2 – 1 copyright © 2010 pearson education, inc. publishing as prentice hall. project management...

2 – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Project ManagementProject Management

Chapter 2


ProjectsProjects

Projects are an interrelated set of activities with a definite starting and ending point, which results in a unique outcome from a specific allocation of resources

The three main goals are to: Complete the project on time Not exceed the budget Meet the specifications to the satisfactions of

the customer


ProjectsProjects

Project management is a systemized, approach to defining, organizing, planning, monitoring, and controlling projects

Projects often require resources from many different parts of the organization

Each project is unique


Planning ProjectsPlanning Projects

There are five steps to planning projects1. Defining the work breakdown structure

2. Diagramming the network

3. Developing the schedule

4. Analyzing the cost-time trade-offs

5. Assessing risks


Work Breakdown StructureWork Breakdown Structure

A statement of all the tasks that must be completed as part of the project

An activity is the smallest unit of work effort consuming both time and resources that the project manager can schedule and control

Each activity must have an owner who is responsible for doing the work


Diagramming the NetworkDiagramming the Network

Network diagrams use nodes and arcs to depict the relationships between activities

Benefits of using networks include1. Networks force project teams to identify and

organize data to identify interrelationships between activities

2. Networks enable the estimation of completion time

3. Crucial activities are highlighted

4. Cost and time trade-offs can be analyzed



Precedent relationships determine the sequence for undertaking activities

Activity times must be estimated using historical information, statistical analysis, learning curves, or informed estimates

In the activity-on-node approach, nodes represent activities and arcs represent the relationships between activities


S T U S precedes T, which precedes U.


AON Activity Relationships

S

T

US and T must be completed before U can be started.

Figure 2.2




T

U

ST and U cannot begin until S has been completed.

S

T

U

V

U and V can’t begin until both S and T have been completed.

Figure 2.2




S

T

U

V

U cannot begin until both S and T have been completed; V cannot begin until T has been completed.

S T V

U

T and U cannot begin until S has been completed and V cannot begin until both T and U have been completed.

Figure 2.2


Developing the ScheduleDeveloping the Schedule

Schedules can help managers achieve the objectives of the project

Managers can1. Estimate the completion time by finding the

critical path

2. Identify start and finish times for each activity

3. Calculate the amount of slack time for each activity


Critical PathCritical Path

The sequence of activities between a project’s start and finish is a path

The critical path is the path that takes the longest time to complete


St. John’s Hospital Project (Modified)St. John’s Hospital Project (Modified)Activity Immediate

PredecessorsActivity Times (wks)

Responsibility

ST. JOHN’S HOSPITAL PROJECT

START

ORGANIZING and SITE PREPARATION

A. Select administrative staff

B. Select site and survey

C. Select medical equipment

D. Prepare final construction plans

E. Bring utilities to site

F. Interview applicants for nursing and support staff

PHYSICAL FACILITIES and INFRASTRUCTURE

G. Purchase and deliver equipment

H. Construct hospital

I. Develop information system

J. Install medical equipment

K. Train nurses and support staff

FINISHExample 2.1

START

START

A

B

B

A

C

D

A

E, G, H

F, I

J, K

0

12

9

10

10

24

10

35

40

15

4

6

0

Kramer

Stewart

Johnson

Taylor

Adams

Taylor

Burton

Johnson

Walker

Sampson

Casey

Murphy

Pike

Ashton

F,I,J in the text


Activity Immediate Predecessors

Activity Times (wks)

Responsibility


START














FINISH

St. John’s Hospital ProjectSt. John’s Hospital Project

Example 2.1

Completion Time

Finish

K6

I15

F10

C10

H40

J4

A12

B9

Figure 2.3

StartG

35

D10

E24

Activity IP Time

A START 12B START 9C A 10D B 10E B 24F A 10G C 35H D 40I A 15J E, G, H 4K F, I 6




Responsibility


START














FINISH


Example 2.1

Completion Time

Finish

K6

I15

F10

C10

D10

H40

J4

A12

B9

Figure 2.3

StartG

35

E24

Path Estimated Time (weeks)

A–I–K 33

A–F–K 28

A–C–G–J 61

B–D–H–J 63

B–E–J 37


Project ScheduleProject Schedule

The project schedule specifies start and finish times for each activity

Managers can use the earliest start and finish times, the latest start and finish times, or any time in between these extremes


Project ScheduleProject Schedule

The earliest start time (ES) for an activity is the latest earliest finish time of any preceding activities

The earliest finish time (EF) is the earliest start time plus its estimated duration

EF = ES + t

The latest finish time (LF) for an activity is the latest start time of any preceding activities

The latest start time (LS) is the latest finish time minus its estimated duration

LS = LF – t


Early Start and Early Finish TimesEarly Start and Early Finish Times

EXAMPLE 2.2

Calculate the ES, EF, LS, and LF times for each activity in the hospital project. Which activity should Kramer start immediately? Figure 2.3 contains the activity times.

SOLUTION

To compute the early start and early finish times, we begin at the start node at time zero. Because activities A and B have no predecessors, the earliest start times for these activities are also zero. The earliest finish times for these activities are

EFA = 0 + 12 = 12 and EFB = 0 + 9 = 9



Because the earliest start time for activities I, F, and C is the earliest finish time of activity A,

ESI = 12, ESF = 12, and ESC = 12

Similarly,

ESD = 9 and ESE = 9

After placing these ES values on the network diagram, we determine the EF times for activities I, F, C, D, and E:

EFI = 12 + 15 = 27, EFF = 12 + 10 = 22, EFC = 12 + 10 = 22,

EFD = 9 + 10 = 19, and EFE = 9 + 24 = 33



The earliest start time for activity G is the latest EF time of all immediately preceding activities. Thus,

ESG = EFC = 22, ESH = EFD = 19

EFG = ESG + t = 22 + 35 = 57, EFH + t = 19 + 40 = 59

Latest finish time

Latest start time

Activity

Duration

Earliest start timeEarliest finish time

0

2

12

14

A

12


Network DiagramNetwork Diagram

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12

F

10

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 27 330 12

Figure 2.4



To compute the latest start and latest finish times, we begin by setting the latest finish activity time of activity K at week 63, which is the earliest project finish time. Thus, the latest start time for activity K is

LSK = LFK – t = 63 – 6 = 57

If activity K is to start no later than week 57, all its predecessors must finish no later than that time. Consequently,

LFI = 57, LFF = 57, and LFJ = 57



The latest start times for these activities are shown in Figure 2.4 as

LSI = 57 – 15 = 42, LFF = 57 – 10 = 47, and

LSj = 63 – 4= 59

After obtaining LSJ, we can calculate the latest start times for the immediate predecessors of activity J:

LSG = 59 – 35 = 24, LSH = 59 – 40 = 19, and LSE = 59 – 24 = 35

Similarly, we can now calculate the latest start times for activities C and D:

LSC = 24 – 10 = 14 and LSD = 19 – 10 = 9



Activity A has more than one immediately following activity: I, F, and C. The earliest of the latest start times is 14 for activity C. Thus,

LSA = 14 – 12 = 2

Similarly, activity B has two immediate followers: D and E. Because the earliest of the latest start times of these activities is 9.

LSB = 9 – 9 = 0



Figure 2.4

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12

F

10

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 27 330 12

42 57

47 57

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

57 63

S = 30

S = 2 S = 35

S = 2

S = 26

S = 2

S = 0S = 0

S = 30S = 30

S = 0S = 0S = 0S = 0S = 0S = 0


Activity SlackActivity Slack

Activity slack is the maximum length of time an activity can be delayed without delaying the entire project

Activities on the critical path have zero slack

Activity slack can be calculated in two ways

S = LS – ES or S = LF – EF

2 – 28

Project CostsProject Costs

The total project costs are the sum of direct costs, indirect costs, penalty cost, and crashing cost.

Direct costs include labor, materials, and any other costs directly related to project activities.

Indirect costs include administration, depreciation, financial, and other variable overhead costs that can be avoided by reducing total project time.

Penalty cost is zero unless the project is completed late

Crashing cost is the cost to decrease the activity time


Cost-Time Trade-OffsCost-Time Trade-Offs

Projects may be crashed to shorten the completion time

Costs to crash

Cost to crash per period =CC – NC

NT – CT

1. Normal time (NT) 3. Crash time (CT)2. Normal cost (NC) 4. Crash cost (CC)

2 – 30

Cost to CrashCost to Crash

To assess the benefit of crashing certain activities, either from a cost or a schedule perspective, the project manager needs to know the following times and costs.

Normal time (NT) is the time necessary to complete and 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-Time RelationshipsCost-Time Relationships

Linear cost assumption

8000 —

7000 —

6000 —

5000 —

4000 —

3000 —

0 —

Dir

ect

cost

(d

oll

ars)

| | | | | |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 2.6


Cost-Time RelationshipsCost-Time Relationships

TABLE 2.1 | DIRECT COST AND TIME DATA FOR THE ST. JOHN’S 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 9 50,000 7 64,000 2 7,000

C 10 4,000 5 7,000 5 600

D 10 16,000 8 20,000 2 2,000

E 24 120,000 14 200,000 10 8,000

F 10 10,000 6 16,000 4 1,500

G 35 500,000 25 530,000 10 3,000

H 40 1,200,000 35 1,260,000 5 12,000

I 15 40,000 10 52,500 5 2,500

J 4 10,000 1 13,000 3 1,000

K 6 30,000 5 34,000 1 4,000

Totals $1,992,000


Assessing RiskAssessing Risk

Risk is the measure of the probability and consequence of not reaching a defined project goal

Risk-management plans are developed to identify key risks and prescribe ways to circumvent them


Simulation and Statistical AnalysisSimulation and Statistical Analysis

When uncertainty is present, simulation can be used to estimate the project completion time

Statistical analysis requires three reasonable estimates of activity times

1. Optimistic time (a)

2. Most likely time (m)

3. Pessimistic time (b)


Statistical AnalysisStatistical Analysis

a m bMean

Time

Beta distribution

a m bMeanTime

3σ 3σ

Area under curve between a and b is 99.74%

Normal distributionFigure 2.7


Statistical AnalysisStatistical Analysis

The mean of the beta distribution can be estimated by

te =a + 4m + b

6

The variance of the beta distribution for each activity is

σ2 =b – a

6

2


Calculating Means and VariancesCalculating Means and Variances

EXAMPLE 2.4

Suppose that the project team has arrived at the following time estimates for activity B (site selection and survey) of the St. John’s Hospital project:

a = 7 weeks, m = 8 weeks, and b = 15 weeks

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

b. Calculate the expected time and variance for the other activities in the project.



SOLUTION

a. The expected time for activity B is

Note that the expected time does not equal the most likely time. These will only be the same only when the most likely time is equidistant from the optimistic and pessimistic times.

The variance for activity B is

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

6

54

6

σ2 = = = 1.7815 – 7

6

28

6

2



b. 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 (σ2)

A 11 12 13 12 0.11

B 7 8 15 9 1.78

C 5 10 15 10 2.78

D 8 9 16 10 1.78

E 14 25 30 24 7.11

F 6 9 18 10 4.00

G 25 36 41 35 7.11

H 35 40 45 40 2.78

I 10 13 28 15 9.00

J 1 2 15 4 5.44

K 5 6 7 6 0.11



Figure 2.4

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12

F

10

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 27 330 12

42 57

47 57

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

57 63

S = 30

S = 2 S = 35

S = 2

S = 26

S = 2

S = 0S = 0

S = 30S = 30

S = 0S = 0S = 0S = 0S = 0S = 0


Analyzing ProbabilitiesAnalyzing Probabilities

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

TE = =Expected activity times

on the critical path Mean of normal distribution

Because the activity times are independent

σ2 = (Variances of activities on the critical path)

z =T – TE

σ2

Using the z-transformation

where

T = due date for the project


Calculating the ProbabilityCalculating the Probability

EXAMPLE

Calculate the probability that St. John’s Hospital will become operational in 65 weeks, using (a) the critical path and (b) path A–C–G–J.

SOLUTION

a. The critical path B–D–H–J has a length of 63 weeks. From slide 50, we obtain the variance of path B–D–H–J: σ2 = 1.78 + 1.78 + 2.78 + 5.44 = 11.78. Next, we calculate the z-value:

0.583.432

2

11.78

6365

z



Using the Normal Distribution appendix, we go down the left-hand column to 0.5 and then across to 0.08. This gives a value of 0.7190. Thus the probability is about 0.719 that the length of path B–D–H–J will be no greater than 65 weeks.

Length of critical path

Probability of meeting the schedule is 0.7190

Normal distribution:Mean = 63 weeks;σ = 3.432 weeks

Probability of exceeding 65 weeks is 0.2810

Project duration (weeks)

63 65

Because this is the critical path, there is a 28.1 percent probability that the project will take longer than 65 weeks.


From the normal table for 98%, z = 2.055

TE = 63 weeks, 3.432 weeks

Critical Path = B - D - H - J

T = TE + z

Expected project completion timeExpected project completion time

EXAMPLE

Determine the completion time of the critical path with 98% probability .

T = 63 + 2.055 (3.432) = 70.05 weeks



SOLUTION

b. From the table in Example 2.4, we determine that the sum of the expected activity times on path A–C–G–J is 12 + 10 + 35 + 4 = 61 weeks and that σ2 = 0.11 + 2.78 + 7.11 + 5.44 = 15.44. The z-value is

1.023.93

4

15.44

6165

z

The probability is about 0.8461 that the length of path A–C–G–J will be no greater than 65 weeks.

EXAMPLE

Calculate the probability that St. John’s Hospital will become operational in 65 weeks, using (a) the critical path and (b) path A–C–G–J.


Near-Critical PathsNear-Critical Paths

Project duration is a function of the critical path

Since activity times vary, paths with nearly the same length can become critical during the project

Project managers can use probability estimates to analyze the chances of near-critical paths delaying the project


Application 2.1Application 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 A

C 4 A

D 4 B, C

E 4 D

F 3 E

G 5 E


Finish

G5

F3

E4

D4


Activity Activity Time (days)Immediate

Predecessor(s)

A 7 —

B 2 A

C 4 A

D 4 B, C

E 4 D

F 3 E

G 5 E

B2

C4

Start A7



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 Yes

B 2

C 4

D 4

E 4

F 3

G 5

The critical path is A–C–D–E–G 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)


Latest Finish (LF)

Slack (LS-ES)


A 7 0 0 7 7 0-0=0 Yes

B 2

C 4

D 4

E 4

F 3

G 5


7 9 9 11 9-7=2 No

7 7 11 11 7-7=0 Yes

19 21 22 24 21-19=2 No

19 19 24 24 19-19=0 Yes

11 11 15 15 11-11=0 Yes

15 15 19 19 15-15=0 Yes



Activity Duration

Earliest Start (ES)

Latest Start (LS)


Latest Finish (LF)

Slack (LS-ES)


A 7 0 0 7 7 0-0=0 Yes

B 2 7 9 9 11 9-7=2 No

C 4 7 7 11 11 7-7=0 Yes

D 4 11 11 15 15 11-11=0 Yes

E 4 15 15 19 19 15-15=0 Yes

F 3 21 21 22 24 21-19=2 No

G 5 19 19 24 24 19-19=0 Yes

Start FinishA7

B2

C4

D4

E4

F3

G5




Indirect project costs = $250 per day and penalty cost = $100 per day for each day the project lasts beyond day 14.

Project Activity and Cost Data

ActivityNormal Time (days)

Normal Cost ($)

Crash Time (days)

Crash Cost ($)

Immediate Predecessor(s)

A 5 1,000 4 1,200 —

B 5 800 3 2,000 —

C 2 600 1 900 A, B

D 3 1,500 2 2,000 B

E 5 900 3 1,200 C, D

F 2 1,300 1 1,400 E

G 3 900 3 900 E

H 5 500 3 900 G



Direct cost and time data for the activities:

Project Activity and Cost Data

Activity Crash Cost/Day Maximum Crash Time (days)

A 200 1

B 600 2

C 300 1

D 500 1

E 150 2

F 100 1

G 0 0

H 200 2

Solution:Original costs:

Normal 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


ActivityImmediate

Predecessor(s)Optimistic

(a)

Most Likely

(m)Pessimistic

(b)Expected Time (t)

Variance (σ)

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


Bluebird University: activity for sales training seminar

6.83 0.25

8.33 1.00

4.00 0.11

17.33 5.44

10.00 0.44

4.00 0.11

7.50 0.69

7.00 0.44

11.50 2.25

4.00 0.00



The director of the continuing education at Bluebird University wants to conduct the seminar in 47 working days from now. What is the probability that everything will be ready in time?

The critical path is

and the expected completion time is

T =

TE is:

A–D–G–I,

43.17 days.

47 days

43.17 days

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


= = = 1.303.83

2.94

47 – 43.17

8.63


T = 47 days

TE = 43.17 days

And the sum of the variances for the critical activities is: 8.63

z = T – TE

σ2

Assuming the normal distribution applies, we use the table for the normal probability distribution. Given z = 1.30, the probability that activities A–D–G–I can be completed in 47 days or less is 0.9032.


Solved Problem 1Solved Problem 1

Your 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?



TABLE 2.2 | ELECTRIC 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



SOLUTION

a. The network diagram is shown in Figure 2.10. Keep the following points in mind while constructing a network diagram.

1. Always have start and finish nodes.

2. Try to avoid crossing paths to keep the diagram simple.

3. Use only one arrow to directly connect any two nodes.

4. Put the activities with no predecessors at the left and point the arrows from left to right.

5. Be prepared to revise the diagram several times before you comeup with a correct and uncluttered diagram.

Start

Finish

A4

B7

C5

D6

E3

F11

G4

H3

Figure 2.10



b. 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 A

Maximum 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



Activity Crash Cost per Day ($) Maximum Time Reduction (days)

A 300 1

B 200 3

C 700 1

D 200 1

E 200 1

F 250 5

G 650 1

H 100 2



Table 2.3 summarizes the analysis and the resultant project duration and total cost. The critical path is C–F–H 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

A–D–G–H: 15 days, B–E–G–H: 15 days, and C–F–H: 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.



An 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?

Figure 2.11

Start

Finish

A

B

C

D

E

F

G



Time Estimate (weeks)

Activity Optimistic Most Likely Pessimistic Immediate Predecessor(s)

A 1 4 7 —

B 2 6 7 —

C 3 3 6 B

D 6 13 14 A

E 3 6 12 A, C

F 6 8 16 B

G 1 5 6 E, F



SOLUTION

a. The expected time and variance for each activity are calculated as follows

te =a + 4m + b

6

Activity Expected Time (weeks) Variance

A 4.0 1.00

B 5.5 0.69

C 3.5 0.25

D 12.0 1.78

E 6.5 2.25

F 9.0 2.78

G 4.5 0.69



SOLUTION

b. 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 = 4.0

B 0 0 + 5.5 = 5.5

C 5.5 5.5 + 3.5 = 9.0

D 4.0 4.0 + 12.0 = 16.0

E 9.0 9.0 + 6.5 = 15.5

F 5.5 5.5 + 9.0 = 14.5

G 15.5 15.5 + 4.5 = 20.0



Based 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 20.0

F 6.5 15.5

E 9.0 15.5

D 8.0 20.0

C 5.5 9.0

B 0.0 5.5

A 4.0 8.0



A

4.0

0.0

4.0

4.0

8.0

D

12.0

4.0

8.0

16.0

20.0

E

6.5

9.0

9.0

15.5

15.5

G

4.5

15.5

15.5

20.0

20.0

C

3.55.5

5.5

9.0

9.0

F

9.05.5

6.5

14.5

15.5

B

5.5

0.0

0.0

5.5

5.5

Finish

Start

Figure 2.12



Start (weeks) Finish (weeks)

Activity Earliest Latest Earliest Latest Slack Critical Path

A 0 4.0 4.0 8.0 4.0 No

B 0 0.0 5.5 5.5 0.0 Yes

C 5.5 5.5 9.0 9.0 0.0 Yes

D 4.0 8.0 16.0 20.0 4.0 No

E 9.0 9.0 15.5 15.5 0.0 Yes

F 5.5 6.5 14.5 15.5 1.0 No

G 15.5 15.5 20.0 20.0 0.0 Yes

Path Total Expected Time (weeks) Total Variance

A–D 4 + 12 = 16 1.00 + 1.78 = 2.78

A–E–G 4 + 6.5 + 4.5 = 15 1.00 + 2.25 + 0.69 = 3.94

B–C–E–G 5.5 + 3.5 + 6.5 + 4.5 = 20 0.69 + 0.25 + 2.25 + 0.69 = 3.88

B–F–G 5.5 + 9 + 4.5 = 19 0.69 + 2.78 + 0.69 = 4.16



So the critical path is B–C–E–G with a total expected time of 20 weeks. However, path B–F–G is 19 weeks and has a large variance.

c. We first calculate the z-value:

z = = = 1.52T – TE

σ223 – 20

3.88

Using 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 B–F–G 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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