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ch 9 pert and cpm Learning objectives: 1- Give a general description of/techniques.

ch 9 pert and cpm Learning objectives: 1- Give a general description of/techniques. 2- Construct simple net work diagrams. 3- Analyze net works that have deterministic time. 4- Describe activity crashing “and solve simple problem. Summary

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ch 9 pert and cpm Learning objectives: 1- Give a general description of/techniques.

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  1. ch 9 pert and cpm Learning objectives: 1- Give a general description of/techniques. 2- Construct simple net work diagrams. 3- Analyze net works that have deterministic time. 4- Describe activity crashing “and solve simple problem.

  2. Summary Projects are comprised f a unique set of activities established to achieve a given set of objectives during a limited life span . PERI/CPM are two commonly used techniques for developing and monitoring projects.

  3. Glossary Activity A task CPM (critical Path method ) Pert/CPM were developed Separately at about the same time in the late 1950s. Both techniques are concerned with integration of a number of different tasks.

  4. Crashing Critical Path Critical Activity Accelerating a project by speeding up those critical- path activities that have lowest ratio of incremental cost to incremental time saved . The path in a network diagram that takes the longest time for completion . An activity on the critical Path . PERT program evaluation and Review Technique) See CPM

  5. Program Evolution Critical path Method and Review Technique Most projects have Certain elements in common 1. The involve Considerable Cost 2. Usually have a long time horizon 3. Involve a large number of activities that must carefully, planned and performance guidelines. Pert / CPM

  6. Planning and Scheduling with Gantt charts : Grantt chart is a popular tool for planning and Scheduling simple projects. This tool enable a manager to initially Schedule project activities and than monitor progress over time by comparing planned progress to actual progress. For example :- Consider the case of bank that planned to great a new direct marketing department . He developed a list of the major activities that would be requires

  7. Activity time to accomplish week after in week project start 1. Locate new facility 8 immediate 2. Interview staff 4 immediate 3.Hire + train staff 9 4 4. Select and order furniture 6 8 5. Remodel and intall phones 11 8 6. Receive Furniture and set up 3 14 7. Move in/ start up 1 19 prepare a Gantt chart for this project .

  8. Gantt chart for bank case : Activity Start 2 4 6 8 10 12 14 16 18 20 1 2 3 4 5 6 7 start 10 20 2 4 6 8 12 14 16 18

  9. PERT and CPM By using pert or cpm , managers are able to obtain : 1. A graphical display of project activities . 2. An estimate of how long the project will take . 3.An indication of which activities are the most critical to timely completion of the project . 4. An indication of how long any activity can be delayed without lengthening the project. Note : pert : stressed probabilistic activity time estimates because the environment in which it was developed was typified by high uncertainty . Cpm: originally made no provision for variable time estimates

  10. The net work diagram Recall the bank case: Nodes: represent both beginning and endings of activities. arrows : represent the project activities that lead from the starting node to the finishing node. 4 2 6 1 5 3

  11. Path: is a sequence of activities that lead from the starting node to the finishing node . from bank 1 - 2 – 4 - 5 – 6 Case digram 1 – 2- 5 – 5 1 – 3 – 5 – 6 Critical path : The longest path which referred to critical activities .

  12. Deterministic time Estimates 1 Deterministic time estimates: Example :- Given the information provided in the a companying net work diagram , determine each of the following 1. The length of each path 2.The critical path 3. The expected length of the project 4. The amount of slack time for each path .

  13. . 6 weeks 2 4 8 weeks 3 weeks 11 weeks start 1 end 9 weeks 6 4 weeks 5 3 1 week

  14. Solution :-(The start and end “activities” have no time they merely serve as reference points ) 1. The path length are A . 1- 2- 4- 5- 6 8 + 6 + 3 +1 = 18 weeks. B . 1 -2 – 5- 6 8 + 11 + 1 = 20 weeks . C . 1 - 3 – 5 - 6 4 + 9 + 1 = 14 weeks . 2 . The longest 20 week (1 -2 -5 -6 ) is the critical path .

  15. 3. The expected length of the project is equal to the length of the critical path 20 weeks . 4. The amount of slack time for each path path length ( week) slack 1-2-4-5-6 8+6+3+1= 18 20-18 = 2 1- 2- 5 – 6 8+ 11+ 1 = 20 20-20 = 0 1- 3 – 5 – 6 4 + 9 +1 = 14 20-14 = 6 Note :- slack time = length of critical path – path length .

  16. A Computing ALogrithm Algorithm is used to develop four pieces of info. About the net work Activities :- Es, earliest time the activity can start , assuming all preceding activities start as early as possible EF , earliest time the activity can finish . LS , latest time the activity can start and not delay the project .

  17. Once these values have been determined , they can be used to find 1. Expected duration of the project 2. Slack time 3. Which activities are on the critical path .

  18. Computing Es and EF Times 14 8 10 10 6 4 2 8 0 8 start 1 3 11 0 6 17 10 20 4 9 3 end 19 5 6 1 4 13

  19. Computation of earliest starting and finished time is aidedby tow simple rules : 1. THE EF is equal to its Es plus it expected duration ,t: ?? EF=Es + t 2. For nodes with one entering arrow : Es for activities at such nodes is equal to EF of entering arrow. For nodes with Multiple entering arrows : equals activities leaving such nodes equals largest EF of entering arrows .

  20. Compute Es and EF for our example : Solution : Assume an Es of o for activities without predecessors. EF 1-2 = 0 +8=8 and EF 1-3=0+4=4 Likewise for the other activities . These results are Summarized in the table below : Activity Duration ,t Es EF 1-2 8 0 8 1-3 4 0 4 2-4 6 8 14 2-5 11 8 19 3-5 9 4 13 4-6 3 14 17 5-6 1 19 20 T+Es

  21. Note : last EF is the project duration , thus ,the expected length of the project is 20 week. ES, EF forward pass??

  22. Computing LS and LF times Computation of the Ls +LF times is aided by the use of tow rules : 1.The L is equal to its LF minus its expected duration : LS= LF -1 2. For nodes with one arrow :LF for arrows entering that node equals the LS of the leaving arrows For nodes with multiple leaving arrows. LS, LF back pass? Compute LF, LS in our examples Set LF of the last activity equal to the EF of that activity . Thus LF 5-6 =EF 5-6 = 20 weeks LS 5-6 =LF 5-6 -1 = 20-1 -19

  23. For LF 4-5 =LF 2-5 = LF 3 -5 =19 For the rest LS 4-5 = 19-3 =16 LS 2-5 = 19 – 11= 18 LS 2-5 = 19 – 9 = 10 There are two arrow leaving Node 2 , 2-4 with LS = 10 , 2-5 Ls = 8 The latest Finish for activity 2-1 thus become 8 which is the smallest Ls for a leaving arrow the LF for 1-3 is equal to the LS for 3-5 . LF 1-3 LS 3-5 =10 The LS for activity 1-3 is LS 1-3 = 10 - 4 =6 The LS for activity 1-2 is LS 1-2 is LS 1-2 =LF1-2 –t = 8 - 8 = 0

  24. The LS , LF computation are Summarized in the table below: Activity Duration LF LS 5-6 1 20 19 4-5 3 19 16 2-5 11 19 8 3-5 9 19 10 2-4 6 16 10 1-2 8 8 0 1-3 4 10 6

  25. Computing Activity slack times Activity slack = LS –ES or LF- ES Activity LS ES slack 1-2 0 0 0 1-3 6 0 6 2-4 10 8 2 2-5 8 8 0 3-5 10 4 6 4-5 16 14 2 5-6 19 19 0 LS -ES Are all Critical Activities

  26. Probabilistic time estimates In the preceding example we assumed that activity times were known and not subject to variation . For other situations we need to use probabilistic approach Probabilistic time estimates use three time estimates for each activity Instead of one : 1- Optimistic time (a) length of time required under optimum condition . 2- Pessimistic time (b) time under worse conditions 3- Most – likely time (m) most probable amount time . Beta distribution is commonly , used to describe inherent Variability in time estimates .

  27. A Beta Distribution Of special interest in network analysis Are the Average or expected time For each activity te and the variance of each activity time. The expected time is computed as a weighted average Of three time estimates : te = a + 4m + b 0 a m te b activity start optimistic time most - likely time ( mode) pessimistic time 6

  28. Standard deviation of each activity Time is estimated 1/6 of difference between the pessimistic and optimistic time estimate .

  29. The variance is found by squiring the standard deviation , thus : 62= 2 or Size of the variance reflects the degree of uncertainty associated with an activity time The larger the variance , the greater the uncertainty It is also desirable to compute the standard deviation of the expected Time for each path . 6path = Go to example

  30. Example: The net work diagram for a shown in the accompanying figure , with three time estimates for each activity, Activity times are in month , do the following : • Compute the expected time for each activity and the expected duration for each path 2. Identify the Critical Path . 3. Compute the variance for each activity and the variance for each path . m a b 2-4-6 2-3-5 b 1-3-4 c end a start 3-6-7 5-7-9 3-4-5 E F d 2-3-6 3-4-6 g i 4-6-8 h

  31. Solution: times te =a+4m=b path1. path activity a m b 6 total a -b - c a 1 3 4 2.83 b 2 4 6 4.00 =10.00 c 2 3 5 3.17d –e -f d 3 4 5 4.00 e 3 5 7 5.00 =16.00 f 5 7 9 7.00 largest critical pathg- h –i g 2 3 6 3.33 h 4 6 8 6.00 13.50 i 3 4 6 4.172.* critical path is d- h -i which has the largest path total 16.00 weeks .Which mean the expected duration of the project .

  32. 3.Variance : Path activity a m b 52 act= (b-a)2 52 path path 36 a-b – c a 1 3 4 (4-1)2/36=9/36 b 2 4 6 (6-2)2/36=16/36 34/63=.944 .97 c 2 3 5 (5-2)2/36= 9/36 D- e-f d 3 4 5 (5-3)2/36= 4/36 e 3 5 7 (7-3)2/36= 16/36 36/36=1.00 1.00 f 5 7 9 (9-5)2/36= 16/36 g- h- i g 2 3 6 (6-2)2/36=16/36 h 4 6 8 (8-4)2/36=16/36 41/36=1.39 1.07 i 3 4 6 (6-3)2/36= 9/36

  33. The previous Solution enable a manager to make probabilistic estimates of the project completion time , such as 1. The probability that the project will completed with months of start ? 2. The probability that the project will take longer than specific months more than it’s completion Go to example

  34. The next example illustrates the use of normal distribution to determine the probabilities for various completion time . Before we look at that example , it is important to make note of two points 1. Related to independence, it is assumed that path duration times are independent of each other 2.A project is not completed until all of it’s activities have been completed (included those on the critical path).

  35. Example :- using the information from the preceding example , Answer the following question : 1. Can the paths be considered independent ? Why ? 2. Determine the probability that the project will be completed within it months of it’s start ? 3.Determine the probability that the project will be completed with 15 months of it’s start. 4. What is the probability that the project will not be completed within 15 months of it’s start ?

  36. Solution :- 1. Yes , the paths can be considered Independent because no activity is on information than on path , and we have information that would suggest that any activity times are interrelated . 2. In order to answer Q’s of this nature , we must take into account the degree to witch the path distribution the specified completion time. See the figures

  37. path 17 months 100% A – b – c 10.0 months D – e – f 16.0 months

  38. 99.95 q – h – i 13.5 months

  39. The shaded portion of each distribution corresponds to the probability the part will be completed within the specified time , observe that paths: A -b - c and g -h -i are far enough to the left of the specified time that is highly likely that both will finished by month 17 but that critical path overlaps the specified completion time. Hence , we need only consider the distribution of path d- e - f in assessing the probability of completion by month 17.to do so, we must first compute the value of Z using the relationship : Z = specified time – Expected time path standard deviation

  40. In our example :- expected time for path d – e – f is 16.0 , Z = 17-16 =+1.00 1.00 From table B with Z = +1.00 Normal distribution .8413 Probability 84% to finish the project within 17 months

  41. 3.The question illustrates how to handle a problem in which more than one of the distribution over laps the specified time See below? 15 months 100% A – b – c 10.0 months .1587 D – e – f 16.0 months

  42. 15 months .9192 Q – h – i 13.5 months

  43. Note that d –e – f and g – h –i paths overlaps month 15. This means that both paths have the potential for delaying the project beyond 15 month. Back to Z equation Path Z= 15-expected path duration from normal path standard deviation distribution table, B1, P(Z) A –b - c 15-10 = +5.15 1.00 .97 D –e - f 15-16 =-1.00 .1587 1.00 G –h - i 15-13.5 =+1.40 .9192 1.07

  44. Note :- Any path with a Z of more than +2.5 is assigned a probability of 1.00 . The joint probability of finishing before month 15 is the product of these probabilities (1.00) ( .1587) (.9192) = .1459 ??

  45. Time – Cost Trade – OFFs – Crashing The desire to shorten the length of a project merely reflects o attempt to reduce the indirect cost associated with running the project . Such as : Facilities of equipment cost , supervision and labor and personal costs. To avoid late penalties. To crashing desirable , a manager needs the following info 1. Regular time and crash time estimates for each activity . 2. Regular cost and crash cost estimates for each activity 3. A list of activities that are on the critical path .

  46. The general procedure for crashing is: 1. Obtain estimates of regular and crash time and costs for each activity . 2.Determine the lengths of all paths and path slack times 3. Determine which activities are on the critical path . 4. Crash critical activities , in order of increasing costs do not exceed benefits .

  47. Example :- Using the info . Below , develop an optimum time- cost solution , Assume that indirect project costs are $1000per day . Cost per day Activity Normal time crash time to crash 1-2 6 6 2-5 10 8 $500 1-3 5 4 300 3-4 4 1 700 4-5 9 7 600 5-6 2 1 800

  48. 2 6 10 2 Solution:- 1.Determine which activities are on the critical path it’s length , and the length of the other path : path length 1-2-5-6 18 1-3-4-5-6 20 critical path End Start 1 6 5 9 5 4 4 3

  49. 2. Rank the critical path activities in order of lost wet crashing cost and determine the umber of days each can be crashed cost per day available activity to crash days 1-3 $300 1 4-5 600 2 3-4 700 3 5-6 800 1

  50. 3.Begin shorting the project , one day at a time , and chek after each reducing to see which path is critical .( After a certain point, another path may equal the length of the shortened critical path ) thus. 4.A shorten Activity 1-3 one day at cost of $ 300, the length of the critical path now become 19 days. b. Activity 1-3 cannot be shortened any more shorten activity 4-5 one day of a cost of $600the length of path 1-3-4-5-6 now become 18 days which is the same as the length of path 1-2-5-6 c. Since the paths are now both critical , further improvements will necessitate shorting one activity on each.

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