Presentation on theme: "pert and cpm ch 9 Learning objectives:"— Presentation transcript:
1 pert and cpm ch 9 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 SummaryProjects 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 developedSeparately at about the same time in the late 1950s. Both techniques are concerned with integration of a number of different tasks.
4 PERT program evaluation and Review Technique) See CPM CrashingCritical PathCritical ActivityAccelerating 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 Pert / CPM Program Evolution Critical path Method and Review Technique Most projects have Certain elements in common1. The involve Considerable Cost2. Usually have a long time horizon3. Involve a large number of activities that must carefully, planned and performance guidelines.
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 immediate2. Interview staff immediate3.Hire + train staff4. Select and order furniture5. Remodel and intall phones6. Receive Furniture and set up7. Move in/ start upprepare a Gantt chart for this project .
8 Gantt chart for bank case : Activity Start 2 4 6 8 10 12 14 16 18 20 34567start2468101214161820
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.426153
11 Path: is a sequence of activities that lead from the starting node to the finishing node .from bank – – 6Case digram – 2- 5 – 51 – 3 – 5 – 6Critical path : The longest path which referred to critical activities .
12 Deterministic time Estimates Example :-Given the information provided in the a companying net work diagram , determine each of the following1. The length of each path2.The critical path3. The expected length of the project4. The amount of slack time for each path .1
14 Solution :- (The start and end “activities” have no time they merely serve as reference points ) 1. The path length areA= 18 weeks.B – 5- 6= 20 weeks .C – 5 - 6= 14 weeks .2 . The longest 20 week ( )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 pathpath length ( week) slack= = 2– = = 01- 3 – 5 – = = 6Note :-slack time = length of critical path – path length .
16 A Computing ALogrithmAlgorithm is used to develop four pieces of info. About the net workActivities :-Es, earliest time the activity can start , assuming all preceding activities start as early as possibleEF , 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 project2. Slack time3. Which activities are on the critical path .
18 Computing Es and EF Times 14810106428start813116101720439end19561413
19 Computation of earliest starting and finished time is aided by tow simple rules : 1. THE EF is equal to its Es plus it expected duration ,t:?? EF=Es + t2. 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= and EF =0+4=4Likewise for the other activities .These results are Summarized in the table below :Activity Duration ,t Es EFT+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 -12. For nodes with one arrow :LF for arrows entering that node equals the LS of the leaving arrowsFor nodes with multiple leaving arrows.LS, LF back pass?Compute LF, LS in our examplesSet LF of the last activity equal to the EF of that activity . ThusLF 5-6 =EF 5-6 = 20 weeksLS 5-6 =LF=
23 For LF 4-5 =LF 2-5 = LF 3 -5 =19 For the rest LS 4-5 = 19-3 =16 There are two arrow leaving Node 2 ,2-4 with LS = 10 , 2-5 Ls = 8The 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 =10The LS for activity 1-3 isLS 1-3 = =6The LS for activity 1-2 is LS 1-2 isLS 1-2 =LF1-2 –t= = 0
24 The LS , LF computation are Summarized in the table below: Activity Duration LF LS
25 Computing Activity slack times Activity slack = LS –ES or LF- ESActivity LS ES slackLS -ESAre allCriticalActivities
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 approachProbabilistic 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 conditions3- Most – likely time (m) most probable amount time .Beta distribution is commonly , used to describe inherentVariability in time estimates .
27 activity start A Beta Distribution Of special interest in network analysisAre the Average or expected timeFor each activity te and the variance of each activity time.The expected time is computed as a weighted averageOf three time estimates :te = a + 4m + bamtebactivitystartoptimistictimemost - likelytime( mode)pessimistictime6
28 Standard deviation of each activity Time is estimated 1/6 of difference betweenthe pessimistic and optimistic time estimate .
29 The variance is found by squiring the standard deviation , thus : 62 = orSize of the variance reflects the degree of uncertainty associated with an activity timeThe larger the variance , the greater the uncertaintyIt is also desirable to compute the standard deviation of the expectedTime for each path .6path =Go to example
30 Activity times are in month , do the following : 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 path2. Identify the Critical Path .3. Compute the variance for each activity and the variance for each path .mab2-4-62-3-5b1-3-4cendastart3-6-75-7-93-4-5EFd2-3-63-4-6gi4-6-8h
31 Solution: times te =a+4m=b path 1 Solution: times te =a+4m=b path 1. path activity a m b total a -b - c a b = c d –e -f d e = f largest critical path g- h –i g h i * critical path is d- h -i which has the largest path total weeks . Which mean the expected duration of the project .
32 Path activity a m b 52 act= (b-a)2 52 path path 36 3.Variance :Path activity a m b act= (b-a) path path36a-b – c a (4-1)2/36=9/36b (6-2)2/36=16/ /63=c (5-2)2/36= 9/36D- e-f d (5-3)2/36= 4/36e (7-3)2/36= 16/ /36=f (9-5)2/36= 16/36g- h- i g (6-2)2/36=16/36h (8-4)2/36=16/ /36=i (6-3)2/36= 9/36
33 The previous Solution enable a manager to make probabilistic estimates of the project completion time , such as1. 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 completionGo 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 points1. Related to independence, it is assumed that path duration times are independent of each other2.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
39 Hence , we need only consider the distribution of path 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 pathd- 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 timepath standard deviation
40 From table B with Z = +1.00 Normal distribution .8413 In our example :-expected time for path d – e – fis ,Z = =+1.001.00From table B with Z = +1.00Normal distribution.8413Probability 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 timeSee below?15 months100%A – b – cmonths.1587D – e – fmonths
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 equationPath Z= 15-expected path duration from normalpath standard deviation distribution table,B1,P(Z)A –b - c =.97D –e - f =1.00G –h - i =1.07
44 Note :-Any path with a Z of more than +2.5 is assigned a probability ofThe joint probability of finishing before month 15 is the product of these probabilities(1.00) ( .1587) (.9192) = ??
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 info1. Regular time and crash time estimates for each activity .2. Regular cost and crash cost estimates for each activity3. 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 times3. 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 dayActivity Normal time crash time to crash$500
48 Solution:-1.Determine which activities are on the critical path it’s length , and the length of the other path :path lengthcritical path26102EndStart16595443
49 2. Rank the critical path activities in order of lost wet crashing cost and determine the umber of days each can be crashedcost per day availableactivity to crash days$
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 $600the length of pathnow become 18 days which is the same as the length of pathc. Since the paths are now both critical , further improvements will necessitate shorting one activity on each.
51 The remaining points for crashing and their costs one: Path activity crash cost per dayno reduction possible$no further reduction possible$700
52 Activity 5-6 would not be advantageous because it has highest crashing cost . How ever activity 5-6 is on paths and he project ?* The option of shortening the least expensive activity on each path would cost $500 for 2-5 and $600 for 4-5 *Thus , shorten Activity 5-6 by one day the project duration is now 17 days .d. At this point , no additional improvement is feasible . The cost to crash Activity 4-5 is $600 for a total of $1.100 and that would exceed the project costs of $1000 per day .E. The crashing sequence is Summarized belowlength afterPath crashing n daysn=Activity crashedCost $ $ $800