weibulltools (version 2.0.0)

dist_mileage: Parameter Estimation of an Annual Mileage Distribution

Description

This function models a mileage random variable on an annual basis with respect to a supposed continuous distribution. First, the distances are calculated for one year (365 days) using a linear relationship between the distance and operating time. Second, the parameter(s) of the assumed distribution are estimated with maximum likelihood. See 'Details' for more information.

Usage

dist_mileage(mileage, time, distribution = c("lognormal", "exponential"))

Value

A list of class mileage_estimation which contains:

• coefficients : A named vector of estimated parameter(s).

• miles_annual : A numeric vector of element-wise computed annual distances using the linear relationship described in 'Details'.

• distribution : Specified distribution.

Arguments

mileage

A numeric vector of distances covered. Use NA for missing elements.

time

A numeric vector of operating times. Use NA for missing elements.

distribution

Supposed distribution of the random variable.

Details

The distribution parameter(s) are determined on the basis of complete cases, i.e. there is no NA in one of the related vector elements c(mileage[i], time[i]). Distances and operating times less than or equal to zero are not considered as well.

Assumption of linear relationship: Imagine a component in a vehicle has endured a distance of 25000 kilometers (km) in 500 days (d), the annual distance of this unit is $$25000 km \cdot (\frac{365 d} {500 d}) = 18250 km$$

Examples

# Data for examples:
date_of_registration <- c("2014-08-17", "2014-03-29", "2014-12-06",
                          "2014-09-09", "2014-05-14", "2014-07-01",
                          "2014-06-16", "2014-04-03", "2014-05-23",
                          "2014-05-09", "2014-05-31", "2014-08-12",
                          "2014-04-13", "2014-02-15", "2014-07-07",
                          "2014-03-12", "2014-05-27", "2014-06-02",
                          "2014-05-20", "2014-03-21", "2014-06-19",
                          "2014-02-12", "2014-03-27")
date_of_repair       <- c(NA, "2014-09-15", "2015-07-04", "2015-04-10", NA,
                          NA, "2015-04-24", NA, "2015-04-25", "2015-04-24",
                          "2015-06-12", NA, "2015-05-04", NA, NA, "2015-05-22",
                          NA, "2015-09-17", NA, "2015-08-15", "2015-11-26",
                          NA, NA)
date_of_analysis     <- "2015-12-31"

## Assume that mileage is only known for units that have failed (date_of_repair != NA).
mileage              <- c(NA, 15655, 13629, 18292, NA, NA, 33555, NA, 21737,
                          29870, 21068, NA, 122283, NA, NA, 36088, NA, 11153,
                          NA, 122842, 20349, NA, NA)

## time in service is the difference between repair and registration for failed
## items and the difference between date of analysis and date of registration
## for intact units.
time_in_service <- difftime(
  as.Date(date_of_repair, format = "%Y-%m-%d"),
  as.Date(date_of_registration, format = "%Y-%m-%d"),
  units = "days"
)
time_in_service[is.na(time_in_service)] <- difftime(
  as.Date(date_of_analysis, format = "%Y-%m-%d"),
  as.Date(date_of_registration[is.na(time_in_service)], format = "%Y-%m-%d"),
  units = "days"
)
time_in_service <- as.numeric(time_in_service)

# Example 1 - Assuming lognormal annual mileage distribution:
params_mileage_annual <- dist_mileage(
  mileage = mileage,
  time = time_in_service,
  distribution = "lognormal"
)

# Example 2 - Assuming exponential annual mileage distribution:
params_mileage_annual_2 <- dist_mileage(
  mileage = mileage,
  time = time_in_service,
  distribution = "exponential"
)