Estimating Life Expectancies of Highway Assets, Volume 1: Guidebook (2012)

Below is the uncorrected machine-read text of this chapter, intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text of each book. Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages.

89 Many of the objectives of life expectancy analysis go beyond the simple calculation of life span. Agencies that gather the necessary data and perform the analysis can benefit in many more ways by constructing useful applications that go farther, to help in developing and selecting policies, planning future work programs, and developing cost-effective designs and projects. This is all a part of the advancements in asset management maturity. This chapter will show how tools built on top of the same building blocks as life expectancy analysis can fill the gap between management needs and data collection. Such applications help to turn data into useful information, which in the hands of proactive management can improve the agency’s efficiency and effectiveness in accomplishing its mission (Figure 5-1). This chapter presents the main building blocks: deterioration models, equivalent age, life extension, asset life, and lifecycle cost. It then presents some sample applications. It concludes with guidance on the process for designing, developing, and refining life expectancy applications. 5.1 Deterioration Models and Life Expectancy Chapter 4 showed the most direct ways to proceed from available data to estimates of life expectancy for the most common types of highway assets. In most cases, agencies will want to go farther, to put their knowledge of life expectancy to work to assist with asset management decision-making, for example, by developing additional tools. Life expectancy is just a part of a larger investigation of deterioration (Figure 5-2). For pavements and bridges, deterioration models have become quite mature, are very widely used, and often form the basis of life expectancy estimates; but deterioration models can be developed for any type of asset, by building on the methods already covered. Deterioration models are used to forecast decline in condition in the absence of corrective action by the agency. More general than life expectancy models, they forecast not only the end-of-life, but all other possible condition levels as well. In many cases, agencies determine life expectancy from their deterioration models. The existence of a deterioration model can improve the accuracy and/or precision of life expectancy estimates. 5.1.1 Regression of Condition Deterministic models are among the oldest techniques in use for deterioration modeling. These models directly predict the most likely value of a condition measure as a function of age and other explanatory variables. This is done by means of a straight or curved line, whose shape and parameters are set by a regression process. C h a p t e r 5 Develop Applications: How to Apply Life Expectancy Models

90 estimating Life expectancies of highway assets Deterministic models were popular before 1960 when they were developed by the AASHO Road Test (before AASHO became AASHTO) for pavements (Patterson 1987). This produced the iconic shape (Figure 5-3) that is still associated with all types of infrastructure deterioration, even though the original AASHO curve is rarely used today. The equation for the AASHO curve is p p p p t t f= − −( )  0 0 ρ α where pt is performance at time t; p0 is initial performance; pf is terminal performance; t is the year of the forecast; r is lifespan; and a is the shaping parameter. Because the basic model lacks explanatory variables, it is easy to develop. The life span esti- mate can be produced by any of the methods discussed so far in this guide and can be a func- tion of explanatory variables or a partitioned data set if desired. The shaping parameter can be determined by linear regression, if life span is known and is not believed to vary much from one agency to another. Subsequent enhancements made the curve “s-shaped” so it would approach the terminal performance level asymptotically. Figure 5-1. Applications put the models to work on day-to-day asset management problems (Patidar et al. 2007). Figure 5-2. Life expectancy as a part of deterioration modeling.

Develop applications: how to apply Life expectancy Models 91 Subsequent research efforts have developed various forms of non-linear deterioration curves that resembled the AASHO curve and could include explanatory variables such as traffic and climate, but which still can be estimated using simpler linear regression techniques. Life expectancy could be read off the curve where it intersected the minimum tolerable condition level or by inverting the equation to make age a function of condition. Despite the apparent simplicity of the deterministic deterioration curve, the model has some drawbacks which limit the ability to improve it or build useful applications with it: • It requires a continuous variable as the condition measure. Many agencies use PCR and/or IRI as condition measures for pavements. A few agencies use retroreflectivity for signs and pavement markings. Other than these examples, the use of continuous condition measures is relatively unusual in asset management, due to the cost of data collection using specialized equipment. • Commonly used regression models assume that variability is constant along the regression line and produce very little useful output about uncertainty. The assumptions about variability are often far off the mark and cause severe inaccuracies in the models. For many of the most useful applications, information about uncertainty is a necessity. • If life expectancy is the main output desired, there are even simpler ways of estimating it that do not require a regression model. Chapter 4 described those methods. A great many regression models for bridge deterioration, based on NBI condition ratings, can be found in the research literature. In the relatively few cases where these models have been tested and validated, they have not performed well because the NBI rating scale is discrete, not continuous. The research work accompanying this guide investigated the issue and found that other types of models produced more accurate forecasts. Even for continuous condition measures, linear regression models can be problematic. For example, a regression model of past pavement performance data from a North Carolina DOT division was developed by the researchers. The dependent variable was PCR, measured on a range of 0–100 where 0 means worst and 100 means best. The resulting equation was PCR Age Jurisdiction A= − + +90 33 2 94 5 37 0 112. . . .  DT Resurfacing_Thick+ 2 27.  where Age = age in years since last resurfacing; Jurisdiction = 1 if sub-divisional; 0 if rural; ADT = average daily traffic in thousands; Resurfacing_Thick = thickness of last resurfacing in inches. 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 Pa ve m en t c on di tio n Age Figure 5-3. AASHO Road Test pavement deterioration model.

92 estimating Life expectancies of highway assets The model’s adjusted R2 was found to be 0.30 with significant autocorrelation problems, suggesting that linear regression is not an appropriate life expectancy method for the given data in the example application. 5.1.2 Markov Models Most common asset management processes use categorical data, which classify condition into a relatively small number of categories. In part, this results from the use of visual inspection tech- niques, which can only discern a few gradations reliably. Another motivation is the common pop- ular use of categorical value measures, such as good-fair-poor, or A-B-C-D-F. The simplest commonly used deterioration modeling technique for this type of data is the Markov model. As explained in Chapter 4, a Markov model requires consistent use of a condition state assess- ment scheme and a uniform time interval between observations and assumes that the probability of making a transition from one condition state to another depends only on the initial state, rather than on age, past conditions, or any other information about the element. Thus, the model is expressed as a simple matrix of probabilities (Table 5-1). A Markov model is a cross-sectional model, able to be developed from a population of assets even if they have not been inspected consistently over their whole lives. This is especially useful for structures whose lives can extend to 50–100 years or more, where a full time series data set is not obtainable. A Markovian transition probability matrix is a special type of matrix with a number of desirable properties that make it easy to process. A well-formed transition probability matrix adheres to the following rules: 1. Square matrix—All transition probability matrices are square, with the number of rows and the number of columns both equal to the number of possible condition states. 2. Upper-right triangular—Only the main diagonal and the upper-right triangle of the matrix are allowed to have non-zero values. This is another way of saying that there can be no movement from any condition state to a better state in a deterioration model. 3. Non-negative—No elements of the matrix may be negative. 4. Positive diagonal—Elements on the diagonal must be non-zero. In other words, there must be a non-zero possibility of remaining in the same condition state from one inspection to the next. 5. Normalized—All rows of the matrix must separately sum to 100%. In other words, the transition probability matrix must account for all possible transitions. 6. Because of the combination of these rules, the lower right corner element must be 100%. Once an asset deteriorates to the worst condition state, it stays there. Probability of each condition state one year later (%) 1 2 3 4 5 Co nd iti on s ta te n o w 1 95.3 4.6 0.1 0 0 2 0 93.2 3.9 1.9 1.0 3 0 0 89.4 7.3 3.3 4 0 0 0 82.8 17.2 5 0 0 0 0 100 Table 5-1. Example Markov deterioration model.

Develop applications: how to apply Life expectancy Models 93 A transition probability matrix can have as few as two condition states, such as pass/fail. It commonly has four or five states for most types of assets. For pavements, there are examples in Arizona, Kansas, and Finland of more than 100 condition states. In those cases, condition is measured on multiple dimensions. For example, if there are five states of roughness, five states of cracking, and five states of rutting, then the deterioration model has 125 rows and 125 columns. Conditions in any future year can be predicted with a Markovian model by simple matrix multiplication. Mathematically, the matrix multiplication for Markovian prediction, when no maintenance action is taken, looks like this: y x p kk j jk j = ∑ for all where xj is the probability of being in condition state j at the beginning of the year; yk is the prob- ability of being in condition state k at the end of the year; and pjk is the transition probability from j to k. This computation can be repeated to extend the forecast for additional years. An example of this computation was shown above in Table 4-28. It is possible to derive transition probabilities if the median number of years between transitions is known. Often this is an appropriate way to develop a deterioration model from expert judgment. It also provides a convenient means of computing, storing, and reporting transition probabilities derived from historical inspection data. If it takes t years for 50% of a population of elements to transition from state j to state k = j + 1, and no other transitions are possible, then the one-year transition probabilities are p p pjj t jk jj= = −( )0 5 11. and So if it takes a median of 10.23 years to transition from state 1 to state 2, then the probabilities after 1 year are 93.4% for state 1 and 6.6% for state 2. 5.1.2.1 Data Preparation The first step in developing a Markov model is to gather a set of inspections for a large group of assets. Each asset must have at least two inspections. It is not necessary to be able to follow any one asset through its whole life, but it is necessary for all possible condition states to be observed somewhere in the data set. Inspections are grouped into pairs, each pair showing the change in condition of an asset (or bundle of assets, such as all the traffic signal heads in one intersection, or all the girders on one bridge) over a period of time. Each inspection can be the beginning of one or more pairs and the end of one or more pairs. The pairs must be uniform in length, commonly either 1 year or 2 years, plus or minus 6 months. If the inspection intervals in the data set are not uniform, it is possible to interleave inspection pairs (Figure 5-4). The deterioration model is intended to describe changes in condition if no agency action is taken to try to improve the condition of the asset. Therefore, it is necessary to remove from the data set any pairs that had agency corrective action between the two inspections. One way to determine this fact is to consult the agency’s information systems where records of past activities are maintained. In practice, this is often an imperfect record of activity. Figure 5-4. Interleaved inspection pairs.

94 estimating Life expectancies of highway assets Another way to detect possible repair activity is to look for improvements in condition. The following formula can be calculated for each inspection pair: IC y xj k k j k k j = −     = = ∑ ∑max 1 1 where IC is the improvement in condition for the inspection pair; j and k are condition states defined for the asset that was inspected (assuming that k = 1 is the best possible condition state); maxj indicates maximization over all possible condition states of the asset; yk is the fraction of the asset in condition state k in the second inspection of the pair; and xk is the fraction of the asset in condition state k in the first inspection of the pair. This equation quantifies improvement as the increase in the fraction at, or better than, any given condition state. Computed over all condition states, the largest increase is selected to represent the inspection pair as its maximum condition improvement. If any one or more of the condition states shows an increase in the fraction at its level or better, then IC is positive. This can indicate either that an error occurred in the inspection process or a preservation activity took place. In the absence of reliable maintenance records, the analyst will often need to assume that all positive IC values indicate repair activity and will remove all such pairs from the data set. 5.1.2.2 Linear Regression One relatively easy way to determine the transition probabilities from the list of inspection pairs is linear regression (Cambridge 2003), using the following steps. Conditions at the beginning of the period: X x x x x xi i i i i[ ] = { }1 2 3 4 5, , , , for all inspection pairs i Conditions at the end of the period: Y y y y y yi i i i i[ ] = { }1 2 3 4 5, , , , for all inspection pairs i These are the known values in the estimation equation. The prediction equation is Y P X[ ] = [ ][ ] where [P] is the transition probability matrix. The unknown transition probabilities can be estimated: P XX XY[ ] = [ ] [ ]−1 Matrix of XX sums: XX x xji ki i [ ] = ∑ for all combinations of j and k Matrix of XY sums: XY x yji ki i [ ] = ∑ for all combinations of j and k The exponent on [XX]-1 indicates matrix inversion. Following the regression computation, the resulting matrix is normalized to ensure that it satisfies the rules of a well-formed transition probability matrix. Any values to the left of the diagonal are set to zero. If any diagonal elements

Develop applications: how to apply Life expectancy Models 95 are less than 0.01, they are changed to 0.01 (or some other small positive value). Negative values to the right of the diagonal are set to zero. Then each row is adjusted to sum to 1.0: ′ = = ∑p p s s pjk jk j j jk k A strong point of the regression method is that it can estimate the probabilities of transition from any starting state to any worse state. The upper-right triangle of the matrix can consist of all positive numbers. This is useful for short-lived assets where a jump of two or more condition states is not unusual between inspections. A weakness of the method is that it is subject to various numerical problems with the matrix inversion step, which can yield incorrect results or fail to produce a result. Thus, the results need careful scrutiny for reasonableness. 5.1.2.3 One-Step Method For long-lived assets, where the inspection interval is short in comparison to the life span, jumps of more than one state at a time may be unusual. In fact, it may be impossible if only two states, such as pass/fail, are used. In this case, the estimation process can be simplified into the one-step method (Thompson and Sobanjo 2010). To set up the estimation of a one-step matrix, the prediction equation is defined as follows, for an example with four condition states: y y y y p p p p p p 1 2 3 4 11 12 22 23 33 3 0 0 0       = 4 44 1 2 3 4p x x x x             Writing out the individual equations necessary to calculate [Y] results in y x p y x p x p y x p x p y x p 1 1 11 2 1 12 2 22 3 2 23 3 33 4 3 3 = = + = + = 4 4 44+ x p Given that the sum of each row in [P] must be 1.0, the following additional equations apply: p p p p p p12 11 23 22 34 331 1 1= − = − = −; ; The vectors [X] and [Y] can be computed from the database of inspection pairs to describe the combined condition of the element before and after: X N x Y N yji i N j i i N[ ] = [ ] = = = ∑ ∑1 1 1 1 , for all condition states j where N is the number of inspection pairs. So the [X] and [Y] vectors are known. Thus, the sys- tem of seven equations and seven unknowns can be solved algebraically for the elements of [P]. This same pattern applies for any number of condition states. Table 5-2 shows an example of the one-step method. The first section is the table of original inspection pairs, showing the data preparation to eliminate pairs that improved in condition.

96 estimating Life expectancies of highway assets Inspection pairs Condition - start of year Condition - end of year Improvement in condition Road Insp Condition state Condition state Condition state segment Year 1 2 3 4 1 2 3 4 1 2 3 4 Screening RS0028 2004 92 8 0 0 82 17 1 0 -10 -1 0 0 RS0028 2005 82 17 1 0 68 27 4 1 -14 -4 -1 0 RS0028 2006 68 27 4 1 58 32 9 1 -10 -5 0 0 RS0028 2007 58 32 9 1 48 37 11 4 -10 -5 -3 0 RS0028 2008 48 37 11 4 46 35 12 7 -2 -4 -3 0 RS0028 2009 46 35 12 7 37 39 14 10 -9 -5 -3 0 RS0028 2010 37 39 14 10 32 37 19 12 -5 -7 -2 0 RS0061 2005 100 0 0 0 84 16 0 0 -16 0 0 0 RS0061 2006 84 16 0 0 78 19 3 0 -6 -3 0 0 RS0061 2007 78 19 3 0 67 27 5 1 -11 -3 -1 0 RS0061 2008 67 27 5 1 65 23 10 2 -2 -6 -1 0 RS0061 2009 65 23 10 2 55 28 12 5 -10 -5 -3 0 RS0061 2010 55 28 12 5 53 24 15 8 -2 -6 -3 0 RS0035 2004 83 10 5 2 75 17 5 3 -8 -1 -1 0 RS0035 2005 75 17 5 3 68 20 7 5 -7 -4 -2 0 RS0035 2006 68 20 7 5 63 24 7 6 -5 -1 -1 0 RS0035 2007 63 24 7 6 52 32 7 9 -11 -3 -3 0 RS0035 2008 52 32 7 9 43 36 10 11 -9 -5 -2 0 RS0035 2009 43 36 10 11 37 39 11 13 -6 -3 -2 0 RS0035 2010 37 39 11 13 33 34 18 15 -4 -9 -2 0 RS0011 2005 29 21 18 32 25 22 17 36 -4 -3 -4 0 RS0011 2006 25 22 17 36 24 18 18 40 -1 -5 -4 0 RS0011 2007 24 18 18 40 100 0 0 0 76 58 40 RS0011 2008 100 0 0 0 83 17 0 0 -17 0 0 0 RS0011 2009 83 17 0 0 77 22 1 0 -6 -1 0 0 RS0011 2010 77 22 1 0 73 23 3 1 -4 -3 -1 0 RS0001 2003 100 0 0 0 86 14 0 0 -14 0 0 0 RS0001 2004 86 14 0 0 75 22 3 0 -11 -3 0 0 RS0001 2005 75 22 3 0 63 31 5 1 -12 -3 -1 0 RS0001 2006 63 31 5 1 62 26 10 2 -1 -6 -1 0 RS0001 2007 62 26 10 2 51 33 12 4 -11 -4 -2 0 RS0001 2008 51 33 12 4 49 33 10 8 -2 -2 -4 0 RS0001 2009 49 33 10 8 42 36 11 11 -7 -4 -3 0 RS0001 2010 42 36 11 11 35 36 15 14 -7 -7 -3 0 RS0004 2006 24 18 15 43 21 18 15 46 -3 -3 -3 0 RS0004 2007 21 18 15 46 18 17 15 50 -3 -4 -4 0 RS0004 2008 18 17 15 50 16 15 15 54 -2 -4 -4 0 RS0004 2009 16 15 15 54 90 10 0 0 74 69 54 0 Delete 0 Delete RS0004 2010 90 10 0 0 79 19 2 0 -11 -2 0 0 RS0016 2006 81 14 4 1 76 18 4 2 -5 -1 -1 0 RS0016 2007 76 18 4 2 62 30 5 3 -14 -2 -1 0 RS0016 2008 62 30 5 3 59 29 7 5 -3 -4 -2 0 RS0016 2009 59 29 7 5 55 31 8 6 -4 -2 -1 0 RS0016 2010 55 31 8 6 51 28 13 8 -4 -7 -2 0 Change in condition for segments where no work done Condition at start Condition at end 1 2 3 4 1 2 3 4 Avg by state 62.6 22.6 7.0 7.9 55.4 26.2 8.8 9.6 Computed transition probabilities using One-Step Method Condition state probabilities 1 2 3 4 Stay in same state 88.5 84.2 74.7 100 Deteriorate one step 11.5 15.8 25.3 0.0 Table 5-2. Example of the one-step method of estimating Markov models.

Develop applications: how to apply Life expectancy Models 97 The second section contains the [X] and [Y] vectors, and the third section shows the results, the non-zero members of the transition matrix. 5.1.2.4 Life Expectancy from Markov Deterioration Chapter 4 showed how any set of condition states can be collapsed into two states, failed and not-failed, after which a version of the one-step method can be used to compute transition prob- abilities and life expectancy, with the formula t pjj = ( ) ( ) log . log 0 5 This method was called “quick and easy” mainly because of the collapsing of condition states, which then requires the assumption that all assets in the not-failed state are equally likely to fail in the next year. A Markov model for the full set of condition states improves on this result because only the assets currently in the second-to-last condition state are in position to possibly reach the worst state in the following year. If the not-failed assets are currently concentrated in the best condi- tion state, it will be many years before very many of them reach the worst state. As a result, life expectancy forecasts made with the help of a fully-developed Markov model can be more accu- rate than the quick-and-easy method. To calculate life expectancy from a Markov transition probability matrix, start with an asset in perfect condition and repeatedly multiply by the transition probability matrix until 50% of the asset is in the worst condition state. Table 4-28 shows an example. 5.1.3 Markov/Weibull Models In Chapter 4, the Weibull survival probability model was used to give an age-dependent probability of failure for the failed/not-failed scenario as an enhancement of the Markov model. The Weibull model can play a similar role in a deterioration model. One useful application for this enhancement is in modeling the onset of deterioration, the transition from the best condition state to the second-best state. This is analogous to the transi- tion from the not-failed state to the failed state and is mathematically the same model. The only difference is that the median state transition time is used instead of the median life expectancy. As shown in Figure 5-5, the Markov model features a rather quick decline in condition, even for a brand new asset, an effect not often observed in practice. The Weibull model can slow the onset of deterioration, making the initial stages of the deterioration model more realistic. Another useful application for the Weibull model in life expectancy analysis is in modeling the transition from the second-worst condition state to the worst (failed) state. The Markov model provides a median transition time, but the Weibull model can refine this estimate and provide a measure of uncertainty in the time to failure. So for assets already in the second-worst state, the Weibull model can provide an estimate of what fraction of them will fail within a defined time period, such as a 2-year budget or a 10-year program horizon. This can help to make budgeting more accurate. The methods for computing these estimates are the same as described in Chapter 4. It is possible to develop a completely age-dependent Weibull survival probability deterioration model if all of the individual state transitions can be analyzed independently, that is, if each asset is in only one condition state at a time and can move to only one other state between inspections. These conditions do not hold true for bridges, where AASHTO CoRe Elements are described as a distribution of members among condition states (with the notable exception of New York in

98 estimating Life expectancies of highway assets Agrawal and Kawaguchi 2009). For other types of assets, tracking each individual piece of equip- ment separately may involve more data collection and management than most agencies would want to undertake. For most cases, securing an age-dependent deterioration model requires more powerful tools. 5.1.4 Ordered Probit Another condition-based approach that could be used by agencies for deterioration predic- tion is the ordered probit model. This model can be used to produce age-dependent perfor- mance curves for assets with discrete, ordered condition states such as the NBI 0–9 rating system. The likelihood of being in any condition state at a time t can be determined as a function of a set of life expectancy factors, an asset’s age, and a set of threshold parameters. These threshold parameters, µ, serve as a sort of boundary between condition states. For instance, consider the pipe culvert 0–3 rating scale discussed in Section 4.1.1.5. Depending on the model sum (Sbx) and the threshold parameters (µ), the probability of being in a condi- tion state will differ according to a normal distribution (Figure 5-6). Mathematically, the exact probability of an asset being in any condition state follows the cumulative standard normal distribution with the variable X taking the following forms: P Condition State x N P Condition =( ) = −[ ] ( )∑0 0 1β ∼ , State x N x N P Co =( ) = −[ ] ( ) − −[ ] ( )∑ ∑1 0 1 0 11µ β β∼ ∼, , ndition State x N x=( ) = −[ ] ( ) − −[ ]∑ ∑2 0 12 1µ β µ β∼ ∼, N P Condition State x N 0 1 3 1 0 12 , , ( ) =( ) = − −[ ] (∑µ β ∼ ) where P(Condition State = i) = Predicted probability of an asset being in condition state i; x = set of independent variables, age, material type, etc.; b = set of parameter estimates corresponding to independent variables; 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 20 25 30 35 Age of element (years) Pr ob ab ili ty o f s ta te 1 Markov (Beta=1) Beta=2 Beta=4 Beta=8 Figure 5-5. Comparison of the Markov and Weibull models.

Develop applications: how to apply Life expectancy Models 99 µ = threshold parameters, which in comparison to parameter estimates and variable values, indicate the likelihood of being in a given con- dition state; [µ - S bx] = X value that can be used to calculate normal distribution test sta- tistic via Z X Mean S dard Deviation = − tan N(0, 1) = indicates the cumulative, standard normal distribution with mean = 0 and standard deviation = 1 By using age as an independent variable in the model, it is possible to make a condition state prediction for each asset across every feasible age while holding all other variables constant. For instance, suppose we calibrated an ordered probit model for pipe culverts with β µ µ µx = = = =∑ 2 444 0 1 116 2 2210 1 2. ; ; . ; . Using the model, the probability of the culvert asset being in each condition state can be determined as follows: P Condition State N=( ) = −[ ] ( )0 2 444 0 1. ,∼ Using Excel, =NORMDIST 0,1,1−( ) =2 444 0 0073. , . P Condition State N N=( ) = −[ ] ( ) − −[ ]1 1 116 2 444 0 1 2 444. . , .∼ ∼ 0 1 1 116 2 444 , . . ( ) −Using Excel, =NORMDIST , 0,1,1( ) − −( ) =NORMDIST , 0,1,12 444 0 0848. . P Condition State N=( ) = −[ ] ( ) − −2 2 221 2 444 0 1 1 116 2 444. . , . .∼ [ ] ( ) − ∼ N 0 1 2 221 2 444 , . .Using Excel, =NORMDIST , 0,1,1 NORMDIST , 0,1,1( ) − −( ) =1 116 2 444 0 3198. . . P Condition State N=( ) = − −[ ] ( )3 1 2 221 2 444 0 1. . ,∼ Using Excel, = NORMDIST , 0,1,11 2 221 2 444 0− −( ) =. . .5881 Figure 5-6. Example illustration of an ordered probit model with µ0 = 0 for pipe culverts (after Washington et al. 2003). P(Condition State 1) P(Condition State 2) P(Condition State 3)P(Condition State 0)

100 estimating Life expectancies of highway assets In other words, the culvert in this specific example is considered to have a 0.73% chance of being in condition state 0, an 8.48% chance of being in condition state 1, a 31.98% chance of being in condition state 2, and a 58.81% chance of being in condition state 3. Therefore, the most likely condition state for this asset is condition state 3. If the same calculations are repeated for different ages resulting in different model sums, then a performance curve like the one in Figure 5-7 can be obtained. 5.1.5 Machine Learning An even more mathematically complex technique for life prediction that has gained popu- larity among some researchers that could be considered by agencies is machine learning. Essentially, this non-linear, adaptive model predicts conditions based on what it has “learned” (pattern identification) from past data. Statistically, an artificial neural network (the most common learning technique) is a non-linear form of 3-Stage Least Squares regression, where “instruments” (variables used to represent relationships between other variables) are estimated to predict future events (Figure 5-8). To facilitate learning, such models are typically Bayesian-based. This approach updates estimates (i.e., posterior means) by applying weighted averages based on previous estimates 0 1 2 3 0 20 40 60 80 Co nd iti on S ta te Age in Years Figure 5-7. Example pipe culvert performance function using the ordered probit model. Input 1 Input 2 Input n Instru- ment 1 Instru- ment n Output Figure 5-8. Example of an artificial neural network.

Develop applications: how to apply Life expectancy Models 101 (i.e., prior means). Typically, these weights are based on the number of observations. Activation functions within the network have included hyperbolic tangent, log-sigmoid, and bipolar- sigmoid functions. Such approaches have been found to work well with noisy data and are rela- tively quick; however, such techniques are better suited for smaller databases (Melhem and Cheng 2003). These models require more sophisticated software to develop and sometimes can be used as a “black box” (i.e., prediction process unknown but assumed appropriate). However, the abil- ity to “learn” makes these models particularly useful to asset managers in applications where it is necessary to adjust predictions in real time in response to new data, such as inputs from moni- toring systems. Such an approach is outside the scope of this guide, but interested managers may want to consider applying machine learning to their databases. 5.1.6 Mechanistic Models This guide emphasizes empirical models; however, some agencies prefer to define life more directly in terms of structural response. For instance, bridge life may be reached at the time the reliability of members to resist shear and strain stresses reaches a threshold level. Another exam- ple would be predicting pipe culvert life based on the time until corrosion based on the resistivity of a material to chloride ions. Such techniques may be difficult to apply at the network level, requiring extensive data on asset dimensions and conditions, and do not account for alternative replacement rationale. Regardless, the condition-based and interval-based methodologies proposed in this guide can still apply to the results of mechanistic models. 5.2 Building Blocks of Life Expectancy Applications The techniques of life expectancy analysis and deterioration models open the door for various useful applications to support asset management decision-making. Before introducing these applications, it is useful to develop a few additional building blocks that make it easier to under- stand and construct these applications. 5.2.1 Equivalent Age Deterioration models are often used to convert from the age of an asset to a forecast of its condition. But many applications also need the opposite capability, namely, to convert from a known condition to an equivalent age. How this is done depends on the type of deterioration model used. 5.2.1.1 Deterministic Models For a deterministic model of condition versus age, such as the AASHO curve, it is usually a simple matter to read the age from the curve for any level of condition. Many functional forms can be inverted to make age the dependent variable of the equation. Even if this is not possible, the equivalent age can be found numerically by iterating through the range of possible ages until the desired condition level is found. 5.2.1.2 Markov Models Converting from a condition state representation to an equivalent age is somewhat more challenging. If an individual asset is rated in just one state or another at a given point in time, then its condition state may correspond to a range of years in the deterioration model. In the more common case where the unit of analysis is a bundle of assets, multiple condition states may be included in the bundle. In that case, the equivalent age depends on the relative fractions in each condition state.

102 estimating Life expectancies of highway assets One way to minimize the complication is to use the Markov prediction formula iteratively until the 50% failure criterion is reached. As long as the asset has not already reached its life expectancy, the remaining life can be determined in this way and then subtracted from the life expectancy to compute equivalent age. To forecast the condition state 1 year following a known condition, the formula is y x p kk j jk j = ∑ for all where xj is the probability of being in condition state j at the beginning of the year; yk is the prob- ability of being in condition state k at the end of the year; and pjk is the transition probability from j to k. This computation can be repeated to extend the forecast for additional years until the failed percentage reaches 50%. (Table 4-28 showed an example.) 5.2.1.3 Weibull Model For a Weibull survival probability model, equivalent age is easily calculated from the inverse of the Weibull prediction formula. The Weibull curve has the following functional form: y gg1 1 0= − ×( )( )exp . α β where y1g is the probability of the not-failed state at age g, if no intervening maintenance action is taken between year 0 and year g; b is the shaping parameter; and a is the scaling parameter, calculated as α β= ( ) t ln2 1 where t is the median life expectancy from the Markov model. This is calculated in the same way as described earlier in this chapter. The inverse of the Weibull formula is ′ = × − ( )( )   ∧g y α β10 1log ln This yields the age that is equivalent to the given non-failed fraction y1. 5.2.1.4 Convert a Markov Model to a Weibull Model Another way to calculate equivalent age for a Markov model is to develop a function to convert a condition state description of condition into a condition index representative of the equivalent point in its life span. Then the inverse Weibull formula, as presented in the preceding section, can be used to estimate the equivalent age based on the condition index at any point in time. This function would be applied to the known asset condition to simplify its representation and would also be applied to conditions forecasted by the deterioration model. In this way, the transition probability matrix is presented in the form of a linear depiction of the change in median condition over time, and any known condition state representation can be converted to a point on that line, making it possible once again to read off the equivalent age directly. The steps to do this are as follows (Table 5-3): 1. Develop the Markovian transition probability matrix using the tools described earlier in this chapter. Either the linear regression method or the one-step method will work.

Develop applications: how to apply Life expectancy Models 103 2. Convert each row of the matrix to median transition time, using the familiar formula t p j jj = ( ) ( ) log . log 0 5 where pjj is the probability of remaining in the same condition state from one year to the next, and tj is the median amount of time spent in condition state j before moving on to condition state j + 1. Table 5-3. Model to estimate equivalent age from a Markov model. Step 1: Transition probabilities (One-Step Method, from previous example) Condition state probabilities 1 2 3 4 Stay in same state 88.5 84.2 74.7 100 Deteriorate one step 11.5 15.8 25.3 0.0 Step 2: Median transition time Sum Median transition time 5.66 4.02 2.38 12.1 t is median transition time p is same-state transition probability Step 3: Condition state weights State weight 1.00 0.53 0.20 0.00 w is weight given to each state t is median transition time N is the total number of condition states Step 4: Condition index forecast Step 5: Equiv age model Coefficient Value Condition state probabilities Cond Predict Square Log Index at life expectancy Year 1 2 3 4 index age deviation likelihood Life expect 16.00 0 100 0.0 0.0 0.0 100.0 0.00 0.0000 1.2688 Predicted life 16.00 1 88.5 11.5 0.0 0.0 94.6 1.26 0.0694 -1.4902 Equiv age parameters 2 78.3 19.9 1.8 0.0 89.2 2.27 0.0731 -1.6377 Scaling (alpha) 13.25 3 69.3 25.8 4.5 0.5 83.8 3.23 0.0538 -0.8676 Shaping (beta) 1.2297 4 61.3 29.7 7.4 1.6 78.5 4.18 0.0327 -0.0288 Std deviation 0.1122 5 54.2 32.0 10.3 3.5 73.2 5.13 0.0168 0.6024 Sum LogLike 19.989 6 48.0 33.2 12.7 6.1 68.1 6.08 0.0070 0.9906 7 42.4 33.5 14.8 9.3 63.1 7.05 0.0021 1.1868 8 37.5 33.1 16.3 13.0 58.3 8.02 0.0002 1.2594 9 33.2 32.2 17.5 17.2 53.7 8.99 0.0000 1.2670 10 29.4 30.9 18.1 21.6 49.4 9.98 0.0005 1.2509 g is equivalent age 11 26.0 29.4 18.4 26.1 45.3 10.97 0.0009 1.2349 y is condition index 12 23.0 27.8 18.4 30.8 41.4 11.97 0.0010 1.2294 13 20.4 26.0 18.2 35.5 37.8 12.97 0.0008 1.2357 14 18.0 24.2 17.7 40.0 34.4 13.98 0.0005 1.2491 15 15.9 22.5 17.1 44.5 31.2 14.99 0.0002 1.2627 16 14.1 20.8 16.3 48.8 28.3 16.00 0.0000 1.2688 17 12.5 19.1 15.5 52.9 25.7 17.01 0.0002 1.2607 18 11.0 17.5 14.6 56.9 23.2 18.03 0.0009 1.2336 19 9.8 16.0 13.7 60.5 21.0 19.05 0.0021 1.1847 20 8.6 14.6 12.8 64.0 18.9 20.06 0.0039 1.1135 21 7.6 13.3 11.8 67.2 17.0 21.08 0.0062 1.0214 22 6.8 12.1 11.0 70.2 15.3 22.09 0.0090 0.9119 23 6.0 10.9 10.1 73.0 13.8 23.11 0.0121 0.7897 24 5.3 9.9 9.3 75.5 12.4 24.12 0.0153 0.6606 25 4.7 8.9 8.5 77.9 11.1 25.14 0.0186 0.5311 )log( )5.0log( jj j p t = ∑j tj 1 wj tk k=j N = CI wj xj j = ( )( )− ×=′ βα 1lnlog ^10 yg

104 estimating Life expectancies of highway assets 3. Each condition state will be allocated a portion of the asset’s life, in proportion to its transition time. So compute the weight of each condition state as w t tj j j k k j N = ∑ ∑= 1 This formula assumes that 1 is the best state and N is the worst. The weight given to the best state is 1.0, and the weight given to the worst state is 0, with all other states having weights in between. 4. Compute the equivalent condition index from the condition state distribution using the formula CI w xj j j = ∑ The forecasts from the transition probability matrix are run through this computation to generate a deterministic time series of condition index, starting at 1.0 for an asset in perfect condition and approaching zero as the asset ages long past its life expectancy. A Weibull model can be developed from this time series, using maximum likelihood estimation constrained so that the equivalent age equals the actual age at the asset life expectancy. Figure 5-9, which is based on the results of Table 5-3, shows that the Weibull model is an excellent fit to data generated in this way. The method is exactly the same as for the calculation of survival probability, but in this case it is used instead on a condition index. Once this model is developed, any condition state vector for the asset can be simplified to an equivalent age, even if the asset has already passed its life expectancy. 5.2.2 Life Extension Benefits of Actions Typically the effect of repair and rehabilitation actions is expressed as an improvement in condition. Once the improved condition is forecast, the methods in the preceding section can be used to calculate equivalent age, before and after the action. The difference in age is one way of expressing the benefit of the action (Figure 5-10). 0 5 10 15 20 25 30 0 20 40 60 80 100 Co nd iti on in de x Equivalent age Actual Predicted Figure 5-9. Condition index versus equivalent age from Table 5-3.

Develop applications: how to apply Life expectancy Models 105 5.2.3 Remaining Life One of the most obvious ways to compute remaining asset life is to subtract the actual age of an asset from its life expectancy. This method is valid if no life extension actions have been performed. However, if an asset has been repaired, or if its maintenance history is unknown, then it is more accurate to use a condition-based approach, taking advantage of the deterioration and equivalent age models presented in this chapter. Assuming that the asset’s life is not limited by impending functional needs or changes in standards, the current condition of the asset can be converted to its equivalent age, essentially finding its most likely place on the deterioration curve. This equivalent age is then subtracted from life expectancy to estimate remaining asset life (Figure 5-11). The equivalent age method works regardless of what preservation work may have been performed in the past, even if the past work is unknown. However, one limitation is that it assumes that no future work will be done. For many applications, this assumption is desirable. However, if the goal is to estimate when the asset will actually be replaced, then the possibility of future life extension actions must be considered. Models of repair feasibility and effectiveness are beyond the scope of this guide, but such models do exist and are widely used for pavements and bridges and could be developed for other assets, based on agency data and experience. If a life extension action is found to be feasible and if its condition benefit can be estimated, then the equivalent age method provides a direct estimate of the added life, as shown in Figure 5-12. Figure 5-11. Remaining asset life from current condition. Figure 5-10. Converting condition improvement to life extension benefit.

106 estimating Life expectancies of highway assets 5.2.4 Lifecycle Cost Models Adding lifecycle cost to life expectancy and deterioration models opens the door to a wealth of useful applications to support transportation asset management decision-making. Among the possible applications are the comparison of design and life extension alternatives, optimiz- ing replacement intervals, optimizing preventive maintenance, evaluating new maintenance materials and techniques, optimizing corridor planning, and responding effectively to funding constraints. Lifecycle cost models are key ingredients in asset management systems, such as for pavements and bridges; but the models are often simple enough that they can be implemented as Microsoft Excel spreadsheets. 5.2.4.1 Time Value of Money One of the key concepts of lifecycle cost analysis is the time value of money. In economic decision-making, people value near-term revenue and near-term costs more highly than money that changes hands years in the future. People are willing to pay interest in order to have access to money today that they might not otherwise see for many years. Agencies issue bonds and pay interest on those bonds. Future needs for money are less certain. Another key aspect of lifecycle cost is the timing of the decisionmaker’s cost and benefit horizon, and the timing of asset life expectancy. State Transportation Improvement Programs (STIPs) and budgets have defined time horizons, where accountability for costs and outputs is increased. Political terms in office are also limited. These factors tend to push costs into the future while concerns for outcomes are more immediate. Figure 5-13 depicts a typical set of cost streams for a bridge, showing how the choice of a discount factor affects the calculation of lifecycle cost. The top and bottom of the figure are two different lifecycle activity profiles (Hawk 2003), sets of agency actions timed according to deterioration, action effectiveness, and cost. Both profiles are feasible for the bridge, each having its own strengths and weaknesses. Alternative 1 features relatively frequent, but small, preventive maintenance activities. With a discount factor of 0.95, the difference in value between future costs and current costs is small, so there is more willingness to spend in the near term to gain long-term benefits, such as extension of the life of the structure. As a result, Alternative 1 has lower overall lifecycle cost at a discount factor of 0.95. At the lower discount factor of 0.90, on the other hand, Alternative 2 becomes more attractive. Figure 5-14 shows another example of lifecycle cost analysis, for replacement of traffic signal lamps. With the shorter time frame measured in weeks, discounting of future costs plays less of a role than for bridges. Yet, the economic considerations are substantial in comparing policies. Figure 5-12. Remaining life with future extension.

Develop applications: how to apply Life expectancy Models 107 Figure 5-13. Example of bridge lifecycle cost alternatives. Alternative #2 Figure 5-14. Example of cost streams for traffic signal lamp replacement.

108 estimating Life expectancies of highway assets The blanket replacement policy saves one million dollars by reducing the mobilization and traffic control costs of unplanned traffic signal failures. The optimal time for replacement depends on the width of the probability distribution, which is the level of uncertainty in the median failure time. If the timing of the blanket replacement policy were set too far to the left or the right, it could end up being more expensive than the response-driven policy. 5.2.4.2 Common Methods The mathematical formulas for computing lifecycle cost are well known in asset management applications. The discount factor is calculated from d i = + 1 1 where i is the real discount rate. Usually agencies set the discount rate for asset management pol- icy to be consistent with the cost of public-sector bond financing. Although inflation can either be included or not included, it is usually much simpler to omit inflation from all lifecycle cost computations. Most published reports about lifecycle cost omit inflation, which is generally the reader’s expectation. The missing inflation, of course, must be added back as a part of discussions of future nominal budgets. The discount rate should be consistent among all asset management applications. The present value of a one-time future cost or benefit is calculated from PV d FVn= × where n is the time interval between the base year of the analysis (usually the current year or the first year of a program), and the year when the cost or benefit is to be realized; and FV is the future value of the cost or benefit estimate for the time that it is realized (again omitting inflation). If a uniform annual series of costs or benefits is expected for an indefinite period of time into the future, this is called a perpetuity. The present value of a perpetuity is PV FV i = where FV is the annual payment, starting 1 year from the present. If the future uniform series is not annual (perhaps it is once every 2 years), it is simplest to change the discount rate to match the desired time interval. First calculate the equivalent dis- count factor, then apply the appropriate exponent for the desired time interval, then convert this back to a discount rate. For example, if i = 5%, then d = 0.9524. For a 42-month (3.5-year) interval, such as the replacement interval for a certain type of street lamp, the full discount is 0.95243.5 = 0.843. The corresponding discount rate i = (1-0.843)/0.843 = 18.624%. For a transportation facility with initial investment, P; compounded amount of all cash flows within a replacement lifecycle, R, and length of replacement lifecycle, N years, the present worth of all payments in perpetuity is given by, PW∞ = P + R/((1 + i)N - 1). However, in cases where the facility already exists, P is a sunk cost, and the present worth is PW∞ = R/((1 + i)N - 1). If a uniform annual stream of costs or benefits has a definite end, then the present value is PV FV i i n = − +( )−1 1

Develop applications: how to apply Life expectancy Models 109 Here FV is again the amount of the future recurring payment, starting 1 year from the present. If the stream of cash flows corresponds to the life span of an asset, then n is typically the median life expectancy of the asset. However, there are applications where n should be some other value, such as a proposed blanket replacement date that might be earlier than the life expectancy. If the uncertainty in the life expectancy is large, or if its variability is asymmetrical (e.g., minimally spread before the median, but widely spread afterward), then it may be more accurate to represent the cash flows individually rather than as a uniform stream. 5.2.4.3 Comparing Alternatives Using Net Present Value Net present value is a term used to describe the sum of all relevant costs and benefits at stake in a decision, with each cash flow discounted to the same year, usually the year of the decision. Lifecycle cost is usually understood to be a type of net present value. It is important to be clear on the definitions of what is and is not included in the computation. In lifecycle cost analysis, generally, two or more specific alternatives are compared, the decision being to select one or the other. One of the decisions might be “do nothing” or “do what we are doing now” or “base case.” If a particular cost has exactly the same amount and timing in both alternatives, then it must either be included in both or excluded from both. If the amount or timing differ, then both should be included. If one alternative includes initial costs and ongoing routine maintenance costs, then the other alternative must also include these costs. Similarly, it is important to include user costs in a con- sistent manner, ensuring that the same types of costs are included or excluded from both alternatives. All costs and benefits that are significant in selecting between the alternatives should be included. Occasionally there can be confusion about whether a cash flow is a cost or a benefit. Whenever possible, it is simpler and less confusing if all cash flows are treated as costs. For user costs, externalities, and costs of other agencies, it is important to be clear and consistent about who is paying the costs. For example, if Public Safety Department costs are included in one alternative, they should be included in both. Sometimes there are large distinctions among alternatives in terms of federal, local, or private cost participation. It is generally necessary for the costs of each alternative to be considered over the same time frame. It is desirable for this time frame to be long enough that all differences between the alternatives can be accounted for. However, for long-lived assets this may be unrealistic. In that case, it is important to consistently account for the long-term residual costs beyond the end of the analysis period. Common ways of doing this include • Computing a salvage value, a hypothetical revenue amount for selling the asset at the end of the period, considering the condition and performance of the asset forecast at that time. • Computing a lump-sum long-term cost representing all future costs beyond the analysis period, sensitive to condition and performance at the end of the period. • Computing the repair cost that would be required at the end of the period to restore the asset to near-new condition (or at least to the same condition) under both alternatives. • Computing lifecycle costs over a long enough period that discounting and/or uncertainty reduces any differences in subsequent costs to irrelevance. • Structuring the analysis as a perpetuity by including recurrent replacement and lifecycle costs extending the total life of the asset and its successors into the indefinite future. Lifecycle cost alternatives are usually compared by selecting the one with the lowest net present cost or highest net present value. However, in asset management often there are relevant costs and benefits that are non-economic or that are experienced by customers and stakeholders rather than by the agency. For these cases, there are more general methods of multi-objective analysis that are appropriate (Patidar et al. 2007).

110 estimating Life expectancies of highway assets 5.2.4.4 Comparing Alternatives Using Equivalent Uniform Annual Cost For certain purposes and certain audiences, it is useful to compare alternatives by converting net present value to equivalent uniform annual cost (EUAC). This calculation is just the inverse of the annuity formula described previously: EUAC NPV i i n = − +( )−1 1 where NPV is the net present value computed as described previously. This method is especially useful when comparing an agency investment against an alternative where the same service is provided by a contractor, where the contractor finances equipment acquisitions and charges the agency an annual amount. EUAC also is used in presenting investment amounts or lifecycle cost analysis to the public, where it might be converted to a cost per person. For example, a proposed sign washing program might be presented as costing 10 cents per taxpayer per year, but which is saving 15 cents due to longer asset life and lower replacement costs. This makes the argument easier to understand for people who do not have an intuitive feel for the much larger amounts that appear in budget documents. Comparisons among alternatives using EUAC should always produce exactly the same results as comparisons using NPV. However, it is very helpful to have tools such as EUAC readily available to help make economic arguments more accessible to the layperson. 5.2.4.5 Comparing Alternatives Using Internal Rate of Return For certain applications, an alternative to NPV is internal rate of return. The rate of return computation still requires computing the NPV of each alternative so in general it uses the same models and principles. The main difference is that the discount rate is considered uncertain and variable. To compute the internal rate of return, the analyst iteratively tries out a range of possible discount rates until finding one that equalizes the NPV between the alternatives. (This process is easily automated in a Microsoft Excel spreadsheet model.) If this rate of return is far from market rates, then one alternative is considered to be far superior to the other. If the discount rate is close to market rates, then the economic analysis might be considered inconclusive. Internal rate of return is useful when the agency is considering creative financing alternatives for a project, where the cost of money may be variable or may be divided between the public and private sectors. It is also useful for communicating with certain audiences that routinely work with discount rates. Sometimes the technique is useful for political decision-making when the difference in NPV among alternatives is small, but it might not be clear to the audience just how small it is. If the rate of return is within a range of familiar market rates, this might provide cover for pursuing an alternative that has greater political appeal in preference to one that strictly minimizes lifecycle costs. 5.2.4.6 Benefit/Cost Ratio There are many applications where it is necessary to compare alternative uses of a fixed amount of money, for example, in setting priorities. For this purpose, benefit/cost analysis is useful. To construct a benefit/cost analysis of asset investments, it is necessary to identify a set of alternatives for each asset and develop a criterion for ordering the alternatives. Usually it is assumed that the assets are independent of each other and that any combination of them can be implemented, subject to a funding constraint. In the simplest and most common case, there are two alternatives: do-nothing and do-something. The do-nothing alternative may have zero cost or may include routine maintenance and operational

Develop applications: how to apply Life expectancy Models 111 costs. In any event, it has a lower cost than do-something. If the decisionmaker is considering spending the additional money needed for the do-something alternative, then there must be a benefit for this expenditure. Often the benefit is calculated by comparing lifecycle costs by subtracting the lifecycle cost of do-something from the lifecycle cost of do-nothing. Lifecycle cost includes the initial cost and is often computed using the NPV method. If this difference in lifecycle costs is positive, then the expenditure is attractive because it saves money in the long term. When there are multiple objectives such as condition, risk, and/or safety to be considered, and not just lifecycle cost, then a utility framework might be used (Patidar et al. 2007) in order to calculate the benefit. A set of investment alternatives is prioritized by sorting the alternatives by the ratio of benefit to cost. When funding is limited, the alternatives with the highest benefit/cost are selected. If a particular asset has more than just the two investment alternatives, a variation on this method is used. The alternatives on the one asset are sorted in order by cost and evaluated by comparing each alternative with the next-less-expensive alternative. The sorting criterion is then the incremental benefit divided by incremental cost, which is called the incremental benefit/cost ratio. 5.2.4.7 Agency Cost and User Cost An important issue in lifecycle cost analysis is the ratio of the values of agency cost to that of user cost. Agency costs are the costs incurred by the transportation facility or service provider and such costs can be incurred during construction, preventive maintenance, rehabilitation, or reconstruction stages and also during normal operation phases. A highway agency responsible for construction and maintenance of highway assets incurs costs including initial costs associated with feasibility studies, engineering design, construction, operation of the facility, maintenance and rehabilitation, and disposal costs. In lifecycle cost analysis for assets such as pavements, preliminary costs such as feasibility and engineering studies are excluded because those are typically common among all alternatives considered. User costs are costs incurred by the highway users over the life of the project and may depend on highway improvements and associated maintenance and rehabilitation strategies over the analysis period. User costs may form a substantial part of total lifecycle costs and can often be the major determining factor in the analysis. User costs can be either work-zone user costs or non-work-zone user costs and components of user costs include vehicle operation cost, travel time cost, safety cost, and the costs from noise and water and air pollution. Agencies sometimes assign a weight less than 1.0 to user costs to reduce their effect on the lifecycle cost analysis. This may be reflective of actual agency decisions that do not give full weight to costs borne by road users and outside the agency’s budget. 5.3 Example Applications With the building blocks discussed in this chapter, it becomes possible to create various useful asset management applications. As the earlier chapters demonstrated, each agency will have its own needs so the applications may differ substantially from one to another. The process of discovering needs and incorporating them and gaining buy-in, interest, and demand may be more important than the sophistication of the applications themselves. The examples in this section are not intended to be full-scale, implementable management systems, bur relatively simple and transparent demonstrations of life expectancy analysis on small but realistic types of problems. These examples can be a source of ideas and clarity for agencies wishing to develop their own decision support tools.

112 estimating Life expectancies of highway assets 5.3.1 Routine Preventive Maintenance A common maintenance planning issue is the question of whether to start routine programs of crew activities that might have life extension benefits. Common examples are sealing of pavement cracks; washing of bridges, signs, pavements, and guiderails; spot painting; concrete patching; and cleaning of equipment enclosures. Here is an example of comparing a preventive maintenance scenario with the do-nothing scenario. Through the application of preventive maintenance, the two scenarios will have different service lives. For comparing asset alternatives that have different service lives, there are at least three approaches: • For each alternative, convert all costs and benefits into EUAC, • For each alternative, compute lifecycle cost over an asset life that is a lowest common denomi- nator of the separate life expectancy estimates, or • For each alternative, find the present worth of periodic payments to perpetuity. In this example, let us make a comparison of pavement management strategies using the EUAC approach comparing the two strategies in Table 5-4. For the routine preventive maintenance strategy, assume crack sealing is performed every 4 years at $400 per lane-mile, resulting in a life extension of 4 years; for the do-nothing option, assume only reconstruction is performed at a cost of $30k for both alternatives. Assume a discount rate of 4%. The EUAC of the two alternatives can be compared as follows: PVRoutine eventive Ma enancePr int = +($ .400 1 1 0 04) + +( ) + +( )     + 4 8 20 1 1 0 04 1 1 0 04 30 000 1 1. . $ , +( )     = 0 04 9 083 24 . $ , lane-mile EUACRoutine Pr inteventive Ma enance = +( ) $ , . . 9 083 0 04 1 0 04 24 1 0 04 1 596 24 +( ) −      =. $ - lane-mile EUACDo Nothing = +( ) −     =$ , . . $30 000 0 04 1 0 04 1 768 20 lane-mile Table 5-4. Example lifecycle activity profiles to be compared. Cost per lane-mile by strategy Year Routine Preventive Maintenance Do-Nothing 1 ... 4 $400 ... 8 $400 ... 12 $400 ... 16 $400 ... 20 $400 $30,000 ... 24 $30,000

Develop applications: how to apply Life expectancy Models 113 With these assumptions, the agency could reduce annual costs by $172 per lane-mile if routine preventive maintenance is completed. 5.3.2 Optimal Replacement Interval Certain types of assets may have various asset life alternatives, depending on different strate- gies for maintenance and life extension. The optimal asset life would be the lifecycle activity profile that can be sustained at minimum lifecycle cost. Here is an example of comparing several alternative profiles. After several decades of service, a railway bridge is slated for reconstruction. The estimated asset life of the structure is 50 years. The reconstruction cost is $600,000. During its replacement cycle, the bridge will require two rehabilitation events, each costing $200,000, at the 25th and 40th years, and the average annual cost of maintenance is $5,000. At the end of the replacement cycle, the bridge will again be reconstructed and the entire cycle is assumed to recur to perpetu- ity. Assuming a discount rate of 5%, the present worth of all bridge agency costs in perpetuity was calculated to be $753.15k. The agency would like to consider a range of potential life extension strategies to deter- mine if they are economical. As a second alternative, it is found that the asset life of the bridge can be extended to 60 years with rehabilitation in the 25th and 45th years, with only minor degradation in the level of service. By adding a third rehabilitation cycle, the agency finds that it can further extend the asset life to 70 or 80 years. Table 5-5 shows all the alternatives. In this example, Option 3 gives the lowest lifecycle cost. In Figure 5-15, the present worth values of the different estimates of lifecycle cost are plotted against the different estimates of asset life of the bridge, using a smoothed trend line. This suggests that the optimum replace- ment cycle is about 64 years. Moreover, the shape of the curve suggests that the present worth of the cost declines rapidly from 50 years to 60 years; but between 60 and 70 years, the curve is relatively flat, indicating that the asset manager has some flexibility in deciding on the replace- ment cycle in this range. Option 1 Option 2 Option 3 Option 4 Replacement Cost 600 600 600 600 Rehabilitation Cost 200 200 200 200 Annual Maintenance Cost 5 5 5 5 Estimated Asset Life 50 60 70 80 Rehabilitation Years 25 25 25 20 40 45 45 40 55 60 Discount Rate 0.05 0.05 0.05 0.05 Compounded Lifecycle Cost 7883.51 12726.51 21145.92 35411.36 Present Worth at Perpetuity 753.15 719.86 718.60 729.21 Table 5-5. Example system replacement interval optimization.

114 estimating Life expectancies of highway assets 5.3.3 Comparing and Optimizing Design Alternatives It is a very common need to compare two products or methods that have different costs, dif- ferent life expectancies, and different life extension possibilities. Here is an example, considering the case of deciding to apply a coating to a pipe culvert. Assume an engineer must decide whether to replace an existing pipe culvert with a coated or a non-coated pipe culvert, provided that a coated culvert is expected to survive 50 years with compounded amount of all cash flows within the replacement lifecycle to be $1200 while a non- coated culvert is expected to survive 40 years and the compounded amount of all cash flows within the replacement lifecycle is $1000. As discussed in Section 5.3.1, three possible ways of making this comparison would be an annual cost basis using EUAC, a least common multiple analysis period consisting of multiple replacement cycles, or a perpetuity of replacement cycles. For this example, a perpetuity is assumed, with a 4% discount rate. The present values of the two options, computed as perpetuities, then are PV no coating ∞ = × +( ) −     =$ . $1000 1 1 0 04 1 263 40 . $ . $ 1 1200 1 1 0 04 1 19 50 PV coating ∞ = × +( ) −     = 6 5. Therefore, the coated design option is preferred. 5.3.4 Comparing and Optimizing Life Extension Alternatives Similar to the previous example is the need to compare two or more life extension alternatives with different costs and effectiveness. Consider the set of alternatives presented in Table 5-6, for a bridge having a do-nothing asset life of 50 years, a replacement cost of $500k, and an interest rate of 4%. In a bridge management system, these types of strategies are typically compared on a NPV basis, and more than one of them may be selected. For the current example, EUAC is used as the selection criterion. Figure 5-15. Smoothed graph of the alternatives in Table 5-5. 700 710 720 730 740 750 760 40 50 60 70 80 90 Pr es en t w or th a t P er pe tu ity ($ 10 00 ) Replacement cycle (year)

Develop applications: how to apply Life expectancy Models 115 EUAC of Deck Overlay = $15k 1 1+0.04 ∗ ( ) + +(20 1 1 0 04. )     + +( )        +40 50 7 500 1 1 0 04 $ . k ∗ ∗ 0 04 1 0 04 1 0 04 1 50 7 50 7 . . . +( ) +( ) − + + = $2.84k EUAC of Deck Patching = $500k 1 1+0.04 $500 50 3 + ( ) +∗         +( ) +( ) − + + ∗ 0 04 1 0 04 1 0 04 50 3 50 3 . . . 1 3 36= $ . k EUAC of Joint Replacement = $500k 1 1+0.04 $300 + ( )∗ 50 2 50 2 5 0 04 1 0 04 1 0 04 + +        +( ) +( )∗ . . . 0 2 1 3 29 + − = $ . k EUAC of Deck Overlay and Joint Replacement = $$100 + 15k 1 1+0.04 1+0.04 $500k∗ + ∗( ) + ( )    20 40 1 1 1+( )         +( ) + + 0 04 0 04 1 0 04 50 9 50 9 . . . ∗ 1 0 04 1 2 74 50 9 +( ) − = + . $ . k EUAC of Deck Patching and Joint Replacement =$700 + $500k 1 1+0.04 ∗ ∗( )         + +50 5 0 04 1 0 04. .( ) +( ) − = + + 50 5 50 5 1 0 04 1 3 32 . $ . k EUAC of Deck Rehabilitation = $200k 1 1+0.04 ∗ ( )  35   + ( )        + $500k 1+0.04 ∗ ∗ 1 0 50 30 . . . $ . 04 1 0 04 1 0 04 1 3 32 50 30 50 30 +( ) +( ) − = + + k EUAC of DoNothing k= +( ) −    $ . . 500 0 04 1 0 04 1 50   = $ .3 28k Activity Frequency Life Extension of Activity at Applied Frequency Activity Cost Deck overlay Every 20 years 7 $15k Deck patching Every year 3 $500 Joint replacement Every year 2 $300 Deck overlay & joint replacement Overlay every 20 years & joint replacement every year 9 $15k for overlay and $100 for joint replacement Deck patching & joint replacement Every year 5 $700 Deck rehabilitation Once at year 35 30 $200k Table 5-6. Example bridge life extension alternatives.

116 estimating Life expectancies of highway assets From this array of activity options, the improvement strategy that minimizes cost under these assumptions is annual deck overlay and joint replacement. It can also be seen that the life extensions from patching, joint replacement, and rehabilitation under these assumptions are not cost-effective. 5.3.5 Pricing Design and Preservation Alternatives Many agencies invest in research and development programs so as to produce practical, cost- effective designs and materials. The primary concern with innovations, however, relates to reli- ability, life extension benefits, and cost of application. To facilitate decisions on whether to apply a new design, agencies often assess break-even points (i.e., the levels at which alternative designs become less costly than the traditional design). Example. For a bridge planned for construction, an agency wishes to assess the feasibility of using solid stainless steel reinforcement bars in place of traditional carbon steel. The bridge length and total deck width (ft) are 148.66 and 49.33, respectively; traffic volume is 8,527 AADT; weight of deck reinforcement is 62,963 lbs; and during the construction, rehabilitation, and deck replacements, work-zone traffic is diverted to a 1.3-mile 30-mph detour. If the lives of two bridges are 75 years and 100 years, respectively, with the activity profiles shown in Figure 5-16A, at what price is the stainless steel alternative preferred? The project durations for initial construction, deck replacement, and deck rehabilitation are 120, 60, and 21, respectively. Values of other analysis variables are as follows: discount rate = 4%; Vehicle occupancy = 1.8; minimum hourly wage = $13.43; average fuel economy = 23 mpg; cost of fuel = $3.75$/gal; traditional carbon steel price = 1.15$/lb; carbon steel service life = 40 yrs; stainless steel service life = 100 yrs. Results. The result of the analysis is shown in Figure 5-16B. This depicts the values of the ratio of the lifecycle cost of the stainless steel option relative to the traditional steel option, at var- ious ratios of the price of stainless steel relative to traditional steel. The differences in the lifecycle costs of stainless and traditional steel arise from their different lifecycle profiles which in turn result from the differences in the deck life (stainless steel decks have been found to have greater longevity (Cope 2009). Figure 5-16B shows that the stainless steel alternative is the superior alternative as long as the stainless steel price is less than 8.7 times the price of traditional steel. This is referred to as the price threshold ratio (PRT) for stainless steel desirability. The higher the PRT, the more favorable is the use of stainless steel. Higher values of the discount rate, vehicle occupancy, minimum hourly wage, fuel cost, and stainless steel service life, and lower values of average fuel economy would cause the Price Ratio function to shift to the right and thus, a higher PRT and consequently, an expanded range of cost-effectiveness for the stainless steel option. 5.3.6 Synchronizing Replacements Along a busy highway corridor, maintenance interventions can often be costly and disrup- tive. In some places, there’s never a good time to close a lane. When an agency has a good set of alternatives for design and life extension, it is useful to see what combination of products and techniques will minimize the required number of traffic control installations. • Year 0: Initial Construction • Year 20: Rehabilitation (deck overlay) • Year 40: Deck Replacement • Year 60: Rehabilitation (deck overlay) • Year 75: End of Life • • • • Year 0: Initial Construction Year 50: Rehabilitation (deck overlay) Year 75: Rehabilitation (deck overlay) Year 100: End of Life Steel Traditional Steel Stainless Figure 5-16A. Example activity profiles for carbon steel and stainless steel options (Cope 2009).

Develop applications: how to apply Life expectancy Models 117 Consider a small system of assets located along the same roadway (Table 5-7). If the location costs (i.e., mobilization, traffic control, and user costs) are estimated to be $7,000 per site visit, then what are the optimal replacement times so as to minimize the present value of costs in perpetuity? Assume assets are to be replaced no later than their remaining asset life. The objective of this problem is to minimize the total lifecycle cost, computed as follows: PV Annual i n n An = +   = Replacement Cost  1 11 alysis Period ∑ where Annual Replacement Cost = Location Cost p S3asset=1xasset Replacement Costasset; x ≡ binary decision variable indicating replacement, 1 = replace, 0 = do-nothing; n ≡ year of potential replacement. The only constraint is that the remaining asset life must be greater than zero, RSL ≥ 0 ∀n. 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 R at io o f E UA C fo r S ta in le ss S te el to T ra di tio na l s te el Ratio of Stainless Steel Price to Traditional Steel Price Price Ratio Threshold (PRT) = 8.7 Stainless steel is more cost-effective Traditional steel is more cost-effective Figure 5-16B. Sensitivity of relative long-term cost-effectiveness of longer life innovative material to the innovative-traditional price ratio. Asset New Construction Asset Life Remaining Life Replacement Cost Pavement Markings 5 3 $200 Traffic Sign 10 4 $300 Traffic Signal 15 5 $500 Table 5-7. Example data for synchronizing replacement intervals.

118 estimating Life expectancies of highway assets This optimization problem can be solved using a solver software package, although it is simple enough to solve by inspection, recognizing that • Ideally an agency would like to coordinate replacements so as to minimize cost. • The new construction asset life estimates have a common multiple of 5 years. Therefore, the optimal solution can be seen to be replace • All assets in year 3. • Pavement markings every 5 years thereafter (i.e., years 8, 13, 18, 23, 28, 33). • Traffic signs every 10 years thereafter (i.e., years 13, 23, 33). • Traffic signals every 15 years thereafter (i.e., years 18, 33). This produces the same lifecycle profile every 30 years with a present value of $26k. Alternatively, if an agency did not coordinate replacement schedules and replaced assets at the time each asset’s full asset life is reached, the optimal solution can be seen to be replace • Pavement markings in year 3 and every 5 years thereafter (i.e., years 3, 8, . . . 33). • Traffic signs in year 4 and every 10 years thereafter (i.e., years 4, 14, 24, 34). • Traffic signals in year 5 and every 15 years thereafter (i.e., years 5, 20, 35). Then a common lifecycle profile every 30 years with a present value of $80k is obtained. This example shows that the strategy of sacrificing 1 year of traffic sign life and 2 years of traf- fic signal life initially, so as to synchronize replacements, ultimately lowers the present value of costs by $54k ($80k–$26k). 5.3.7 Effect of Funding Constraints Agencies are constantly faced with the need to do more with less. Decision support tools based on life expectancy and lifecycle cost can help. Following is an example of working around time and budget constraints to maximize the benefit from a limited budget. Assume an agency has calculated the utility of a set of projects with respect to life expec- tancy, deterioration, lifecycle cost, and estimated project cost (Table 5-8). Assume a budget of $2.75M. To select a set of projects, optimization techniques can be applied to the problem: Maximize ProgramUtility Utility Subje = ∑xi i i m ct to ProgramCost Cost≤ ∑xi i i m where x ≡ binary decision variable with 1 = program, 0 = do not program; m ≡ number of potential projects. This simple example can be readily solved in Microsoft Excel for a small sample size. In this case, the optimal solution would be to replace bridge A, rehabilitate bridge B, replace pipe culvert A, and patch bridge C yielding a total utility of 242 at a cost of $2.675M. The remaining $75k could be carried over to the next planning cycle.

Develop applications: how to apply Life expectancy Models 119 5.3.8 Value of Life Expectancy Information For some of the asset types described in this guide, an agency might not have any data collection processes at all and no way to implement a condition-responsive replacement or life extension program to optimize life expectancy. Usually the cost of data collection is a major barrier to improvement. Here is an example showing the potential cost savings of using life expectancy analysis to design and implement a maintenance program. 5.3.8.1 Value of Quantifying Life Extension Suppose a life expectancy model predicts a box culvert life of 60 years. If an asset is 45 years old and expert opinion puts the asset life at 50 years, then a replacement project is likely to be programmed within 5 years. However, statistical evidence would suggest this project should not be programmed for another 15 years. The consequences of this can be quantified via lifecycle analysis. Assume the cost of replacement is $100k at a discount rate of 4%. Remaining EUAC of replacement, as scheduled by expert opinion $ . . $ .100 0 04 1 0 04 1 18 46 5 k k +( ) − = Remaining EUAC of replacement, as scheduled by life expectancy modeling $ . . $ .100 0 04 1 0 04 1 4 99 15 k k +( ) − = Based on this analysis, reliance on expert opinion may cost an additional $13.47k over the asset’s life depending on the accuracy of the life expectancy model. Thus, reliable life estimates can benefit agencies in setting financial needs and effectively spending taxpayer funds. 5.3.8.2 Value of Additional Explanatory Variables Life expectancy models can be made more accurate and realistic by the addition of more explanatory variables; but, agencies may be reluctant to add variables because of the implied addition of costs for data collection and/or quality assurance that come with a new data-based application. The following hypothetical example shows how to structure an analysis of the potential benefits of additional data, using a lifecycle cost framework. In Table 5-9, a number of statistical models were developed to predict the asset life of a highway asset. The series of statistical models employs an increasing number of variables. Each additional variable implies added costs as given in the table. The cost of data collection was then combined with the cost of replacement, which was constant for the particular asset, and the total cost was turned into present worth at perpetuity for the sake of comparison of different models’ lifecycle costs. Activity Utility Cost Bridge A replacement 100 $2400k Bridge B rehabilitation 75 $250k Box Culvert A replacement 55 $100k Pipe Culvert A replacement 35 $5k Bridge C deck patching 32 $20k Table 5-8. Example ranked projects with associated utility and cost.

120 estimating Life expectancies of highway assets Once the lifecycle costs of different statistical models, as well as the expert opinion, were converted into present worth at perpetuity, those results were plotted against the number of variables in Figure 5-17. The plot suggests that the total cost declines with an increasing number of variables used in the performance model to predict an asset’s life, provided that the added variables enabled an extension of asset life for selected assets as shown in Table 5-9. Such an analysis could motivate road agencies to collect data and improve the calculation of the life expectancies of highway assets. Some data items, such as the weather data used in some of the example models earlier in this guide, are widely available free of charge. There may be opportunities to spread the cost of certain types of data, such as traffic data, over many asset types. The type of analysis shown in the example can help the agency to optimize its data investment. 5.3.9 Highway Asset Valuation Reliable estimation of asset life helps improve the accuracy of asset valuation. Most approaches of asset valuation, including the GASB 34 approach, use asset life as a critical variable. The following No. Of Variables in Model Asset Life (yr) Cost for Data Cost of Replacement Total Cost Present Worth at Perpetuity Practice (rule of thumb) 0 10 0 1000 1000 1590 Statistical Model 1 1 12 100 1000 1100 1382 Statistical Model 2 2 13 100 1000 1100 1242 Statistical Model 3 3 14.5 110 1000 1110 1079 Statistical Model 4 4 15.5 120 1000 1120 991 Statistical Model 5 5 17 140 1000 1140 882 Statistical Model 6 6 18 150 1000 1150 818 Statistical Model 7 7 18.5 155 1000 1155 788 Statistical Model 8 8 18.5 160 1000 1160 791 Statistical Model 9 9 18.5 180 1000 1180 805 Discount rate = 0.05 Increasing number of variables Table 5-9. Examples of lifecycle cost including data cost. 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 0 2 4 6 8 10 Pr es en t W or th a t P er pe tu ity No. of variables in the performance model Figure 5-17. Example cost savings due to employing increasingly better performance models.

Develop applications: how to apply Life expectancy Models 121 example involves a box culvert built in 2002 and having a culvert condition rating of 8. The culvert is 8 years of age. Its replacement cost is $123,752 and salvage is assumed to be $0. All costs and values are in 2010 dollars and the analysis year is 2010. The following equations show how asset life is used to find the asset value. Using the Sum-of-Years Digits (SOYD) functional form for asset depreciation, for example, and assuming a 57-year service life, the asset value at any year can be found. The depreciation at any year is calculated as: SOYD N t N N HC St = − +   +( ) −( )1 2 1  Thus, the asset value at any year, Vt, is found as follows V HC SOYDt t t= −∑ 1 where N - t + 1 is the useful remaining life at beginning of year t; N is the planning period or service life; t is the given year; HC is historical cost adjusted to 2010$ using FHWA Construction Price Index; S is the salvage value. Thus, the asset value at the 4th year, for example, is found as follows: Vt = − =$ , $ , $ , .123 752 4 562 104 923 It can be seen that the asset value, Vt, can be heavily influenced by the service life N. 5.4 Role of a User Group Earlier chapters showed how to build a constituency for life expectancy analysis that makes it more likely that the necessary data collection and analysis will get done and that the results will be put to work productively. Members of this constituency can do more than make information requests and provide data and resources. If stakeholders are to feel confident that their needs will be met and if the not-invented-here syndrome is to be avoided, stakeholders need to take an active role in application development and subsequent enhancement. One of the best ways to create involvement and buy-in is to form a user group for the applications to be developed (Figure 5-18). A user group should consist of people who will be hands-on users of the applications, as well as people who may receive and act on the information. Ideally, some of the applications will be of use to the units that collect the necessary data (e.g., workflow management and quality assurance) so representatives of these units can also be user group members. A user group has the following tasks at different stages of the application lifecycle: Planning • Ensure that the user group includes the necessary stakeholders and that all prospective applications are represented. • Perform or review the asset management self-assessment, specifically concerned with life expectancy analysis and its potential uses (Gordon 2010). • Review and perform or update, as necessary, the planning steps described in Chapters 1 through 3 of this guide.

122 estimating Life expectancies of highway assets • Become familiar with available methods and tools as described in Chapters 4 and 5 of this guide. • Evaluate possible additional applications and recruit users who may want to see such applications developed. If an application idea has no interested users, this indicates that either the application was not such a good idea after all, the agency already has the tools it needs, the agency is not yet at a maturity level where it can use the application, or some form of organizational change may be necessary first. • Ensure that senior managers and outside stakeholders are asking questions about how the proposed applications can answer the agency’s mission and seeing the possibilities. In other words, make sure there is a demand for the information to be produced. • Ensure that senior managers and outside stakeholders understand the kinds of information to be provided and the boundaries on coverage, quality, and timeliness that will become possible. In other words, make sure they understand the potential supply of information. • Review and refine definitions and mockups for compatibility with agency business processes, related information systems, and available data. Development • Ensure that in-house and/or consultant labor and resources are made available to develop the applications. Oversee letting and procurement activities as needed. If a consultant is to be hired, members of the user group should select a single author for the Request for Proposals and should review the draft of the document. • Review the prototypes and documentation developed. • While prototype development and refinement are underway, resolve issues of terminology, procedures, and data standards. Be prepared to refine and modify these over time, learning from experience with the prototypes. Create and maintain a working document to describe the user group’s decisions and recommendations on these matters. Ensure that the developers of the system have input and access to this document and can raise new issues through an organized process. • Communicate progress to stakeholders, and show results early and often. Convey a constructive and upbeat attitude about the applications. • Coordinate with committees involved with other aspects of asset management in the agency. • Assist in maintaining the flow of time and resources necessary to see the application through to completion. Figure 5-18. Example user group structure.

Develop applications: how to apply Life expectancy Models 123 Production • Oversee and attend training classes, for new users and applications, and refresher courses for existing users. • Provide constructive input on new functionality that may be needed. • Report problems and follow up on solutions. • Through an organized process, such as voting, advise on priorities for new enhancements. • Use the products and promote the results to stakeholders. • Attend conferences and share ideas with other agencies. • Ensure that the applications contribute to implementation of the Transportation Asset Management Plan (TAMP) (Gordon 2010). Use what is learned from the applications to improve the TAMP and to advance the agency’s state of asset management maturity. Often a user group will be large and may expand over time to include all hands-on users and many indirect users of the applications. Once the group reaches a certain size, it should create sub-groups to whom it delegates many of the tasks above. 5.5 Development of Applications With so many useful applications, it may be tempting to launch a large system development effort to implement them. Although that has been done, and has often been successful, it is not the only way to proceed. Another alternative is to select a relatively small subset of applications at first (often just one), and develop a working prototype that addresses the core functions throughout the process— from data collection to analysis to reports. This should be conceived as the smallest possible system that can produce useful outputs and should work from existing data if possible. Review the prototype first, then gradually expand it to cover more applications and add more features. As part of this review and expansion, identify the data gaps, procedures, and standards that are required in the context of a working application. Having a simple useful program in place works wonders for focusing a development effort, avoiding peripheral features that might or might not eventually be needed, and streamlining the implementation. Priority setting is more natural and harder to avoid if users are ready and eager to put the system to work. This incremental prototyping style of development is often given the name “agile development” or “extreme programming.” Even though it has been styled as a cultural theme for programmers, this type of development is actually driven more by the hands-on users of the systems. It gives users more day-to-day control and involves them more deeply in the creation of the tools they will use. Even if the actual concept, design, and programming are done by an outside consultant, there will not be a “not-invented-here” syndrome if the agency owns the concepts, requirements, and design of their one-of-a-kind product.