This document is part of a larger demonstration illustrating the utility of using large amounts of data and processing in a cloud environment to derive value from these activities. The overall project simulates the operations of a fictional corporation that includes gas stations in 20 cities across the US. Each of these cities has a number of gas stations that were built or acquired over time and each station has a number of fuel pumps used to generate revenue. The demonstration project can be viewed at: http://carma.demo.ociweb.com/.

Here we describe the analytics involved in establishing a maintenance interval for the fuel pumps that minimizes the total cost of repairs. This is a demonstration of using cloud based processing to gather and ingest large amounts of data, present them in a useful fashion, and utilize analytic methods to generate value from the gathered data.

Overview

In this document we derive a suitable model to use for determining when the simulated pumps will fail. We then generate some simulated data with the date of inception – when the pumps were last new or repaired – and the age at which they will be simulated as failed. After that we illustrate how we will estimate the model parameters from data and show the accuracy of this estimation process using the generated simulation data.

Once we have the model and derived parameters from simulated data, we then determine the maintenance costs for repairing the pumps assuming that we do this at an interval where most of the maintenance can be scheduled rather than reacting to the actual failure events. We find the minimum of this cost over the different maintenance intervals and schedule the maintenance accordingly. This effectively reduces the amount of data that we have for estimating the parameters, so once we start the new maintenance intervals we will need to spend a longer amount of time before reaching the same confidence in our estimated parameters.

Once we have examined the costs without using scheduled maintenance and then optimized the interval for scheduled maintenance, we obtain results that show a positive value for implementing the scheduled maintenance:

After we obtain these results, we discuss some of the issues that might arise as well as some cautions about use of the model and subsequent parameter estimation.

Fuel Pump Reliability

In order to determine maintenance costs for the fuel pumps, the reliability of the pumps must be modeled. Since this is a demonstration project, we can make many simplifying assumptions without losing the ability to illustrate the techniques involved in doing this. Some conditions that we ignore include the specifics of how failures occur. We also can characterize the types of failure as an aggregate average cost for maintenance actions rather than attempting to identify each failure case or repair result. We can also lump all of the different component lifetimes into a single time to failure value that represents the aggregate average over all of the components.

Pump Survival

Pump survival to a specific age is defined as the probability that a pump has not failed when it achieves that age. Each pump will fail with a probability defined by a probability density function (PDF) characteristic of the pumps. Using different density functions will result in different survival curves for the pumps.

Modeling pump failures will consist of assigning an initial age (date of installation or repair) to each pump, then determining failure points using the model.

Note that once a model has been selected, processing of the inbound actual data will then consist of estimating the model parameters from the observed data. Predictions can then be made using the model with the best fit parameters. We will consider two models here: a constant failure rate model, and a log-normal failure rate model.

Constant Failure Rate Model

The simplest failure density function is the constant failure rate, where survival is characterized by the exponential density function. A survival curve for this density function is:

We expect fuel pumps to last for a time, then fail after they reach an age where components start to wear out. The constant failure rate model assumes that failures occur regularly through the lifetime at a regular rate. Clearly the constant failure rate model does not reflect the expected failure behaviors for the pumps.

Log-normal Failure Rate Model

The log-normal probability distribution can be used to create a failure rate model that more closely resembles the right side of the typical failure rate “bathtub” characteristic curve describing hardware failures. The relevant formulas describing the model include:

\[\begin{aligned} (1)\quad & f(t) = \frac{\phi \left(\frac{\mathrm{ln}(t) - \mu}{\sigma} \right)}{ \sigma t } & \quad\text{Probability density function} \\ (2)\quad & F(t) = \Phi \left( \frac{\mathrm{ln}(t) - \mu}{\sigma} \right) & \quad\text{Cumulative density function} \\ (3)\quad & S(t) = 1 - F(t) = 1 - \Phi \left( \frac{\mathrm{ln}(t) - \mu}{\sigma} \right) & \quad\text{Survival function} \\ (4)\quad & h(t) = \frac{f(t)}{S(t)} = \frac{ \phi \left( \frac{ \mathrm{ln}(t) - \mu }{ \sigma} \right) }{ \sigma t \left[ 1 - \Phi \left( \frac{\mathrm{ln}(t) - \mu}{\sigma} \right) \right] } & \quad\text{Hazard function} \\ \end{aligned}\]

where

\[\begin{aligned} (5)\quad & \phi(u) = \frac{1}{\sqrt{2\pi}} \, \mathrm{e}^{^{-u^2}/_2} & \quad\text{Standard normal probability density function} \\ (6)\quad & \Phi(u) = \int\limits_{-\infty}^u \phi(z)\,\mathrm{d}z & \quad\text{Standard normal cumulative density function} \\ \end{aligned}\]

The log-normal density function results in a survival curve that more closely resembles actual failure characteristics:

The hazard curve can be derived as the ratio of the failure probability to the survival probability. This curve can be used to understand the probabilty of failure at a particular age. It represents the probability that a pump that has survived to an age will then fail at that age.

The hazard rate in this case is the tail end of the typical “bathtub” curve expected for pumps. If we assume the early failures are not present due to burn in prior to purchase, the log-normal model is suitable for modeling the pump failures.

Generating Simulated Failures

As part of the demonstration simulation, we need to generate simulated failures. We can also use these simulated failures to test the parameter estimation formulas that we will use to derive the parameters from data.

Simulating the hardware failures consists of determining the inception date for each pump and then determining a simulated age at failure for that pump. The inception date is the time at which the pump is “new”; that is, the starting point of the aging calculations.

Simulating Age at Failure

We can generate a simulated age at failure for all of our pumps using random variates from the log-normal distribution. To find the age at the first failure for each of the pumps in the population, we generate one random value for each pump. Sorting these ages and plotting the portion of surviving population against the age at failure results in the following chart:

Note that this generated simulation data resembles the theoretical survival curve very well. This gives us confidence that we are generating the age at failure correctly.

Assigning Inception Dates

The inception date for each pump will be simulated as if each pump was last replaced at a random time in relation to any other pump. This will be done by allowing the inception date for a pump range from MTTF months prior to now up to now. For any pump that fails prior to now, the inception date is moved to that failure date and a new age to failure generated. This results in random ages for all of the pumps. The date of a simulated failure is the sum of the inception date and the age to failure. Since this procedure utilizes the age at failure to possibly adjust the inception date, these dates are derived after the age at failure data is available.

The following histogram illustrates the distribution of inception times for the population prior to adjusting for failures earlier than now. The boxplot at the bottom shows that the distribution is uniform across the entire time period.

Estimating Model Parameters

Once we have simulated data, we can use that as input data to simulate actual measurements and determine if we will be able to estimate the model parameters closely enough to be useful.

Parameter Estimation Formulas

Estimating the actual parameter values for the log-normal model from measured sample data will be done using a maximum likelihood estimator for each of the two parameters. We will estimate the location (\(\hat{\mu}\)) and spread (\(\hat{\sigma}\)) parameters for the model. The exponentiation of these parameters is measured in time and can be used to characterise the age at failure for the pumps.

The values that are being used as input to the estimation formulas are the age at failure for each detected failure. This is the time at which a failure was detected to occur minus the time where the pump was considered ‘new’. New is the condition a pump is in when it was last either replaced or repaired.

\[\begin{aligned} (7)\quad & \hat{\mu} = \frac{1}{n} \displaystyle\sum_{k=0}^n \mathrm{ln}(t_{fail,k} - t_{0,k}) & \quad\text{location estimation} \\ (8)\quad & \hat{\sigma} = \sqrt { \frac{1}{n} \displaystyle\sum_{k=0}^n \big( \mathrm{ln}(t_{fail,k} - t_{0,k}) - \hat{\mu} \big)^2 } & \quad\text{spread estimation} \\ \end{aligned}\]

Note that the input data to these estimation formulas are not directly available from actual input data. The simulated data provides this information directly. When receiving actual measurement data, failures need to be detected first. This will be done by applying a threshold to the amount of time between sensor readings and when a reading has not been received from a sensor within this time, the associated pump will be considered to have failed. Once a failure has been detected in this way, then the inception age (last repair or replace date) will be subtracted from the current date, when the failure was detected, and the result used as the age at failure in the parameter estimation formulas.

Parameter Estimation Performance

We examine the estimated values of the parameters, referenced back to the non-log-normal values for generated simulation data. To do so, we generate a rolling window of the observed input data for the parameters as the data is received. This gives an intuition of how the values will behave when collecting actual measurement data.

The plot below shows the estimated mu parameter value (translated back into months) over the first 7,000 generated data points using a 50 sample window. During this time, the estimate ranges from a low of 31.2 to a high of 36.7 months. The largest error from the actual value is 2.81 (8.26%) months.

The plot below shows the estimated sigma parameter value (translated back into months) over the first 7,000 generated data points using a 50 sample window. During this time, the estimate ranges from a low of 1.14 to a high of 1.26 months. The largest error from the actual value is 0.0634 (5.28%) months.

Comparing the survival curve of the actual age profile with the profiles derived from the estimated parameters, we can see in the chart below that the shape of the survival curve fits well, and the estimation errors bound the derived curve such that the actual curve lies within the estimated ones.

Predictions using the model

Now that we have confidence in the simulation data, and the processes used to generate it, we can use the parameters estimated from that data to make some predictions. The obvious useful prediction to make would be to estimate the age at failure, which the log-normal model estimates. So once we have the model parameters, we can estimate when pumps will fail and perform preventative maintenance - such as replacing aging pumps - prior to the actual failure.

To utilize the prediction of the age at which pumps will fail, we will need to determine the costs of repairing or replacing the pumps. Then we will determine when to schedule this maintenance so that the total cost is minimized. We will perform this minimization with respect to the maintenance interval, so that the independent variable for minimizing the cost will be the age at which we schedule pumps to be maintained.

Once we select a maintenance interval, we can determine the per pump cost for the maintenance and compare that to the case where we do no schedule maintenance. This will allow us to determine if, and if so by how much, scheduling maintenance on the pumps will reduce overall costs.

When we schedule maintenance for pumps at a certain age, it is likely that some of the pumps will fail prior to this scheduled time. There will be no reduction in costs for these pumps. Likewise there will be no increase in their maintenance cost either.

Also, preventative maintenance prior to an actual failure will result in useful life of replaced components not being utilized, which is a cost as well. These costs must be balanced by any gains realized through scheduled maintenance compared to unscheduled, emergency, repairs.

To start the prediction process, we can observe the model characteristic that allows us to determine the age at which a proportion of the entire population, or quantile, will have failed. This is the inverse of the cumulative probability function:

\[\begin{aligned} (9)\quad & t = F^{-1}(q) & \quad\text{Inverse cumulative density function} \\ \end{aligned}\]

Using this formula allows us to determine the age to schedule preventative maintenance, allowing a known portion of pumps to fail prior to this interval. The x-axis in the chart below is the quantile, or proportion of failed pumps, for which we are finding the predicted age.

Maintenance Costs

In order to determine when to schedule maintenance actions we need to consider costs for the maintenance. Once we can determine what the total cost of maintenance actions are at various possible maintenance intervals, we can minimize the cost to find the interval to actually schedule.

Costs involved can be averaged over the entire population of pumps in order to determine what the cost of an individual repair is. Costs to consider include: cost of the repair or replacement of a pump; cost of lost revenue during the time it takes to repair a pump; and, the cost of not utilizing the full available lifetime of a pump by repairing or replacing it prior to it failing.

It may be simpler to understand the cost components by analogy. The predictions that we are making are similar to scheduling when to replace tires. The mileage on any given tire is specific to that tire – tires are replaced as necessary and not as a group; this is similar to the age at which we will be performing maintenance on the pumps. If we wait until they fail, that corresponds to not replacing a tire until it goes flat. Clearly not a desireable event. If we schedule a tire to be replaced at a particular mileage, that corresponds to performing scheduled maintenance at a certain pump age. If we do that we have the cost of the replacement tire and labor for doing so. We also have any lost revenue from use of the tire while we wait for the tire to be replaced or repaired. This corresponds to the lost revenue while a pump is not working. Finally we have the cost of replacing a tire prior to it going flat; corresponding to the time we do not use the pump when it would not have failed. For the tire this is represented by the miles not driven by the tire, and for the pumps it is the time between when a repair of replacement was performed and when the pump would have failed.

For all of the costs, there are some assumptions that we can make. These assumptions bound the costs involved and provide a structure for determining how those costs vary with the scheduled maintenance interval. These include:

  • \(C_{\mathrm{pump}} \, = \, \$2,000.00 \,\) – cost of a new pump (replacement cost)

If a maintenance action requires a pump to be replaced, this is the maximum cost for the maintenance action.

  • \(C_{\mathrm{repair}} \, = \, 60\% \, C_{\mathrm{pump}} \, = \, \$1,200.00\) – average non-replacement maintenance cost

This is the aggregate average over the entire pump population for the cost of maintenance that does not require the pump to be replaced. This proportion holds for both scheduled and unscheduled maintenance actions.

  • \(p_{\mathrm{replace}} \, = \, 20\%\) – proportion of maintenance actions that require the pump to be replaced

This is the proportion of all repairs that require the entire pump to be replaced.

  • \(p_{\mathrm{discount}} \, = \, 80\%\) – scheduled maintenance cost as a proportion of unscheduled maintenance costs

This is a reduction in maintenance costs due to the ability to leverage supplier buying power by bulk and scheduled purchases as well as the ability to schedule labor costs rather than paying a premium for unscheduled, emergency, labor.

  • \(mttr_{\mathrm{unscheduled}} \, = \, 7 \, \mathrm{days}\) – mean time to repair (MTTR) for unscheduled maintenance

This is the amount of time from when a pump actually fails to when that pump is returned to operation. This includes detection time, the time to schedule an emergency repair, and the actual time to repair.

  • \(mttr_{\mathrm{scheduled}} \, = \, 2 \, \mathrm{days}\) – mean time to repair (MTTR) for scheduled maintenance

This is the amount of time from when a scheduled maintenace action starts to when that maintenance action is complete and the pump being maintained is returned to operation.

Repair costs

The cost of a repair for any pump will include the parts and labor for the repair. If the repair is done on a scheduled basis, then the cost should be adjusted to account for the shorter interval (on average) requiring more maintenance (on average) than would be incurred for repairs only performed on pumps which had failed.

Parts and labor costs

An unscheduled maintenance action cost will be the cost of a new pump times the number of new pumps needed plus the cost of repair times the number of pumps repaired:

\[\begin{aligned} (10)\quad & C_{\mathrm{unscheduled}} = p_{\mathrm{replace}} \, C_{\mathrm{pump}} \, + \, ( 1 - p_{\mathrm{replace}} ) \, C_{\mathrm{repair}} & \quad\text{Cost of unscheduled maintenance action} \\ & = 0.2 \, \times \, $2,000.00 \, + 0.8 \, \times \, $1,200.00 & \\ & = $1,360.00 & \\ \end{aligned}\]

Scheduled maintenance costs are similar to the unscheduled costs, but have a lower cost for parts and labor:

\[\begin{aligned} (11)\quad & C_{\mathrm{scheduled}} = p_{\mathrm{discount}} \, C_{\mathrm{unscheduled}} & \quad\text{Cost of scheduled maintenance action} \\ & = 0.8 \, \times \, $1,360.00 & \\ & = $1,088.00 & \\ \end{aligned}\]

With these two repair costs available, the repair cost as a function of the quantile of failed pumps can be derived, and from there the repair cost as a function of the maintenance interval.

\[\begin{aligned} (12)\quad & C_{\mathrm{maintenance}} = q \, C_{\mathrm{unscheduled}} \, + \, ( 1 - q ) \, C_{\mathrm{scheduled}} & \quad\text{Cost of maintenance action} \\ & \, = F(t) \, C_{\mathrm{unscheduled}} \, + \, S(t) \, C_{\mathrm{scheduled}} & \\ & \, = C_{\mathrm{unscheduled}} \, \left( p_{\mathrm{discount}} \, + ( 1 - p_{\mathrm{discount}} ) \, F(t) \right) & \\ & \, = \$1,360.00 \, \times \, \big( 0.8 \, + 0.2 \, \times \, F(t) \big) \, = \, \$272.00 \, \times \, F(t) \, + \, \$1,088.00 & \\ \end{aligned}\]

This cost as a function of the maintenance interval for the theoretical model as well as the bounds of estimated model parameters is shown in the chart below:

Adjustment due to shorter intervals

Since we are proposing to perform maintenance actions – and incur maintenance costs – prior to pumps actually failing, we need to account for the loss of use of the un-failed portion of the pumps lifetime. For example if we simply repaired each pump each week, we would expect to not see any failures, but we would be paying for 147 weeks (34 months), on average, of repairs where we would have payed only once for the repair if we waited for it to fail.

Remembering the tire analogy, we can think of this as the per-mile cost of the tire times the number of miles not driven. For the pumps, this is the per-month cost of the maintenance times the amount of time that the pump is not used. The amount of time the pump is not used can be estimated by the MTTF value. The per-month cost for the pump is simply the cost of a maintenace action, which is identified in equation (12) as \(C_{\mathrm{maintenance}}\) divided by the interval at which this cost is incurred. We can either compare the per-month costs for all maintenance actions or use some other common time frame. Since companies typically account for costs anually or quarterly we can use either of these time frames as a reference. For here we will use annual costs, so we can simply multiply the monthly cost (maintenance costs divided by the age in months at which the maintenance is performec) by 12 (months in a year).

\[\begin{aligned} (13)\quad & C_{\mathrm{adjusted}} = & 12 \, \times \, \left( \frac{C_\mathrm{maintenance}(t_{\mathrm{interval}})}{t_{\mathrm{interval}}} \right) & \quad\text{Annualized cost of early repair} \\ \end{aligned}\]

The chart below illustrates this cost and shows that this is the element of cost that penalizes early maintenance with a very high cost for short intervals.

Lost Revenue

The time when a pump is out of service – either for a scheduled maintenance action or for an emergency repair after failure – represents the loss of revenue from that pump. This is considered a cost and must be accounted for when determining the cost of scheduling maintenance for pumps.

In the assumptions, we can see that scheduled outages are much smaller (2 days) than outages due to failures (7 days). This is due to the additional time to detect the failure and schedule the repair. The actual repair time will be the same for both. The revenue lost during an outage is:

\[\begin{aligned} (14)\quad & C_{\mathrm{repair time}}(t) = \frac{\mathrm{Revenue}}{n_{\mathrm{total}}} \, \times \, t & \quad\text{Lost revenue for a repair} \\ & = \, \$782.78 \, \times \, t & \text{[per pump per day]} \\ \end{aligned}\]

The cost of lost revenue for repairing or maintaining a pump depends on the number of repairs after failures and the number of maintenance actions prior to failures. This proportion is the quantile that we have used before.

\[\begin{aligned} (15)\quad & C_{\mathrm{lost revenue}} & = q \, C_{\mathrm{repair time}} \left( mttr_{\mathrm{unscheduled}} \right) & \quad\text{Cost due to lost revenue} \\ & & + \left( 1 - q \right) \, C_{\mathrm{repair time}} \left( mttr_{\mathrm{scheduled}} \right) & \\ & & = \$782.78 \, \times \big( 2 \, {\mathrm{days}} \, + \, q \, \left( 5 \, {\mathrm{days}} \right) \big) & \\ \end{aligned}\]

And now we can use the cumulative distribution to translate the formula from the quantiles to the maintenance interval. Since we want to be able to combine this cost with the maintenance cost calculated above, we go ahead and annualize the cost here as well.

\[\begin{aligned} (16)\quad & C_{\mathrm{lost revenue}} & = \frac{12}{t} \, \$782.78 \, \times \big( 2 \, {\mathrm{days}} \, + \, 5 \, {\mathrm{days}} \, \times \, F(t) \big) & \quad\text{Cost due to lost revenue} \\ \end{aligned}\]

Total Maintenance Cost

Once we have determined what the annualized cost is to perform the maintenance and the cost for not having the pumps available during maintenance, we can simply combine those costs to determine what the annual cost per pump is. This can be compared with the cost of not performing scheduled maintenance in order to determine what the benefit is from scheduling the maintenance. The total cost is the sum of the individual costs:

\[\begin{aligned} (17)\quad & C_{\mathrm{total}} \, = \, & C_{\mathrm{adjusted}} \, + \, C_{\mathrm{lost revenue}} & \quad\text{Total cost per pump} \\ \end{aligned}\]

We can determine the maintenance cost that are incurred in the absence of any scheduled repairs by simply setting the maintenance interval to MTTF and the percentage of failed pumps to 100% in the above formulas. The degenerate formulas for no scheduled maintenance are:

\[\begin{aligned} (13a)\quad & C_{\mathrm{adjusted}} \, = \, \$1,360.00 \, \times \, \left( \frac{12}{\mathrm{MTTF}} \right) \, = \, \$480.00 & \quad\text{Annualized cost of repair} \\ \end{aligned}\]

\[\begin{aligned} (14a)\quad & C_{\mathrm{lost revenue}} \, = \, \$782.78 \, \times \, 7 \, = \, \$5,479.46 & \quad\text{Per pump per repair} \\ & \$5,479.46 \, \times \, \left( \frac{12}{\mathrm{MTTF}} \right) \, = \, \$1,933.93 & \quad\text{Annualized} \\ \end{aligned}\]

Resulting in a total cost prior to scheduling maintenance of:

\[\begin{aligned} (18)\quad & C_{\mathrm{total}} \, = & \, \$480.00 \, + \, \$1,933.93 \, = \, \$2,413.03 & \quad\text{Annual cost per pump} \\ & C_{\mathrm{entire}} \, = & \, n_{\mathrm{total}} \, \times \, \$2,413.03 \, = \$16,897,510.00 & \quad\text{Total annualized revenue for repairs} \\ \end{aligned}\]

Now that we have a cost for repairs where we do not schedule maintenance, we can determine the cost when we do and find the optimum maintenance age for the pumps as well as the total annualized cost of repairs for comparison with the non-scheduled option.

In the plot below, we see the total cost curve for the nominal case and for the extreme conditions of the estimated model parameters. The minimum cost has been identified for each of the curves as well.

The annualized cost per pump per year for the nominal case is $1360.55. The maintenance interval at the minimum cost (prescribed repair age) is 25.8 months. Annualized and summed over all of the pumps, the cost becomes:

\[\begin{aligned} (19)\quad & C_{\mathrm{total}} \, = & \, \$1,360.55 & \quad\text{Annual cost per pump} \\ & C_{\mathrm{entire}} \, = & \, n_{\mathrm{total}} \, \times \, \$1,360.55 \, = \$9,523,850.00 & \quad\text{Total annualized revenue for repairs} \\ \end{aligned}\]

Discussion and Cautions

To summarize the results:

  • When not performing scheduled maintenance (repairing pumps only after failure), the annual cost is $2,413.03 per pump, or $16.9 MM for the entire corporation.

  • When performing scheduled maintenance at an interval of 25.8 months, the annual cost is $1,360.55 per pump, or $9.5 MM for the entire corporation.

  • Implementing scheduled maintenance results in a savings of 43.6% (costs are 56.4% of the base case).

  • Implementing scheduled maintenance saves $1,052.48 per pump annually, or $7.4 MM for the corporation.

When we created our model, we made several simplifying assumptions. Among them were that the pumps would fail with a probability described by a log-normal distribution. This may or may not be correct; the actual failure mechanisms that contribute to the failures might generate a different distribution. We did not account for the entire observed failure behavior of real components: which exhibit many failures at an early age - which we ignored assuming the pumps and parts we used were “burned-in”; they also exhibit a small but non-zero failure rate during the majority of their lifetime. We did account for the “end-of-life” failures that are typically observed, but used a log-normal distribution. This is likely to be acceptable, but more detailed analysis is warranted.

Prior to instituting the scheduled maintenance, we would expect to see many failures at a steady rate over time. This would be expected to be near \(\frac{n_{\mathrm{total}}}{\mathrm{MTTF}}\), or 206 failures per month. This would mean that the 50 sample window we assumed for estimating the parameters would span a time of about 1 week.

After adopting the scheduled maintance, we would expect to observe a much smaller number of failures since we are deliberately attempting to perform maintenance prior to failures in most cases. The number of failures that we would expect to see when using scheduled maintenance is \(\frac{n_{\mathrm{total}}}{\mathrm{MTTF}} \, \times \, F(t_{\mathrm{interval}})\) or 13.5 failures per month. This is a much smaller observation rate and a 50 sample window will span a time of 16 weeks. This means that any updates to the maintenance interval will be much slower to adapt to any changes in the actual failure rates if they occur.

From the estimator performance we can see that it is possible to diverge quite far from the actual model parameters for short times. It would be prudent to implement a limit to changes in the maintenance interval in an attempt to smooth any excursions due to estimation error.

From the total cost curve plot, we can see that the area where the optimal maintenance interval occurs is a shallow minimum. This is encouraging since it means that any error in the implemented interval from the actual optimal interval will be close in cost to the desired minimum cost. This means that even if we selected a probability density function that was not accurate, it is likely that a more accurate density function would result in a similar interval selection.

Well. Optimistically anyway.