Graduating from college and applying for grad school
is a "piece of cake" using CPM and PERT.
By Duber Gomez-Fonseca
Fall 2012
In Partial Fulfillment of
CS (or Math or Stat) 4395-Senior Project
Department of Computer and Mathematical Sciences
University of Houston-Downtown
Faculty Advisor:
Dr. Timothy Redl:
_________________________________
Committee Member:
Dr. Yunwei Cui:
_________________________________
Committee Member:
Ms. Sangeeta Gad:
_________________________________
Department Chairman:
Dr. Shishen Xie
_________________________________
I
Table of contents
Abstract
Acknowledgements
Section 1 Introduction: History and background information
1.1 Overview of Project management
1.2 CPM: Critical path method
1.2.1 Description of CPM
1.2.2 Examples
1.2.3 Computations
1.2.4 Construction of the time schedule
1.3 PERT: Program Evaluation and Review Technique
1.3.1 Description of PERT
Section 2: My Projects
2.1 Making a cake
2.1.1 Description of the project
2.1.2 Construction of the network
2.1.3 Solution using CPM
2.1.4 Solution using PERT
2.1.5 Analysis of the solution methods
2.1.6 Addressing possible project variations: “What if?”
2.2 How to graduate from college
2.2.1 Description of the project
2.2.2 Construction of the network
2.2.3 Solution using CPM
2.2.4 Solution using PERT
2.2.5 Analysis of the solution methods
2.2.6 Addressing possible project variations: “What if?”
2.3 Applying for grad school
2.3.1 Description of the project
2.3.2 Construction of the network
2.3.3 Solution using CPM
2.3.4 Solution using PERT
2.3.5 Analysis of the solution methods
2.3.6 Addressing possible project variations: “What if?”
Section 3: Summary, conclusions, and future work
3.1 Summary and conclusions
3.2 Future work
References.
ll
lll
1
1
2
2
6
7
10
13
13
15
15
15
16
18
21
22
23
24
24
26
26
29
30
31
32
32
33
34
37
38
39
40
40
41
42
II
Table of figures
Figure 1.1.How not to be late to work?
Figure 1.2 Representation of an arc
Figure 1.3 Representation of an event (node)
Figure 1.4 Representation of node-arc activity
Figure 1.5 An incorrect network
Figure 1.6 the correct use of a dummy variable
Figure 1.2.1 Project network for example 1
Figure 1.2.2Critical path network
Figure 1.2.3 Phases for project planning with CPM-PERT
Figure 1.2.4 Critical path
Figure 1.2.5 Preliminary schedule for the project of Example 1
Figure 1.2.6 Determination of the Total and Free Floats
Figure 1.2.7 FF and TF computations example one
Figure 1.2.8 the mean and the variance computations for example one
Figure 2.1.1list of activities for Betty’s cake
Figure 2.2.2 Betty’s Cake Project Network
Figure 2.3.3 Preliminary schedule for the project of section 2.2.
Figure 2.1.4 FF & TF computations
Figure 2.1.5 PERT computations for problem on
Figure 2.1.6 the average estimation time and variance for problem on
Figure 2.1.7 the probabilistic times computations problem one
Figure 2.2.1 list of activities for an average person go to school
Figure 2.2.2 Critical Path Network, Project Two
Figure 2.2.3 Preliminary Schedule for the Project of the problem 2.2
Figure 2.2.4 FF and TF Computations for problem
Figure 2.2.5 the mean and the variance computations fro problem two
Figure 2.2.6 the probabilities of the project network for problem two
Figure 2.3.1 List of Activities for problem three
Figure 2.3.2 Project Network for problem three
Figure 2.3.3 preliminary schedule for problem three
Fugue 2.3.4 FF and TF calculations for problem three
Figure 2.3.5 mean computations for problem three
Figure 2.3.6 the mean and the variance computations for problem three
Figure 2.3.7 the probabilistic computations for problem three
3
4
4
5
5
6
6
7
7
10
11
12
13
15
16
17
20
21
22
22
22
25
26
28
29
30
30
33
34
36
37
38
38
38
III
Abstract
The focus of this project is to demonstrate that simple project management and complex
project management have many management elements in common when the concern is time
management. Being able to schedule the non-critical activities and manage slack time gives the
manager complete control over any project. Three projects are considered and two statistical tools,
PERT and CPM, applied to these projects demonstrate that an average person can use these tools
in their normal life just as well as large companies can use them in more complex projects. PERT
stands for Project Evaluation and Review Technique. It is a decision-making tool designed to save
time in achieving end-objectives, and is of particular interest to those engaged in research and
development programs for which time is a critical factor. It is designed to analyze and represent
the tasks in a given project. It is normally used in conjunction with the Critical Path Method
(CPM), which is an algorithm for scheduling a set of project activities. It is an important tool for
effective project management. Being able to schedule the noncritical activities and manage the
slack time, gives the CPM and PERT user complete control over any project. In addition, CPM and
PERT give the user the capability, through the ‘red flags’ process, to understand and predict the
known time based on the given time completion.
IV
ACKNOWLEDGMENTS
Primarily I wish to acknowledge my Advisor Dr. Timothy Redl, University of Houston
Downtown for the importance of his help when it was needed. He helped me in shaping my ideas
and thoughts. He influenced my thinking the first time I took a class with him as my Professor
(Operation research) and specifically when he gave me the opportunity to do my own research
related with operation research. The topic I chose was the critical Path Method CPM since I
already had some knowledge acquire in my country (Colombia) when studying Economics as my
bachelor I had to take “Programacion Matematica” Mathematical Programing. Back then I could
not understand why an economics student had to take such class; of course, my interest was not
much for that class. In fact, it was the strongest motivation I had in order to pick CPM as the
subject of my Senior Project.
Also I which to acknowledge my dear friend Robert Lewis who helped me in the
development of my project, gave me his unconditional support, and helped me with his expertise.
I want also to acknowledge to Dr. Richard A. Alo, Jackson State University for the golden
opportunity he gave me when he was part of the University of Houston Downtown.
I want to acknowledge my committee member Dr. Yunwei Cui, University of Houston
Downtown. Ms. Sangeeta Gad, University of Houston Downtown and all the people that in some
way gave me ideas how to achieve my goal.
V
1. Introduction
The world has been rapidly expanding in different areas such as technology, architecture,
finance, and so on. At the same time, the rapid growth of the cities due to family displacements
from rural to urban areas has led authorities to change the way cities are structured. Today it is
common to see big construction projects such as buildings, bridges, tunnels, and a big variety of
smaller projects that allow people to have a better way of life. In the United States for example,
there are some major cities where the technology in those construction areas are as simple as they
are breath taking. Chicago is the third largest city in U.S and is home of 1,209 high-rises in which
73 stand taller than 550 feet (168m). But there is yet a taller building in Chicago, the 108 story
Willis Tower which rises 1,451 feet (442m). The Willis Tower was the tallest building for a while,
until the International Commerce Center building in Hong Kong was built. This building is 118
floors tall and now the Burj Khalifa building in Dubai, with more than 163 floors (which was
finished in 2010) are the second and first tallest buildings in the world. But there must be a
question, how much time does the construction companies spend in planning, drawing, building,
and finishing those mega constructions? There must be a tool that allows engineers, architects, and
executives to plan this type of fabrication.
1.1 Overview of Project management
In order to manage, control, and direct a project, it is necessary to know several things. For
example, how much time does the construction company have in order to start, build, and finish a
project? How much money can the construction company earn as well as lose if they finish on
time, earlier, or later than the time estimated at the beginning of the project? How much labor does
the company need in order to accomplish their goal? Also, the company must know about the
machinery, tools, and material necessary to start and finish on time. In starting a project such as the
Willis Tower, the architect must know that there are some events that have to start before others.
At the same time, there are some events during the process of constructing a house or a building
that can be delayed without affecting the normal development of the whole project.
It is known that some hospitals, construction companies, IT, banks, and other type of
companies offering services need to improve and innovate the way they offer their services
in order to be more competitive in the market. By improving their services, these
1
companies have to come up with new strategies. In the case of hospitals, they need to track
patients and their treatment. Construction companies have to improve time management in
order to increase their profit. In the case of IT, they need to know how to manage different
activities; otherwise they may end up losing a contract. The secret of the companies in the
twenty-first century is to succeed based on the economic crisis that is plaguing global
economies.
1.2 CPM: Critical path method
CPM stands for Critical Path Method. CPM is one of the more effective ways companies
can manage, formulate and build a project. Even though there is a broad spectrum of Project
Management software in the market, CPM has become a very effective tool in project planning
and decision making. But what is CPM? Who created it?
1.2.1 Description of CPM
Introduction to CPM
The Critical Path Method (CPM) is a project modeling technique developed in the late
1950s. There were two people involved in the development of the algorithm, Morgan R. Walker of
DuPont and James E. Kelly, Jr. of Remington Rand. The Critical Path name was attributed by
Kelly based on the developers of the Program Evaluation and Review Technique which was
developed about the same time by Booz Allen Hamilton and the US Navy. CPM is commonly used
in all kinds of projects such as construction, aerospace, defense, software development, research
projects, product development, plant maintenance, among others. CPM is a widely used technique
for analyzing and managing the sequence in which a project has to be managed. The essence of
CPM is based on calculations in order to determine how long it will take to complete important
steps during the process and also to be able to analyze how those steps are related to each other.
Also this method calculates the minimum completion time for a project as well as the start and
finish of each one of the project activities. CPM is a mathematical technique that gives the
manager the capability to effectively plan, schedule and implement each one of the activities
within the project.
2
Benefits of Using CPM
There are several reasons why companies’ managers want to use CPM, due to the benefits
they can obtain CPM can:
•
Provide a graphical view of the project
•
Predict the time require to complete the project
•
Show which activities are critical to maintain the schedule and which are not
•
Calculate the schedule starts and finish dates for each task pertaining to the project
completion.
•
Determine the tasks that are critical to the project and must be completed exactly as
scheduled.
•
Calculate the slack time available in non-critical tasks, as well as how long they can be
delayed before they affect the project finish date.
All of the above reasons give a manager a reliability of doing a good job of being able to
control all of the activities. There are several remarkable benefits to using CPM
•
Planning
•
Analyzing and scheduling
•
Controlling project task
Now that we know what CPM is and the benefits that it provides to CPM users, let’s to
construct a CPM model. In order to construct a CPM model it is necessary to know the entire list
of tasks that a project requires. Once the list of tasks is completed, it is necessary to list each one of
them individually (e.g. A, B, C, D, E, F). One more thing that is necessary to do in order to get the
CPM project started, is to know the time each individual activity is going to take. Here is a small
example of how to start a CPM project. In this case each activity is labeled. There are three tasks
identified, and finally the time for each task is given
Activities
Description Activities
Time/minutes
A
Wake up
15
B
Take a shower
10
C
Getting dressed
15
Figure 1.1. How not to be late to work?
3
Not all the projects are as small as the previous example. But it is very important to specify
the individual activities. For example, a listing can be made of all possible events of all the
activities in the project. Once this first step is done, it is necessary to determine the sequence of
each of the activities. For example, some activities are dependent on the completion of others. A
listing of the immediate predecessors of each activity is useful in order to construct a CPM
network diagram. This gives the user the ability to identify which activity is required to be done
first before starting or continuing to the next task. Once the listing of tasks is done, it is necessary
to move to the second part of the CPM project, which is the construction of the network.
CPM Network Diagram
A network diagram represents a project that shows the precedence relationships of the
activities of the project along with activity times. The activity is said to consume time and
resources. They are represented by arrows (arcs).
Figure 1.2 Representation of an activity (arc)
The precedence relationships of the activities are indicated through events. In this case
those events are represented by nodes. Events are just points in time, represented by circles. Nodes
do not consume any resources. They represent the beginning of some activities and the ending of
some other activities.
Figure 1.3 Representation of an event (node)
Now it is time to see how the activities and events are related to each other. Consider the
following diagram in which an activity (i, j) with duration Di,j is represented by an arrow between
two events or nodes. In this example we have events і (tail) and j (head). Most of the time the
activities are named by letters (such as A, B, C, etc.) while the events are denoted by numbers
(such as 1, 2, 3, etc.)
4
i
j
Di, j
1
D1, 2
2
Figure 1.4 Representation of node-arc activity
Now it is time to introduce a Dummy activity. Some times in a network diagram it is
necessary to use a dummy activity. The dummy activity does not consume time or resources. There
are two potential problems that can occur when building a network diagram. Since an arrow serves
to represent both the sequence and duration of the event, there are cases when showing sequences
with a single arrow is inadequate. Here is an example of the need for a dummy activity.
Activity
Predecessor
A
None
B
None
C
A
D
A, B
Activity A
Activity B
Activity C
Activity B does
not precede
Activity C!
Activity D
Figure 1.5 an incorrect network
Based on the previous example, is it possible to see activity B does not precede activity C. So it is
necessary to introduce a dummy variable. The dummy variable does not consume time or
resources. It is just an indicator that two activities were completed.
5
Activity C
Activity A
Dummy variable
Activity B
Activity D
Figure 1.6 the correct use of a dummy variable
Section 1.2.2 First CPM Example
In this section we show some examples of how a CPM project looks.
Example 1
A publisher has a contract with an author to publish a textbook. The activities associated
with the production on the textbook are given below. The author is required to submit to the
publisher a hard copy and a computer file of the manuscript. Developed the associated network for
the project.
Activity
Predecessors
(s)
A: Manuscript proofreading by editor
B: Sample pages separation
C: Book cover design
D: Artwork preparation
E: Author’s approval of edited manuscript and sample pages A, B
F: Book formatting
E
G: Author’s review of artwork
F
H: Author’s review of work
D
I: Production of printing plates
G, H
J: Book production and binding
C, I
Figure 1.2.1 Project network for example 1
Duration
(weeks)
3
2
4
3
2
4
2
1
2
4
6
F-2
E-2
6
4
3
G-2
2
7
H-1
A-3
B-2
D-3
I-2
5
J-4
C-4
1
9
8
Figure 1.2.2Critical path network
Section 1.2.3 Critical path (CPM) Computations
Now that the network is completed, it is a good moment to introduce the CPM
computations. The end result in CPM is the construction of the time schedule for the project, as
shown below.
Network
Project activities
Network
Calculations
Time Schedule
Figure 1.2.3 Phases for project planning with CPM-PERT
The two techniques, CPM and PERT, which were developed independently, differs in that
CPM assumes deterministic activities durations and PERT assumes probabilistic durations. The
order in which this presentation is going to be introduced is CPM first and then proceeds with the
detail of PERT.
To achieve this objective properly, we carry out special computations that produce the following
information:
1. Total duration needed to complete the project.
2. Classification of the activities of the project as a critical and noncritical.
7
An activity is said to be critical if there is not “leeway” in determining its start and finish time.
A noncritical activity allows some scheduling slack, so that the start time of the activity can be
advanced or delayed within limits without affecting the completion date if the entire project.
To carry out the necessary computations, we define an event as a point in the time at which
activities are terminated and others are started. In terms of the network, an event corresponds to a
node. Define
□ j = Earliest occurrence time of event j
∆j = latest occurrence time on event j
Dij = Duration of activity (i , j)
The definitions of the earliest and latest start occurrences of event j are specified relative to the
stat and the complexion dares if the entire project.
The critical path calculations involve two passes: the forward pass determines the earliest
occurrence time of the events, and the backward pass calculate their latest occurrence times.
Forward pass (Earliest Occurrence times, ▭). The computations start at node 1 and advance to
node n.
Initial step. Set□1 = 0 to indicate that the project starts at a time 0.
General step j. given that nodes p, q …, and v are linked directly to the node j by incoming
activities
(p, j), (q, j),… and (v, j ) and that the earliest occurrence times of events p, q, …, and v
have already been computed, then the earliest occurrence time of event j is computed as
□j = max {□p + Dpj, □q + Dqj,… □v + Dvj}
The forward pass is complete when ▭n at a node n has been computed. By definition ▭j
represents the longest path (duration) to node j.
Backward Pass (Latest Occurrence Times, ∆). Following the completion of the forward pass, the
backward pass computations start at node n and end at node 1.
Initial step. Set ∆n = □n to indicate that the earliest and latest occurrences of the last node of the
project are the same.
General step j. Given the nodes p, q, …, and v are linked directly to node j by outgoing activities
(j, p), (j, q), …, and (j, v) and that the latest occurrence time of node j is computed as
∆j = min {∆p - Djp, ∆q - Djq,…, ∆v - Djv}
8
The backward pass is completed when ∆₁ at a node 1 is computed. At this point, ∆₁ = □₁ (=
0).
Based on those preceding calculations, an activity (i, j) will be critical if it satisfies three
conditions.
1. ∆i = □i
2. ∆j = □j
3. ∆j - ∆i = □j - □i = Dij
There three conditions state that the earliest and that the latest occurrence times of end nodes i
and j are equal and the duration Dij fits “tightly” in the specified time span. An activity that does
not satisfy all three conditions is thus noncritical.
By definition, the critical activities of a network must constitute an uninterrupted path that
spans the entire network from start to finish.
At this point, all this information can be confusing. The best way to see how CPM
calculations work in a real life example is by going through a complete exercise. Let’s determine
the critical path for the project in Figure 1.2.1. All the durations are in weeks.
Forward pass
Node 1. Set □₁ = 0
Node 2. □₂ = □₁ + D₁₂ = 0 + 2 = 2
Node 3. □₃ = Max {□₁ + D₁₃, □₂ + D₂₃} = Max {0 + 2, 3 + 0} =3
Node 4. □₄ = □₄ D₃₄ = 3 + 2 = 5
Node 5. □₅ = □₅ + D₁₅ = 0 + 3 = 3
Node 6. □₆ = □₆ + D₄₆ = 5 + 2 = 7
Node7. □₇ = □₇ + Max {□₆ + D₆₇, □₅ + D₅₇} = max {7 + 2, 3 + 1} = 9
Node 8. □₈ = □₈ + max {□₇ + D₇₈, □₁ + D₁₈} Max {9 + 2, 0 + 4} = 11
Node 9. □₉ = □₉ + D₈₉ = 11 + 4 = 15
Backward pass
Node 9. Set ∆₉ = 15
Node 8. ∆₈ = ∆₉ - D₈₉ = 15 – 4 = 11
Node 7. ∆₇ = ∆₈ - D₇₈ = 11 – 2 = 9
9
Node 6. ∆₆ = ∆₇ - D₆₇ = 9 – 2 – 7
Node 5. ∆₅ = ∆₇ - D₅₇ = 9 – 1 = 8
Node 4. ∆₄ = ∆₆ - D₄₆ = 7 – 2 = 5
Node 3. ∆₃ = ∆₄- D₃₄ = 5 – 2 = 3
Node 2. ∆₂ = ∆₃ - D₂₃ = 3 – 0 = 3
Node 1. ∆₁ = min {∆₈ - D₁₈, ∆₅ - D₁₅, ∆₃ - D₁₃, ∆₂ - D₁₂} {9 - 4, 8 - 3, 3 - 2, 3 - 3}
= {5, 5, 1, 0} = 0
Correct computations always will end with ∆₁ = 0
The forward and the backward pass computations can be made directly on the network as
show in the figure 1.2.3. Applying the rules for determining the critical activities, the critical path
is 1→ 2→ 4 → 5 →6, which as should be expected, spans the network form start (mode 1) to
finish (mode 6). The sum of the durations of the critical activities [(1,2, (2,3), (3,4), (4,6), (6,7),
(7,8), and (8, 9)] equal the duration of the project (15 weeks).
Figure 1.2.4 CPM calculations
Section 1.2.4 Construction of the Time Schedule
Now that the activities, project, and network are done, it is time for the construction of the
time schedule. This section shows how the information obtained for the calculation in section 1.2.3
can be used to develop the time schedule. It is important to keep in mind that for an activity (i, j),
□I represents the earliest start time, and ∆j represents the latest completion time. This means that
the interval (□i, ∆j) delineates the maximum span during which activity (i, j) may be schedule
without delaying the entire project.
Construction of the preliminary schedule. The method for constructing a preliminary
schedule is going to be illustrated in the next examples; so it is important to follow two steps.
1. The critical activities (shown by solid lines) must be staked on the right of another to
ensure that the project is complete within the specified 15 weeks.
10
2. The noncritical activities (shown by dashed lines) have time spans that are larger than their
respective durations, thus allowing slack (or leeway) in scheduling within their allotted
time intervals.
How should we schedule the noncritical activities within their respective span? Normally, it
is preferable to start each noncritical activity as early as possible. In this manner, slack time it
will remain opportunely available at the end of the allotted span where they can be used to
absorb unexpected delays in the execution of the activity. It may be necessary, however, to
delay the start of a noncritical activity past its earliest start time. For example, in the figure
1.2.5 suppose that each of the noncritical activities C and D requires doing two different
activities that requires the use of the same computer, and that computer is the only one.
Scheduling both C and D as early as possible requires two computers between time 0 and 3. It
is possible to overlap by starting D at time 0 and pushing the start time of D somewhere
between times 3 and 4.
If all the noncritical activities can be schedule as early as possible, the resulting schedule
automatically is feasible. Otherwise, some precedence relationship may be violated if
noncritical activities are delayed past their earliest start time. Take for example activities D and
H in the figure 1.2.5. In the project network (figure 1.2.4) though D must be complete before
H, the spans of D and H in the figure 1.2.5 allow us to schedule D
Figure 1.2.5 Preliminary schedule for the project of Example 1
11
Critical path calculations including floats
Floats are the slack times available within the allotted span of the noncritical activity. The
most common are the total float and the free float. Figure 1.2.6 gives a convenient summary
for computing the total float (TFij) and the free float (FFij) for an activity (i, j). The total flow
is the excess of the time span defined from the earliest occurrence of event i to the latest
occurrence of event j over the duration of (i, j) that is
TFij = ∆j - □i – Dij
The free float is the excess of the occurrence of the time span defined from the earliest
occurrence of event i to the earliest occurrence of event j over the duration of (i, j) that it
FFij = □j - □i – Dij
By definition FFij ≤ TFij
Red-Flagging Rules. For a noncritical activity (i, j)
a) If FFij = TFij then the activity can be schedule anywhere within its (□j, ∆j) span without
causing schedule to conflict.
b) If FFij < TFij, then the start of the activity can be delayed by at most FFij relative to its
earliest start time (□i) without causing conflict. Any delay larger than FFij (but not more
than TFij) muct be coupled with an equal delay relative to □j in the start time of all the
activities leaving node j.
The implication of the rules is that noncritical activity (i, j) will be red-flagged if its FFij <
TFij. This red flag is important only if we decide to delay to start of the activity past its earliest
start time, □i, in which case we must pay attention to the start times of the activities leaving
node j to avoid schedule conflicts.
Figure 1.2.6 Determination of the Total and Free Floats
12
The following is an example of how the total float and the free float look like in a real
problem. Form the figure 1.2.4 it is possible to compute the floats for noncritical activities of
the network. The following table summarizes the computation of the total and free floats
Noncritical activity
Duration
Total float (TF)
B (1, 3)
2
3–0–2=1
D (1, 5)
3
8–0–3=5
C (1, 8)
4
11 – 0 – 4 = 7
H (5, 7)
1
9–3–1=5
Figure 1.2.7 FF and TF computations example one
Free float (FF)
3–0–2=1
3–0–3=0
11 – 0 – 4 = 7
9–3–1=5
The computations of the red-flag activity D because their FF < TF. The remaining activities
(B, C, and H) have a FF = TF, and hence may be schedule anywhere between their earliest start
and latest completion time.
To investigate the significance of the red-flagged activities, consider activity C. It is possible to
see that its FF = 0. This means that any delay in staring C past its earliest time (= 0) must be
coupled with an equal delay in the start of its successor activity J.
1.3 PERT Network
1.3.1 Description of PERT
PERT differs from CPM in that it bases the duration of an activity on three estimates:
1. Optimistic time, a, which occurs when executions are extremely well.
2. Most likely time, m, which occurs when execution is done under normal conditions.
3. Pessimistic time, b, which occurs when execution goes extremely poorly.
The range (a, b) encloses all possible estimates of the duration of an activity. The estimate m
lies somewhere on the range (a, b). Based on the estimates, the average duration time D̅ , and the
variance, v, are approximate as:
D̅ =
V=
^2
CPM calculations given in figure 1.2.4 may be applied directly, with D̅ replacing the single
estimated D.
It is possible now to estimate the probability that a node j in the network will occur by a
prespecified schedule, Sj. Let ℮j be the earlier occurrence time of the node. Because of the
durations of the activities leading from the start node to node j are random variables. Assuming
that all the activities in the network are statistically independent is possible to determine the mean,
13
E {℮j}, and variance, v {℮}, in the following manner. If there is only one path from the start node
to node j, then the mean is the sum of the expected durations, D̅ , for all the activities along this
path and the variance is the sum of the variances, v, of the same activities. On the other hand, if
more than one path leads to the node j, then it is necessary first to determine the statistical
distribution of the duration of the longest path. This problem is rather difficult because it is
equivalent to determine the distribution of the maximum of two or more random variables. A
simplifying assumption this calls for computing the mean and variance, F{℮j} and var {℮j}, as
those of the path to node j that has the largest sum of the expected activity durations. If two or
more path have the same mean, the one with the larger variance is selected because it reflects the
most uncertainty and, hence, leads to a more conservative estimate of probabilities.
Once the mean and the variance of the path to node j, E{℮j} and var {℮j} have been
computed, the probability that node j will be realized by a present time Sj is calculated using the
following formula;
P{℮j ≤ Sj} = P ℮
℮
℮
≤
℮
℮
= { ≤ }
Where z = Standard normal random variable
Kj =
℮
{℮}
The standard normal variable z has mean 0 and standard deviation 1. The justification for
the use of the normal distribution is that ℮j is the sum of independent random variables. According
to the central limit theorem ℮j is approximately normally distributed.
Consider the project of the example 1.2.4. To avoid repeating critical path calculations, the values
of a, m, and b in the table below are selected such that D̅ ij = Dij for all i and j in the example 1.2.4
Activity
A
B
C
D
E
i–j
1–2
1–3
1–8
1–5
3–4
(a, m, b)
(1, 3, 5)
(1, 2, 3)
(2, 4, 6)
(1, 3, 5)
(1, 2, 3)
Activity
F
G
H
I
J
i–j
4–6
6–7
5–7
7–8
8-9
(a, m, b)
(1, 2, 3)
(1, 2, 3)
(.5, 1, 1.5)
(1, 2, 3)
(2, 4, 6)
The mean D̅ ij and the variance vij for the different activities are given in the flowing table. Noted
that for the dummy activity (a, m , b) = (0, 0, 0) hence its mean and variance also are equal to zero.
14
Activity
i–j
D̅ ij
vij
Activity
i–j
A
1–2
3
.444
F
4–6
B
1–3
2
.111
G
6–7
C
1–8
4
.444
H
5–7
D
1–5
3
.444
I
7–8
E
3–4
2
.111
J
8-9
Figure 1.2.8 the mean and the variance computations for example one
D̅ ij
2
1
1
2
4
Vij
.111
.111
1.000
.111
.444
The next table gives the longest path from the node 1 to the different nodes, together with their
associated mean and standard deviation.
Section 2: My Projects
Section 2.1 Making a cake
In this part of my project I am going to introduce three different situations in which it is necessary
to use CPM. As mentioned before, CPM can be applied in small projects such as the one
mentioned in problem one of section 2. CPM also can be applied in bigger management problems
such as problem 2 and 3 of the section 2. In this type of problems it will be shown how CPM can
easily help people to organize construction projects as well as organize their lives in terms of
major events such as graduating from college or applying to grad school.
Section 2.1.1 description of the project
Problem 1.
Betty is planning to surprise her boyfriend by baking a cake for him. The following table provides
the associated activities and their duration. Do the following:
A. Construct the project network
B. Explain the time schedule
C. Calculate and explain the free float and total float
D. Calculate the PERT network
15
Making a cake with CPM
Activitie
s
A
B
C
D
E
F
G
H
I
J
K
M
Description
Decide what type I will do
Prepare the cake pans
The ingredients
Preheat the oven
Measuring the ingredients
Mixing the ingredients
Poor the butter evenly into the prepared
pans
Place the cake on the preheat oven
Close the oven and set the timer
Check the doneness
Cool the cake
Decorating
no
no
no
no
yes
yes
type
of
predecesso
r
0
0
0
0
c
e
yes
yes
yes
yes
yes
yes
b, f
d, g
h
h, I
j,
a, k
Require
predecessor
Duration
Minutes
10
10
10
2
8
5
3
3
2
2
30
10
95
Figure 2.1.1list of activities for Betty’s cake
Betty is not an expert baking cakes. She knows that the best way to optimize time is following a CPM
schedule, where she can track each one of the activities she is going to do in order to get the cake baked on
time.
The first thing Betty does is to write all the possible activities she is going to use when baking the cake.
She has to determine which one of those activities is more critical and requires starting with no delay. The
table above shows all the possible activities Betty will need in order to achieve her goal. After she listed
all the activities, she realized that she may need 95 minutes before she gets her cake baked. She needs to
optimize this time, otherwise her boyfriend is going to arrive and the cake is not done.
2.1.2 Construction of the Network
The network below shows the critical path that Betty will need to follow if she wants to get her task done
on time. She realized that, by following this critical path method she will not need 95 minutes. Instead she
is going to spend 70 minutes in the process of making and baking her cake.
16
Figure 2.2.2 Betty’s Cake Project Network
CPM calculations for Betty’s cake
The following calculations show how Betty utilized the critical path method. First, she listed all the
possible activities that she is going to use in order to surprise her boyfriend. Since she realized
there are some activities that do not have predecessor, she set a node 1 as a starting node where the
value of that node was set equal to zero. Betty has to follow the CPM rules, which means, she has
to find the forward and the backward pass computations. It is important to recall that the forward
pass will give Betty the earliest start time of the project, which means the exact time in which each
activity has to start. Also Betty has to keep in mind that the way to calculate the forward pass is by
choosing the largest duration of an activity. For example, in the figure 2.2, node 2, 3, 4, and 5,
Betty has to choose node 4. But here we raise a question, why did Betty choose node 4 if the nodes
2, 3, And 4 has the same duration time? The answer is simple; Betty has to follow CPM rules.
When Betty calculates the earliest time in node 13, she has to pick the earliest time coming from
the node 12 which is equal to 60. On the other hand, Betty cannot pick the earliest time coming
from the node 2 since it is equal to 10 and this will be against the CPM rules. Here Betty’s forward
pass calculations.
17
Section 2.1.3 Solution with CPM
Forward Pass
Node 1. Set □1 = 0
Node 2. □2 = □1 + D12 = 0 + 10 = 10*
Node 3. □3 = □1 + D13 = 0 + 2 = 2
Node 4. □4 = □2 + D24 = 10 + 8 = 18*
Node 5. Max {□4 + D45, □1 + D15} = Max {18 + 5, 0 + 10} = 23*
Node 6. □6 = □5 +D56 = 23 + 3 = 26*
Node 7. □7 = Max {□3 + D37, □6 + D67} = Max {2 + 3, 26 + 0} = 26*
Node 8. □8 = □7 + D78 = 26 + 2 = 28*
Node 9. □9 = □8 + D89 = 28 + 2 = 30*
Node 10. □10 = □9 + D910 = 30 + 30 = 60*
Node 11. Max {□1 + D111, □10 + D1011} = Max {10 + 0, 60 +10} = 70*
By the end of the forward computations Betty realized she will need 70 minutes to achieve her
goal. Notice that the critical path are marked with a (*) at the end of each computation. This is an
easy way for the reader to identify and keep in mind the path. Now it is time for Betty to calculate
the Backward Pass. It is important to recall that by the end of the backward path the last node has
to be equal to zero. Betty is going to obtain the latest start time, which means, the slack time in
which Betty can delay a non-critical activity without delay the entire project. Here the backward
pass computation.
Backward pass
Node 11. Set ∆11 = □11 = 70
Node 10. = ∆11 – D1011 = 70 – 10 = 60
Node 9. = ∆10 – D 910 = 60 – 30 = 30
Node 8. = ∆9 – D89 = 30 – 2 = 28
Node 7. = ∆8 – D78 = 28 – 2 = 26
Node 6. = ∆7 – D67 = 26 – 0 = 26
Node 5. = ∆6 – D56 = 26 – 3 = 23
Node 4. = ∆5 – D45 = 23 – 5 = 18
Node 3. = ∆7 – D37 = 26 – 3 = 23
Node 2. = ∆4 – D24 = 18 – 8 = 10
Node 1. = min {∆3 – D13, ∆2 – D1 2, ∆5 – D15, ∆11 – D111} = min {23 – 2, 10 – 10, 23 – 10,
70 – 10} = 0
Notice that the correct computations will ∆1 = 0
18
The forward and the backward computations can be made directly on the network as show in the
figure 2.2. Applying the rules for determining the critical activities, the critical path is
1
2
4
5
6
7
8
9
10
11 which should be expected, spans
the network form start (node 1) to the finish (node 11)
Time Schedule
In this section it is shown how the information obtained from the calculations in section 2.2.3 can
be used to develop the time schedule. We recognize that for an activity (i, j) □i represents the
earliest start time, and ∆j represents the latest completion time. This means the interval (□i, ∆j)
delineates the maximum span during which activity (i, j) may be schedule without delaying the
project.
The method in the construction of the preliminary schedule is illustrated by an example. This is an
important step that Betty has to do if she really wants to be on top of her project. From the figure
2.3 Betty can get the preliminary time schedule for the different activities of her project by
delineating their respective time spans as shown in the figure below. Betty has to do two
observations:
1. The critical activities shown in red lines must be stacked one right after another to ensure
that the project is complete within the specified 70-minutes time.
2. The noncritical activities, shown in white, have time spans that are larger than their
respective durations, thus allowing slack (or “leeway”) in scheduling them within their
allotted time intervals.
How should Betty schedule the noncritical activities within their respective spans? Normally, it
is preferable to start each activity as early as possible. In this manner, slack periods will remain
opportunely available at the end of the assigned time where they can be used to absorb
unexpected delays in the execution of the activity. In some time it is necessary to delay some
activities as much as is possible within the slack time in order to keep the project on going.
This sometime happens where there are two different activities that require the use of the same
machine or maybe the use of the same pot, as is in the case of Betty’s cake. On the other hand,
it is necessary to check the precedence relationship between activities and see if they do not
violate any of the precedence relationship rules. If all noncritical activities can be schedule as
early as possible, the resulting schedule is automatically feasible. Otherwise some precedence
relationship may be violated if noncritical activities are delayed past their earliest times. In the
case of Betty’s cake the precedence relationship is well kept for all the activities since there are
only four noncritical activities. We may thing for example that the activities D and H could be
violating the precedence relationship, but since D has a slack of 21minutes and H has a slack of
23 minutes will tell us that D has to be done first before H starts.
19
Figure 2.3.3 Preliminary schedule for the project of section 2.2.3
Determination of the floats.
Floats are slack of times available within the allotted span of the noncritical activity. The most
common is the total float and the free float. As mentioned in section 1.2.4 the total floats and free
floats is the excess of time span define from earliest occurrence of event I to the latest occurrence j
over the duration of (I, j).
One of the things Betty has to look at is the Red-Flagging Rule for a noncritical activity.
a. If FFij = TFij, then the activity can be schedule anywhere within its (□j, ∆j) span
without causing schedule conflict.
b. If FFij < TFij, then the start of the activity can be delayed by at most Ffij relatively
to its earliest start time (□i) without causing schedule conflict. Any delay larger
than Ffij(but no more than Tfij) must be coupled with an equal delay relatively to
□j in the start time of all the activities leaving the node j.
With this in mind Betty is going to proceed and calculate the FF and the TF. The graph below
shows Betty’s calculations.
Noncritical activity
Duration
A (1, 11)
10
B (1, 5)
10
D (1, 3)
2
H (4, 7)
3
Figure 2.1.4 FF & TF computations
Total float (TF)
70 – 0 – 10 = 60
23 – 0 – 10 = 13
23 – 0 – 2 = 21
26 – 2 – 3 = 21
Free float (FF)
70 – 0 – 10 = 60
23 – 0 – 10 = 13
2–0–2=0
26 – 2 – 3 = 21
20
Betty’s computations for red-flag activities D because their FF < TF. The remaining
activities (A, B, and H) have FF = TF, and hence may be scheduled anywhere between their
earliest start and latest completion time.
To investigate the significance of the red-flagged activities, consider activity D, Betty noticed that
activity D, its FF = 0. This means that any delay in starting D past its earliest start time (= 0) and
because its delay time exceeds free float by 2, thus succeeding activities (H) cannot start any
earlier than time 4.
Section 2.1.4 Solution using PERT
In this section dedicated exclusive to PERT where there are going to be some differences from
section 2.2.3 which talks about the CPM solutions. As mentioned in section 1.2.4 PERT differs
from CPM in that it bases the duration of an activity on three estimates, optimistic time (a), most
likely time (m), and pessimistic time (b). Noticed that the optimistic time, a, which occurs when
execution goes extremely well, the most likely time, m, which occurs when execution is done
under normal conditions, and pessimistic time, b, which occurs when the execution goes extremely
poorly.
Going back to our first problem, Betty’s cake, the table below shows Betty’s calculations
regarding each one of the activities. Following the formula given in the section 1.2.4, Betty could
calculate each one of the optimistic, most likely, and pessimistic times.
Hint: if the formulas were forgotten, here they are:
D̅ =
V=
^2
Activity
i–j
(a, m, b)
A
1 – 11
7, 10, 12
B
1–5
9, 10, 11
C
1–2
8, 10, 12
D
1–3
1, 2, 3
E
2–4
6, 8, 10
F
4–5
3, 5, 7
Figure 2.1.5 PERT computations for problem one
Activity
G
H
I
J
K
L
i–j
5–6
3–7
7–8
8–9
9 – 10
10 – 11
(a, m, b)
1, 3, 5
2, 3, 4
1, 2, 3
1, 2, 3
25, 30, 35
8, 10, 12
21
After Betty applied the formulas she got the results shown in the table below. What she did was
replace D̅ by the single D (estimate time). With this in mind now is possible for Betty to calculate
the mean and the standard deviation of her project.
Activity
i–j
D̅ ij
vij
Activity
i–j
A
1 – 11
10
.694
G
5–6
B
1–5
10
.111
H
3–7
C
1–2
10
.444
I
7–8
D
1–3
2
.111
J
8–9
E
2–4
8
.444
K
9 – 10
F
4–5
5
.444
M
10 – 11
Figure 2.1.6 the average estimation time and variance for problem one
D̅ ij
3
3
2
2
30
10
Node
Kj
Longest path
Path mean
Path standard
deviation
2
1-2
10
0.83
3
1-3
2
0.33
4
1-2-4
18
0.94
5
1-2-4-5
23
1.15
6
1-2-4-5-6
26
1.33
7
1-2-4-5-6-7
26
1.33
8
1-2-4-5-6-7-8
28
1.37
9
1-2-4-5-6-7-8-9
30
1.41
10
1-2-4-5-6-7-8-9-10
60
2.18
11
1-2-4-5-6-7-8-9-10-11
70
2.30
12
1-2-4-5-6-7-8-9-10-11
70
2.30
Figure 2.1.7 the probabilistic times computations problem one
Sj
10
3
15
20
30
28
24
32
65
68
75
0
3.03
-3.19
-2.60
3.00
1.50
-2.91
1.42
2.29
-0.87
2.17
Vij
.444
.111
.111
.111
2.777
.444
P{z ≤ kj}
.5000
.9988
.0007
.0047
.9987
.9332
.0018
.9222
.9890
.1922
.9850
2.1.5Analsis of the solution methods
In this section, the solution given by the two methods of CPM and the PERT will be shown. By
following the solution of the project network, it was possible to have done the cake in 70 minutes.
In the figure 2 where all the possible activities were listed, there was a time line of 95 minutes to
achieve Betty’s goal. Once the network was done, it was possible to achieve the whole project in
70 minutes. It was also possible to construct the time schedule, an important tool, that allowed the
CPM user to manage each minute during the project. There is an important feature on the time
schedule and that is red-flagging. CPM has rules that have to be kept in order to maintain the
project working properly. That rules have been explained in section one. One particular rule in
CPM is to preserve the precedence relationship between activities. For example, if activity j
precede activity i, it is necessary for the activity j to be completed before activity i starts. If
Activity j is delay and overlaps the activity i per most of its earliest time, some precedence
relationship will be violated. In section 2.4.3 graph 2.4 it is possible to observe the computation
22
red-flag activity D because its FF < TF the remaining activities can be schedule anywhere between
their earliest start and latest completion time.
On the other hand, the diagram for CPM and PERT are identical except for the activity time. The
time estimates for CPM is deterministic, while those for PERT is probabilistic. As mentioned in
section 2.2.4 in PERT, each activity has three times:
a = optimistic time under the best of conditions
b = pessimistic time under the worse of the conditions
m = most likely (Probable) time estimates under normal conditions
The solution obtained by PERT will allow us to identify what happens if the project is finished
before the deterministic time. In this case what happens if the cake is baked in 68 or 65 minutes?
But also will allow us to know what happens if the cake is baked in 75 minutes. All of these
questions are going to be answered in the next section.
2.2.6 Addressing Possible Variations “What if”
This section will focus on the analysis of the possible “what if” events that can happen to Betty
during the time she is running her project. Betty is a very focus person. She is not easily distracted
by things that could possibly take her away from her obligations. She decided to give her boyfriend
a surprise.
She started making the cake and she realized there were some activities that needed to be done
with no delay. She had completed all the activities A, B, C, E, F, G, and she was working on the
activity H when she got a call. Her boyfriend called her and they talked for 10 minutes. Now let’s
take a look at what could possibly happen to the entire project if she talked with her boyfriend for
10 minutes? To complete activity H, Betty needs 3 minutes, but she spent 10 minutes talking with
her boyfriend. At the same time, activity H has a FF = TF and a slack time of 21 minutes. Which
means there is not going to be any delay in the entire project. First of all, H is a noncritical activity
and its FFij = TFij, then this activity can be scheduled anywhere within its earliest start and its
latest completion time.
On the other hand, what, if that event would had happened during the activity J? First of all j is a
critical activity and it has duration of 2. Since j’s FF = TF = 0 means that that activity cannot be
delay, it has to be done in the allotted time. By spending 10 minutes talking with her boyfriend by
phone, Betty had delay the entire project 10 minutes. Now the estimated completion time for the
entire project is no going to be 70 minutes, it will be 80 minutes. Adding to this, there will be other
consequences, For example, since activity j has duration of 2, but Betty talked by phone per 10
minutes, she could possible forget to run that activity on time. She could end up burning the cake,
since in the activity j all she has to do is to check the doneness.
23
In the PERT case, Betty will be able to calculate the probabilistic time. The probabilistic time will
allow Betty to know the normal distribution of the whole project, if it is delayed more than the
initial 70 minutes or if the project is finished before the 70 minutes. For example, in Betty’s cake,
the critical path is C, E, F, G, I, J, K, M
(a) Expected project completion time = E(e) = E(ec) + E(eE) + E(eF) + E(eG) + E(eI) + E(eJ) +
E(ek) + E(eM) = 10 + 8 + 5 + 3 + 2 + 2 + 30 + 10 = 70
Project variance = σ ² = .444 + .444 + .444 + .444 + .111 + .111 + 2.777 + .444 = 5.219
(b) probability that the project completion time e ≤ 68
Sj = 68
E(e) = 70
σ = √5.33 = 2.30
c=
()
=
.
= −0.87
P(e ≤ 68) = P(z ≤ c ) = P(z ≤ -0.876) = .1922 (from normal distribution table)
Now probability that the project completion time e ≤ 75
Sj = 75
E(e) = 70
σ = 2.284
c=
()
=
.
= 2.17p
P(e ≤ 75) = P(z ≤ c) = P(z ≤ 2.17) = .9850 (from distribution table)
The probability that Betty will finish 2 minutes earlier is .1922, and the probability that she finish
the cake 5 minutes late, 75, is .9850.
It is possible to talk more about the “what if” events that can happens at the moment of running a
project. The best advice for a CPM user is to avoid any distraction and keep tracking each one of
the activities on time. This will reduce the risk of taking more time that the allotted time to run the
entire project.
2.2 How to graduate from college
2.2.1 Description of the project
The second part of this project will introduce a second CPM and PERT management problem.
This involves a much bigger problem than the first one. This does not mean that the previous
problem was easy or less important, just that in this problem the variable time will be given in
months and also involve a longer period of time. As mentioned in section one, CPM and PERT
24
facilitates users to manage simple problem to very complex problem with many more
variables.
The second problem will show how an average person can control, manage, and plan his/her
studies. The theme of this second problem is how to graduate from high school by using CPM
and PERT analysis. In the table below there is a list with all possible items a person will need
in order to start and finish his/her college. It is important to mention that some high school
students do not wait until they finish high school in order to start looking for a college that
satisfies their needs.
How to graduate from undergrad school with CPM analysis
Activitie
s
Description
Require
predecessor
Duration/Month
s
Select a college based on
A
cost
no
3
B
Find a job
no
4
C
Enroll in college
no
3
D
Decide a degree
no
16
E
get an advisor
yes
1
F
Apply for a loan
yes
2
Apply for scholarships and
G
grants
yes
3
H
Be a freshman
yes
10
I
Be softmore
yes
10
J
Be a junior
yes
10
K
Be a senior
yes
10
Complete
the
L
requirements
yes
5
M
Fill out Graduation form
yes
1
Pay for the graduations
1
N
rights
yes
O
Graduation day
yes
1
Figure 2.2.1 list of activities for an average person go to school
type
of
predecessor
0
0
0
0
D
A, D
C
C
H
I
J
K
L
B, M
N
Now that all the activities have been listed and we are ready to start the project management
network, it is important to list the steps we have to follow in order to achieve our goal. For
example,
A. Construct the project network
B. Explain the time schedule
C. Calculate and explain the free float and total float
25
D. Calculate the PERT network
2.2.2 Construction of the network.
In the construction of the network there are two steps involved: the forward path and the backward
path. In the calculation in the forward part in is necessary to follow the CPM rules in turns of activity
relationship. From the graph above, it is possible to observe that the completion of the project will take 80
months. However, after the critical path network is done, it is possible com complete the whole project in
51 months.
Fi
Figure 2.2.2 Critical Path Network, Project Two
Section 2.2.3 Solution with CPM
Forward Pass
Node 1. Set □1 = 0
Node 2. Max {□1 + D12, □4 + D42} = {0 + 3, 16 + 0} = 16
Node 3. □3 = □1 + D13= 0 + 3 = 3*
Node 4. □4 = □1+D14=0+16=16
26
Node 5. □5 □3 + D35 = 3 + 10 = 13*
Node 6. □6 = □5 + D56 = 13 + 10 = 23*
Node 7. □7 = □6 + D67 = 23 + 10 = 33*
Node 8. □8 = □7 + D78 = 33 + 10 = 43*
Node 9. □9 = □8+D89 = 43 + 5 = 48*
Node 10. =Max{□1 + D110, □9 + D910} = {0 + 4,48 + 1} = 49*
Node 11. □11 = □10 +D1011 = 49 + 1 = 50*
Node 12. = Max {□2 + D212, □4 + D412, □3 + D312, □11 + D1112} ={16 + 2, 16 + 1, 3 + 3, 50 +
1}
= 51*
Now that the forward computation had been calculated, it is possible to see that the whole
project will take 51 months and not 80 as it was when the activities were listed. At this point the
whole project makes sense, since for a regular person it will take 48 months to graduate from
college; But then why 51 months and not 48 in the forward computations? The reason why there
are three more months involved in the completion of the project is because some people do not
wait until they finish high school to start with the whole process of searching, applying, and
enrolling into college.
Now the backward computations.
Node 12. Set ∆12 = □12 = 51
Node 11. ∆11 - ∆12 - D1112 = 51 – 1 = 50
Node 10. ∆10 = ∆11 – D1011 = 50 – 1 = 49
Node 9. ∆9 = ∆10- D910 = 48 – 1 = 48
Node 8. ∆8 = ∆9 – D89 – 48 – 5 = 43
Node 7. ∆7 = ∆8 - D78 = 43 – 10 = 33
Node 6. ∆6 = ∆7 – D67 = 33 – 10 = 23
Node 5. ∆5 = ∆6 – D56 = 23 - 10 = 13
Node 4. ∆4 = Min {∆12 - D412, ∆2 - D42} = {51 - 1, 49 - 0} = 49
Node 3. ∆3 = Min {∆5 – D35, ∆12 - D312} = {13 - 10, 51 - 3} = 3
Node 2. ∆2 = ∆12 – D212 = 51 – 2 = 49
Node 1. ∆1 = Min {∆10 – D110, ∆4 - D14, ∆3-D13, ∆2-D12} = {49 + 4, 3 - 3, 46 – 16, 49 - 3} = 0
Notice that the correct computations will ∆1 = 0
The forward and the backward computations can be made directly on the network as show in the
figure 2.2. Applying the rules for determining the critical activities, the critical path is
1
3
5
6
7
8
9
10
11
12 which should be expected,
spans the network form start (node 1) to the finish (node 12)
27
Time Schedule
As mentioned in section 2.1.3 the time schedule for the different activities of the figure 2.2 allows
us to find the respective time span, as shown in the figure 2.2.3. The critical activities are shown in
red and it is shown that the whole project can be completed in 51 months instead of the initial 80
months shown in the figure 2.2.1. Also the noncritical activities are shown in blue and the span
times are shown in white. One way to keep in mind what a noncritical activity is, is that the
noncritical activities have time spans that are larger than their respective duration in schedule them
within their allotted time intervals.
Figure 2.2.3 Preliminary Schedule for the Project of the problem 2.2
Notice that the easiest way to manage noncritical activities is scheduling them as early as
possible. Keep in mind that this will save slack time by the end of the project so that if something
unpredicted occurs it can absorb any delay in the execution of the an activity. However, it may be
necessary to delay some of the activities because some activities are runing at the same time. For
example, in the figure 2.2.3 to run activities A and B require to find a job first then select the
college. So it is necessary to delay activity A until activity B can be completed.
It is necessary to keep in mind that is all the activities can be scheduled on time; the
resulting schedule will be feasible. Otherwise, some relationship precedence will be violated if
noncritical activities are delayed past their earliest start time.
28
Total float
As mentioned in section 2.1.3, the determination of the floats will allow the user to identify
red flags on the project. Just as remainder, the red flags are raise when the FF < TF then the start
of the activity can be delayed by at most FFij relatively to its earliest start time. On the other hand,
if the FF = TF then the activity can be schedule anywhere within its latest start and latest
completion times.
The graph below shows the calculations of the TF and FF floats calculation from the
critical path network figure 2.2.2. It is possible to observe that from the graph 2.2.2 two red flag
were raised. In the activity A the FF < TF so activity A can be delayed relatively to its earliest start
time if the manager does not want to affect the entire project. The activity D, on the other hand, its
TF = 45 and the FF = 0. This means that any delay in D must be coupled with at least an equal
delay in the start of its successor activities, in this case activity E.
Noncritical activity
Duration
Total float (TF)
A (1, 2)
3
49 – 0 – 3 = 46
B (1, 10)
4
49 – 0 – 4 = 45
D (1, 4)
16
49 – 0 – 16 = 33
E (4, 12)
1
51 – 16 – 1 = 34
F (2, 12)
2
51 – 16 – 2 = 33
G (3, 12)
3
51 – 3 – 3 = 45
Figure 2.2.4 FF and TF Computations for problem two
Free float (FF)
16 – 0 – 3 = 13
49 – 0 – 4 = 45
16 – 0 – 16 = 0
51 – 16 – 1 = 34
51 – 16 – 2 = 33
51 – 3 – 3 = 45
Section 2.2.4 Solution Using PERT
This section is dedicated to the PERT solution. As mentioned before, the differences
between CPM and PERT are not many. PERT differs from CPM just on three different times;
optimistic (a), pessimistic (b), and most likely (m). Just a remainder, the pessimistic time occurs
when the execution of the project goes extremely well, most likely time imply that the project is
going under normal conditions, and the pessimistic time occurs when the project execution goes
extremely poor.
D̅ =
V=
^2
After the formulas above have been applied it is easy to find the mean ad the standard deviation.
Basically what was done here was to replace D by D̅ . D is just the time duration of the activities ij.
29
Activity
i–j
D̅ ij
vij
Activity
i–j
A
1–2
3
.111
I
5–6
B
1 – 10
4
.444
J
6–7
C
1–3
3
.25
K
7–8
D
1–4
16
.444
L
8–9
E
4 – 12
1
.174
M
9 – 10
F
2 – 12
2
.25
N
10 – 11
G
3 – 12
3
.25
O
11 – 12
H
3–5
10
.694
Figure 2.2.5 the mean and the variance computations fro problem two
Node
Longest path
2
3
4
5
6
7
8
9
10
11
Path
mean
Path
standard
deviation
0.33
0.5
0.66
0.97
1.39
1.48
1.62
1.69
2.15
2.23
1–4–2
16
1–3
3
1–4
16
1–3–5
13
1–3–5–6
23
1–3–5–6–7
33
1–3–5–6–7–8
43
1–3–4–5–6–7–8–9
48
1 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10
49
1–3–4–5–6–7–8–9–1050
11
12
1 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10 51
2.27
11 – 12
Figure 2.2.6 the probabilities of the project network for problem two
D̅ ij
10
10
10
5
1
1
1
Sj
Vij
1
.25
.444
.25
.174
.340
.174
Kj
P{z ≤ kj}
16
2
14
15
20
30
45
50
52
48
0
-2
-3.03
2.06
-2.15
-2.02
1.23
-1.18
1.39
-0.89
.5000
.0228
.0012
.9803
.0158
.0217
.8907
.1190
.9177
.1867
55
1.76
.9608
2.2.5 Analysis of the Solution Methods
This section is dedicated to the analysis of the methods. When project two started, it was
possible to see, after the activities were listed, that graduating from college will take 80 months.
Once the project network was done, the entire project of graduating from college will take only 51
months. Now we can raise a question, will graduating from college necessarily take more than 48
months? The answer is straight forward. It is not necessary for an average high school student to
finish high school in order to start the process of selecting a college. CPM offers the opportunity of
managing every activity within the model in order to control each one of the events. Sometimes the
events or activities cannot be controlled or something unexpected happens delaying some
activities. CPM offers the possibility of managing the slack time within the noncritical activities.
For example, running the activity H (be a freshman) a student may fail a class. Retaking this class
can delay the student at least one more semester. But these type of events are going to be analyzed
in the next section.
30
2.2.6 Addressing Possible Variations “What is”
This section describes those events that are not planned. For example, a student who went
to college had an amazing performance in his/her classes, got a decent GPA. But what would
happen if in the very last semester the student failed a class? Let’s look at example activity K (be a
senior). Since activity k is a critical activity, which means it has to be done right way and cannot
have any delay. If the student fails a class at this point of the project, the whole project is going to
be delayed for at least six months.
On the other hand, what if the student takes the failed class in summer? Taking a summer
class takes at least three weeks, so the student is going to delay the entire project three more weeks
than was expected. Something that is very important to keep in mind is the noncritical activities
cannot be delayed. Critical activities do not have time span. So it is not possible to reschedule that
single activity without affecting the entire project
Similarly, a student how are planning to graduate from college must know the possibilities
of finishing after or before than the estimated time. With PERT is possible to calculate this type of
probabilities, so that student will be aware of losing or gaining time. For example, in project two,
the initial time is 51 months. But what if the entire project finishes after 51 months, For instance,
55 months instead? For the project two figure 2.2.2 the critical path is C, H, I, J, K, L, M, N, O and
so now if time to calculate the;
(a) Expected project completion time = E(e) = E(ec) + E(eh) + E(ei) + E(ej) + E(ek) + E(el) +
E(em) + E(em) + E(eo) = 3 + 10 + 10 + 10 + 10 + 5 + 1 + 1 + 1 = 51
Project variance = σ ² = .25 + .694 + 1 + .25 + .444 + .25 + 1.74 + .340 + .174 = 5.142
(b) Probability that the project completion time e ≤ 55
Sj = 55
E(e) = 51
σ = √5.142 = 2.26
c=
()
=
.
= 1.77
P(e ≤ 55) = P(z ≤ c) = P(z ≤ 1.77) = .9616 (From the normal distribution table)
So the probability of finishing the entire project five days later is .9616
So now what is the probability of finishing the project five days before?
31
(a) Expected project completion time = E(e) = E(ec) + E(eh) + E(ei) + E(ej) + E(ek) + E(el) +
E(em) + E(em) + E(eo) = 3 + 10 + 10 + 10 + 10 + 5 + 1 + 1 + 1 = 51
Project variance = σ ² = .25 + .694 + 1 + .25 + .444 + .25 + 1.74 + .340 + .174 = 5.142
(b) Probability that the project completion time e ≤ 46
Sj = 46
E(e) = 51
σ = √5.142 = 2.26
c=
()
=
.
= −2.21
P(e ≤ 46) = P(z ≤ c) = P(z ≤ -2.21) = .0080
The probability of finishing the project five days earlier is .0080
Though PERT it is possible to obtain the optimistic (a) time = 46 months = .0080, the most likely
time (m) = 51 months = .9608, and the pessimistic time (b) 55 months = .9616. with this in mind, it
is possible to track each one of the activities and events knowing that the best idea is to finish the
project as earlier as possible.
2.3 Applying for grad school
2.3.1 Description of the Project
In the third part of project I want to introduce a different problem with different activities and
different time. In section 2.1 the duration’s activities were given in minutes, in section 2.2 the
duration’s activities were given in months, in section 2.3 the duration’s activities are given in
weeks. The purpose of using different time is to show one of the greatest tools that CPM has. It
allows the user to manage anytime of type that the project may require. It does not matter if it is a
big project such the one shown in section 2.2 which involve 51 months, or a small one shown in
section 2.1 in which are necessary 70 minutes.
The purpose of the third project is to inspire people to use CPM and PERT in order to organize
their lives. At this time I am going to be using CPM and PERT to create a project that involve
people applying for grad school. Some time, it is not easy to be a college student and being
thinking what school should apply to. In this project, a student can actually be in the last two
semester of college and start making preparations to go to grade school. The table below shows all
the possible activity that a senior student should do if he/she is trying to get into grad school.
A senior student wants to apply for grad school, the table below shows the associated activities
and their durations, construct the network and provide the analysis.
32
Applying for grad school based on CPM analysis
Activities
A
B
C
D
E
F
G
H
I
J
K
L
M
N
Description
To be a senior
Graduate from undergrad school
Define the grad program
Decide on at least 9 schools
To visit some schools
To satisfy the
graduate program
requirements
To take the GRE
Write a personal statement
Letters of recommendation
Transcripts
Example of the work Done
To complete all the paperwork required
Submit applications
Hearing back from the interviewers
Require
predecessor
no
no
no
yes
yes
no
yes
yes
yes
yes
yes
yes
yes
yes
Duration/weeks
40
8
16
4
8
4
8
1
1
4
4
2
4
3
107
type
of
predecessor
0
0
0
C
D
D, E
A
G
H
B
A
H, I , J
L
M
Figure 2.3.1 List of Activities for problem three
2.3.2 Construction of the Network
After listing all possible activities and considering all the possible options, a college
student can build a project network. This project network will help the student to determine if
really 107 weeks are needed in order to apply for grad school. Notice that 107 weeks are 26.3
months, around 2.2 years. Then it will raise the question, does a student really need that amount of
time in order to apply for grad school? The answer is simple. This project shows the preparations a
student must have such as; finishing undergrad school, preparing and taking the GRE test, getting
documents ready for application, and obviously searching for grade schools. The figure 2.3.2
shows the project management network from figure 2.3.1.
33
Figure 2.3.2 Project Network for problem three
Once all the possible activities were listed, in the figure 2.3.1. it is possible to see that the
initial time given to complete all the activities was 107 weeks. After the project management
network was built, there were only needed 59 weeks, which save us 48 weeks, speeding up the
entire project.
Section 2.3.3 Solution with CPM
As explained in section 1.2.3 the critical path method implies two calculations, the forward
path computations, and the backward path computations. The following computation shows that
forward and the backward critical path computation for the figure 2.3.2.
34
Forward Pass
Node 1. Set □1 = 0
Node 2. □2 = □1 + D12 = 0 + 40 = 40*
Node 3 □3 = □1 + D13 = 0 + 8 = 8
Node 4. □4 = □1 + D14 = 0 + 16 = 16
Node 5. □5 = □4 + D45 = 16 + 4 = 20
Node 6. □6 = □5 + D56 = 20 + 8 = 28
Node 7. □7 = □2 + D27 = 40 + 8 = 48*
Node 8 □8 = □7 + D78 = 48 + 1 = 49*
Node 9. Max { □3 + D39, □8 + D89} = {8 + 4, 49 + 1} = 50*
Node 10. □10 = □9 + D910 = 50 + 2 = 52*
Node 11. □11 = □10 + D1012 = 52 + 4 = 56*
Node 12. □12 Max{ □2 + D212, □6 + D612, □11 + D1112} = {40 + 8, 28 + 4, 56 + 3} = 59*
Now that the forward computation path is done it is possible to observe that the time
needed to complete the project is almost twice less than the initial time.
Backward path calculation.
Node 12. set ∆12 = □12 = 59
Node 11. ∆11 = ∆12 – D1112 = 59 – 3 = 56
Node 10. ∆10 = ∆11 – D1011 = 56 – 4 = 52
Node 9. ∆9 = ∆10 – D910 = 52 – 2 = 50
Node 8. ∆8 = ∆9 – D89 = 50 – 1 = 49
Node 7. ∆7 = ∆8 – D78 = 49 – 1 = 48
Node 6. ∆6 = ∆12 – D612 = 59 – 4 = 55
Node 5. ∆5 = ∆6 – D56 = 55 – 8 = 47
Node 4. ∆4 = ∆4 – D45 = 47 – 4 = 43
Node 3∆3 = ∆9 – D39 = 50 – 4 = 46
Node 2. ∆2 = Min {∆12 – D212, ∆7 – D27} = {59 – 8, 48 - 8} = 40
Node 1. ∆1 = Min {∆4 – D14, ∆3 – D13, ∆2 – D12} = {43 – 16, 46 - 8, 40 - 40} = 0
Notice that the correct computations will ∆1 = 0
The forward and the backward computations can be made directly on the network as show in the
figure 2.2. Applying the rules for determining the critical activities, the critical path is
1
2
7
8
9
10
11
12 which should be expected, spans the
network form start (node 1) to the finish (node 12)
35
Time Schedule
As mentioned in section 2.1.3 the time schedule for the different activities of the figure 2.2
allows us to find the respective time span. Now that that the time schedule had been found, it is
time to go over it and see if is possible to schedule all the noncritical activities as earlier as
possible. As shown in the sections 2.1 and 2.2, the critical activities are shown in red; the
noncritical activities are shown in blue and white colors.
But how should the noncritical activities be schedule with their respective time spans?
Look for instance activity B and C both activities start at zero time. For a student to run activity B
and C at the same, it will be necessary to have a twin brother, since it is necessary for the student
to do both activities himself, nobody else can do it for him. So the student can schedule activity B
between zero and 8 while activity C can be pushed the start time somewhere between 8 and 11
without affecting the entire project, this is possible thanks to the time spans of the activities.
Figure 2.3.3 preliminary schedule for problem three
On the other hand, as mentioned before, if all noncritical activities can be scheduled on
time, the resulting schedule will be automatically feasible. Otherwise some precedence relationship
will be violated if noncritical activities are delayed past their earlier start times. Take for example,
activities D and E from the figure 2.3.3 in the project network figure 2.3.2. Even though the
36
activity D has to be completed before E, the spans of D and E will allow us to scheduled D
between 18 and 22, and E between 20 and 28 which violated the requirement that D must precede
E.
Total float
From the time schedule in figure 2.3.3, it is possible to observe that when some precedence
relationships are violated. It is necessary then to apply the red flag rule. The red flag rules will help
us to schedule those activities that overlap without affecting the entire project. In the table below
shows the computation red flag for the activities B, C, D, and E because their FF < TF, the
remaining activities F, J, and K have FF = TF and hence can schedule anywhere between the
earliest start time and the and the latest completion time.
Noncritical activity
Duration
Total float (TF)
B (1, 3)
8
46 – 1 – 8 = 37
C (1, 4)
16
43 – 0 – 16 = 27
D (4, 5)
4
47 – 16 – 4 = 27
E (5, 6)
8
55 – 20 – 8 = 37
F (6, 12)
4
59 – 28 – 4 = 27
J (3, 9)
4
50 – 8 – 4 = 38
K (2, 12)
8
59 – 40 – 8 = 11
Fugue 2.3.4 FF and TF calculations for problem three
Free float (FF)
8–0–8=0
16 – 0 – 16 = 0
20 – 16 – 4 = 0
28 – 20 – 8 = 0
59 – 28 – 4 = 27
50 – 8 – 4 = 38
59 – 40 – 8 = 11
Section 2.3.4 Solution Using PERT
This section is dedicated to the PERT solution. As we know from section 2.1 and 2.2 the
differences between CPM and PERT vary just on three different times; optimistic (a), pessimistic
(b), and most likely (m). Just a remainder, the pessimistic time occurs when the execution of the
project goes extremely well, most likely time imply that the project is going under normal
conditions, and the pessimistic time occurs when the project execution goes extremely poor.
D̅ =
V=
^2
As section 2.1 and 2.2 with PERT is easy to find the mean ad the standard deviation.
Basically what was done here was to replace D by D̅ . D is just the time duration of the activities ij.
37
Activity
i–j
(a, m, b)
Activity
A
1–2
45, 40, 50
H
B
1–3
6, 8, 11
I
C
1–4
12, 16, 20
J
D
4–5
1, 4, 5
K
E
5–6
6, 8, 10
L
F
6 – 12
2, 4, 8
M
G
2–7
4, 8, 10
N
Figure 2.3.5 mean computations for problem three
i–j
7–8
8–9
3–9
2 – 12
9 – 10
10 – 11
11 – 12
Activity
i–j
D̅ ij
vij
Activity
i–j
A
1–2
40
.694
H
7–8
B
1–3
8
.694
I
8–9
C
1–4
16
1.77
J
3–9
D
4–5
4
.444
K
2 – 12
E
5–6
8
.444
L
9 – 10
F
6 – 12
4
1
M
10 – 11
G
2–7
8
1
N
11 – 12
Figure 2.3.6 the mean and the variance computations for problem three
Path
standard
deviation
2
1–2
40
0.83
3
1–3
8
0.83
4
1–4
16
1.33
5
1-2–7
48
1.30
6
1–4–5
20
1.48
7
1–4–5–6
28
1.63
8
1–2–7–8
49
1.32
9
1–2–7–8–9
50
1.46
10
1 – 2 – 7 – 8 – 9 - 10
52
1.54
11
1 – 2 – 7 – 8 – 9 – 10 - 11
56
1.62
12
1 – 2 – 7 – 8 – 9 – 10 – 11- 12
59
1.69
12
1 – 2 – 7 – 8 – 9 – 10 – 11- 12
59
1.69
Figure 2.3.7 the probabilistic computations for problem three
Node
Longest path
Path
mean
(a, m, b)
.5, 1, 2
.3, 1, 4
2, 4, 6
6, 8, 13
1, 2, 4
2, 4, 5
1, 3 ,4
D̅ ij
1
1
4
8
2
4
3
Sj
40
6
20
45
25
25
52
55
48
58
55
64
Vij
.06
.38
.44
1.36
.25
.25
.25
Kj
0
-2.40
3
-2.30
3.37
-1.84
2.72
3.42
-2.59
1.23
-2.36
2.95
P{z ≤ kj}
.5000
.0082
.9987
.0107
.9996
.0329
.9967
.9997
.0048
.8907
.0091
.9984
2.3.5 Analysis of the Solutions Methods
For section 2.3 the solutions of CPM and PERT are interesting. In the CPM solution it was
possible to reduce the initial time given, after the activities were listed, from 107 weeks to 59
weeks. As mentioned before 59 weeks are around 2.3 years, and it is pretty much the time that a
student, who is interested in going to grad school, should spend in getting everything ready to go.
38
Another important tool from CPM is that allow managing the time between activities even though
those activities overlap each other as shown in section 2.3.3.
On the other hand, the solution obtained from PERT comes in handy when red flags are
identified. Also, PERT allows the user to determine the best, more likely, and worse time of the
project.
2.3.6 Addressing possible project variations: “what if”
As a student wants to be sure he/she will be running on time during all the activities
completions, it is necessary to watch out for possible delays during the project. For instance, what
would happen is a student running activity G (to take the GRE) the student does not get a high
score in the GRE test. It is known that grad schools are interested on students with a good grade
performance. It will be necessary for the student to reschedule a new GRE test, study even harder
for it, and retake it again. Since the activity time is given in weeks, the student can spend at least
20 more weeks just by retaking the GRE test. So now the completion of the whole project will not
be complete on time since activity G is a critical activity, with not time slack, that needs to be done
on time, the whole project will end up spending 79 weeks.
Now let’s look at the same situation but with different activity. For example, activity E (to
visit some schools) to complete activity E it is necessary eight weeks. However, because the slack
time of the activity E = 27, then the activity can be schedule somewhere between week 20 and
week 47 without delaying the entire project.
On the other hand, using PERT will allow us to know what would be the probability of
finishing the project earlier or later than the expected given time. Let’s take a look, for example,
what will be the probability that the project finish on week 55 and not in the 59th week as is
expected?
(a) Expected project completion time = E(e) = E(ea) + E(eg) + E(eh) + E(ei) + E(el) + E(em) +
E(en) = 40 + 8 + 1 + 1 + 2 + 4 + 3 = 59
Project variance = σ ² = 0.694 + 1 + .06 + .38 + .25 + .25 + .25 = 2.88
(b) Probability that the project completion time e ≤ 55
Sj = 55
E(e) = 59
σ = √2.88 = 1.69
c=
()
=
.
= −2.36
P(e ≤ 55) = P(z ≤ c) = P(z ≤ -2.36) = .0091 (From the normal distribution table)
39
So the probability of finishing the entire project four days earlier is .0091
So now what is the probability of finishing the project five days later?
(a) Expected project completion time = E(e) = E(ea) + E(eg) + E(eg) + E(ei) + E(el) + E(em) +
E(en) = 40 + 8 + 1 + 1 + 2 + 4 + 3 = 59
Project variance = σ ² = 0.694 + 1 + .06 + .38 + .25 + .25 + .25 = 2.88
(b) Probability that the project completion time e ≤ 64
Sj = 64
E(e) = 51
σ = √2.88 = 1.69
c=
()
=
.
= 2.95
P(e ≤ 64) = P(z ≤ c) = P(z ≤ 2.95) = .9984 (From the normal distribution table)
So the probability of finishing the entire project five days later is .9616
Section 3
3.1Conclusion and conclusions
As the rapid growth of the human population continues, it is also necessary to rapidly to
grow infrastructure and technology. Now it is possible to see scientist walking on the moon’s
surface, airplanes breaking the sound barrier, and small robots used in medicine capable of making
reparations at the cellular level. All of these giant projects, at some point, started with a simple
idea. They require planning, managing, and implementation and that is where CPM and PERT are
very appropriate. Both of them allow the user to manage small projects and huge projects where
the time completion given can be years. The purpose of this study was to show the powerful tool
that CPM and PERT are when managing daily life activities such as baking a cake, graduating
from college, or applying to grad school. Some people may think that those projects are simple
decision making, but when something is not planned, the expected result can easily be a failure.
40
3.2 Future Work
As part of the future work, I will be interesting to include the variable cost to the already
given variable time. For example, the normal time given for an activity can be reduced by using
increased resources. The limit beyond which an activity cannot be shortened will be known as the
crash limit.
41
References
Bronson , Richard , and Govindasami Naadimuthu. Operation Research . Second edition . 2nd . the United
Statesof Americ : McGraw-Hill, 1997. Print.
Taha, Hamdy A. Operation Research An Introduction . Eight Edition. 8th . Upper Saddle River, New
Jersey : Pearson Education, Inc, 2007. Print.
http://hadm.sph.sc.edu/courses/J716/CPM/CPM.html
http://www.netmba.com/operations/project/cpm/
http://www.mindtools.com/critpath.html
42