Establishing a Project Management Framework for Internet Architecture

Overview of PERT and CPM

One advantage of project management is cost-effective, on-time delivery. This can be guaranteed by implementing project management techniques like PERT and CPM. When planning, carrying out, and overseeing a project, PERT and CPM help management find the most time-consuming or consuming way through a network of tasks or activities. In order to reduce the overall cost and time of the project, managers can use these approaches to optimize the longest time duration. By assisting the project manager at each step of the process, these tools are beneficial. The methods are highly helpful in keeping an eye on project expenses, offer project documentation, and provide critical path and slack time through mathematical simplification. But when weighing the benefits of both approaches to project management, PERT is more appropriate for projects with an unclear time horizon since it is event-oriented, probabilistic, and only focused on time. CPM, on the other hand, is a deterministic, activity-oriented paradigm that is employed for small-scale, recurring tasks [7].

The fundamental tasks of scheduling, planning, and controlling are made easier by the PERT and CPM models. The planning stage divides a project’s overall requirements for supplies, labour, and machinery into clear categories. Its focus is on allocating clearly defined tasks to a chronological order of completion. An operational network that links every activity to a time dimension, resource requirements, and a way to distinguish between important and non-essential tasks are all provided to management by PERT and CPM.

PERT is frequently used in construction projects to allocate resources, determine the critical path, and schedule and monitor activities. Office buildings, roads, swimming pools, and the development of a countdown and ”hold” mechanism for space mission launch are a few examples of these projects. It can be used by construction managers to identify possible delays and make the necessary adjustments to keep the project on time. PERT is used in the manufacturing industry to assess production procedures, pinpoint areas for enhancement, and finalize corporate mergers, including ship design and the creation and promotion of a completely new product. The method is applied to optimize resource utilization, develop production schedules, and evaluate production capability. PERT is widely used in information technology (IT) projects to assign resources, identify critical routes, and manage complex operations [11].

Method of Data Collection

Data was gathered through interviews with subject- matter experts. The researcher conducts an in-person interview with the expert to obtain the necessary data. The most likely time, the optimistic time, and the pessimistic time are among the details that were gleaned from the interview, together with the activity description and its predecessors. Details of the data are shown in Tables 1 & 2.

Network Diagram

For a graphical depiction of the software, similar to that seen in Figure 1, the project can then be transformed into a network diagram. Each task originates at a starting node and ends at a subsequent node, represented by the arrows.

Probabilities of Completing the Project

The total of all the task durations on the critical route equals the project length X. PERT makes the assumption that every activity’s time is independent and identically distributed. Therefore, X has a normal distribution with mean EX(X) and variance VAR(X), according to the central limit theorem [12].

$$ Z = \frac {\mathrm {X} - \mathrm {E X} (\mathrm {X})}{\sqrt {\mathrm {V A R} (\mathrm {X})}} $$ Let X stands for the length of the project. Next, the project’s anticipated duration is expected to be as ( ) EX X Sum of the expected times of task = ( ) ( ) ( ) ( ) ( ) ( ) ( ) EX a EX b EX d EX e EX f EX i EX j $$ = E X (a) + E X (b) + E X (d) + E X (e) + E X (f) + E X (i) + I $$

7 6 14 13.5 4.33 3 8.17

$$ = 7 + 6 + 1 4 + 1 3. 5 + 4. 3 3 + 3 + 8 $$

56 =

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The variance of the project duration is ( )

VAR X Sum of the variances of task =

2 2 2 2 2 2 2 2 12 23 35 12 56 67 78 89

$$ = \sigma_ {1 2} ^ {2} + \sigma_ {2 3} ^ {2} + \sigma_ {3 5} ^ {2} + \sigma_ {1 2} ^ {2} + \sigma_ {5 6} ^ {2} + \sigma_ {6 7} ^ {2} + \sigma_ {7 8} ^ {2} + \alpha $$ σ σ σ σ σ σ σ σ

0.11 0.11 5.44 3.36 0.44 0.11 0.25

$$ = 0. 1 1 + 0. 1 1 + 5. 4 4 + 3. 3 6 + 0. 4 4 + 0. 1 1 + 0 $$

9.82 = The standard deviation of the project duration is ( )

$$ S T D = \sqrt {V A R (X)} $$

9.82 3.13 = = A normal distribution with mean µ and variance σ2 is shown in Figure 2. With a mean of 56 and a standard deviation of 3.13, X has a normal distribution. For every given normal distribution, the probability that a random variable would lie between one standard deviation and the mean is 0.68. Consequently, the project has a 68% chance of taking 50 to 62 days to finish. In contrast, there is a 95% chance that X will fall within two standard deviations, or between 44 and

68 days. Moreover, the X has a 99.7% chance of falling within the range of 38 to 74 days, or three standard deviations.

Additionally, we are able to compute the likelihood of completing the project by the deadline. For example, the client would like to know the likelihood that the project would be finished three days earlier than anticipated. The calculation looks like this:

( ) 53 56 53 Z 3.13 (Z 0.96) 0.3315

$$ \begin{array}{l} (Y \leq 5 3) = \operatorname {P r o b a b i l i t y} \left(Z \leq \frac {5 3 - 5 6}{3 . 1 3}\right) \\ = \operatorname {P r o b a b i l i t y} (Z \leq - 0. 9 6) \\ = 0. 3 3 1 5 \\ \end{array} $$ Probability X Probability Probability There is a 33.15% chance that the project will be finished three days sooner than anticipated.

( ) 59 56 59 Z 3.13 (Z 0.96) 0.8315

$$ \begin{array}{l} (Y \leq 5 9) = \operatorname {P r o b a b i l i t y} \left(Z \leq \frac {5 9 - 5 6}{3. 1 3}\right) \\ = \operatorname {P r o b a b i l i t y} (Z \leq - 0. 9 6) \\ = 0. 8 3 1 5 \\ \end{array} $$ Probability X Probability Probability The project has an 83.15% chance of being finished in less than 59 days. The following formula determines the likelihood that the project will be finished in 53 to 59 days.

53 56 59 56 Z ( 0.96 Z 0.96) 3.13 3.13 0.3315 0.3315

$$ \begin{array}{l} \operatorname {P r o b a b i l i t y} \left(\frac {5 3 - 5 6}{3. 1 3} < Z < \frac {5 9 - 5 6}{3. 1 3}\right) = \operatorname {P r o b a b i l i t y} (- 0. 9 6 < Z < 0 \\ = 0. 3 3 1 5 + 0. 3 3 1 5 \\ \end{array} $$

0.663 =

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There is a 66.3% chance that the project will be completed in 53-59 days [13].

Confidence Interval for the Project Completing Time If X represents the actual project completion time, then calculate an approximate 99% confidence interval for the project completion time.

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As a result, based on ten tasks, we can be 99% positive that the completion time will fall between 53 and 59 days.

The project network’s activity duration means and variations are displayed in Table 3.

Table 4 shows the longest path mean and standard deviation for the project.

For every task, the latest start and earliest completion times are specified as follows: LSij = LCj − Dij ECij = ESi + Dij TFij = LCj − ESi − Dij = LCj − ECij = LSij − ESi FFij = ESj − ESi − Dij ESj − ESi = LCj − LCi = Dij Where: Dij = Activity duration LSij = Latest Start time LCij = Latest completion time ECij = Earliest completion time TFij = Total float FFij = Free float

*Non -critical activity Table 5: Summary

Table 5 provides a handy summary of the critical path estimates along with the floats for the non-critical activities. The network computations are the source of columns (1), (2), (3), (4), (5), (6), (7) and (8). The algorithms above can be used to determine the remaining data. A representative summary of the critical path calculations is shown in Table 4. Observe that the total float of a crucial activity (and only a critical activity) must be zero. When the total float is zero, the free float likewise needs to be zero. However, the opposite is not true in that a non-critical activity can have zero free float [14].