Project Management and Scheduling - Research Paper Example

August 17, Project Management Scheduling Outline The Critical Path (CPM) method and the Program Evaluation Technique (PERT) are two vital toolsof project management and scheduling. While CPM is a deterministic model used for scheduling of projects activities of known time durations, PERT is a probabilistic model generally used to schedule activities with uncertain time durations. a) Scheduling the constructing a new building for AXBC PLC. Activity Optimistic time (a) Most likely (m) Pessimistic (b) Predecessors A 1 2 4 B 3 6 12 C 1 2 3 A B D 2 4 5 A E 3 6 10 A F 2 5 7 G 1 3 5 B D H 1 3 4 D E F I 1 2 6 K J 2 5 6 G F K 1 2 3 H L 2 5 6 G H M 3 4 7 K L N 5 8 12 J k L O 4 8 10 D b) Expected Time duration (te), the Standard Deviation (σ) and the Variance (σ 2) Since the time duration for the associated activities in constructing a new building for AXBC PLC cannot be estimated very precisely, the degree of uncertainty in the duration of an activity is handled by using the probabilistic technique of Program Evaluation and Review Technique (PERT). The time is expressed in the form of three time estimates namely, the Optimistic Time (represented as a): The time by which an activity will be completed if everything went well Most Likely Time: (represented as m): The time which is taken most frequently by an activity Pessimistic Time: (represented as b):The time by which an activity will be completed even under adverse conditions The expected time for an activity represents the average time it would take if the activity is performed over and over again. It is known empirically that the probability density function of activity duration closely follows a Beta distribution, which defines the following relationship (Heldman & Baca, 267) Expected duration of an activity = {Optimistic + 4 * (Most likely) +Pessimistic time} / 6 te (A) = (1+4*2+4) / 6 = 13 / 6 = 2.5 Activity Optimistic(a) Most Likely (m) Pessimistic(b) Expected Time duration (te) A 1 2 4 2.500 B 3 6 12 6.500 C 1 2 3 2.000 D 2 4 5 3.833 E 3 6 10 6.166 F 2 5 7 4.833 G 1 3 5 3.000 H 1 3 4 2.833 I 1 2 6 2.500 J 2 5 6 4.666 K 1 2 3 2.000 L 2 5 6 4.666 M 3 4 7 4.333 N 5 8 12 8.166 O 4 8 10 7.666 Standard Deviation is used for calculating the variability associated with the high degree of uncertainty in estimated time durations. The standard deviation is represented by sigma symbol ‘σ’ and is calculated as σ = (Pessimistic– Optimistic) / 6 σ = (b – a) / 6 σ (A) = (4 – 1) / 6 = 3 / 6 = 0.5 Variance is defined as the square of Standard Deviation and is denoted by the symbol ‘σ2’. For Example, Variance = σ2 (A) = [Standard Deviation of A] 2 = (0.5)2 = 0.25 The Standard Deviation (σ) and the Variance (σ2) for each activity is calculated Activity a m B (te) Standard Deviation (σ) Variance (σ2) A 1 2 4 2.500 0.500 0.250 B 3 6 12 6.500 1.500 2.25 C 1 2 3 2.000 0.333 0.110 D 2 4 5 3.833 0.500 0.250 E 3 6 10 6.166 1.116 1.245 F 2 5 7 4.833 0.833 0.693 G 1 3 5 3.000 0.666 0.443 H 1 3 4 2.833 0.500 0.250 I 1 2 6 2.500 0.833 0.693 J 2 5 6 4.666 0.666 0.443 K 1 2 3 2.000 0.333 0.110 L 2 5 6 4.666 0.666 0.443 M 3 4 7 4.333 0.666 0.443 N 5 8 12 8.166 1.166 1.359 O 4 8 10 7.666 1.000 1.000 c) Early Start, Early Finish, Latest Start, Latest Finish and Slack The time analysis elements such as the forward pass (Early Start time ES and Early Finish time EF) and the Backward Pass (Late Start time LS and Late Finish time LF) are used to find the Critical Path. These times are calculated using the Expected Time (te) for the respective activity relative to the zero date (date when the project clock starts ticking) of the project. 1. Forward Pass The Forward Pass provides the details about the earliest time by which an activity can be started (ES) and the earliest time by which it can be finished (EF) (Chase, 82). ES (A) = ES (B) ES (F) = 0 since these activities have no predecessors. ES of activities with predecessors = max (EF of the predecessor activities) EF of an activity = ES of the activity +Expected Time Duration of the respective activity (te) Example, ES (C) = max {EF (A), EF (B)} ES (C) = max (2.500, 6.500) = 6.500 EF (C) = 6.500 + 2 = 8.500 The ES and EF times for all activities are calculated. ES (A) = 0 EF (A) = ES (A) + te (A) = 0 +2.500 = 2.500 ES (B) = 0 EF (B) = 0 +6.5 = 6.500 ES (C) = max {EF (A), EF (B)} = 6.500 EF (C) = 6.500 + 2 = 8.500 ES (D) = max {EF (A)} = 2.500 EF (D) = 2.500 +3.833 = 6.333 ES (E) = max {EF (A)} = 2.500 EF (E) = 2.500 +6.166 = 8.666 ES (F) = 0 EF (F) = 0 +4.833 = 4.833 ES (G) = max {EF (B), EF (D)} = 6.500 EF (G) = 6.500 +3 = 9.500 ES (H) = max {EF (D), EF (E), EF (F)} = 8.666 EF (H) = 8.666 +2.833 = 11.499 ES (I) = max {EF (K)} = 13.499 EF (I) = 13.499 +2.500 = 15.999 ES (J) = max {EF (F), EF (G)} = 9.500 EF (J) = 9.500 +4.666 = 14.166 ES (K) = max {EF (H)} = 11.499 EF (K) = 11.499 +2 = 13.499 ES (L) = max {EF (G), EF (H)} = 11.499 EF (L) = 11.499 +4.666 = 16.165 ES (M) = max {EF (L), EF (K)} = 16.165 EF (M) = 16.165 + 4.333 = 20.498 ES (N) = max {EF (J), EF (K), EF (L)} = 16.165 EF (N) = 16.165 +8.166 = 24.331 ES (O) = max {EF (D)} = 6.333 EF (O) = 6.333 +1 = 7.333 ES (FINISH) = max { EF (C), EF (O), EF (I), EF (M), EF (N)} = 24.331. Finish is a milestone so it has duration of 0. Therefore, EF (FINISH) = 24.331 + 0 = 24.331 2. Backward Pass The Backward Pass Late Start (LS) and Late Finish (LF) values are calculated considering that the earliest completion time of the project for the last activity and then working backwards towards the predecessors. For all the last activities, the LF will be equal to the respective EF value. Therefore, values of LF for the activities O, N, M, I and C are equal to the respective values of EF. LF (FINISH) = EF (FINISH) = 24.331 LS (FINISH) = 24.331 – 0 =24.331 LF (O) = EF (O) = 7.333 LS (O) = 7.333 – 1 = 6.333 LF (N) = EF (N) = 24.331 LS (N) = 24.331 – 8.166 = 16.165 LF (M) = EF (M) = 20.498 LS (M) = 20.498 – 4.333 = 16.165 LF (L) = min {LS (M), LS (N)} = 16.165 LS (L) = 16.165 – 4.666 = 11.449 LF (K) = min {LS (I), LS (M), LS (N)} = 13.499 LS (K) = 13.499 – 2.000 = 11.499 LF (J) = min {LS (N)} = 16.165 LS (J) = 16.165 – 4.666 = 11.499 LF (I) = EF (I) 15.999 LS (I) = 15.999 – 2.500= 13.499 LF (H) = min {LS (L), LS (K)} = 11.499 LS (H) = 11.499 – 2.833 = = 8.666 LF (G) = min {LS (L), LS (J)} = 11.499 LS (G) = 11.499 – 3.000 = 8.499 LF (F) = min {LS (H), LS (J)} = 8.666 LS (F) = 8.666 – 4.833 = 3.833 LF (E) =min {LS (H)} = 8.666 LS (E) = 8.666 – 6.166 = 2.500 LF (D) = min {LS (H), LS (O), LS (G)} = 6.333 LS (D) = 6.333 – 3.833 = 2.500 LF (C) = EF (C) = 8.500 LS (C) = 8.500 – 2.000 = 6.500 LF (B) =min {LS (C), LS (G)} = 6.500 LS (B) = 6.500 – 6.500 = 0 LF (A) =min {LS (C), LS (D), LS (E)} = 2.500 LS (A) = 2.500 – 2.500 = 0 LF (START) =min {LS (A), LS (B), LS (F)} = 0 LS (A) = 0 The Critical Path can be determined by finding the Total Slack for each activity. Total Slack for any activity is the maximum time by which the start of an activity can be delayed without affecting the critical path. Total Slack = LF – EF or LS – ES Total Slack (A) = LF (A) – EF (A) or LS (A) – ES (A) = 2.500 – 2.500 = 0 The value for each activity is tabulated below. Activity (te) Early Start (ES) Early Finish (EF) Late Finish (LF) Late Start (LS) Slack Critical activity A 2.500 0 2.500 2.500 0 0 YES B 6.500 0 6.500 6.500 0 0 YES C 2.000 6.500 8.500 8.500 6.500 0 YES D 3.833 2.500 6.333 6.333 2.500 0 YES E 6.166 2.500 8.666 8.666 2.500 0 YES F 4.833 0 4.833 8.666 3.833 3.833 NO G 3.000 6.500 9.500 11.499 8.499 1.999 NO H 2.833 8.666 11.499 11.499 8.666 0 YES I 2.500 13.499 15.999 15.999 13.499 0 YES J 4.666 9.500 14.166 16.165 11.499 1.999 NO K 2.000 11.499 13.499 13.499 11.499 0 YES L 4.666 11.499 16.165 16.165 11.449 0 YES M 4.333 16.165 20.498 20.498 16.165 0 YES N 8.166 16.165 24.331 24.331 16.165 0 YES O 7.666 6.333 7.333 7.333 6.333 0 YES d) AON (Activity-On-Node) Diagram are a type of precedence network diagrams that use circles (nodes) representing activities and the arrows representing the precedence relationship between the nodes. e) Paths: Critical and Non-Critical The path with the longest Expected time duration is the critical path. Activities on the critical path have slack as 0, since these activities cannot be delayed at all. Each path from the Start to the Finish node is defined along with the respective time duration. The Expected Completion Time (Et) for each path is equal to the expected duration of the constituent activities. For Example, A-E-H-K-I is calculated as: Et (A-E-H-K-I) = [te (A) + te (E) + te (H) + te (K) + te (I)] = = 2.500 + 6.166 +2.833 + 2.000 + 2.500 = 15.999 All the paths (both critical and non critical paths) starting from the start node going till termination and the Finish node are enlisted below. A-E-H-K-I = 15.999 A-D-H-K-I = 13.666 A-C = 4.5 A-D-O = 13.999 A-D-G-L-M = 18.332 A-D-G-L-N = 22.165 A-D-G-J-N = 22.165 A-E-H-L-N = 24.331 A-E-H-L-M = 20.498 A-D-H-L-M = 18.165 A-D-H-L-N = 21.998 B-C = 8.5 B-G-L-M = 18.499 B-G-J-N = 22.332 B-G-L-N = 22.332 F-H-K-I = 12.166 F-J-N = 17.665 F-H-L-M = 16.665 F-H-L-N = 20.498 F-H-K-M = 13.999 F-H-K-N = 17.832 Therefore, the longest path is A-E-H-L-N with the Expected duration of 24.331 days. This longest path is also the Critical path since all the activities A, E, H, L, and N have slack as 0. Variance of the duration of the critical path A-E-H-L-N (σ2) = [σ2 (A) + σ2 (E) + σ2 (H) + σ2 (L) + σ2 (N)]] = 0.250 + 1.245 +0.250 + 0.443 + 1.359 Variance (σ2)= 3.547 days2 Standard Deviation of the the critical path A-E-H-L-N (σ) = (3.547)1/2 = 1.88 f) Probability of completing this project in 22 days Pr [days Read More