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I inspect a road network's condition every three years:

+-----+-----------+
| age | condition |
+-----+-----------+
|   0 |        20 |
|   3 |        19 |
|   6 |        19 |
|   9 |        18 |
|  12 |        18 |
|  15 |        17 |
|  18 |        16 |
+-----+-----------+

The rating system is from 0-20, 20 being perfect condition.


I have an exponential regression function that I can use to generate a deterioration curve: condition = 21 - exp(0.06 * age). The deterioration curve is a revised & simplified version of the the original (page 53).

As far as I can tell, the deterioration curves (both the original and the revised) are primarily intended to be used to project the condition of a road right from it's initial construction.

However, I'm wondering if I can use the deterioration curve for a different purpose. Rather than predict the condition of a new road, I want to predict the condition of an existing road, at a known condition.

The documentation of the revised deterioration curve hints that I can do this sort of thing:

The future condition can be estimated by considering the condition information available from inspections as well as the rate of deterioration provided by the models.

Although a pavement might, in reality, be a particular age based on its last rehabilitation, it may not be performing at that age. For example, a pavement that was reconstructed five years ago might be expected to be in good condition, but due to some unknown variable (e.g. construction errors), it might actually be in poor condition. This pavement would be performing as if it were an older age than it actually is.

Thus, those sections that have not been maintained or rehabilitated since their last inspection could be assigned a pseudo-age based on their condition at last inspection. This pseudo age can be easily calculated by rearranging the equation to solve for age given condition.

For example, a road section that [text removed] has been inspected and assigned a grade of 13 would have the following pseudo age using the new exponential model:

13 = 21 - exp(0.06*age)
ln8 = 0.06 * age
age = 35 years

Once this pseudo age has been calculated, the predicted condition score at a particular point in the future can be predicted based on it. For example, if the condition at 20 years from the inspection was needed, it could be calculated by inputting an assumed pseudo age of 55 years into the model.

The Question:

I can create a table of ages from 0-50, and use the function to predict condition:

+-----+-----------+
| age | condition | <--predicted
+-----+-----------+
|   0 |      20.0 |
|   1 |      19.9 |
|   2 |      19.9 |
|   3 |      19.8 |
|   4 |      19.7 |
|   5 |      19.7 |
|   6 |      19.6 |
|   7 |      19.5 |
|   8 |      19.4 |
|   9 |      19.3 |
|  10 |      19.2 |
|  11 |      19.1 |
|  12 |      18.9 |
|  13 |      18.8 |
|  14 |      18.7 |
|  15 |      18.5 |
|  16 |      18.4 |
|  17 |      18.2 |
|  18 |      18.1 |
|  19 |      17.9 |
|  20 |      17.7 |
|  21 |      17.5 |
|  22 |      17.3 |
|  23 |      17.0 |
|  24 |      16.8 |
|  25 |      16.5 |
|  26 |      16.2 |
|  27 |      15.9 |
|  28 |      15.6 |
|  29 |      15.3 |
|  30 |      15.0 |
|  31 |      14.6 |
|  32 |      14.2 |
|  33 |      13.8 |
|  34 |      13.3 |
|  35 |      12.8 |
|  36 |      12.3 |
|  37 |      11.8 |
|  38 |      11.2 |
|  39 |      10.6 |
|  40 |      10.0 |
|  41 |       9.3 |
|  42 |       8.6 |
|  43 |       7.8 |
|  44 |       7.0 |
|  45 |       6.1 |
|  46 |       5.2 |
|  47 |       4.2 |
|  48 |       3.2 |
|  49 |       2.1 |
|  50 |       0.9 |
+-----+-----------+

I want to predict the future condition of my road from it's last known condition--going forward.

I could do this by simply appending the condition values from the deterioration curve, that are less than the last known condition, to my existing inspection values.

The age would not be from the deterioration curve, but instead is just a sequential list of numbers, starting at the last inspection.

+-----+-----------+
| age | condition |
+-----+-----------+
|   0 |        20 |
|   3 |        19 |
|   6 |        19 |
|   9 |        18 |
|  12 |        18 |
|  15 |        17 |
|  18 |        16 |
+-----+-----------+  
|  19 |      15.9 | <-- Appended.
|  20 |      15.6 |
|  21 |      15.3 |
|  22 |      15.0 |
|  23 |      14.6 |
|  24 |      14.2 |
|  25 |      13.8 |
|  26 |      13.3 |
|  27 |      12.8 |
|  28 |      12.3 |
|  29 |      11.8 |
|  30 |      11.2 |
|  31 |      10.6 |
|  32 |      10.0 |
|  33 |       9.3 |
|  34 |       8.6 |
|  35 |       7.8 |
|  36 |       7.0 |
|  37 |       6.1 |
|  38 |       5.2 |
|  39 |       4.2 |
|  40 |       3.2 |
|  41 |       2.1 |
|  42 |       0.9 |
|  43 |           |
|  44 |           |
|  45 |           |
|  46 |           |
|  47 |           |
|  48 |           |
|  49 |           |
|  50 |           |
+-----+-----------+

From there, I can do analysis on the projected condition of the road. For example I can graph it:

enter image description here

However, while I'm tempted to just say "job well done!" and go have a celebratory pint, I suspect that this might be a classic example of a non-stats guy completely butchering the numbers.


Is appending a portion of a deterioration curve to existing values appropriate/correct, according to stats theory?

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This seems more like an algebra problem than a statistics problem. I think there is an interesting statistics problem lurking, though---I'm interested to know how the 0.06 was estimated from data for example.

The problem is, given your formula above, how would you forecast condition in the future given condition today. They key to your formula is the exponential growth. Exponential growth is very easy to project forward from any given level. So, to begin, let's re-write the formula to put the exponential growth by itself on the right-hand-side. Let's also add a time subscript to make it easy to keep track of:

\begin{align} \text{condition}_t &= 21 - exp(0.06 \cdot \text{age}_t)\\ (21-\text{condition}_t) &= exp(0.06 \cdot \text{age}_t) \end{align}

Now, suppose we had the slightly easier task of forecasting future $(21-\text{condition}_t)$ given current $(21-\text{condition}_t)$. Let $\tau > t$ be a future time period. This means that $\text{age}_\tau-\text{age}_t=\tau-t$:

\begin{align} (21-\text{condition}_\tau) &= exp(0.06 \cdot \text{age}_\tau)\\ &= exp \left(0.06 (\text{age}_t + \text{age}_\tau-\text{age}_t) \right)\\ &= exp \left(0.06 \cdot \text{age}_t \right) \cdot exp \left(0.06 (\text{age}_\tau-\text{age}_t) \right)\\ &= (21-\text{condition}_t) \cdot exp \left(0.06 (\text{age}_\tau-\text{age}_t) \right)\\ &= (21-\text{condition}_t) \cdot exp \left(0.06 (\tau-t) \right) \end{align}

So, this is easy. To forecast $(21-\text{condition}_t)$ into the future, just take the current value of it and multiply by $e$ to the power 0.06 times the number of years into the future. Finally, to forecast condition itself into the future, we just solve the above equation for it:

\begin{align} (21-\text{condition}_\tau) &= (21-\text{condition}_t) \cdot exp \left(0.06 (\tau-t) \right)\\ \text{condition}_\tau &= 21 - (21-\text{condition}_t) \cdot exp \left(0.06 (\tau-t) \right) \end{align}

And that's it.

If you wanted to put some probability and statistics into the question, there are two ways you could go. First, as I said above, we could try to account for the uncertainty in estimating the coefficient on age, i.e. the rate of decay. Another way we could go is we could hypothesize that the condition variable in the first equation you gave is actually the expected value of condition. That is that the equation you gave is not quite correct. The correct equation is:

\begin{align} E \left\{ \text{condition}_t \right\} &= 21 - exp(0.06 \cdot \text{age}_t) \end{align}

With an additional assumption (called "independent increments") the final equation we wrote above could be re-derived as:

\begin{align} E \left\{ \text{condition}_\tau | \text{condition}_t \right\} &= 21 - (21-\text{condition}_t) \cdot exp \left(0.06 \cdot (\tau-t) \right) \end{align}

That is, that the equation is describing the expected future condition of the road. The actual future condition of the road could be better or worse than this, depending on unmodeled, random (at least from our point of view) factors---weather, traffic, quality of maintenance, &c.

So, you might ask yourself, can I make a confidence interval for future condition? Can I say something like, "my expectation is that future condition at some specified future date will be 17, and I am 90% sure that future condition will be between 16 and 18." To do this, we need a model of the evolving random variable $\text{condition}_t$. Models like this are called "stochastic processes."

I don't think this is actually what you are after, so I will stop here, but what you would probably want to do to go further is to assume that $ln \left( 21 - \text{condition}_t \right)$ is a Poisson process (the most famous stochastic process). Then you could derive the confidence interval I mention above.

As an aside, the original formula has the unfortunate property that it produces negative numbers for $\text{condition}_t$ after a while. So, I'm wondering if that is really the right formula to use.

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  • $\begingroup$ Thanks for the excellent answer Bill. A question: I'm afraid I don''t understand the time subscript part. Can you recommend a resource that explains this concept? $\endgroup$ – Wilson Dec 6 '17 at 1:58
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    $\begingroup$ Sure, $\text{condition}_t$ just means the road condition in year $t$. Like $\text{condition}_{2015}$ would mean the road condition in 2015. $\text{condition}_\tau$ just means condition in a different year, which I am calling $\tau$, said "tau," like towel except without the "el." $\endgroup$ – Bill Dec 6 '17 at 21:56
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    $\begingroup$ Try googling "understanding subscript notation" without the quotes. I found several nice explanations that way. $\endgroup$ – Bill Dec 6 '17 at 21:59
  • $\begingroup$ Thanks Bill. This related question might be of interest: Vertically translated depreciation curve: Update the exponential regression coefficient $\endgroup$ – Wilson Dec 13 '17 at 1:10

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