project management chapter 2 copyright ©2013 pearson education, inc. publishing as prentice hall02...
TRANSCRIPT
Project Management
Chapter 2Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 -0 1
What is a Project?
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 -0 2
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?
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 03
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 -0 4
Planning Projects
• Defining the Work Breakdown Structure
• Diagramming the network
• Developing the schedule• Analyzing cost-time
trade-offs
• Assessing risks
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 -0 5
Work Breakdown Structure
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 -06
Work Breakdown Structure
A statement of all the work that has to be completed.
Activity
The 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 fornursing and support staff
Organizing and Site Preparation Physical Facilities and Infrastructure Level 1
Level 0
Level 2
Relocation of St. John’s Hospital
02 -07
Diagramming the Network
• Program Evaluation and Review Technique (PERT)
• Critical Path Method (CPM)
• Activity-on-Node approach
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 08
Establishing Precedence Relationships
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 09
Precedence RelationshipsDetermining the sequence for undertaking activities.
FinishStart
A
B
C
D
E
F
G
H
I
J
K
A
—B
—C
AD
BE
BF
AG
CH
DI
AJ
E,G,HK
F,I,J
Immediate Predecessor
Example 2.1
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 11
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 2.1
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 12
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
StartA7
Application 2.1
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 13
Developing the Schedule
• Path–The sequence of activities between a
project’s 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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 15
• 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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 16
Latest finish time
Latest start time
Activity
Duration
Earliest start timeEarliest 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 project’s start and finish.
Example 2.2
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 17
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 18
K
6
C
10
G
35
J
4
H
40
B
9
D
10
E
24
I
15
FinishStart
A
12
F
10
0
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 19
Example 2.2
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 63 690 12
The Critical Path takes 69 weeks
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 20
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 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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 21
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 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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 22
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 DurationEarliest 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 23
Calculate the four times for each activity in order to determine the critical path and project duration.
Activity DurationEarliest 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
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
Application 2.2
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 24
Application 2.2
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 25
The critical path is A–C–D–E–G with a project duration of 24 days.
Start FinishA7
B2
C4
D4
E4
F3
G5
Analyzing Cost-Time Trade-Offs
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 26
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 27
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 – NCNT – CT
Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall.
Cost-Time Relationships
Linear cost assumption
8000 —
7000 —
6000 —
5000 —
4000 —
3000 —
0 —
Dire
ct c
ost (
dolla
rs)
| | | | | |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-Offs
Determining the Minimum Cost Schedule:1. Determine the project’s 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.Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 29
Example 2.3 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 ($)
A12 $12,000 11 $13,000 1 1,000
B9 50,000
764,000 2 7,000
C10 4,000
57,000 5 600
D10 16,000
820,000 2 2,000
E24 120,000 14 200,000 10 8,000
F10 10,000
616,000 4 1,500
G35 500,000 25 530,000 10 3,000
H40 1,200,000 35 1,260,000 5 12,000
I15 40,000 10 52,500 5 2,500
J4 10,000
113,000 3 1,000
K6 30,000
534,000 1 4,000
Totals$1,992,000 $2,209,500
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 30
Example 2.3Determine 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) = $80,000A–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 weeksCopyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 31
Example 2.3
STAGE 1Step 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 32
Example 2.3
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 33
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 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 34
Example 2.3
Finish
K6
I15
F10
C10
D8
H40
J 1
A12
B9
StartG
35
E24
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 35
STAGE 2
Example 2.3
STAGE 3Step 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 36
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.
Example 2.3
Finish
K5
I15
F10
C10
D8
H40
J1
A12
B9
StartG
35
E24
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 37
STAGE 3
Example 2.3
STAGE 4Step 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 38
Example 2.3
Finish
K5
I15
F10
C8
D8
H40
J1
A12
B7
StartG
35
E24
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 39
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-K
A-C-G-J-K
64 1,995.04.0
512.00.0
2,511.0
3 K 1 B-D-H-J-K
A-C-G-J-K
63 1,999.04.0
504.00.0
2,507.0
4 B, C 2 B-D-H-J-K
A-C-G-J-K
61 2,003.015.2
488.00.0
2,506.2
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 40
Indirect project costs = $250 per day Penalty cost = $100 per day past day 14.
Project Activity and Cost Data
ActivityNormal Time (days)
Normal Cost ($)
Crash Time (days)
Crash Cost ($)
Immediate Predecessor(s)
A 51,000
41,200
—
B 5800
32,000
—
C 2600
1900
A, B
D 31,500
22,000
B
E 5900
31,200
C, D
F 21,300
11,400
E
G 3900
3900
E
H 5500
3900
G
Application 2.3
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 41
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 42
13
19
F
2
15
21
13
13
G
3
16
16
5
5
D
3
8
8
8
8
E
5
13
13
5
6
C
2
7
8
0
0
B
5
5
5
0
1
A
5
5
6
Start
ES
LS
ID
DUR
EF
LF
16
16
H
5
21
21
Finish
Application 2.3
Project Activity and Cost DataActivity Crash Cost/Day Maximum Crash Time (days)
A 200 1B 600 2C 300 1D 500 1E 150 2F 100 1G 0 0H 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 2.3
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 43
Step 1: The critical path is , and the project duration is
B–D–E–G–H21 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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 44
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 45
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 46
The recommended completion date is day 17.
TrialCrash
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 2.3
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 47
Assessing Risks
• Risk-management Plans– Strategic Fit
– Service/Product Attributes
– Project Team Capability
– Operations
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 48
Assessing Risks
• Statistical Analysis
– Optimistic time (a)
– Most likely time (m)
– Pessimistic time (b)
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 49
Statistical Analysis
a m bMean
Time
Beta distribution
a m bMeanTime
3σ 3σ
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 by
te =a + 4m + b
6
• The variance of the beta distribution for each activity is
σ2 =b – a
6
2
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 51
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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 52
Example 2.4
a. The expected time for activity B is
The variance for activity B is
te = = = 9 weeks7 + 4(8) + 15
6
54
6
σ2 = = = 1.7815 – 7
6
28
6
2
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 53
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 (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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 54
Application 2.4
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 55
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 56
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 (σ)
A —5 7 8
B —6 8 12
C —3 4 5
D A11 17 25
E B8 10 12
F C, E3 4 5
G D4 8 9
H F5 7 9
I G, H8 11 17
J G4 4 4
For the Bluebird University sales training seminar activities, calculate the means and variances for each activity.
Application 2.4
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 57
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 A11 17 25
E B8 10 12
F C, E3 4 5
G D4 8 9
H F5 7 9
I G, H8 11 17
J G4 4 4
For the Bluebird University sales training seminar activities, calculate the means and variances for each activity.
6.830.25
8.331.00
4.000.11
17.335.44
10.000.44
4.000.11
7.500.69
7.000.44
11.502.25
4.000.00
Application 2.4
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 58
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 59
• 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. 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: σ2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89 Next, we calculate the z-value:
0.873.45
311.89
6972 z
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 60
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 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;σ = 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 61
Example 2.5
b. The sum of the expected activity times on path A–C–G–J–K is 67 weeks and that σ2 = 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55. The z-value is
1.273.94
515.55
6772 z
The probability is about 0.90 that the length of path A–C–G–J–K will be no greater than 72 weeks.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 62
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:
A–D–G–I,
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 63
Application 2.5
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 64
Start
A
B
C
D
FinishE
F
G J
H
I
The Network Diagram is:
The critical path is A-D-G-I
= = = 1.313.84
2.94
47 – 43.16
8.63
T = 47 days
TE = 43.16 daysAnd 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 0.9049.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 65
Application 2.5
Monitoring and Controlling Projects
• Monitoring Project Status– Open Issues and Risks– Schedule Status
• Monitoring Project Resources
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 66
Solved Problem 2.1
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 $200 per day. The data on direct costs and activity precedent relationships are given in Table 2.2.
a. Draw the project network diagram.b. What completion date would you recommend?
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 67
Solved Problem 2.1
ELECTRIC MOTOR PROJECT DATAActivity
Normal Time (days)
Normal Cost ($)
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 68
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 69
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 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 – $1,0004 days – 3 days
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 70
Solved Problem 2.1
Activity Crash Cost per Day ($)
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 71
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 72
Solved Problem 2.1The 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.
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 73
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 74
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 B
D 6 13 14 A
E 3 6 12 A, C
F 6 8 16 B
G 1 5 6 E, F
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 75
Solved Problem 2.2
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 76
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 =4.0
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 77
Solved Problem 2.2
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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 78
Solved Problem 2.2
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
3.5
5.5
5.5
9.0
9.0
F
9.0
5.5
6.5
14.5
15.5
B
5.5
0.0
0.0
5.5
5.5
Finish
Start
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 79
Solved Problem 2.2Start (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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 80
Solved Problem 2.2The 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
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 81
Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 02 - 82
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any
means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.
Printed in the United States of America.