Why do we need to know how to schedule the activities in a project? Even more important - why do we need to know how to do it by hand??? The answer on both questions are quite simple: Firstly we need to know how to schedule the activities in a project to know what must happen when and in which order to ensure that we complete the project in time. Secondly we need to know how to do it by hand in order to understand what a network diagram is and how it can support us in managing the project. 

I know you most probably use a software scheduling tool to do the scheduling for you, but if you do not understand the principles of scheduling how will you ever know if the answer you get from the software package is representative of how you envisage the project to be done? Remember the software scheduling package will do all the calculations based on the information provided by you - therefore you need to understand what information the package needs and what it is going to do with it. Another important aspect to keep in mind is that the software package will use defaults or tables set up by you, the PMO or somebody else and therefore the resulting schedule presented by the software is not necessarily always a true reflection of how you visualized the project. Understanding how scheduling is done will help you to identify if the schedule presented by the software is realistic and a true representation of the project schedule to be implemented.

Please bear in mind that this discussion is based on the activity-on-node method (AON) as normally used in the Critical Path Method (CPM). Also please note that this is a discussion of the basic principles of project scheduling and will therefore not address the more complicated scheduling scenarios.

Before we look at what scheduling is all about there are a few rules that we should keep in mind. Adherence to these rules can save you a lot of sweat and tears, whether you do the scheduling by hand or using a software scheduling tool. A network diagram should have:

1. One and only one start activity.
2. One and only one end activity.
3. All activities except the first one should have at least on predecessor. (Predecessor:- a task whose start or finish date determines the start or finish date of its successor task.) 
4. All activities except the last one should have at least one successor. (Successor:- a task whose start or finish date is driven by its predecessor task.)

See the figure 1a below for an example of how the network should look:

A network diagram (or "project schedule network diagram" as per the PMBOK® Guide) consists of boxes representing the activities (or tasks) linked together with arrows representing the logical relationship (dependency) that exists between the activities. The logical relationships can include:

The finish-to-start dependency is the most commonly used relationship in network diagramming.
A finish-to-start relationship is used when the predecessor activity or activities (Activities C, D and E) can not start unless the predecessor activity (Activity A) is completed.

In the example on figure 4 activity B can not stop until activity C is completed and activity C can not stop before activity D is completed - it is important to note the direction of the arrow. Bear in mind that there is no direct relationship between activity E and activities B and D, the relationship is between E and C.

From the example network diagram in Figure 5 it can be seen that activity E (successor) can not stop before Activity F (predecessor) has started. Please note the definitions of a predecessor and successor. Predecessor:- a task whose start or finish date determines the start or finish date of its successor task. Successor:- a task whose start or finish date is driven by its predecessor task.

The lead relationships indicated in the example as per Figure 6 represent the following:

• The SS-1 day relationship between activities C and B indicates that activity B cannot start until activity C has started minus one day. In other words activity B can start one day before activity C starts.
• The FS-2 days relationship between activities B and D indicate that activity D cannot start until activity B is completed minus two days. In other words activity D can start 2 days before activity B is completed.
• The SF-3 days relationship between activities E and C indicate that activity C can not finish until activity E has started minus 3 days. This means that activity C can finish 3 days before activity E starts.
• The FF-4 days relationship between activities E and D indicate that activity D can not finish until activity E is completed minus 4 days. Therefore activity D can finish 4 days before activity E finishes.

The network diagram represents the scheduling logic of the project. This scheduling logic consists of the activities and the logical relationships or dependencies that exists between the activities. We need to determine what the logical relationship between the activities are and then use these relationships to create the network diagram for our project. 

Some relationships are fairly obvious to determine for instance you can not paint the wall before the wall is built, assuming we are referring to a bricks and mortar wall. These dependencies that are legally or contractually required or inherent in the nature of the work are referred to as  Mandatory dependencies and normally make up the bulk of the dependencies in the network diagram.

Some relationships are not that obvious to determine. If for instance you need to tile the floor and paint the walls of a room - which should be done first? If you first paint the wall and then tile the floor the tiler may get tile glue and grout on the walls. If you first tile the floor and then paint the walls the painter may spill paint on the floor or the tiles may be damaged with the scaffolding. These dependencies are called discretionary dependencies, preferred logic or soft logic. Many of the dependencies prescribed by a project management methodology are discretionary dependencies.

We must also remember that some dependencies involve a relationship between project activities and non-project activities, usually outside the project team’s control. These are referred to as External dependencies. An example of an external dependency can be the obtaining of a environmental impact assessment done by a third party.

Internal dependencies involve a precedence relationship between project activities and are generally inside the project team’s control. For example, if the team cannot test a machine until they assemble it, this is an internal mandatory dependency.

The minimum information required to build the network logic diagram is the activity (number and/or description), the activity duration and the predecessor - see Figure 8.

From the information displayed in the table with the scheduling data (Figure 8) it is easy to compile the network diagram (Figure 9). We must keep in mind that the durations allocated to the activities at this stage may be fairly high level estimates as the resources may not necessarily been allocated to the activities yet and as such the durations may be subject to adjustment once the resources are assigned to the activities. See my post "From Stakeholder to Budget" for more information in this regard.

Once the network logic is sorted out we can start to calculate the early start and early finish dates for every activity - and this exercise we call a forward pass. But what do we mean by "forward pass", "Early Start Date" and "Early Finish Date"?

Forward pass:- A critical path method technique for calculating the early start and early finish dates by working forward through the network diagram from the project start date or a given point in time.
Early Start date (ES):- the earliest possible point in time when the uncompleted portions of a schedule activity can start based on the schedule network logic. The early start date of the first activity in the network diagram is set as equal to zero. In a FS relationship as is the case with the example in Figure 10, The early start date of the successor(s) activitie(s) are set as equal to the early finish date of the predecessor activity.
Early Finish date (EF):- the earliest possible point in time when the uncompleted portions of a schedule activity can finish based on the schedule network logic. The early finish date of an activity is calculated by adding the activity duration to the early start date. (EF = ES + Duration).

Let's look at a simple example of a forward pass:

In order to calculate the early start and early finish for every activity we start at the beginning (start activity) and work from left to right until we reach the end activity. There are more than one method that can be used to calculate the early start and early finish dates for each activity - we will here only discuss the most commonly used method that also happens to be the easiest method to use.

• Start:-The start activity kicks of the project. In this example the start activity is a milestone with a duration of zero days. The project starts on day zero and due to the fact that the start activity has a duration of zero, the early start and early finish for the start activity will both be zero.
• Act A:- This activity has a FS relationship to the start activity meaning that as soon as the go-ahead for the project is obtained, this activity can start. If the approval is obtained on day zero then the earliest that act A can start (ES) is on day zero. Using the formula EF = ES + duration we can calculate the EF date for this activity. EF = 0 + 2. The EF is therefore day 2.
• Act B:-This activity also has a FS relationship to the start activity meaning that as soon as the go-ahead for the project is obtained, this activity can also start. If the approval is obtained on day zero then the earliest that act B can start (ES) is on day zero. Using the formula EF = ES + Duration we can calculate the EF date for this activity. EF = 0 + 5. The EF is therefore day 5.
• Act C:-This activity has a FS relationship with act A. Therefore if the earliest act A can finish is on day 2, then the earliest act C can start is on day 2. The EF is calculated as 2+6=8 using the formula (EF=ES+Duration).
• Act D:-This activity has a FS relationship with act A. Therefore if the earliest act A can finish is on day 2, then the earliest act C can start is on day 2. The EF is calculated as 2+7=9 using the formula (EF=ES+Duration).
• Act E:-This activity has a FS relationship with act A as well as with act B. The FS relationship stipulates that act E can not start unless both the predecessors (A and B) have been completed. Act A can be completed at the earliest on day 2 and act B can be completed at the earliest on day 5. Therefore due to the fact that act E can not start before both A and B are completed, the earliest act E can start is on day 5. The EF is calculated as 5+5=10 using the formula (EF=ES+Duration).
• Act F:-This activity has a FS relationship with act B. Therefore if the earliest act B can finish is on day 5, then the earliest act F can start is on day 5. The EF is calculated as 5+2=7 using the formula (EF=ES+Duration).
• Act G:-This activity has a FS relationship with act B. Therefore if the earliest act B can finish is on day 5, then the earliest act G can start is on day 5. The EF is calculated as 5+3=8 using the formula (EF=ES+Duration).
• Act H:-This activity has a FS relationship with activities C, D and E. The FS relationship stipulates that act H can not start unless all the predecessors (C,D and E) have been completed. Act C can be completed at the earliest on day 8, act D on day 9 and act E can be completed at the earliest on day 10. Therefore due to the fact that act H can not start before C,D and E are completed, the earliest act H can start is on day 10. The EF is calculated as 10+2=12 using the formula (EF=ES+Duration).
• Act I:-This activity has a FS relationship with activities E, F and G. The FS relationship stipulates that act I can not start unless all the predecessors E, F and G) have been completed. Act E can be completed at the earliest on day 10, act F on day 7 and act G can be completed at the earliest on day 8. Therefore due to the fact that act I can not start before all three the predecessors (E, F and G) are completed, the earliest act I can start is on day 10. The EF is calculated as 10+3=13 using the formula (EF=ES+Duration).
• Finish:- The finish activity is, just like the start activity, a milestone with zero duration. The finish activity is dependent FS on act H and act I. The earliest H can be completed is on day 12 and for I it is day 13. The finish milestone therefore falls on day 13.

We have now completed the forward pass of this example. What information did it make available to us?

• The earliest day on which every activity can start and finish.
• The shortest time span within which the project can be completed namely 13 days.

Once we have completed the forward pass we can do the backward pass to calculate the latest day on which every activity can start and finish. Let's look at what we mean by:

Backward Pass:- A critical path method technique for calculating the late start and late finish dates by working backward through the schedule model from the project end date.

Late Finish date (LF):- the latest possible point in time when the uncompleted portions of a schedule activity can finish based on the schedule network logic. The late finish of the last activity (Finish activity) is set as equal to the early finish of that activity as calculated by the forward pass. In determining the late finish dates of every other activity we simply ask the question "If the latest the successor can start is day X and the FS relationship stipulates that it can not start until all of the predecessors are completed, then what is the latest date on which the predecessor(s) can be completed?"

Late Start date (LS):- the latest possible point in time when the uncompleted portions of a schedule activity can start based on the schedule network logic. The late start of every activity is calculated by the formula LS = LF - Duration.

Let's look at how we do the backward pass. In order to calculating the late start and late finish dates for every activity we start at the end (finish activity) and work from right to left until we reach the start activity:

• Finish:- As stated, the LF of the last activity in the network diagram is set as equal to the EF of that activity. In this case (Figure 11) it is day 13. As this finish activity is a milestone with zero duration the LS will be the same as the LF as calculated by the formula LS = LF - Duration.
  
• Act H:- If the latest date of the finish activity is day 13, then the latest act H can finish will be day 13 as there is a FS relationship between act H and the finish activity that stipulates that the finish activity can not start until act H is completed. The LS = LF - Duration, that is 13 - 2 = 11. The latest act H can therefore start is day 11.
• Act I:- If the latest date of the finish activity is day 13, then the latest act I can finish will be day 13 as there is a FS relationship between act I and the finish activity that stipulates that the finish activity can not start until act I is completed. The LS = LF - Duration, that is 13 - 3 = 10. The latest act I can therefore start is day 10.
• Act C:- If the LS date of act H is day 11, then the latest act C can finish will be day 11. The LS= LF-Duration, that is 11 - 6 = 5. 
• Act D:- If the LS date of act H is day 11, then the latest act D can finish will be day 11. The LS= LF-Duration, that is 11 - 7 = 4. 
• Act E:- There is a FS relationship between act H and act E as well as between act I and act E. Therefore act E can not finish later than the start dates of either act H or act I. The LS of act H is day 11 and that of act I is day 10. Therefore the latest act E can finish will be day 10. Finishing act E on day 10 will provide for act I to start on day 10 and will not be a problem for act H starting on day 11. The LS= LF-Duration, that is 10 - 5 = 5. 
• Act F:- If the LS date of act I is day 10, then the latest act F can finish will be day 10. The LS= LF-Duration, that is 10 - 2 = 8. 
• Act G:- If the LS date of act I is day 10, then the latest act G can finish will be day 10. The LS= LF-Duration, that is 10 - 3 = 7.

• Act A:- There is a FS relationship between act A and activities C, D and E. This means that act A can not finish later than the late start dates of any of these three successor activities. The late start dates are C equals day 5, D equals day 4 and  E equals day 5. Therefore the latest act A can finish will be day 4. The LS = LF-Duration, that is 4 - 2 = 2.
• Act B:- There is a FS relationship between act B and activities E, F and G. This means that act B can not finish later than the late start dates of any of these three successor activities. The late start dates are E equals day 5, F equals day 8 and  G equals day 7. Therefore the latest act B can finish will be day 5. The LS = LF-Duration, that is 5 - 5 = 0.
• Start:- There is a FS relationship between the start milestone and activities A and B. That means that the start activity can not occur later than the latest of the start dates of either act A or B. The LS for A equals day 2 and that for B equals day 0. Therefore the latest date for the start activity is day 0.

We have now completed the backward pass and in doing so calculated the latest dates on which every activity can start or end.

The Critical Path is generally the sequence of schedule activities determining the duration of the project and is:

• The longest path through the project. 
  
• The path with the least float.

See figure 14 for the critical path of our example project:

On this project it would be a good idea to watch carefully over Act B, Act E and Act I. If there is any slippage on any one of these three activities it will delay the project end date.

A project can have multiple critical paths and critical paths can also change due to progress made or not made on the project. If for instance in the above example act A that has initially had 2 days of float, for some reason is only completed on day 6 and not on the latest day 4 as originally planned, the critical path will then be as follows (see figure 15).

How many days float does activity E have?

a) 6 days
b) 10 days
c) 0 days
d) 9 days

For the model solution please go to model answer

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The objective of this presentation is to discuss the compressing of a project’s schedule in as simple and easy a way as possible.

What is schedule compression?

Whenever we find ourselves with a project schedule that indicates that the estimated project finish date is later than the finish date required by the project stakeholders we need to find ways to compress the schedule. In other words we need to find ways to be able to complete the project within the required time frame. There are numerous ways or strategies that can be followed to do this, of which the most commonly used techniques are:

• Fast tracking.
• Crashing.
  
• Resource reallocation.
  

What is fast tracking?
Fast tracking is a schedule compression technique where we change the logic of the project schedule by overlapping critical activities rather than working them strictly in sequence. For instance, if we have a project scheduled as follows:

According to the schedule it will take 18 days to complete this project. If it was a requirement that the project must be completed within 15 days then we should find a way to compress the schedule from 18 to 15 days – that means we must somehow save 3 days.

Fast tracking is a technique where we run two or more activities concurrently and not in sequence as currently scheduled. There are some aspects that need to be kept in mind such as:

• In order to shorten the project (to save the 3 days of the duration) we will have to shorten the longest string of activities, called the critical path (indicated in red). That means that two or more activities on the critical path will have to be overlapped.
• Activities selected for fast tracking should have a duration longer than the number of days that is required to shorten the schedule. It will not be sufficient to run two activities, one with a duration of two days, concurrently and hope to save three days off the schedule. The most that the schedule can be shortened will be two days and therefore another or an additional activity should be selected for compression.
• Care should be taken if the activities selected for fast tracking share the same resources. If it is the same resource working on both activities then that resources may become scheduled to work more hours than what is humanly possible for the period of the overlap.
  

In the example schedule above the best candidates for fast tracking would therefore be activities A, B, C and D. Some of the possible scenarios could include:

• Executing activity A and activity B concurrently. That will save a maximum of two days.
  
• Executing activity B and activity C at the same time. That will save a maximum of five days.
  
• Executing activity C and activity D simultaneously. That will save a maximum of four days.
  
• Overlapping activity A and activity B with one day, activity B and activity C with one day and activity C and activity D with one day.
  

The most important aspect that should be kept in mind is that fast tracking can introduce more risk into the project. If the project was originally scheduled in such a way that activity B cannot or should not start before activity A is completed, then it is safe to assume that there must have been a very good reason for this to be the case. For instance if activity A is the unloading and safeguarding of the rifle on the shooting range and activity B is handing it back to the shooting instructor I am sure you will appreciate the risk if you are the shooting instructor and I am attempting to do both activities at the same time!

The bottom line is fast tracking should only be done if the risk introduced by changing the order in which activities are executed, can be managed.

Above is an example of the same project fast tracked by overlapping activity B and activity C by three days. The project is now scheduled to be completed in 15 and no longer in 18 days. Bear in mind this is only valid if:

• The nature of activities A and B are such that the overlap is physically possible.
  
• The risk can be managed.
  
• The overlap is not causing an unmanageable schedule overload for the resources involved.
  

What is crashing?
Crashing is a technique used to shorten the estimated schedule duration for the least incremental cost by adding more resources. Crashing works only for activities on the critical path where additional resources will shorten the activity’s duration. Bear in mind that not all activities can be completed in a shorter timeframe if more resources are added. Activities where a lot of communication and creativity is involved (for instance creating the WBS) will most probably take longer if more resources are added. On the other hand some activities are excellent candidates for crashing for instance an activity where one labourer is scheduled to dig a trench of 100 meter long in four days. Using 2 labourers could possible shorten this activity to 3 days. Using 4 labourers could possibly bring it down to 2 days etc.

Care should be taken when estimating what impact the adding of resources on the duration will have as it is not just a mathematical calculation where we can assume by doubling the number of resources we will halve the duration. The measure of compression can for example be impacted by:

• Whether the nature of the activity lends it to be compressed by adding more resources.
  
• The type and number of resources added.
  
• How much of a learning curve the new resources will have as things can at first get extremely slow before it starts to speed up. 
  

Let’s look at an example of a project estimated to take 18 days to complete:

If in the above example the project needed to be completed in 15 days and we decide to use crashing, how could we do it? When selecting the best activities for crashing we should keep in mind that:

• As with fast tracking, only shortening activities on the critical path will result in compressing the duration of the project.
  
• Again, the longer the activity the bigger the potential for compressing it.
  
• The cost vs. duration compression ratio will not be the same for all activities. The objective is to implement the most economical alternative.
  

If for instance in the example above, we decide to assign two additional resources to activity B. We also estimate that if there are three resources assigned to activity B then the activity can be completed in 4 days as compared to 7 days if only one resource is assigned. This would mean that we now estimate to complete the project in 15 days as per figure D:

When selecting activities for resource reallocation we should take into consideration:

• The activity to which the resources are moved must be on the critical path.
  
• The activity from which the resources are moved must not be on the critical path.
  
• The resources moved must be appropriately skilled to complete the activity to which they are reallocated to.
  
• There may also be a learning curve for them on the new activity.
  

If in the above example (Figure E) we have decided to move 2 resources from activity G to activity B then the schedule could typically look as follows:

In the above case Activity B is compressed from 7 to 4 days due to 2 additional resources assigned to this critical activity. Activity G will now take 4 and not only 2 days to complete due to two of the resources taken away from this activity. The net result is that the duration of the critical path (and therefore for the project) is reduced from 18 to 15 days. The duration of the non-critical path F, G, H and J is extended with 3 days with the result that the amount of float on this path is reduced from an original of 9 days to 4 days after the resource reallocation.

Any one or a combination of any of these techniques can be used to do schedule compression – keep in mind however that if a second round of compression is required, the first step is always to confirm what the latest critical path is as the previous compression could have created a new critical path. 

The objective of this presentation is to make Earned Value Management (EVM) as simple and easy as possible, therefore we will only discuss the basic and most commonly used formulas. So let’s have a look what EVM is all about.

What is Earned Value Management?

EVM is a method for measuring project performance. It compares what is actually happening on the project against the approved baselines and helps us to answer questions such as:

• Where are we on the project money and time wise?
  
• Will we make it within budget and schedule?
  
• What do we need to do to get back on track?

It is important to note that EVM does not measure in days, weeks or months - It measures in monetary terms. 

How do we determine Planned Value (PV)?

We need to know what portion of the project is scheduled to be completed by the reporting date – normally referred to as “today”. 

 As time moves on we will see that the PV becomes more. Let’s now assume that the timeline has moved on and that by today we have scheduled 50% of the activity to be done. The PV will now be 50% of the BAC, which is $20K. At 75% scheduled the PV will be $30K. Once the reporting date is the same or later than the scheduled finish date of the activity it means that the total activity was scheduled to be completed by this date. Therefore the PV now becomes and stays the same as the BAC = $40K. The planned value is therefore that portion of the BAC estimate of the activity that is scheduled to be done by the reporting date.

How do we determine Earned Value (EV)?

EV is also based on the BAC of the activity. EV is determined by calculating what portion of the BAC has been done, irrespective of what portion was scheduled to be done. 

Let’s assume that we have an activity that we estimate will cost $40K to complete. (BAC). Let’s also assume that we have already completed 25% of the activity by the reporting date. The EV will then be 25% of the BAC, which is 25% of $40K = $10K. The passage of time does not change the EV – only the amount of progress made in completing the activity. Let’s now assume that we have completed 50% of the activity. The EV will now be 50% of the BAC, which is $20K. At 75% completed the EV will be $30K. Once the activity has been 100% completed it means the EV now becomes and stays the same as the BAC, in this example $40K. The earned value is therefore that portion of the BAC of the activity that is completed, whether it was scheduled to be done or not.

How do we determine Actual Cost?

How do we determine the variances?

How do we calculate the variances on the project? In other words how do we determine how our time and cost performance measures against our baselines? Let’s assume that we have a project with the following measurements: We have planned to do $40K worth of work (PV), we have done $40K worth of work (EV), and it actually cost us $40K to do the work (AC).

Even without the use of formulas we can see that in this example the project is exactly on schedule and on budget, but let us look at the variance formulas.

 What will the schedule variance (SV) be?  SV is determined by comparing what we have done (EV) with what we have planned to do (PV). Therefore SV = EV – PV. In this example SV = $40K -$40K and that equals 0. In other words no schedule variance as we are on schedule.

How would we determine the cost variance? CV is determined by comparing what we have done (EV) with what it cost us to do it (AC). Therefore CV = EV – AC. In this example CV = $40K - $40K and that equals zero indicating that we have no cost variance as we are exactly on budget. 

How do we determine the Indices?

Let’s assume again we have a project where the PV, EV and AC are exactly the same on $40K. From the variance analysis we saw that the project was on schedule and on budget. For calculating indices we again only use PV, EV and AC – but as we are now looking at the relationship between them and not the variance, we divide EV by PV for the Schedule Performance Index (SPI) and divide EV by AC for the Cost Performance Index (CPI). Here we see that SPI = EV / PV and that gives us $40K / $40K = 1. A SPI of 1 means that our schedule progress on the project is exactly as we have planned. Therefore we are progressing at 100% of the rate scheduled. CPI = EV / AC and that gives us $40K / $40K = 1. A CPI of 1 means that for every $1 we invest in the project we are making $1 worth of progress. Therefore we are exactly on budget

How can we use EVM to do forecasting?

The most commonly used forecasts are:

• ETC – how much money do we need to complete the rest of the project?
  
• EAC – how much will the project cost us in total?
  
• VAC – how much will we be over or under budget at the end of the project?
  

How do we determine the Estimate to Completion?

We see the EAC is higher as our previous forecast. The reason for this is that up to now our cost performance was not too good (CPI = 0.91) and therefore if we continue this trend we will need a larger budget for the project.

If we know what ETC is we can add what we have already spent (AC) to the ETC to arrive at a forecasted EAC. Let us assume that we re-estimated the ETC to be $35K and AC is $33K. Therefore we will forecast an EAC of $68K for the project.

As previously stated re-estimating the ETC can be more accurate but also more time consuming. This method takes our experiences on the project to date into account and the forecasted ETC also takes applicable risk responses and corrective actions into consideration.

How much will the budget to be over or under (VAC)?

Variance at Completion, what we normally refer to as how much will we be over or under budget at completion, is calculated by subtracting EAC from BAC. Therefore VAC = BAC – EAC. The forecasted value of the VAC will obviously depend on our forecasted value of EAC. If for example BAC = $60K and EAC = $68K then VAC = $60K (BAC) - $68K (EAC) indicating that we forecast the project to exceed the approved budget by $8K. Using one of the other methods to calculate the EAC, we will obviously calculate a different value for VAC.

Please take note that there are much more to EVM – this discussion only covered the most basic aspects.