The Critical Path Method for the PMP Exam and the CAPM Exam by Aileen Ellis, AME Group Inc.

here we have another sample question related to network diagrams you are a project manager for a new road work project you have determined the following dependencies when I look at the rest of the question I can see it lists activities relationships and durations from this I know I need to build a network diagram I begin with the start the question says activity a is the first activity with a duration of five days when I continue to skim the question I see that both activity B and activity C our successors to activity a and I fill in their durations activity D is a successor to activity B with a duration of two days activity E is a successor to both B and C with a duration of six days F is a successor to D and E with a duration of nine days G is a successor to E with a duration of eight days H is a successor to both F and G with a duration of four days the project is finished once activity H is completed I now have my Network diagram the first question I see what is the project duration another question what is the duration of the critical path to solve both of these questions I do the same thing I do what's called a forward pass I begin with activity a I ask what is that earliest activity a may begin since activity a is the first activity the earliest it may start is day one if it runs five days day one day two day three day four the earliest it may finish is day five activity be made again the day after activity a ends therefore B may begin on day six the earliest B may finish is day 8 how did I get that it begins on day 6 it runs day 6 day 7 day 8 activity C may begin after activity a finishes the earliest C could begin is day 6 if it runs seven days the earliest it could finish is day 12 now how did I get 12 I take 6 plus the duration of 7 subtract 1 I get day 12 the earliest activity D may begin is the day after activity B ends therefore the early start of activity D is 9 if it runs 2 days day 9 day 10 the earliest it may finish is day 10 activity E is a little trickier activity II may begin after activity B and C have both completed I look at the early finish of B which is day 8 the early finish of C which is day 12 12 is higher therefore the earliest emai begin is day 13 if it's 6 days long 13 plus 6 minus one day 18 F may begin after both D and D have completed I look at the early finish of D and E 10 and 18 18 is the larger number therefore the earliest F could begin would be the following day day 19 if F runs nine days 19 plus nine minus one day 27 the earliest G may begin is the day after Aoife Nicias G the earliest may begin is day 19 if it runs 8 days 19 plus 8 minus one day 26 is the earliest G may finish for each to begin both F and G need to be finished I look at the early finish of F day 27 the early finish of G day 26 day 27 is a higher or a bigger number therefore the earliest age could begin would be day 28 if H runs four days 28 plus four minus one the earliest each could finish is day 31 since H is the last activity for the project the project duration is 31 days the next question what is the duration of the critical path it's the same answer 31 days the critical path is what drives the duration of the project the next question we see what is the float of activity B first let's ask what does the word float really mean for me float means flexibility when we think of the critical path method float is the difference between the late numbers and the early numbers therefore to calculate float we need to do a backward pass at least as far as activity B to begin the backward pass I start with my last activity each and I bring down my early finish of 31 days unless that as my late finish now I ask what is the latest H may start with a duration of 4 days to finish by day 31 mathematically I say 31 minus 4 plus 1 gives me day 28 this means activity H the latest it may begin is day 28 if it will fit finish by day 31 let's now look at activity F I'm asking what's the latest F may finish and H begin by day 28 that would be one day earlier day 27 if F is nine days long what's the latest F may start I take 27 minus 9 and 1 the latest F may start is day 19 if it has a duration of nine days and it's late finishes day 27 I move on to activity G the latest G may finish is day 27 if H will start by day 28 what's the latest G may start 27 minus 8 plus 1 it would be day 20 the latest G may start is day 20 if it has a duration of 8 and a late finish of day 27 you can see I'm working in columns I go to activity D and I ask what's the latest DM a finish an activity F begin on day 19 the late finish of activity D would be day 18 if activity D is two days long the latest it could start would be day 17 mathematically 18 minus two plus one day 17 activity e is a little trickier I'm asking what's the latest II may finish and F may start by day 19 and G start by day 20 I look at the number 19 in the number 20 I take the smaller of these two numbers 19 and subtract one the latest emai is day 18 if F were to begin on day 19 or G by day 20 let's go back to activity e what's the latest he could start I take 18 subtract 6 add 1 I get day 13 what does this really mean it means the latest e may start is day 13 if it has a duration of six and a late finish of day 18 activity B is a little trickier I ask what's the latest beam a finish for D to have a late start of day 17 in e a late start of day 13 I look at the smaller number day 13 subtract one the latest beam a finish is day 12 if D is to start by day 17 and ebuy day 13 looking back at activity B what's the latest beam a start 13 minus 3 plus 1 is 10 what does this really mean it means the latest beam a start this day 10 if it has a duration of 3 and a late finish of day 12 now I ask what's the latest seem a finish I look at e he has a late start of day 13 meaning the latest C can finish is day 12 what's the latest seem a start I say 12 minus 7 plus 1 day 6 activity a again a little trickier what's the latest a may finish for B to start by day 10 and C by day 6 I look at the smaller number day 6 the latest a may finish is day 5 if a is 5 days long 5 minus 5 plus 1 the latest a may start is day 1 now really to answer the question what is the float of activity B I didn't need to figure out the lead start and finish of a or C I just needed to go back as far as B the float of activity B is the flexibility of activity B it's how much flexibility it has between its early and late number the equation float is the late finish of activity B which is day 12 - the early finish of activity B which is day 8 12 minus 8 equals 4 days what does this really mean it means B may slip up to 4 days from its early numbers before it would hit or affect the critical path or the project duration of 31 days let's say we get another question it asks what is the free flow of activity B free flow is how much an activity can slip from its early finish before it affects the earliest start of any successor if we visualize the free flow of activity B is how much B may slip from its early finish of day 8 before it would affect that earliest start of activity D day 9 if we look at the equation it would be nine minus eight minus one equals zero what does this really mean it means if B slips at all B will start to eat into DS float let's say we get another question the question asks which activities are critical what the question is really asking which activities are on the critical path how do we know an activity is on the critical path we know this if it's late numbers are the same as its early numbers here we can see activity a c e F and H are all on the critical path because their late finish numbers are the same as their early finish numbers the last question we'll look at today for this example if management says to end the project by day 25 what is the project float the way I like to visualize this management is saying activity each needs to end by day 25 to calculate float late finish - early finish 25 days - 31 days equals a negative 6 days why is the answer negative because the plan we created shows the project finishing in 31 days management says to finish no later than 25 days therefore if we stay on the plan we have created we will be six days leads a negative six days of flute I hope this example helped you as you're preparing for the PMP or the CAPM exam check back later for more examples thank you

Channel: Aileen Ellis

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Keywords: Project Management Professional (Profession), pmbok guide 6th Edition, capm exam, pmp exam, pmi, AME Group Inc., pmp, pmp exam prep 6th Edition, PMP Exam Sample Questions, pmp exam questions, capm, the project management institute, capm exam prep, CAPM Exam Sample Questions, pmp exam prep, pmbok6, pmp training videos, aileen ellis, aileen, the critical path method, CPM, forward pass, backward pass, float, free float, total float, project float

Id: Ifr5x3sfRBE

Length: 14min 29sec (869 seconds)

Published: Sun Feb 23 2014