PMGT3623 Scheduling Week 5 Lecture Plan PDF

PMGT3623 Scheduling Week 05: Probabilistic Approach to Project Network Diagram – Part II 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 05: Probabilistic Approach to Project Network Diagram – Part II 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 05: Probabilistic Approach to Project Network Diagram – Part II Topics Covered - Uniform Distribution - Normal Distribution: Z-score and Z-table - Project Variance and Standard Deviation 5 Quick Overview (Last two weeks) ❖ A deterministic approach for task duration is okay for simple projects (week 3). ❖ In the last week, we learned the probabilistic approach for task duration allocation since, in most projects, a deterministic approach would lead to biased estimation. ❖ In that same week, we learned the basic approach and concept of probability and probability theory. ❖ We learned two probabilistic approaches for estimation: Expected Value and PERT (Program Evaluation and Review Technique). ❖ PERT is more complex than the EV and is based on the Beta distribution. ❖ This week, we will emphasise Uniform distribution, Normal distribution and Z- score concept. 6 Uniform Distribution ❖ Possible outcomes are equally likely: A uniform distribution is a probability distribution in which all outcomes are equally likely. ❖ This means that any value within the specified range has an equal chance of occurring. ❖ A simple example of the discrete uniform distribution is throwing a fair dice. The possible values are 1, 2, 3, 4, 5, and 6, and each time the dice is thrown, the probability of a given score is 1/6. ❖ There are two main types of uniform distributions: discrete and Figure: Uniform Distribution continuous. In project scheduling, we typically refer to the continuous uniform distribution. 7 Uniform Distribution (cont.…) ❖ If the value of a variable is uniformly distributed between a minimum value (a) and a maximum value (b), then the expected value of that distribution is - 𝑎+𝑏 𝐸= 2 ❖ The variance of a uniform distribution is given by: (𝑏 − 𝑎)2 𝑉𝑎𝑟 = 12 ❖ The probability of that variable taking a value less than a specific value is given by : (𝑥 − 𝑎) 𝑃 𝑋≤𝑥 = (𝑏 − 𝑎) 𝑤ℎ𝑒𝑟𝑒, 𝑎 ≤ 𝑥 ≤ 𝑏 8 Uniform Distribution (cont.…) Only for demonstration (in case you are curious about how to prove the formulas in the previous slide. These derivations (this slide and the next two) are not for assessment. ❖ The probability density function (pdf) of the random variable X (follows a uniform distribution) is given by (from the second last slide): 1 𝑓 𝑥 = 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏 … … … … ….. (𝑖) 𝑏−𝑎 𝑏 𝑇ℎ𝑢𝑠, 𝐸 𝑋 = න 𝑥. 𝑓 𝑥 𝑑𝑥 𝑎 9 Uniform Distribution (cont.…) ❖ The probability density function (pdf) of the random variable X (follows a uniform distribution) is given by: 1 𝑓 𝑥 = 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏 … … … … ….. (𝑖) 𝑏−𝑎 𝑏 𝑇ℎ𝑢𝑠, 𝐸 𝑋2 = න 𝑥 2. 𝑓 𝑥 𝑑𝑥 𝑎 10 Uniform Distribution (cont.…) ❖ The probability density function (pdf) of the random variable X (follows a uniform distribution) is given by: 1 𝑓 𝑥 = 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏 … … … … ….. (𝑖) 𝑏−𝑎 𝑇ℎ𝑢𝑠, 𝑉𝑎𝑟 𝑋 = 𝐸 𝑋 2 − 𝐸 𝑋 2 11 Uniform Distribution (cont.…) An example of its application in Scheduling You are managing a project with a single task (Task A). The duration of Task A is uniformly distributed between a minimum duration of 10 days and a maximum duration of 20 days. You need to calculate – (a) The expected duration, (b) The variance of the task duration, and (c) Determine the probability that the task will be completed within 15 days. (d) Determine the probability that the task will be completed within five days. (e) Determine the probability that the task will be completed within 30 days. Solution: 10 + 20 𝑎 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑑𝑢𝑟𝑎𝑡𝑖𝑜𝑛 = = 15 𝑑𝑎𝑦𝑠 2 (20 − 10)2 25 𝑏 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = = 𝑑𝑎𝑦𝑠 2 12 3 (15 − 10) 𝑐 𝑃 𝑋 ≤ 15 = = 0.50 (20 − 10) 𝑑 0 (𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒) 𝑒 1 (𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 ℎ𝑖𝑔ℎ𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒) 12 Variance and Standard Deviation For PERT (from the last week) 𝑃−𝑂 𝜎 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 6 2 𝑃−𝑂 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 6 For Uniform Distribution (this week) (𝑏 − 𝑎)2 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 12 (𝑏 − 𝑎) 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 12 So far, we have not seen the applications of these measures in project scheduling. 13  Normal Distribution Normal distribution ❖ Often called “bell curve”. 100 data points ❖ All values are plotted in a symmetrical fashion. ❖ Most results clustered around the sample’s mean. ❖ The frequency of other values reduces 1,000 data points symmetrically toward either extreme. Standard Deviation By how much the members of a group differ from the mean value for the group. 10,000 data points Further from the “mean”, the higher the deviation is. 14 Normal Distribution (cont.…) ❖ Most of the outcomes are close to the mean value ❖ Symmetrical and extend from - ∞ to + ∞ ❖ It is defined by two parameters: the mean (μ) and the standard deviation (σ). ❖ The mean determines the centre of the distribution, while the standard deviation determines the spread or width of the distribution. Mean: μ (mu) Standard deviation: σ (sigma) 68.2% numbers will be within 1 standard deviation distance of the mean 95.4% numbers will be within 2 standard deviation distance of the mean 15 Normal Distribution (cont.…) An example You are the project manager of a company. Consider a project with a task whose duration is normally distributed with a mean (μ) of 10 days and a standard deviation (σ) of 2 days. What will be your estimated duration for the task (i.e., for the project since the project consists of only this task) if (i) your company can tolerate up to 15.9% risk and (ii) your company cannot afford a risk over 2.3% 97.7% 84.1% 50% 50% (i) 10 + 2 = 12 days Since 100 - 15.9 = 84.1 (ii) 10 + 2*2 = 14 days Since 100 - 2.3 = 97.7 16 Normal Distribution (cont.…) An example You are the project manager of a company. Consider a project with a task whose duration is normally distributed with a mean (μ) of 10 days and a standard deviation (σ) of 2 days. What will be your estimated duration for the task (i.e., for the project since the project consists of only this task) if (i) your company can tolerate up to 15.9% risk and (ii) your company cannot afford a risk over 2.3% Can we find the estimated duration for the task for which your company can tolerate up to 30% risk by following the same process? Solution: Z-score? NO 17 Z-score: Universal Currency Which students got a better mark? o 16 out of 20 (student 1, S1) o 30 out of 40 (student 2, S2) If you consider the raw marks, then S2 got a higher mark than S2 (since 30>16). However, this is not a fair comparison. So, what can we do for a fair comparison? One way is to convert both marks into a per cent value. Then o S1: 80% (since 16 out of 20) o S2: 75% (since 30 out of 40) This is a fair and correct comparison and represents the actual comparative results. So, how to compare two numbers from two different normal distributions? 18 Z-score: Universal Currency (cont.…) Random Number Generator (RNG1): ❖ It generates random numbers that follow a normal distribution with µ = 50, Std = 15 and range = [1 to 100]. ❖ You give it a command to get a random number. Then RNG1 generates the number 80. Random Number Generator (RNG2): ❖ It generates random numbers that follow a normal distribution with µ = 130, Std = 20 and range = [60 to 200]. ❖ You give it a command to get a random number. Then RNG2 generates the number 90. Now, the question is: How do you compare these numbers (i.e., 80 and 90)? One way to answer this question is to visualise these numbers in the normal distribution curve 19  Z-score: Universal Currency (cont.…) 68.2% of data 80 90 68.2% of data 95.4% of data 95.4% of data 99.6% of the data 99.6% of the data 97.7% of the numbers generated by 2.3% of the numbers generated by RNG2 RNG1 are less than or equal to 80 are less than or equal to 90 RNG1: RNG2: µ = 50 and Std = 15 µ = 130 and Std = 20 range = [1 to 100]. range = [60 to 200]. Although 80