# How to predict failure of machinery using data science

**R | Asitav Sen**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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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.

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, 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.

### Advantages for machine owners

- Reduce time to repair
- Avoid unplanned maintenance
- Reduce inventory cost
- Increase bottom line

# 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.

### Advantages for OEMS

- Avoid unplanned breakdowns
- Increase customer satisfaction (lower repair time)
- Optimize inventory
- Improve top line
- Improve product

# 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.

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, 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 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.^{1} 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

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!

For any questions and/or clarifications, please do not hesitate to contact me.

# Notes

The data has been sourced from [Deepti Chevvuri’s Github] (https://github.com/DeeptiChevvuri/Predictive-Maintenance-Modelling-Datasets). The data contains error logs as well, which has not been used in this analysis.↩︎

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**R | Asitav Sen**.

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