Application 2.4


ActivityImmediate

Predecessor(s)Optimistic

(a)

Most Likely

(m)Pessimistic

(b)Expected Time (t)

Variance ()

A 5 7 8

B 6 8 12

C 3 4 5

D A 11 17 25

E B 8 10 12

F C, E 3 4 5

G D 4 8 9

H F 5 7 9

I G, H 8 11 17

J G 4 4 4

For the Bluebird University sales training seminar activities, calculate the means and variances for each activity.

6.83 0.258.33 1.004.00 0.11

17.335.44

10.000.44

4.00 0.117.50 0.697.00 0.44

11.502.25

4.00 0.00

Application 2.4


Analyzing Probabilities

TE = =Expected activity times on the critical path Mean of normal distribution Because the activity times are

independentp2 = (Variances of activities on the critical path)

z =T TE

p

Using the z-transformationwhere

T = due date for the project


Because the central limit theorem can be applied, the mean of the distribution is the earliest expected finish time for the project

Example 2.5Calculate the probability that St. Johns Hospital will become operational in 72 weeks, using (a) the critical path and (b) path ACGJK.a. The critical path BDHJK has a

length of 69 weeks. From the table in Example 2.4, we obtain the variance of path BDHJK: 2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89 Next, we calculate the z-value: 0.87

3.453

11.896972

==

=z


Example 2.5Using the Normal Distribution appendix, we find a value of 0.8078. Thus the probability is about 0.81 the length of path BDHJK will be no greater than 72 weeks. Length of critical

path

Probability of meeting the schedule is 0.8078

Normal distribution:Mean = 69 weeks; = 3.45 weeks

Probability of exceeding 72 weeks is 0.1922

Project duration (weeks)69 72

Because this is the critical path, there is a 19 percent probability that the project will take longer than 72 weeks.


Example 2.5

b. The sum of the expected activity times on path ACGJK is 67 weeks and that 2 = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55. The z-value is 1.27

3.945

15.556772

==

=z

The probability is about 0.90 that the length of path ACGJK will be no greater than 72 weeks.


The director of the continuing education at Bluebird University wants to conduct the seminar in 47 working days from now. Using the activity data from Application 2.4, what is the probability that everything will be ready in time?

The critical path isand the expected completion time is

T =

TE is:

ADGI, 43.16 days.

47 days

43.16 days

(0.25 + 5.44 + 0.69 + 2.25) = 8.63And the sum of the variances for the critical activities is:

Application 2.5


Application 2.5


Start

A

B

C

D

FinishE

F

G J

H

I

The Network Diagram is:

The critical path is A-D-G-I

= = = 1.313.842.94

47 43.16 8.63

T = 47 daysTE = 43.16 daysAnd the sum of the variances for the critical activities is: 8.63

z = T TE 2

Assuming the normal distribution applies, we use the table for the normal probability distribution. Given z = 1.30, the probability that activities ADGI can be completed in 47 days or less is 0.9049.


Application 2.5

Monitoring and Controlling Projects Monitoring Project Status

Open Issues and Risks Schedule Status

Monitoring Project Resources


Solved Problem 2.1Your company has just received an order from a good customer for a specially designed electric motor. The contract states that, starting on the thirteenth day from now, your firm will experience a penalty of $100 per day until the job is completed. Indirect project costs amount to $200 per day. The data on direct costs and activity precedent relationships are given in Table 2.2.

a. Draw the project network diagram.b. What completion date would you recommend?


Solved Problem 2.1ELECTRIC MOTOR PROJECT DATA

Activity

Normal Time (days)

Normal Cost ($)

Crash Time (days)

Crash Cost ($)

Immediate Predecessor(s)

A 4 1,000 3 1,300 None

B 7 1,400 4 2,000 None

C 5 2,000 4 2,700 None

D 6 1,200 5 1,400 A

E 3 900 2 1,100 B

F 11 2,500 6 3,750 C

G 4 800 3 1,450 D, E

H 3 300 1 500 F, G


Solved Problem 2.1a.The network diagram is shown in Figure 2.10.

Keep the following points in mind while constructing a network diagram.

Start

Finish

A4

B7

C5

D6

E3

F11

G4

H3


Solved Problem 2.1b. With these activity times, the project will be

completed in 19 days and incur a $700 penalty. Using the data in Table 2.2, you can determine the maximum crash-time reduction and crash cost per day for each activity. For activity AMaximum crash time = Normal time Crash time =

4 days 3 days = 1 day

Crash cost per

day= = Crash cost Normal costNormal time Crash time

CC NCNT CT

= = $300$1,300 $1,0004 days 3 days


Solved Problem 2.1

Activity Crash Cost per Day ($)Maximum Time Reduction

(days)A 300 1B 200 3C 700 1D 200 1E 200 1F 250 5G 650 1H 100 2


Solved Problem 2.1

PROJECT COST ANALYSISStage Crash

ActivityTime Reduction (days)

Resulting Critical Path(s)

Project Duration (days)

Project Direct Costs, Last Trial ($)

Crash Cost Added ($)

Total Indirect Costs ($)

Total Penalty Costs ($)

Total Project Costs ($)

0 C-F-H 19 10,100 3,800 700 14,600

1 H 2 C-F-H 17 10,100 200 3,400 500 14,200

2 F 2 A-D-G-H 15 10,300 500 3,000 300 14,100

B-E-G-H

C-F-H


Solved Problem 2.1The critical path is CFH at 19 days, which is the longest path in the network.

The cheapest activity to crash is H which, when combined with reduced penalty costs, saves $300 per day.

Crashing this activity for two days gives

ADGH: 15 days, BEGH: 15 days, and CFH: 17 days

Crash activity F next. This makes all activities critical and no more crashing should be done as the cost of crashing exceeds the savings.


Solved Problem 2.2An advertising project manager developed the network diagram in Figure 2.11 for a new advertising campaign. In addition, the manager gathered the time information for each activity, as shown in the accompanying table.a. Calculate the expected time and variance for each activity.

b.Calculate the activity slacks and determine the critical path, using the expected activity times.

c. What is the probability of completing the project within 23 weeks?

Start

Finish

A

B

C

D

E

F

G


Solved Problem 2.2

Time Estimate (weeks)Activity Optimistic Most Likely Pessimistic Immediate Predecessor(s)

A 1 4 7 B 2 6 7 C 3 3 6 BD 6 13 14 AE 3 6 12 A, CF 6 8 16 BG 1 5 6 E, F


Solved Problem 2.2a. The expected time and variance for each

activity are calculated as followste = a + 4m + b

6

Activity Expected Time (weeks) VarianceA 4.0 1.00B 5.5 0.69C 3.5 0.25D 12.0 1.78E 6.5 2.25F 9.0 2.78G 4.5 0.69


Solved Problem 2.2b. We need to calculate the earliest start, latest start,

earliest finish, and latest finish times for each activity. Starting with activities A and B, we proceed from the beginning of the network and move to the end, calculating the earliest start and finish times.

Activity Earliest Start (weeks) Earliest Finish (weeks)A 0 0 + 4.0 = 4.0B 0 0 + 5.5 = 5.5C 5.5 5.5 + 3.5 = 9.0D 4.0 4.0 + 12.0 =16.0E 9.0 9.0 + 6.5 = 15.5F 5.5 5.5 + 9.0 = 14.5G 15.5 15.5 + 4.5 =20.0


Solved Problem 2.2Based on expected times, the earliest finish date for the project is week 20, when activity G has been completed. Using that as a target date, we can work backward through the network, calculating the latest start and finish times

Activity Latest Start (weeks) Latest Finish (weeks)G 15.5 20.0F 6.5 15.5E 9.0 15.5D 8.0 20.0C 5.5 9.0B 0.0 5.5A 4.0 8.0


Solved Problem 2.2

A

4.0

0.04.0

4.08.0

D

12.0

4.08.0

16.020.0

E

6.5

9.09.0

15.515.5

G

4.5

15.515.5

20.020.0

C

3.55.55.5

9.09.0

F

9.05.56.5

14.515.5

B

5.5

0.00.0

5.55.5

Finish

Start


Solved Problem 2.2Start (weeks) Finish (weeks)

Activity Earliest Latest Earliest Latest Slack Critical PathA 0 4.0 4.0 8.0 4.0 NoB 0 0.0 5.5 5.5 0.0 YesC 5.5 5.5 9.0 9.0 0.0 YesD 4.0 8.0 16.0 20.0 4.0 NoE 9.0 9.0 15.5 15.5 0.0 YesF 5.5 6.5 14.5 15.5 1.0 NoG 15.5 15.5 20.0 20.0 0.0 Yes

Path Total Expected Time (weeks) Total VarianceAD 4 + 12 = 16 1.00 + 1.78 = 2.78AEG 4 + 6.5 + 4.5 = 15 1.00 + 2.25 + 0.69 = 3.94

BCEG 5.5 + 3.5 + 6.5 + 4.5 = 20 0.69 + 0.25 + 2.25 + 0.69 = 3.88BFG 5.5 + 9 + 4.5 = 19 0.69 + 2.78 + 0.69 = 4.16


Solved Problem 2.2The critical path is BCEG with a total expected time of 20 weeks. However, path BFG is 19 weeks and has a large variance.

c. We first calculate the z-value:

z = = = 1.52T TE

223 20

3.88Using the Normal Distribution Appendix, we find the probability of completing the project in 23 weeks or less is 0.9357. Because the length of path BFG is close to that of the critical path and has a large variance, it might well become the critical path during the project



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Slide 1What is a Project?What is Project Management?Defining and Organizing ProjectsPlanning ProjectsWork Breakdown StructureWork Breakdown StructureDiagramming the NetworkEstablishing Precedence RelationshipsExample 2.1Example 2.1Application 2.1Application 2.1Developing the ScheduleDeveloping the ScheduleDeveloping the ScheduleExample 2.2Slide 18Slide 19Slide 20Slide 21Developing a ScheduleApplication 2.2Application 2.2Application 2.2Analyzing Cost-Time Trade-OffsAnalyzing Cost-Time Trade-OffsCost-Time RelationshipsAnalyzing Cost-Time Trade-OffsExample 2.3Example 2.3Example 2.3Example 2.3Example 2.3Example 2.3Example 2.3Example 2.3Example 2.3Example 2.3Example 2.3Application 2.3Slide 42Application 2.3Application 2.3Application 2.3Application 2.3Application 2.3Assessing RisksAssessing RisksStatistical AnalysisStatistical AnalysisExample 2.4Example 2.4Example 2.4Application 2.4Application 2.4Application 2.4Application 2.4Analyzing ProbabilitiesExample 2.5Example 2.5Example 2.5Application 2.5Application 2.5Application 2.5Monitoring and Controlling ProjectsSolved Problem 2.1Solved Problem 2.1Solved Problem 2.1Solved Problem 2.1Solved Problem 2.1Solved Problem 2.1Solved Problem 2.1Solved Problem 2.2Solved Problem 2.2Solved Problem 2.2Solved Problem 2.2Solved Problem 2.2Solved Problem 2.2Solved Problem 2.2Solved Problem 2.2Slide 82

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