Concept: Jointly, R_{Tot} and I/C$_{Tot} offer insights into schedule performance not otherwise available. Combining them, however, introduces limitations not otherwise visible. The previous post detailed the new insights. This post sets out the limitations.
Practice: When used together, R_{Tot} and I/C$_{Tot} face some limitations. The limitations and reasons for them are described below, along with examples. Finally, one key implication of the limitations is mentioned and then, in the next post, it is elaborated.
*Limitation: The period-to-period rise and fall of R*_{Tot} and I/C$_{Tot} is always synchronous.
*Reason: *The total value associated with R-tasks and with I/C-tasks must be equal. So, rise-and-fall calculations based on the totals will reflect that fact.
- The synchronous behaviour is easy to see in an example. Figure 1 charts the R
_{Tot} and I/C$_{Tot} from a sample project.
*Proof:* Given that the total value earned at the Actual Time equals the total value planned at the ES time (i.e., EV@AT=PV@ES [1]), the difference between the two must be 0.
Next, to calculate P, each task (i) is evaluated by subtracting its PVi@ES from its EVi@AT. The differences must be positive, negative, or zero. If a difference is zero, then PVi@ES = EVi@AT. Positive differences are associated exclusively with R-tasks; negative ones with I/C-tasks (see R-tasks and I/C-tasks).
Non-zero differences of individual tasks (i.e., EVi@AT - PVi@ES) must sum to 0—otherwise, the total EV@AT ≠ total PV@ES. So, the value of R-tasks equals the (absolute) value of I/C-tasks. (No, this is not a circular argument [2].)
Figure 1
*Limitation: Although synchronous, the amount of rise and fall differs for non-zero values of Rework and of Impediments/Constraints. *
*Reason:* Rework is always a fraction of the total value of R-tasks because it is set by a mathematical model. By contrast, the full value of I/C-tasks is directly observed. So, the two values can and do differ.
- Again, Figure 1 illustrates the difference in values. Note how I/C$
_{Tot} seems to stabilize between August and October while R_{Tot} continues to decline. It should also be mentioned that the scale of Cost $ obscures non-zero differences early in the project.
*Proof:*Like Rework, Impediments/Constraints must end at 0 because, ultimately, there are no further tasks to run late. I/C’s progress to that end is not, however, as regulated as Rework’s decline to 0. Indeed, sometimes the I/C amount seems to stall at a particular level.
*Limitation: Although initial values of the two are often close, they separate over time. *
*Reason:* At the start, the mathematical model ensures that a large fraction of the value in R-tasks is liable for Rework. So, there is little difference between the value associated with Rework and the value associated with I/C. Over time, the fractional amount falls according to the mathematical model, but the I/C amount does not have the same regulated decline. Consequently, the two costs tend to separate over time.
- Note in Figure 1 how the gap grows as the project proceeds.
*Implication: The tight mathematical relationship between RTot and I/C$Tot narrows how much the combination can tell us. *
- For instance, one might expect the combination of R
_{Tot }and I/C$_{Tot} to illuminate trade-offs between working ahead (and facing rework) and not starting work (and facing Impediments/Constraints).
- But, you cannot trade off one over the other if they must always remain in sync—any trade-off ends in a balance between the two metrics.
- Intuitively, it seems that the two metrics should be more independent of one another. There’s much more to be said on this point—see the next post.
Notes:
[1] Arguments for EV@AT=PV@ES are laid out in the next post.
[2] Although this claim follows from the principle that the total values are equal, it is not intended to prove that proposition. Instead, it is derived as a corollary: if EV@AT=PV@ES, the sum of differences is zero. If the sum of differences is zero, the sum of negative differences must equal the sum of positive differences. So, if EV@AT=PV@ES, the sum of negative differences (i.e., I/C-tasks) must equal the sum of positive differences (i.e., R-tasks). |