CRITICAL PATH

Published on
Scene 1 (0s)

Project Scheduling :. CRITICAL PATH METHOD. Presented by: Sharymae Faith Cabuenas BS Mangement Accounting.

Scene 2 (24s)

Critical Path Method (CPM). is a statistical technique of project management that manages well defined activities of a project. It is an algorithm used for planning, scheduling, coordination and control of activities in a project. it is assumed that the activity duration is fixed and certain. CPM is used to compute the earliest and latest possible start time for each activity..

Scene 3 (1m 10s)

Calculation of the Critical Path. These are the following key elements into your schedule: All project tasks with start and end dates Task durations Task relationships or dependencies.

Scene 4 (2m 13s)

How does the Critical Path Method calculate the project’s finish date?.

Scene 5 (3m 7s)

Early Start (ES) is the earliest date a task can start Early Finish (EF) is the earliest date a task can be completed Late Start (LS) is the latest date a task can start without delaying the project end date Late Finish (LF) is the latest date a task can finish without delaying the project end date.

Scene 6 (4m 1s)

CALCULATION OF CRITICAL PATH. EARLIEST LATEST EARLIEST START START FINISH SLACK CRITICAL ACTIVITY (ES) (LS) (EF) (LS-ES) PATH? A 0 0 6 0 YES B 0 7 2 7 C 6 10 9 4 D 6 7 11 1 E 6 6 9 0 YES F 9 13 11 4 G 11 12 14 1 H 9 9 13 0 YES I 13 13 15 0 YES J 15 15 17 0 YES.

Scene 7 (5m 9s)

START. FINISH. FORWARD PASS. C 6 9 3 10 13. F 9 11 2 13 15.

Scene 8 (10m 9s)

VARIABILITY IN PROJECT COMPLETION TIME. Variation in activity can causes variation in the project completion time. Variation in non-critical activities ordinarily has no effect on the project completion time because of slack time associated with these activities . Variability leading to a longer than the expected total time for the critical activities will always extend the project completion time, and conversely, variability that results than the expected total time for the critical activities will reduce the project completion time, unless other activity become critical.

Scene 9 (10m 56s)

GIVEN EXAMPLE : PORTA-VAC PROJECT NETWORK. THERE ARE FOUR PATHS: PATH 1 = A-E-H-I-J PATH 2 =A-C-F-J PATH 3 = A-D-G-J PATH 4 = B-H-I-J Let the random variable T i - denote the total time to complete path ( i ) T i = to the sum of expected times of the activities along path (I ) Computation: E(T i ) = t A + t E + t H + t i + t j = 6+3+4+2+2 =17 weeks.

Scene 10 (11m 53s)

The variance of T i , is the sum of the variances of the activities along path i . for the path 1 (the critical path) the variance in completion is σ 2/1 = σ 2/ A + σ 2/ E + σ 2/ H + σ 2/ I + σ 2/ J = 1.78+0.11+0.69+0.03+0.11 = 2.72 weeks 2 where σ 2/ A , σ 2/ E , σ 2/ H , σ 2/ I , σ 2/ J , and of are the variance of the activities A, E, H, I and J. The formula of σ 2/1 is based on the assumption that the activity times are independent. If two or more activities are dependent, the formula provides only an approximately of the variance of the path completion time. The closer the activities are to being independent the better the approximation. Standard deviation is the square root of the variance, as σ for the path 1 completion time is: σ = √ σ 2/1 = √2.72 = 1. 65.

Scene 11 (13m 50s)

NORMAL DISTRIBUTION OF THE CRITICAL PATH COMPLETION TIME.

Scene 12 (14m 54s)

. Example: 20 weeks allotted time for the PORTA-VAC Project so, probability for path 1 for 20 weeks is T 1 ≤ 20. The z-score for the normal probability distribution at T 1 = 20 is: z 1 = 20-17 = 1.82 1.65 -- so, Using z= 1.82 and the normal distribution and the table of normal distribution; the probability of path 1 meeting the 20 weeks deadline is 0.9656..

Scene 13 (16m 42s)

COMPUTING THE PROBABILITY OF EACH PROJECT PATH METHOD THE 20-WEEK DEADLINE EXPECTED PATH COMPLETION TIME STANDARD DEVIATION OF PATH COMPLETION TIME Z SCORE PROBABILITY OF MEETING DEADLINE E(T 1 ) = 6 + 3 + 4 + 2 + 2 = 17 σ2/1 = 1.78 + 0.11 + 0.69 + 0.03 + 0.11 =2.72 Z 1 = 20-17/√2.72 = 1.82 0.9656 E(T 2 ) = 6 + 3 + 2 + 2 = 13 σ2/2 = 1.78 + 0.11 + 0.03 + 0.11 =2.03 Z 2 = 20-13/√2.03 = 4.91 > 0.9999 E(T 3 ) = 6 + 5 + 3 + 2 = 16 σ2/3 = 1.78 + 1.78 + 0.25 + 0.11 = 3.92 Z 3 = 20-16/√3.92 =2.02 0.9783 E(T 4 ) = 2 + 4 + 2 + 2 = 10 σ2/4 = 0.44 + 0.69 + 0.03 + 0.11 =1.27 Z 4 = 20-10/√1.72 = 7.02 > 0.9999.

Scene 14 (20m 34s)

THANK YOU!. THE END. 3,000+ Free Background &amp; Pattern Vectors.