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Project Management

Chapter 7

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What is a Project?

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.

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What is Project Management?

Project Management

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

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Defining and Organizing Projects

Defining the Scope and Objectives of a Project

Selecting the Project Manager and Team

Recognizing Organizational Structure

Could be a TQM project/team

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Constructing Project Networks

Defining the Work Breakdown Structure

Diagramming the Network

Developing the Project Schedule

Analyzing Cost-Time Trade-offs

Assessing and Analyzing Risks

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Defining the Work Breakdown Structure

Work Breakdown Structure

A statement of all work that has to be completed. (more than a statement, but an actual breakdown of all tasks that need to be performed in order to accomplish the objective)

Activity

The smallest unit of work effort consuming both time and resources that the project manager can schedule and control. (A task)

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Purchase and deliver equipment

Construct hospital

Develop information system

Install medical equipment

Train nurses and support staff

Work Breakdown Structure

Select administration staff

Select site and survey

Select medical equipment

Prepare final construction plans

Bring utilities to site

Interview applicants for nursing and support staff

Organizing and Site Preparation

Physical Facilities and Infrastructure

Level 1

Level 0

Level 2

Relocation of St. John’s Hospital

Figure 7.1

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Diagramming the Network

Network Diagram – A network planning method designed to depict the relationships between activities, that consist of nodes (circles) and arcs (arrows)

Program Evaluation and Review Technique (PERT)

Critical Path Method (CPM)

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Diagramming the Network

Precedence relationship

A relationship that determines a sequence for undertaking activities; it specifies that one activity cannot start until a preceding activity has been completed.

Estimating Activity Times

Statistical methods

Learning curve models

Managerial opinions

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Example 7.1

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Finish

Start

A —

B —

C A

D B

E B

F A

G C

H D

I A

J E,G,H

K F,I,J

Immediate Predecessor

Example 7.1

Figure 7.2

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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

Application 7.1

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Finish

Start

Application 7.1

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

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Developing the Project Schedule

Path

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

Critical Path –

The sequence of activities between a project’s start and finish that takes the longest time to complete.

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Developing the Schedule

Earliest start time (ES) - The earliest finish time of the immediately preceding activity.

Earliest finish time (EF) - An activity’s earliest start time plus its estimated duration (t). EF = ES + t

Latest finish time (LF) - The latest start times of the activity that immediately follows.

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

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

S = LS – ES or S = LF – EF

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Developing the Schedule

Latest finish time

Latest start time

Activity Name

Estimated time

Earliest start time

Earliest finish time

Figure 7.3

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Finish

Start

Path Time (wks)

A-I-K 33 A-F-K 28

A-C-G-J-K 67

B-D-H-J-K 69

B-E-J-K 43

Paths are the sequence of activities between a project’s start and finish.

Example 7.2

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Finish

Start

Earliest start time

Earliest finish time

0 9

9 33

9 19

19 59

22 57

12 22

59 63

12 27

12 22

63 69

Example 7.2

Figure 7.3

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Example 7.2

Finish

Start

0 9

9 33

9 19

19 59

22 57

12 22

59 63

12 27

12 22

63 69

0 12

The Critical Path takes 69 weeks

Figure 7.3

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Finish

Start

0 9

9 33

9 19

19 59

22 57

12 22

59 63

12 27

12 22

63 69

0 12

48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

63 69

Example 7.2

Latest start time

Latest finish time

Figure 7.3

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Finish

Start

0 9

9 33

9 19

19 59

22 57

12 22

59 63

12 27

12 22

63 69

0 12

48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

63 69

Example 7.2

S = 0

S = 2

S = 26

S = 0

S = 36

S = 2

S = 41

S = 0

Figure 7.3

Critical Path

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Developing a Schedule (Pert or Gantt)

Gantt chart

Figure 7.4

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Application 7.2

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

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.

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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

9-7=2

7-7=0

Yes

21-19=2

19-19=0

Yes

11-11=0

Yes

15-15=0

Yes

Application 7.2

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Application 7.2

The critical path is A–C–D–E–G with a project duration of 24 days.

Start

Finish

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Analyzing Cost-Time Trade-Offs

Project Crashing

Shortening (or expediting) some activities within a project to reduce overall project completion time and total project costs

Project Costs

Direct Costs

Indirect Costs

Penalty Costs

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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 —

Direct cost (dollars)

| | | | | |

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 7.5

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Analyzing Cost-Time Trade-Offs

Determining the Minimum Cost Schedule:

Determine the project’s critical path(s).

Find the activity or activities on the critical path(s) with the lowest cost of crashing per week.

Reduce the time for this activity until…

It cannot be further reduced or

Another path becomes critical, or

The increase in direct costs exceeds the indirect and penalty cost savings that result from shortening the project.

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

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Example 7.3

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		$2,209,500

Table 7.1

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Example 7.3

Determine the minimum-cost schedule for the St. John’s 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

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

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Example 7.3

STAGE 1

Step 1. The critical path is B–D–H–J–K.

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

A–C–G–J–K: 64 weeks

B–D–H–J–K: 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.

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Example 7.3

Finish

Start

STAGE 1

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Example 7.3

STAGE 2

Step 1. The critical path is still B–D–H–J–K.

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 are

A–C–G–J–K: 64 weeks and B–D–H–J–K: 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.

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Example 7.3

Finish

Start

STAGE 2

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Example 7.3

STAGE 3

Step 1. The critical paths are B–D–H–J–K 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: A–C–G–J–K: 63 weeks and B–D–H–J–K: 63 weeks

Net savings are $8,000 – $4,000 = $4,000.

Total project costs are $2,511,000 – $4,000 = $2,507,000.

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Example 7.3

Finish

Start

STAGE 3

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Example 7.3

STAGE 4

Step 1. The critical paths are still B–D–H–J–K and A–C–G–J–K.

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

A–C–G–J–K: 61 weeks and B–D–H–J–K: 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.

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Example 7.3

Finish

Start

STAGE 4

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Example 7.3

Stage	Crash Activity	Time 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	—	552.0	80.0	2,624.0
1	J	3	B-D-H-J-K	66	1,992.0	3.0	528.0	20.0	2,543.0
2	D	2	B-D-H-J-K A-C-G-J-K	64	1,995.0	4.0	512.0	0.0	2,511.0
3	K	1	B-D-H-J-K A-C-G-J-K	63	1,999.0	4.0	504.0	0.0	2,507.0
4	B, C	2	B-D-H-J-K A-C-G-J-K	61	2,003.0	15.2	488.0	0.0	2,506.2

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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	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

Application 7.3

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Application 7.3

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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

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

Application 7.3

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

B–D–E–G–H

21 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 7.3

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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 7.3

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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 7.3

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The recommended completion date is day 17.

Trial	Crash Activity	Resulting Critical Paths	Reduction (days)	Project Duration (days)	Costs Last Trial	Crash Cost Added	Total Indirect Costs	Total Penalty Costs	Total Project 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	$12,750

Further 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 7.3

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Assessing and Analyzing Risks

Risk-management Plans

Strategic Fit (linkage to Chapter 1) – if doesn’t fit need to ask yourself why doing

Service/Product Attributes (requirements)

Project Team Capability (capable?)

Operations (impact on)

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Assessing Risks

Statistical Analysis

Optimistic time (a)

Most likely time (m)

Pessimistic time (b)

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Statistical Analysis

Mean

Time

Beta distribution

Mean

Time

3σ

Area under curve between a and b is 99.74%

Normal distribution

Figure 7.6

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Statistical Analysis

The mean of the beta distribution can be estimated by

te =

a + 4m + b

The variance of the beta distribution for each activity is

σ2 =

b – a

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Example 7.4

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.

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Example 7.4

a. The expected time for activity B is

The variance for activity B is

te = = = 9 weeks

7 + 4(8) + 15

σ2 = = = 1.78

15 – 7

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Example 7.4

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

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Application 7.4

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:

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Application 7.4

Start

Finish

The Network Diagram is:

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Activity	Immediate 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

For the Bluebird University sales training seminar activities,

calculate the means and variances for each activity.

Application 7.4

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Activity	Immediate 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

For the Bluebird University sales training seminar activities,

calculate the means and variances for each activity.

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

Application 7.4

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Analyzing Probabilities

TE = =

Expected activity times on the critical path



Mean of normal distribution

Because the activity times are independent

σp2 =  (Variances of activities on the critical path)

z =

T – TE

σp

Using the z-transformation

where 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

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Example 7.5

Calculate the probability that St. John’s Hospital will become operational in 72 weeks, using (a) the critical path and (b) path A–C–G–J–K.

a. The critical path B–D–H–J–K has a length of 69 weeks. From the table in Example 2.4, we obtain the variance of path B–D–H–J–K: σ2p = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89 Next, we calculate the z-value:

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Example 7.5

Using the Normal Distribution appendix, we find a value of 0.8078. Thus the probability is about 0.81 the length of path B–D–H–J–K will be no greater than 72 weeks.

Length of critical path

Probability of meeting the schedule is 0.8078

Normal distribution:

Mean = 69 weeks;

σ2p = 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.

Figure 7.7

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Example 7.5

b. The sum of the expected activity times on path A–C–G–J–K is 67 weeks and that σ2p = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55. The z-value is

The probability is about 0.90 that the length of path A–C–G–J–K will be no greater than 72 weeks.

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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 7.4, 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

43.16 days.

47 days

43.16 days

(0.25 + 5.44 + 0.69 + 2.25) = 8.63

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

Application 7.5

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Application 7.5

Start

Finish

The Network Diagram is:

The critical path is A-D-G-I

7-‹#›

= = = 1.31

3.84

2.94

47 – 43.16

8.63



T = 47 days

TE = 43.16 days

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

z =

T – TE

σ2p



Assuming the normal distribution applies, we use the table for the normal probability distribution. Given z = 1.31, the probability that activities A–D–G–I can be completed in 47 days or less is 0.9049.

Application 7.5

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Monitoring and Controlling Projects

Monitoring Project Status

Open Issues and Risks

Schedule Status

Monitoring Project Resources

Figure 7.8

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Monitoring and Controlling Projects

Controlling Projects

Project Close Out – An activity that includes writing final reports, completing remaining deliverables, and compiling the team’s recommendations for improving the project process.

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Solved Problem 7.1

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 7.2.

a. Draw the project network diagram.

b. What completion date would you recommend?

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Solved Problem 7.1

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

Table 7.2

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Solved Problem 7.1

The network diagram is shown in Figure 7.9. Keep the following points in mind while constructing a network diagram.

Start

Finish

Figure 7.9

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Solved Problem 7.1

With these activity times, the project will be completed in 19 days and incur a $700 penalty. Using the data in Table 7.2, 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 cost

Normal time – Crash time

CC – NC

NT – CT

= = $300

$1,300 – $1,000

4 days – 3 days

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Solved Problem 7.1

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

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Solved Problem 7.1

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.

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Solved Problem 7.1

PROJECT COST ANALYSIS
Stage	Crash Activity	Time 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

Table 7.3

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Solved Problem 7.2

An advertising project manager developed the network diagram in Figure 7.10 for a new advertising campaign. In addition, the manager gathered the time information for each activity, as shown in the accompanying table.

Calculate the expected time and variance for each activity.

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

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

Start

Finish

Figure 7.10

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Solved Problem 7.2

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

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Solved Problem 7.2

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

te =

a + 4m + b

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

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Solved Problem 7.2

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

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Solved Problem 7.2

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

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Solved Problem 7.2

4.0

0.0

4.0

8.0

12.0

4.0

8.0

16.0

20.0

6.5

9.0

15.5

4.5

15.5

20.0

3.5

5.5

9.0

5.5

6.5

14.5

15.5

5.5

0.0

5.5

Finish

Start

Figure 7.11

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Solved Problem 7.2

	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

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Solved Problem 7.2

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.

We first calculate the z-value:

z = = = 1.52

T – TE

σ2



23 – 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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Project Management Introduction

Most realistic projects are large and complex

Tens of thousands of steps and millions of dollars may be involved

Managing large-scale, complicated projects effectively is a difficult problem and the stakes are high

The first step in planning and scheduling a project is to develop the work breakdown structure

Time, cost, resource requirements, predecessors, and people required are identified for each activity

Then a schedule for the project can be developed

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Planning and Scheduling Project Costs: Budgeting Process

The overall approach in the budgeting process of a project is to determine how much is to be spent every week or month

This can be accomplished in four basic budgeting steps

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Four Steps of the Budgeting Process

Identify all costs associated with each of the activities then add these costs together to get one estimated cost or budget for each activity

In large projects, activities can be combined into larger work packages. A work package is simply a logical collection of activities.

Convert the budgeted cost per activity into a cost per time period by assuming that the cost of completing any activity is spent at a uniform rate over time

Using the ES and LS times, find out how much money should be spent during each week or month to finish the project by the date desired

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Monitoring and Controlling Project Costs

Costs are monitored and controlled to ensure the project is progressing on schedule and that cost overruns are kept to a minimum

The status of the entire project should be checked periodically

It can be used the answer questions about the schedule and costs so far

It can be used to make adjustments to a project or kill the project altogether

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