PMGT3623 Scheduling Lecture Notes PDF

PMGT3623 Scheduling Week 04: Probabilistic Approach to Project Network Diagram – Part I Dr Shahadat Uddin The University of Sydney Page 1 Acknowledgement of Country I would like to acknowledge the Traditional Owners of Australia and recognise their continuing connection...

PMGT3623 Scheduling Week 04: Probabilistic Approach to Project Network Diagram – Part I Dr Shahadat Uddin The University of Sydney Page 1 Acknowledgement of Country I would like to acknowledge the Traditional Owners of Australia and recognise their continuing connection to land, water and culture. I am currently on the land of the Gadigal people of the Eora Nation and pay my respects to their Elders, past, present and emerging. PMGT3623 Scheduling 2 PMGT3623 Overview Week Topic Week 01 Introduction to Scheduling, Course Resources and Assessment Components Week 02 Define and Sequence Project Tasks Week 03 Project Network Diagram (Discrete Approach) Week 04 Probabilistic Approach to Project Network Diagram – Part I Week 05 Probabilistic Approach to Project Network Diagram – Part II Week 06 Confidence Analysis of Project Network Diagram Week 07 Knowledge Test Week 08 Implementation of Project Network Diagram using Microsoft Project Week 09 Simple Task Allocation Approach Mid-Semester Break Week 10 Complex Task Allocation Approach Week 11 Progress Reporting and Earned Value Analysis Week 12 Group Assignment Presentation Week 13 Review PMGT3623: Scheduling 3 PMGT3623 Assessments No Assessment Name Weight Due date Comment 1 Weekly Participation 10% W2-W6; W9-W10 Best 6 (out of 7) 2 Knowledge Test 20% W7 3 Group Assignment Presentation (Part A) 10% W12 and W13 4 Group Assignment Report Submission (Part B) 20% Friday of W13 By 11:59 pm 5 Final Exam 40% Exam Week PMGT3623 Scheduling 4 Week 04: Probabilistic Approach to Project Network Diagram – Part I Topics Covered - Why do we need the Probabilistic Approach to Project Network Diagram? - Probability and Probability Distribution o Expected Value o Beta Distribution - Program Evaluation and Review Technique (PERT) 5 Why do we need the probabilistic approach? From the last week 4 7 7 9 9 12 C (3) F (2) G (3) 0 4 4 7 7 9 9 12 A (4) 6 9 12 0 4 H (3) 5 6 9 12 Start End D (1) 6 11 6 7 I (5) 0 5 7 12 B (5) 5 9 9 11 1 6 E (4) J (2) 6 10 10 12 ❖ We considered a deterministic approach for task duration in the last week. ❖ Such simple approximation could be okay for simple projects. However, for complex project we need to follow a probabilistic approach for task duration allocation. 6 Why do we need the probabilistic approach? (cont.…) How confident shall we be regarding our earlier critical path results (from the last week) if we knew the estimates were only probable (not certain)? 4 7 7 9 9 12 C (3), 89% F (2), 99% G (3), 95% 0 4 4 7 7 9 9 12 A (4), 95% 6 9 12 0 4 H (3), 85% 5 6 9 12 Start End D (1), 79% 6 11 6 7 I (5), 99% 0 5 7 12 B (5), 98% 5 9 9 11 1 6 E (4), 88% J (2), 87% 6 10 10 12 To answer the above question, project managers need to know probability, probability distribution and their principles? 7 Probability and Probability Theory Project managers must make probability-guided decisions in many project scheduling contexts. Here are a few of such contexts. o What is the probability of completing the project on time? o What is the probability that a specific task will finish on time? o What is the expected project duration? o What is the probability of completing the project within a certain number of days? o What is the probability of delay due to the identified risks? o What is the variance of the project completion time? o What is the probability distribution of project completion times? o How do changes in task durations affect the project completion probability? o And many more questions like these…. Understanding probability would help in making informed decisions in such contexts (this week and the next week) 8 Probability and Probability Theory (cont.…) Event 1: Toss a coin Possible outcomes = {head, tail} Set P (head) = 0.5 P (tail) = 0.5 P(head or tail) = 1 P (not head and not tail) = 0 P (head) = 1 – P (tail) Event 2: Roll a regular dice Possible outcomes =? Possible outcomes = {1, 2, 3, 4, 5, 6} Sample space P (1) = P (2) = P(3) = P(4) = P(5) = P(6) = ? 𝟏 P (1) = P (2) = P(3) = P(4) = P(5) = P(6) = 𝟔 P(2 or 3) = P(2) U P(3) = ? 𝟏 𝟏 𝟐 P(2 or 3) = P(2) U P(3) = + = Addition rule 𝟔 𝟔 𝟔 𝟏 P (1) = 𝟔 𝟓 Or, P(1) = 1 – 𝟔 Or, P(1) = 1 - P (2 or 3 or 4 or 5 or 6) ഥ) Thus, 𝐏 𝐀 = 𝟏 − 𝐏(𝑨 Complementary rule 9 Probability and Probability Theory (cont.…) Set and Probability Set: A set is a collection of well-defined and distinct objects. For example, if A is a set of numbers from 1 to 10, then A={1,2,3,4,5,6,7,8,9,10} Subsets: If every member of set A is also a member of set B, then A is said to be a subset of B, written A ⊆ B. For example, {1, 3} ⊆ {1, 2, 3, 4}. Basic Set Operation Unions: Two sets can be "added" together (but not duplicate), denoted by ‘∪’ Examples: {1, 2} ∪ {red, white} ={1, 2, red, white}. Intersections: A new set can also be constructed by determining which member(s) two sets have "in common“, denoted by ‘∩’. Examples: {1, 2} ∩ {red, white} = ∅; and {1, 2, green} ∩ {red, green} = {green}. 10 Probability and Probability Theory (cont.…) Set and Probability (cont.…) Given that, Z = {1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4}, B = {2,5,6,7} and C = {4,8,9,10}. Find out A∩B = ? A∩B = {2} A∩C = ? A∩C = {4} B∩C = ? B∩C = {ø} AUB = ? AUB = {1,2,3,4,5,6,7} AUC = ? AUC = {1,2,3,4,8,9,10} BUC = ? BUC = {2,4,5,6,7,8,9,10} Compare to Z: Compare to Z: P(A) = ? P(A) = 4/10 P(B) = ? P(B) = 4/10 P(C) = ? P(C) = 4/10 P(AUB) = ? P(AUB) = 7/10 P(AUC) = ? P(AUC) = 7/10 P(BUC) = ? P(BUC) = 8/10 P(A∩B) = ? P(A∩B) = 1/10 P(A∩C) = ? P(A∩C) = 1/10 P(B∩C)= ? P(B∩C)= 0/10 11 Probability and Probability Theory (cont.…) Formal Definition The probability of an event “A” is the value that ranges between “0” and “1”. It is denoted as P(A) and defined as 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑃(𝐴) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 The sum of the probabilities of “n” mutually exclusive and collectively exhaustive events Ai equals “1”. That is n A  p( A ) = 1 i =1 i B C The three events of this circle are ❖ mutually exclusive (as they do not overlap), and ❖ collectively exhaustive as they all three form the circle Examples: dice roll and coin toss 12 Probability and Probability Theory (cont.…) Probability example Experiment: Rolling 2 regular die [dice] and summing 2 numbers on top (each die labelled with 1, 2, 3 …..6). Sample Space: S = {2, 3, …, 12} Dice2 outcome Probability Examples: Dice1 outcome P(2) = ? 1/36 1 2 3 4 5 6 1 2 3 4 5 6 7 P(7) = ? 6/36 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 P(10) = ? 3/36 5 6 7 8 9 10 11 6 7 8 9 10 11 12 13 Probability and Probability Theory (cont.…) Probability Theory: Complementary Rule 𝑃 𝐴ҧ = 1 − 𝑃(𝐴) Example: A single dice is rolled once. (i) What is the probability of turning up of “1 or 2 or 3 or 4” (say, this is event A) in this roll? (ii) What is the probability of turning up of “5 or 6” in this roll (say this is event B)? Solution: P (A)=4/6 P(B) = 2/6 Sample space, S = { 1, 2, 3, 4, 5, 6} It is clear that B is the complement of A since B = S – A Thus, P (complement of A) = 1 – P (A) 2 4 That is, = 1 − 6 6 14 Probability and Probability Theory (cont.…) Probability Theory: The multiplication rule for independent events P( A B) = P( A).P( B) Example: A single dice is rolled twice. What is the probability of turning up “not 1” in the first roll and “2” in the second roll? Solution: 5 𝑃 𝑛𝑜𝑡 1 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑙𝑙 = 6 1 𝑃 2 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑙𝑙 = 6 𝑃 "𝑛𝑜𝑡 1 𝑖𝑛 1𝑠𝑡 𝑟𝑜𝑙𝑙" 𝑎𝑛𝑑 "2 𝑖𝑛 2𝑛𝑑 𝑟𝑜𝑙𝑙" = 𝑃 𝑛𝑜𝑡 1 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑙𝑒 × 𝑃 2 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑟𝑜𝑙𝑙 5 1 = × 6 6 5 = 36 15 1. Exercise on ‘Multiplication Rule’ You are managing a project that consists of three sequential tasks, Task A, Task B and Task C. The completion of the project on time depends on the successful and timely completion of these three tasks. You have the following probabilities: Task P (On-time completion) A 0.80 B 0.90 C 0.80 Assuming the completion of Task A, Task B and Task C are independent events. Calculate the probability that the project will be completed on time. Solution: 𝑃 𝑇𝑖𝑚𝑒𝑙𝑦 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑜𝑛 = 𝑃 𝐴 𝑎𝑛𝑑 𝐵 𝑎𝑛𝑑 𝐶 = 0.80 × 0.90 × 0.80 = 0.576 16 Probability and Probability Theory (cont.…) Expected Value ❖ The expected value E(x) or the expected monetary value is calculated by multiplying each “ n “ outcome by its probability and then adding the results. ❖ The following equation can represent it: n E ( x) =  x P( x ) i =1 i i ❖ The sum of all P-values must be 1 ❖ The expected value approach is considered appropriate for rational decision- makers when the probabilities of all outcomes can be assessed. 17 Probability and Probability Theory (cont.…) Expected Value Example: Application in Scheduling Being the project manager, you have been given a job of allocating the during for Task A. You then analyse the available historical data (similar to Task A). From this analysis, you came to know that there is a 70% chance that the project team could complete this task in 20 days, 20% chance in 15 days and 10% chance in 12 days. Based on this information, what will be the expected duration for this task? Solution: First, you need to check the sum of p-values = 0.70 + 0.20 + 0.10 = 1.00 (okay) 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 = 0.70 × 20 + 0.20 × 15 + 0.10 × 12 = 14 + 3 + 1.2 = 18.2 𝑑𝑎𝑦𝑠 18 Probability and Probability Theory (cont.…) Expected value - Further exercise What is the expected value if you roll (a) a single dice; and (b) four dices? n E ( x) =  x P( x ) i =1 i i E(Rolling one dice) = 1 ∗ 𝑝(1) + 2 ∗ 𝑝(2) + 3 ∗ 𝑝(3) + 4 ∗ 𝑝(4) + 5 ∗ 𝑝(5) + 6 ∗ 𝑝(6) = 1 ∗ (1/6) + 2 ∗ (1/6) + 3 ∗ (1/6) + 4 ∗ (1/6) + 5 ∗ (1/6) + 6 ∗ (1/6) = (1/6)(1 + 2 + 3 + 4 + 5 + 6) = (1/6) ∗ 21 = 3.5 Thus, expected value for rolling four dices = 3.5 × 4 = 14 What is the expected value of rolling a dice if the probability of getting an odd number is twice the probability of getting an even number? 19 Probability and Probability Theory (cont.…) Expected Value and Mean/Average Average of 6 numbers: 1, 2, 3, 4, 5 and 6 1+2+3+4+5+6 1 1 1 1 1 1 Average = = 1× + 2× + 3× + 4× + 5× + 6× = 3.5 6 6 6 6 6 6 6 If you roll a dice, what will be the expected value? 1 Sample outcome = {1, 2, 3, 4, 5 and 6}, with each has equal probability of 6 1 1 1 1 1 1 Expected Value (EV) = 1 × + 2× + 3× + 4× + 5× + 6× = 3.5 6 6 6 6 6 6 Thus, Expected value and Average will be same only if all outcomes will have the same probability to appear/occur 20 Probability and Probability Theory (cont.…) Expected Value and Risk (Project-level decision-making) Scenario: Your company has $2M capital. Which investment option will you choose from the following? Option 1: Project A o 90% probability of successful completion with a profit of $400,000 o 10% chance of failure. In that case, your company will lose $200,000 Option 2: Project B o 70% probability of successful completion with a profit of $900,000 o 30% chance of failure. In that case, your company will lose $500,000 Option 3: Bank Deposit o Deposit money in a bank with a guaranteed return of $150,000 21 Scenario: Your company has $2M capital. Which investment option will you choose from the following? Option 1: Project A o 90% probability of successful completion with a profit of $400,000 o 10% chance of failure. In that case, your company will lose $200,000 Expected value =? 90%*$400,000 – 10%*$200,000 = $340,000 Option 2: Project B o 70% probability of successful completion with a profit of $900,000 o 30% chance of failure. In that case, your company will lose $500,000 Expected value =? 70%*$900,000 – 30%*$500,000 = $480,000 Option 3: Bank Deposit o Deposit money in a bank with a guaranteed return of $150,000 Expected value = $150,000 22 2. Exercise on Expected Value (generated by ChatGPT) Imagine you are a project manager responsible for a critical software development task that has three potential completion scenarios depending on various risk factors like technical challenges or team availability. Estimates and Probabilities: Scenario 1: Fast Completion - Duration: 10 days; Probability: 25% Scenario 2: Expected Completion - Duration: 15 days; Probability: 60% Scenario 3: Delayed Completion - Duration: 20 days; Probability: 10% Calculate the expected duration for this project using the Expected Value approach. Solution: It is not possible to calculate the expected duration using the EV approach since the sum of p-values does not equal 1. 23 Probability Distribution: Beta Distribution What is it? ❖ The beta distribution is a family of continuous probability distributions defined on the interval [0, 1], and parameterised by two positive shape parameters, denoted by α and β, that appear as exponents of the random variable and control the shape of the distribution. (Source: Wiki) ❖ The beta distribution has been applied to model the behavior of random variables limited to intervals of finite length in a wide variety of disciplines. ❖ It enables better management of projects by easing improved estimation of different project activities (PERT is used for cost and schedule estimation) 24 Probability Distribution: Beta Distribution (cont.…) Why is it important? ❖ It has a finite range from 0 to 1, making it useful for modeling phenomena that cannot be above or below given values. ❖ Estimation/ prediction using the beta distribution is more accurate compared to the similar estimation/ prediction using the normal distribution ❖ The distribution has two parameters:  and β, that determine its shape. When  and β are equal, the distribution is symmetric. ❖ Increasing the values of the parameters decreases the variance 25 Probability Distribution: Beta Distribution (cont.…) Right-skewed beta distribution Left-skewed beta distribution 26 Beta Distribution and PERT (Program Evaluation and Review Technique) ❖ PERT is based on beta distribution; also known as three-point estimation ❖ A network analysis technique used when task durations have a high degree of uncertainty 27 Beta Distribution and PERT (cont.…) (Program Evaluation and Review Technique) Approximations for the mean and the standard deviation of the completion time of a single activity are based on the Beta distribution. 𝑂 + 4𝑀𝐿 + 𝑃 𝜇 = 𝑀𝑒𝑎𝑛 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 = 6 𝑃−𝑂 𝜎 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 6 Where, O: Optimistic estimation, ML: Most-likely estimation, and P: Pessimistic estimation Malcolm, D. G., et. al. Application of a technique for research and development program evaluation. Operations research 7, 646-669 (1959). 28 Beta Distribution and PERT (cont.…) (Program Evaluation and Review Technique) An example For a task (say updating a computer software to comply with the new system) of a project, the following information is given. Find the expected completion time and standard deviation of this task using PERT. Optimistic (time) = 2 days, Pessimistic (time) = 6 days, and Most likely (time) = 4 days) Solution: 2 + 4(4) + 6 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 = = 4 𝑑𝑎𝑦𝑠 6 6−2 2 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = = 𝑑𝑎𝑦𝑠 6 3 29 Expected Value versus PERT Expected Value ❖ Expected Value provides a simple average outcome based on probabilities and is used in a wide range of fields for quick estimates. ❖ It is a general statistical concept used in a wide range of contexts including in project management (scheduling) ❖ There is no limitations of the total number of parameters 𝐸𝑉 = ෍(𝑥𝑖 × 𝑝𝑖 ) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡, ෍ 𝑝𝑖 = 1 PERT ❖ PERT uses three estimates to account for uncertainty in task durations and helps in detailed project planning and scheduling. ❖ It is more complex than the EV and is based on the beta distribution ❖ Total number of parameters is three 𝑂 + 4𝑀𝐿 + 𝑃 𝜇= 6 30 3. Exercise on PERT You are managing a project that consists of three sequential tasks, Task A, Task B and Task C. The completion of the project on time depends on the successful and timely completion of these three tasks. You have the following information (in days): Task O ML P A 3 4 5 B 6 8 10 C 9 12 15 Calculate the expected completion time of this project. Solution: 3 + 4(4) + 5 𝐸 (𝐴) = = 4 𝑑𝑎𝑦𝑠 6 6 + 4(8) + 10 𝐸 (𝐵) = = 8 𝑑𝑎𝑦𝑠 6 9 + 4(12) + 15 𝐸 (𝐶) = = 12 𝑑𝑎𝑦𝑠 6 𝐸𝑥𝑝𝑙𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑗𝑒𝑐𝑡 = 4 + 8 + 12 = 24 𝑑𝑎𝑦𝑠 31 Review Questions a) What is a probability distribution? How can you use it for project scheduling? b) What are the advantages of using PERT for task estimation? c) What are the differences between PERT and Expected Value regarding their applications for task duration estimation? d) What is probability? Why does a project manager need to know about this concept? e) What are the importance of beta distribution? f) What is the expected value of rolling a dice if the probability of getting an odd number is twice the probability of getting an even number? 32