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How To Predict Machine Failure Using Data Science

How To Predict Machine Failure Using Data Science

Asitav Sen
W3Schools

It is well known, how annoying a machine breakdown can be. Production takes a direct hit because of equipment failures. A great deal of money is lost by the time production restarts. It also impacts OEMs and dealers in terms of lost reputation and business opportunity. Fortunately, these issue can be tackled to a major extent by using data science.

Opportunity for machine owners

Cost of breakdown is not just the opportunity loss (of potential profit from production), but it also includes fixed cost of the machine. Further, delay in production can attract penalties and lost orders. At times, when other machines also depend on the failed machinery, the cost escalates through the roof. The cost of single breakdown can easily exceed thousands of dollars. The worst part is, this loss can hardly ever be recovered.

A close up of a device

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The figure above shows some of the components of cost of downtime.



Using predictive models, one can now estimate failure probability. This gives us two abilities. First, the ability to plan maintenance in a manner to minimize loss. Second, to optimize inventory better. Instead of keeping a lot of spare parts in inventory, it becomes possible to keep only the ones that will be required in near future.

Opportunity for manufacturers (OEMs)

Breakdowns may not directly impact an OEM, but it harms the reputation and may also end up in lost business. If a critical item is not available at the nearest point, customers may not hesitate to procure the item from local market. Further, manpower may not be available to repair the machine immediately.

These problems can be avoided if you already has an idea about possible breakdowns. OEMs/Dealers can then, either plan a maintenance and replace the parts or immediately support in case of breakdown. Furthermore, it can help OEMs launch new revenue models of maintenance contracts. This can also ensure that customers do not buy spare parts from local markets.

Perhaps the biggest advantage these models give an OEM is the ability to improve their products. The models indicate which are the factors that impact a component failure, giving them a direction on how to improve component life.

How to implement

The process starts with identifying the problems to solve. Usually the problem should be broad and broken down into specifics. For e.g. objective could be to reduce cost of operations. One of the several ways to achieve it is by reducing downtime and optimizing spare parts inventory.A screenshot of a cell phone

Description automatically generated

Once the problems are identified, the data needs to be collected for analysis. Often the data will not be available. In such case, infrastructure to collect data needs to be built. While building the infrastructure and processes, attempt should be made to enhance the utility of such infrastructure/processes. This can be done by assessing other possible utilities of the data collected. If the marginal cost of adding more data to solve a significant problem is low, it should be pursued. 

Once the data starts getting collected, it needs to be cleaned and visualized. This is not just for the data science team but also for other business stakeholders. If possible and feasible, the dashboard(s) may be built for different stakeholders, depending on the need. The predictive aspects of analyses is a natural progression from exploratory analyses. The dashboards and visualizations are to be followed by the creation of predictive model(s), which are to be tested, reviewed and deployed.

Example using real life data

Code heavy!!! Skip if you are not interested in the codes. However, you may like to understand how well the below model performed in results

To illustrate a possible model, real life (normalized) data has been used. The data contains log of certain parameters, taken hourly. In addition, log of failures and maintenance will also be used. The parameters recorded hourly are voltage, vibration, rotation and pressure. Data from 100 machines of 4 types are recorded. Failure of 4 components are recorded. For the sake of simplification, we will analyze failure of ‘comp2’.

Several exploratory steps have been skipped to keep it short and focus on the predictive model.

# Following data are available
# 1. Telemetry – Logs hourly parameters (Voltage, Pressure, Rotation and Vibration) for each machine
# 2. Failures – Log of component failures. Contains the time slot (that matches the telemetry log) and machineID
# 3. Maint – Log of maintainance. Contains time slot, machineID and the component replaced
# 4. Assets – Information about the machines – machineID, Model and age

# Function to calculate number of periods since last maintenance of a component
timeslm <- function(k) {
  output <- c()
  output[1] = 0
  for (i in 2:length(k)) {
    if (k[i 1] == 1) {
      output[i] = 1
    } else {
      output[i] = output[i 1] + 1
    }
  }
  return(output)
}

# For the sake of simplicity, failures of ‘comp2’ will be analysed

#preparing data
failures2 <-
  failures %>% filter(failure == “comp2”) #new data frame with failure of comp2
maint2 <-
  maint %>% filter(comp == “comp2”) #new data frame with maintenance of comp2

df <-
  telemetry %>% left_join(failures2, by = c(“machineID”, “datetime”)) %>% #Joining log with failure
  mutate(failed = ifelse(is.na(failure), 0, 1)) %>%  #creating column with binary (failed or not)
  left_join(maint2, by = c(“machineID” = “machineID”, “datetime” = “datetime”)) %>% #Joining maintenance data
  mutate(maint = ifelse(is.na(comp), 0, 1)) %>% #New column with binary (maintained or not)
  inner_join(assets, by = “machineID”) #Joining machine details
df$datetime <-
  parse_date_time(df$datetime, “mdy HMS p”) #Changing column type to date time
df$machineID <- as.factor(df$machineID)
df$model <- as.factor(df$model)

timesm <- timeslm(df$maint) #Calculating periods since maintenance
df$timesm <- timesm

df <-
  df %>% mutate(t = ifelse(is.na((
    as.numeric(datetime lag(datetime, 1))
  )), 0, (as.numeric(
    datetime lag(datetime, 1)
  )))) %>%
  mutate(tim = cumsum(t)) #Adding time columns

df <- df[, c(1, 14, 15, 13, 11, 12, 2:6, 8, 10)]
#Column details – datetime = event log time
#                 machineID = Machine Identification number
#                 volt, rotate, pressure, vibration are some of the parameters that are measured
#                 failed and maint indicate if the component (comp2) failed. 1 indicates true.
#                 age is the age of the machine
#                 timesm is the time since maintenance
#                 tim is the time since the beginning of event logging


#removing dfs that are not required
rm(telemetry)
rm(assets)
rm(failures)
rm(maint)
rm(timesm)

# adding grouping to calculate cumulative parameter values.
g <- c()
g[1] = 1
for (i in 2:nrow(df)) {
  if (df$timesm[i] < df$timesm[i 1]) {
    g[i] = g[i 1] + 1
  } else {
    g[i] = g[i 1]
  }
}

df$group <- g

# removing unwanted data
rm(g)


#Will add new columns with cumulative parameters – sum, mean

df1 <-
  df %>% group_by(group) %>% mutate(
    volt.cum = cumsum(volt),
    vib.cum = cumsum(vibration),
    pres.cum = cumsum(pressure),
    rot.cum = cumsum(rotate),
    volt.mean = cumsum(volt) / seq_along(volt),
    vib.mean = cumsum(vibration) / seq_along(vibration),
    pres.mean = cumsum(pressure) / seq_along(pressure),
    rot.mean = cumsum(rotate) / seq_along(rotate)
  ) %>%
  filter(maint == 1)   # filtered required data

# In case you are interested to know about the transformations in detail please get in touch.

#preparing test and train data
df.train <- df1 %>% filter(datetime < “2015-11-15 06:00:00 UTC”)
df.test <- df1 %>% filter(datetime > “2015-11-15 06:00:00 UTC”)

#removing unwanted data
rm(df1)
rm(df)

# Fitting Kaplan Meier

kap.fit<-survfit(Surv(timesm,failed)~model, data=df.train)

#Plotting

fig<-ggsurvplot(
  kap.fit,
  pval = F, # show p-value
  break.time.by = 1000, #break X axis by 25 periods
  #risk.table = “abs_pct”, # absolute number and percentage at risk
  #risk.table.y.text = FALSE,# show bars instead of names in text annotations
  linetype = “strata”,
  # Change line type by groups
  conf.int = T,
  # show confidence intervals for
  #conf.int.style = “step”,  # customize style of confidence intervals
  #surv.median.line = “hv”,
  # Specify median survival
  ggtheme = theme_minimal(),
  # Change ggplot2 theme
  legend.labs =
    c(“Model 1”, “Model 2”, “Model 3”, “Model 4”),
  # change legend labels
  ncensor.plot = F,
  # plot the number of censored subjects (outs) at time t
  #palette = c(“#000000”, “#2E9FDF”,”#FF0000″)

)+
  labs(x=”Hours”)
fig

Kaplan Meier model predicts survival chances by 3000 hours of operation (post maintenance) reduces significantly. But there is a substantial amount of uncertainty in prediction, except in case of model 4. The model also shows that it is almost certain that there will be no failures till about 700 hours of operation, post maintenance.

Summary of the fit is as shown below.

summary(kap.fit)$table

##              records n.max n.start events   *rmean *se(rmean) median 0.95LCL
## model=model1     103   103     103     37 2502.051   260.3187   1800    1440
## model=model2     112   112     112     32 2625.123   291.1468   2160    1489
## model=model3     257   257     257     78 2634.175   210.9060   1800    1440
## model=model4     191   191     191     70 2502.417   177.2059   2160    1800
##              0.95UCL
## model=model1    3240
## model=model2      NA
## model=model3    2880
## model=model4    2880

See Also
Flow battery

Since the uncertainty is high, I will not use this model and try Cox regression.

Cox regression model shows that among all the parameters used, cumulative rotation is the one that significantly impacts failure of ‘comp2’.

With this insight, OEM can think of ways to improve component quality or find out ways to keep the rotation low.

# Cox regression
cox.fit<-coxph(Surv(timesm,failed) ~ volt.cum+vib.cum+pres.cum+rot.cum+volt.mean+vib.mean+pres.mean+rot.mean+factor(model), data=df.train)
summary(cox.fit)

## Call:
## coxph(formula = Surv(timesm, failed) ~ volt.cum + vib.cum + pres.cum +
##     rot.cum + volt.mean + vib.mean + pres.mean + rot.mean + factor(model),
##     data = df.train)
##
##   n= 663, number of events= 217
##
##                           coef  exp(coef)   se(coef)      z Pr(>|z|)   
## volt.cum            -2.389e-04  9.998e-01  1.416e-04 -1.687 0.091529 . 
## vib.cum             -3.233e-04  9.997e-01  2.849e-04 -1.135 0.256382   
## pres.cum             2.164e-05  1.000e+00  1.557e-04  0.139 0.889459   
## rot.cum             -1.496e-04  9.999e-01  4.161e-05 -3.594 0.000326 ***
## volt.mean            2.527e-01  1.287e+00  1.687e-01  1.498 0.134178   
## vib.mean             5.436e-01  1.722e+00  3.291e-01  1.652 0.098603 . 
## pres.mean            1.520e-01  1.164e+00  1.851e-01  0.821 0.411487   
## rot.mean            -4.229e-02  9.586e-01  4.338e-02 -0.975 0.329642   
## factor(model)model2 -4.183e-01  6.582e-01  2.786e-01 -1.501 0.133273   
## factor(model)model3 -1.932e-01  8.243e-01  2.322e-01 -0.832 0.405304   
## factor(model)model4  9.385e-02  1.098e+00  2.397e-01  0.391 0.695429   
## —
## Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ‘ 1
##
##                     exp(coef) exp(-coef) lower .95 upper .95
## volt.cum               0.9998     1.0002    0.9995    1.0000
## vib.cum                0.9997     1.0003    0.9991    1.0002
## pres.cum               1.0000     1.0000    0.9997    1.0003
## rot.cum                0.9999     1.0001    0.9998    0.9999
## volt.mean              1.2875     0.7767    0.9250    1.7920
## vib.mean               1.7222     0.5807    0.9035    3.2826
## pres.mean              1.1641     0.8590    0.8100    1.6731
## rot.mean               0.9586     1.0432    0.8804    1.0437
## factor(model)model2    0.6582     1.5193    0.3813    1.1363
## factor(model)model3    0.8243     1.2132    0.5229    1.2994
## factor(model)model4    1.0984     0.9104    0.6866    1.7571
##
## Concordance= 0.98  (se = 0.004 )
## Likelihood ratio test= 1060  on 11 df,   p=<2e-16
## Wald test            = 135.2  on 11 df,   p=<2e-16
## Score (logrank) test = 483.9  on 11 df,   p=<2e-16

So, a new model was built to include only cumulative rotation as parameter. The model was used to predict outcome on test data. Confusion matrix was built to analyze accuracy.

# Revised cox model
cox.fit<-coxph(Surv(timesm,failed) ~ rot.cum, data=df.train)

# Predicted probabilities
pred<-predict(cox.fit,newdata=filter(df.test), type = “expected”)

d<-df.test%>%cbind(pred=pred)%>%select(12,14,23)%>%mutate(pred=ifelse(pred>0.5,1,0))

table(act=d$failed,pred=d$pred)

##    pred
## act  0  1
##   0 49 12
##   1  7 32


Results

  • 81 % accuracy was achieved
  • 87 % accurate when did not predict failure
  • 72 % accurate when predicted failure

Wondering what does it mean in terms of saved cash? Or what are all possible ways to create value out of this? One needs more detail of the business and operations to answer those questions. If you are really interested, you know what to do!

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