7.1 Introduction
• A project is a collection of tasks that must be
completed in minimum time or at minimal cost.
• Tasks are called “activities.”
– Estimated completion time (and sometimes costs) are
associated with each activity.
– Activity completion time is related to the amount of
resources committed to it.
– The degree of activity details depends on the
application and the level of specificity of data.
1
• Determining precedence relations among the
activities is crucial to developing optimal schedules.
• Objective of Project Scheduling.
– Project Scheduling is used to plan and control a project
efficiently and economically.
– With Project Scheduling we can:
• characterize the activities
• monitor the project progress.
2
• Project Scheduling provides us with the following
information:
–
–
–
–
–
–
–
–
The minimal expected completion time for a project.
The critical activities.
The earliest and latest time to start and finish an activity.
The amount of slack time for each activity.
The most cost effective completion alternatives.
The activities on which extra resources should be spent.
Whether or not a project is on schedule or within budget.
A schedule that offers a constant level of resources
utilization.
– A schedule that completes a project in minimum time give
3
limited resources.
7.2 Identifying the Activities of a
Project
KLONE COMPUTERS, INC.
• KLONE Computers manufactures personal
computers.
• It is about to design, manufacture, and market the
Klonepalm 2000 palmbook computer.
4
• There are three major tasks to perform:
– Manufacture the new computer.
– Train staff and vendor representatives.
– Advertise the new computer.
• KLONE needs to develop a precedence relations
chart.
• The chart gives a concise set of tasks and their
immediate predecessors.
5
Activity Description
Activity
Description
Manufacturing
activities
Training activities
Advertising activities
A
B
C
D
E
Prototype model design
Purchase of materials
Manufacture of prototype model
Revision of design
Initial production run
F
G
H
Staff training
Staff input on prototype models
Sales training
I
J
Preproduction advertising campaign
Production advertising campaign
6
From the activity description chart, we
can determine immediate predecessors
for each activity.
7
Precedence Relationships Chart
AActivity
ctivity
AA
BB
CC
DD
EE
FF
G
G
HH
II
JJ
Im
Imm
med
ediate
iate
PPred
redecesso
ecessorr
NNone
one
AA
BB
G
G
DD
AA
CC,F
,F
DD
AA
DD,I,I
EEstim
stimated
ated
CCoom
mppletio
letionn TTim
imee
90
90
15
15
55
20
20
21
21
25
25
14
14
28
28
30
30
45
45
8
7.4
Building PERT/CPM Networks
• The PERT/CPM approach to project
scheduling
– A network representation of the project that reflects
activity precedence relationships.
– It is designed to perform a thorough analysis of
possible project schedules.
– Primary objectives of PERT/CPM
• To determine the minimal possible completion time for the
project.
• To determine a range of start and finish time for each
activity, so that the project can be completed in minimal 9
time.
• PERT is a scheduling method in which activity
completion times are treated as a random variable.
• CPM is a scheduling method which assumes the
completion time for an activity is deterministic,
dependent only on the amount of money spent to
complete it.
• Both methods require one to identify the activities
and the precedence relationships between
10
activities.
7.5 The PERT/CPM Approach for
Project Scheduling
KLONE COMPUTERS, INC. - Continued
• Management at KLONE would like to schedule
the activities so that the project is completed in
minimal time.
11
• Management wishes to know:
– The earliest completion date for the project.
– The earliest and latest start times for each activity which will
not alter this date.
– The earliest finish times for each activity which will not alter
this date.
– Activities with rigid schedule and activities that have slack in
their schedules.
12
An Activity On Node (AON) Network Representation
of the Klonepalm 2000 Computer Project
B
15
C
5
Immediate Estimated
Activity PredecessorE Completion Time
21
None
A
None
90
B
A
15
C
5
BB
D
20
D G
H
E
21
20 D
28
F
A
25
G
C,F
14
H
D
28
I
A
30
J
J
D,I 45
45
A
A
90
F
25
I
30
G
14
A
A
13
• Earliest start time / Earliest finish time
– Make a forward pass through the network as follows:
• Evaluate all the activities which have no immediate
predecessors.
– The earliest start for such an activity is zero ES = 0.
– The earliest finish is the activity duration EF = Activity duration.
• Evaluate the ES of all the nodes for which EF of all the
immediate predecessor has been determined.
– ES = Max EF of all its immediate predecessors.
– EF = ES + Activity duration.
• Repeat this process until all nodes have been evaluated
– EF of the finish node is the earliest finish time of the project.
14
Earliest Start / Earliest finish
Evaluate the ES of all the nodes for which EF has been determined
Evaluate all the activities which have no immediate predecessor
90,105
B
B
15
A
A
90
90,115
F
F
25
90,120
I
I
30
149,170
E
E
21
110,124
115,129
G
G
14
170
129,149
D
D
20
149,177
H
H
28
177
194
194
120,165
149,194
J
J
45
EARLIEST FINISH TIME
FINISH
0,90
105,110
C
C
5
15
• Latest start time / Latest finish time
– Make a backward pass through the network as follows:
• Evaluate all the activities that immediately precede the finish
node.
– The latest finish for such an activity is LF = minimal project
completion time.
– The latest start for such an activity is LS = LF - activity duration.
• Evaluate the LF of all the nodes for which LS of all the
immediate successors has been determined.
– LF = Min LS of all its immediate successors.
– LS = LF - Activity duration.
• Repeat this process backward until all nodes have been
evaluated.
16
Latest Start / Latest finish
90,105
95,110
0,90
0,90
C
C
5
90,115
90, 115
F
F
25
29,119
90,120
119,149
I
I
30
149,170
173,194
E
E
21
129,149
149,177
115,129
129,149
115,129 129,149
153,173
166,194
129,149 D 146,166 H
G
129,149 D
H
G
129,149 20
28
14
129,149
129,149
129,149
149,194
149,194
FINISH
5,95
A
A
90
B
B
15
105,110
110,115
194
J
J
45
17
• The Critical Path and Slack Times
– Activity start time and completion time may be delayed by
planned reasons as well as by unforeseen reasons.
– Some of these delays may affect the overall completion date.
– To learn about the effects of these delays, we calculate
the slack time, and form the critical path.
18
– Slack time is the amount of time an activity can be delayed
without delaying the project completion date, assuming no
other delays are taking place in the project.
Slack Time = LS - ES = LF - EF
19
Slack time in the Klonepalm 2000 Project
A c tiv ity
A
B
C
C
D
E
F
G
H
I
LS - E S
0 -0
95 - 90
110 - 105
119 - 119
173 - 149
90 - 90
115 - 115
166 - 149
119 - 90
149 - 149
S la c k
0
5
5
0
24
0
0
17
29
0
Critical activities
must be rigidly
scheduled
20
-
The Critical Path
• The critical path is a set of activities that have no slack,
connecting the START node with the FINISH node.
• The critical activities (activities with 0 slack) form
at least one critical path in the network.
• A critical path is the longest path in the network.
• The sum of the completion times for the activities
on the critical path is the minimal completion time
of the project.
21
90,105
95,110
0,90
0,90
C
C
5
90,115
90, 115
F
F
25
90,120
119,149
I
I
30
115,129
115,129
G
G
14
149,170
173,194
129,149
129,149
D
D
20
THE CRITICAL PATH
E
E
21
149,177
166,194
H
H
28
149,194
149,194
J
J
45
The forward and backward passes provided the
earliest and latest times to start and finish each22
activity.
FINISH
A
A
90
B
B
15
105,110
110,115
194
• Analysis of possible delays
– We observe two different types of delays:
• Single delays.
• Multiple delays.
– In some cases the overall project completion time will be
delayed.
23
– The conditions that specify each case are presented next.
– Single delays
• A delay of a certain amount in a critical activity, causes the entire
project to be delayed by the same amount.
• A delay of a certain amount in a noncritical activity will delay the
project by the amount the delay exceeds the slack time.
• When the delay is less than the slack, the entire project is not
delayed.
24
– Multiple delays
• Observe three cases of delays in two noncritical activities.
– Case 1: There is no path linking the two non-critical activities.
– Case 2: The non-critical activities are on the same path, separated
by a critical activity.
– Case 3: The two non-critical activities are on the same path, with no
critical activity separating them.
• In each one of the above cases, the amount specified will
cause no delay in the entire project.
25
Case 1: Activity E and I are each delayed 15 days.
ES=149
THE PROJECT COMPLETION TIME
DELAYED START=149+15=164
IS NOT DELAYED
LS=173
C
5
B
15
A
90
F
25
E
21
G
14
I
30
ES=90
DELAYED START=90+15 =105
LS =119
D
20
H
28
J
45
26
Case 2: Activity B is delayed 4 days;
Activity E is delayed 15 days.
ES=90
DELAYED START =94
LS =95
B
15
A
90
ES=149
DELAYED START=149+15 =164
LS =173
C
5
F
25
I
30
Activity B and E are separated
by the critical activities G and D.
E
21
G
14
D
20
H
28
J
45
THE PROJECT COMPLETION TIME IS NOT DELAYED
27
• Case 3: Activity B is delayed 4 days;
Activity C is delayed 4 days.
ES= 90
DELAYED START =94
DELAYED FINISH =
94+15=109
B
15
A
90
DELAYED START=
109 + 4 =113;
C LS =110
3 DAYS DELAY
IN THE ENTIRE
PROJECT
E
21
5
F
25
I
30
G
14
D
20
H
28
J
45
THE PROJECT COMPLETION TIME IS DELAYED 3 DAYS
28
7.7 The Probability Approach to
Project Scheduling
• Activity completion times are seldom known with
100% accuracy.
• PERT is a technique that treats activity completion
times as random variables.
• Completion time estimates are obtained by the
29
Three Time Estimate approach
• The three time estimate approach provides
completion time estimate for each activity.
• We use the notation:
a = an optimistic time to perform the activity.
m = the most likely time to perform the activity.
b = a pessimistic time to perform the activity.
• Approximations for the mean and the standard
deviation of activity completion time are based on
the Beta distribution.
a + 4m + b
 = the mean completion time =
6
b -a
 = the standard deviation =
6
30
– Assumption 1
• A critical path can be determined by using the mean
completion times for the activities.
• The project mean completion time is determined solely by the
completion time of the activities on the critical path.
To calculate the mean and standard
deviation
of the project completion time
– Assumption
2
• The time
complete
one activity
is independent of the time
we to
make
some
simplifying
to complete
any other activity.
assumptions.
– Assumption 3
• There are enough activities on the critical path so that the
distribution of the overall project completion time can be
approximated by the normal distribution.
31
The three assumptions imply that the overall
project completion time is normally distributed,
with
Mean = Sum of mean completion times along
the critical path.
Variance = Sum of completion time variances
along the critical path.
32
KLONE COMPUTERS - revised
• Three time estimates were given for each activity
A c tiv ity
O p tim is tic M o s t L ik e ly P e s s im is tic
A
76
86
120
B
12
4
15
5
18
6
15
18
18
21
33
24
16
10
26
13
30
22
I
24
22
18
27
32
50
J
38
43
60
C
D
E
F
G
H
33
• Management at KLONE is interested in
– The probability that the project will be completed within
194 days.
– An interval estimate of the project completion time.
– The probability that the project will be completed within
180 days.
– The probability that the project will take longer than
210 days.
– An upper limit for the number of days within which it can
34
virtually be sure the project will be completed.
• Calculation of the means and the variances
A = [76+4(86)+120]/6 = 90
A = (120 - 76)/6 = 7.33 A2 = (7.33)2 = 53.78
– For the rest of the activities we have
Activity

A
B
C
D
E
F
G
H
I
J
90
15
5
20
21
25
14
28
30
45

7.33
1.00
0.33
3.00
1.00
2.33
2.00
1.33
4.67
3.67
2
53.78
1.00
0.11
9.00
1.00
5.44
4.00
1.78
21.78
13.44
35
• The mean times are the same as in the CPM
problem, previously solved for KLONE.
• Thus, the critical path is A - F- G - D - J.
– Expected completion time = A + F + G + D + J = 194
– The project variance =A2 +F2 +G2 +D2 +J2 = 85.66
– The standard deviation =
2 = 9.255
Under the assumptions stated above the completion time of
the project is normally distributed with = 194 days and
= 9.255 days.
36
• Now the quantities of concern for management can
be calculated.
– The probability of completion in 194 days =
194 -194
P(X  194) = P(Z 
)  P(Z  0)  0.5
9.255
• An interval in which we are reasonably sure the
completion date lies is
  z 0.025 
– The interval is = 194  1.96(9.255)  {175, 213} days.
• There is 95% confidence that the real completion time lies within
175 days and 213 days.
37
The probability of completion in 180 days =
P(X  180) = P(Z -1.51) = 0.5 - 0.4345 = 0.0655
0.0655
0.99
0.0418
.4345 .4582
0.01
180
-1.51
194
0
210
1.73
x X
2.33 Z
– The probability that the completion time is longer
than 210 days = P(X  210) = P(Z  1.73) = 0.5 - 0.458 = 0.0418
– Assume a 99% certain completion date is acceptable
P(Z  2.33) = 0.99; x=+ z =194 + 2.33(9.255) = 215.56 days.
38
7.8 Computer Solution of
PERT/CPM
• A linear programming approach
– Variables
• Xi = The start time of the activities for i=A, B, C, …,J
• X(FIN) = Finish time of the project
– Constraints
• Each immediate predecessor relationship has one constraint
• The constraints state that the start time of an activity must not
occur before the finish time of the immediate predecessor.
39
– The Model
C
5
Minimize X(FIN)
ST
X(FIN)  X + 21
E
X(FIN) XH + 28
X(FIN) XJ + 45
XE  XD + 20
XH  XD + 20
XJ  XD + 20
XJ  XI + 30
All X’s are nonnegative
F
25
XD XG +
XG XC +
XG XF +
XI  XD +
XF  XA +
XC XB +
XD XG +
XB  XA +
G
14
5
25
90
90
15
14
90
40
– An Excel solution provides:
• A schedule of (not necessarily the earliest)
start time that will complete the project in minimal time.
• The earliest starts, returned when using the following objective
function
Minimize XA + XB + XC +…+ XJ + X(FIN)
• The latest starts that can be obtained (after solving for the
project earliest finish) as follows
– Add the constraint X(FIN) = 194
– Change the objective function to
Maximize XA + XB + XC +…+ XJ + X(FIN).
41
• A Network Approach
– WINQSB provides
•
•
•
•
Activity schedule chart (ES, EF, LS, LF).
Mean and standard deviation of the critical path.
Optional: Earliest and latest time Gantt chart.
Optional: The probability of completing the project within
a specified time.
– No programming is needed.
– Input data consists of
• Immediate predecessors for each activity.
• Completion time (for a, m, and b).
42
43
Critical Activities
WINQSB Scheduling Solution
44
7.9 Cost Analysis Using the
Expected Value Approach
• Spending extra money, in general should
decrease project duration.
• Is this operation cost effective?
• The expected value criterion is used to answer
this question.
45
KLONE COMPUTERS Cost Analysis Using Probabilities
• Analysis indicated:
– Completion time within 180 days yields an additional profit
of $1 million.
– Completion time between 180 days and 200 days, yields
an additional profit of $400,000.
46
KLONE COMPUTERS Cost Analysis Using Probabilities
• Completion time reduction can be achieved by
additional training.
Which option should be
• Two possible activities
are
considered
for
pursued?
training.
– Sales personnel training:
• Cost $200,000;
• New time estimates are a = 19, m= 21, and b = 23 days.
– Technical staff training:
• Cost $250,000;
47
•
Evaluation of spending on sales personnel training.
– This activity (H) is not critical.
– Under the assumption that the project completion time is
determined solely by critical activities, this option
should not be considered further.
• Evaluation of spending on technical staff training.
– This activity (F) is critical.
– This option is further studied as follows:
• Calculate expected profit when not spending $250,000.
• Calculate expected profit when spending $250,000.
• Select the decision with a higher expected profit.
48
– Case 1: Do not spend $250,000 on training.
This case represents the current situation.
Expected gross additional profit =
P(completion time is less than 180)($1 million) +
P(180<Completion time<200)($400,000) +
P(Completion time> 200)(0) =
P(X<180)($1 million) + P(180<X<200)($400,000) + P(X>200)(0)
180-194
[P(Z< 9.255 )=P(Z< -1.51) = .0655; P(-1.51<Z<.65) = .6767]
Expected gross additional profit = .06555(1M)+.6767(400K)=
= $366,180.
49
– Case 2: Spend $250,000 on training.
• The new timeExpected
estimates change
expected time
and=standard
Gross the
Additional
Profit
deviation ofP(
a critical
activity. time<180)(1M)+
completion
P(180<completion time<200)(400K)+
P(completion
time>200)(0) =
= (12 + 4 (14)
+ 16)/6 = 14
= (16 -12)/60.1611(1M)+0.7277(400K)
=0.67
= $452,180
• The new estimates are
2= 0(.67)2 =0.44
• The project has a new critical path (A-B-C-G-D-J), with a mean time =
189 days, and a standard deviation of = 9.0185 days.
P(Z< 180-189 ) = P(Z< -.99) = 0.1611;
9.0185
P( - 0.99<Z< 1.22) = 0.7277
50
– The expected net additional profit =
452,180-250,000 = $202,180 < $336,180
Expected additional net profit when
spending $250,000 on training
Expected profit without
spending $250,000 on training
Conclusion: Management should not spend money on
additional training of technical personnel
51
7.10 The Critical Path Method
(CPM)
• The critical path method (CPM) is a deterministic
approach to project planning.
• Completion time depends only on the amount of
money allocated to the activity.
• Reducing an activity’s completion time is called
“crashing.”
52
• There are two crucial completion times to
consider for each activity.
– Normal completion time (NT).
– Crash completion time (CT).
– NT is achieved when a usual or normal cost (NC)
is spent to complete the activity.
– CT is achieved when a maximum crash cost (CC)
is spent to complete the activity.
53
The Linearity Assumption
[Normal Time - Crash Time] [Crash Cost - Normal Cost]
=
[Normal Cost]
[Normal Time]
54
Normal
NC = $2000
NT = 20 days
A demonstration
Total Cost = $2600
of the
Job time = 18 days
Linearity assumption
Time
20 …and save on
18 completion time
…and save more on
16
completion time
Add to the
14
Add more to the
normal cost...
12
normal cost...
Save 25% of the
10
max. time reduction
Add 25% of the extra
cost to the normal cost
8
6
4
2
Marginal Cost =
=
Crashing
CC = $4400
CT = 12 days
Additional Cost to get Max. Time Reduction
Maximum Time reduction
(4400 - 2000)/(20 - 12) = $300 per day
5 10 15 20 25 30 35 40 45
Cost ($100)
55
• Meetings a Deadline at Minimum Cost
– Let D be the deadline date to complete a project.
– If D cannot be met using normal times, additional resources
must be spent on crashing activities.
– The objective is to meet the deadline D at minimal
additional cost.
– Tom Larkin’s political campaign problem illustrates the
56
concept.
TOM LARKIN’S POLITICAL CAMPAIGN
• Tom Larkin has to plan 26 weeks of his mayoral
campaign activities.
• The campaign consists of the following activities
Activity
Activity
A.
A. Hire
Hire cam
camppain
ain staff
staff
Im
No
rm
al
chSSch
ed
chuued
le
Red
ch ed
Imm
med
ediate
iate No
Norm
rm
Noal
al
rmSSch
al
ed
ed
leuule
le
Reduuced
ced SSch
eduule
le
PPred
red ecesso
Co
TTim
Co
ecessorr TTim
imee
Cost
st
imee
Cost
st
None
44
2.0K
22
5.0K
None
2.0K
5.0K
B.
rep are
B. PPrep
are ppoositio
sitionn ppap
aper
er
C.
C. Recru
Recruitit vo
volu
lunnteers
teers
None
None
AA
66
44
3.0
3.0
4.5
4.5
33
22
99
10
10
D.
D. Raise
Raise fu
funnddss
EE.. FFile
ile can
canddid
idacy
acy ppap
apers
ers
AA,B
,B
DD
66
22
2.5
2.5
0.5
0.5
44
11
10
10
11
FF.. PPrep
rep are
are cam
camppaig
aignn m
material
aterial
G
G.. LLoocate/staff
cate/staff hhead
eadqquuarters
arters
EE
EE
13
13
11
13.0
13.0
1.5
1.5
88
11
25
25
1.5
1.5
H.
H. Ru
Runn pperso
ersonnal
al cam
camppaig
aignn
I.I. Ru
Runn m
med
edia
ia cam
camppaig
aignn
C,G
C,G
FF
20
20
99
6.0
6.0
7.0
7.0
10
10
55
23.5
23.5
16
16 57
NETWORK PRESENTATION
A
C
H
G
B
D
E
FINISH
F
I
Project completion (normal) time = 36 weeks
58
To meet the deadline date
of 26 weeks some activities
must be crashed.
Mayoral Campaign Crash Schedule
AA cc ttiivv iittyy
AA
BB
CC
DD
EE
FF
GG
HH
II
NN TT
44
66
44
66
22
11 33
11
22 00
99
NN CC (($$ ))
22 00 00 00
33 00 00 00
44 55 00 00
22 55 00 00
55 00 00
11 33 00 00 00
11 55 00 00
66 00 00 00
77 00 00 00
CC TT
22
33
22
44
11
88
11
11 00
55
CC CC
55 00 00 00
99 00 00 00
11 00 00 00 00
11 00 00 00 00
11 00 00 00
22 55 00 00 00
11 55 00 00
22 33 55 00 00
11 66 00 00 00
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SOLUTION
• Linear Programming Approach
– Variables
Xj = start time for activity j.
Yj = the amount of crash in activity j.
– Objective Function
Minimize the total additional funds spent on crashing activities.
– Constraints
• The project must be completed by the deadline date D.
• No activity can be reduced more than its Max. time reductioN.
• Start time of an activity takes place not before the Finish time of an
immediate predecessor.
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Minimize total crashing costs
Min1500YA  2000YB  2750YC  3750YD  500YE  2400YF  17500YH  2250YJ
ST
Meet the deadline
X ( FIN )  X I  9  YI
X ( FIN )  26
X ( FIN )  X H  20  YH
X I  X F  13  YF
YA  2
XH  XG 1
YB  3
Maximum
time-reduction
constraints
X H  X C  4  YC
YC  2
X G  X E  2  YE
YD  2
X F  X E  2  YE
YE  1
X E  X D  6  YD
YF  5
A
YH  10
C
B
D
E
X D  X B  6  YB
H
G
X D  X A  4  YA
FINISH
F
Activity can
start only after
all the
predecessors
are completed.
X C  X A  4  YA
I
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• Other cases of project crashing
– Operating optimally within a given budget.
• When a budget is given, minimizing crashing costs is a constraint,
not an objective.
• In this case the objective is to minimize the completion time.
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TOM LARKIN - Continued
• The budget is $75,000.
The objective function becomes a constraint
Minimize X(FIN)
1500 YA+ 2000 YB + 2750 YC + 3750 YD + 500 YE + 2400 YF
+1750 YH + 2250 YJ
This constraint becomes the objective function
X(FIN)  26
1500 YA+ 2000 YB + 2750 YC + 3750 YD + 500 YE + 2400 YF
+1750 YH + 2250 YJ  75,000 - 40,000 = 35,000
The rest of other crashing model constraints
remain the same.
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