PMGT3623 Scheduling Lecture Notes PDF

PMGT3623 Scheduling Week 06: Confidence Analysis of Project Network Diagram 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...

PMGT3623 Scheduling Week 06: Confidence Analysis of Project Network Diagram 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 06: Confidence Analysis of Project Network Diagram Topics Covered - Confidence calculation for Project Scheduling - Probabilistic estimation for Project Scheduling - Task Crashing 5 Overview of the Content of this Week ❖ We will explore two case studies in this week. ❖ All materials for this week are based on weeks 2-5. ❖ These two case studies will cover – o Network Diagram o Probability and Probability Distribution o Normal Distribution o Z-score o Beta Distribution and PERT o Task Crashing 6 Probabilistic PERT and Confidence Analysis: KLONE Computers We covered this case study in brief in Week 2 Activity Description A Prototype model design B Purchase of materials Manufacturing C Manufacture of prototype model activities D Revision of design E Initial production run F Staff training Training activities G Staff input on prototype models H Sales training Advertising activities I Pre-production advertising campaign J Post-redesign advertising campaign 7 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) Precedence Relationships Duration information for each activity (days) Activity Predecessor Activity Optimistic Most Likely Pessimistic A 76 86 120 A None B 12 15 18 B A C 4 5 6 C B D 15 18 33 D G E 18 21 24 F 16 26 30 E D G 10 13 22 F A H 24 28 32 G C,F I 22 27 50 J 38 43 60 H D I A J D,I 𝑂𝑝𝑡𝑖𝑚𝑖𝑠𝑡𝑖𝑐 + 4 × 𝑀𝑜𝑠𝑡 𝐿𝑖𝑘𝑒𝑙𝑦 + 𝑃𝑒𝑠𝑠𝑖𝑚𝑖𝑠𝑡𝑖𝑐 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = 6 2 𝑃𝑒𝑠𝑠𝑖𝑚𝑖𝑠𝑡𝑖𝑐 − 𝑂𝑝𝑡𝑚𝑖𝑠𝑡𝑖𝑐 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 6 8 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) Now we can calculate the expected value and variance for each activity (PERT): 2 120 − 76 76 + 4 × 86 + 120 = 53.78 = 90 6 6 Activity Optimistic Most Likely Pessimistic µ (E) Var (s2) A 76 86 120 90 53.78 B 12 15 18 15 1.00 C 4 5 6 5 0.11 D 15 18 33 ? 20 ? 9.00 E 18 21 24 21 1.00 F 16 26 30 25 5.44 G 10 13 22 14 4.00 H 24 28 32 28 1.78 I 22 27 50 30 21.78 J 38 43 60 45 13.44 9 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) By applying the CPM techniques (learned in week1), we can find out the critical path and slack time for each activity 90,105 105,110 149,170 95,110 B C 110,115 173,194 E B C E 15 5 21 0,90 90,115 115,129 129,149 149,177 0,90 90, 115 115,129 129,149 166,194 A F G D H A F G D H 90 25 14 20 28 149,194 90,120 149,194 119,149 I J I J 45 30 Critical path: A → F → G → D → J 10 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) Finding mean and variance for the critical path ….. Activity µ (E) Var (s2) A 90 53.78 Facts B 15 1.00 C 5 0.11 ❖ A critical path can be determined using the activities' mean D 20 9.00 completion times. E 21 1.00 ❖ The project mean completion time is determined solely by the F 25 5.44 completion time of the activities on the critical path. G 14 4.00 H 28 1.78 I 30 21.78 ❖ The critical path is A → F → G → D → J. J 45 13.44 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑃𝑟𝑜𝑒𝑗𝑐𝑡 𝐶𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑜𝑛 𝑇𝑖𝑚𝑒 = 𝐸𝐴 + 𝐸𝐹 + 𝐸𝐺 + 𝐸𝐷 + 𝐸𝐽 𝑃𝑟𝑜𝑗𝑒𝑐𝑡 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝑉𝑎𝑟 𝐴 + 𝑉𝑎𝑟 𝐹 + 𝑉𝑎𝑟 𝐺 + 𝑉𝑎𝑟 𝐷 + 𝑉𝑎𝑟 𝐽 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑟𝑜𝑗𝑒𝑐𝑡 = 85.66 = 9.255 Management at KLONE is interested in information regarding the probabilistic nature of the completion time of the project. 11 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) What is the probability of the completion in 194 days? P(X ≤ 194) Mean 194 Variance 85.66 Std 9.255 𝑋−𝜇 𝑍= 𝜎 12 Z-table Z=0 13 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) What is the probability of the completion in 194 days? Mean 194 P(X ≤ 194) Variance 85.66 Std 9.255 𝑋−𝜇 𝑍= 𝜎 194 14 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) An interval in which the project team is 95% sure that the completion time lies. o Consider the middle 95% of the normal distribution curve o Take out 2.5% (or 0.025) from both sides o The boundary is (1 - 0.025) = 0.9750 (on both sides) Mean 194 Variance 85.66 Std 9.255 𝑋−𝜇 𝑍= 𝜎 0.025 0.025.95 m 15 Z-table 95% 16 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) An interval in which the project team is 95% (i.e., µ ± Z0.025 × s ) confirmed that the completion time lies. Mean 194 Variance 85.66 Std 9.255 X−μ Z= σ 𝑋−𝜇 𝑍=.95 𝜎 m Thus, the probability that the completion time lies in the interval [176, 212] is 0.95. 17 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) What is the probability of the completion in 180 days? Mean 194 Variance 85.66 P(X ≤ 180) Std 9.255 𝑋−𝜇 𝑍= 𝜎 18 Z-table Z = -1.51 19 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) What is the probability of the completion in 180 days? Mean 194 Variance 85.66 P(X ≤ 180) Std 9.255 180−194 =P Z≤ 9.255 = P Z ≤ −1.51 = 1 − 𝑃 𝑍 ≤ 1.51 20 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) What is the probability that the completion time is longer than 210 days? P(X ≥ 210) Mean 194 Variance 85.66 Std 9.255 𝑋−𝜇 𝑍= 𝜎 21 Z-table Z = 1.73 22 Probabilistic PERT and Confidence Analysis: KLONE Computers (cont.…) What is the probability that the completion time is longer that 210 days? Mean 194 P(X ≥ 210) Variance 85.66 210−194 Std 9.255 =P Z≥ 9.255 = P Z ≥ 1.73 = 1 − 𝑃(𝑍 < 1.73) 0.4582 ? 0.0418 194 210 X 0 1.73 Z 23 1. Exercise on Confidence Analysis For the KLONE Computers example (consider the same network Mean 194 mean, variance and standard deviation), find out a completion time Variance 85.66 that has only a 1% chance to exceed. Std 9.255 Solution For p