Risk Analysis of the Space Shuttle: Pre-Challenger Prediction of Failure
Tegegne
June 16, 2020
library(ggplot2)
sessionInfo()## R version 4.0.0 (2020-04-24)
## Platform: x86_64-w64-mingw32/x64 (64-bit)
## Running under: Windows 10 x64 (build 18363)
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## Matrix products: default
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## locale:
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## [3] LC_MONETARY=English_United States.1252
## [4] LC_NUMERIC=C
## [5] LC_TIME=English_United States.1252
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## attached base packages:
## [1] stats graphics grDevices utils datasets methods base
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## other attached packages:
## [1] ggplot2_3.3.1
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## loaded via a namespace (and not attached):
## [1] Rcpp_1.0.4.6 knitr_1.28 magrittr_1.5 munsell_0.5.0
## [5] colorspace_1.4-1 R6_2.4.1 rlang_0.4.6 stringr_1.4.0
## [9] tools_4.0.0 grid_4.0.0 gtable_0.3.0 xfun_0.14
## [13] withr_2.2.0 htmltools_0.4.0 ellipsis_0.3.1 yaml_2.2.1
## [17] digest_0.6.25 tibble_3.0.1 lifecycle_0.2.0 crayon_1.3.4
## [21] vctrs_0.3.1 glue_1.4.1 evaluate_0.14 rmarkdown_2.2
## [25] stringi_1.4.6 compiler_4.0.0 pillar_1.4.4 scales_1.1.1
## [29] pkgconfig_2.0.3Here are the available libraries
devtools::session_info()## - Session info ---------------------------------------------------------------
## setting value
## version R version 4.0.0 (2020-04-24)
## os Windows 10 x64
## system x86_64, mingw32
## ui RTerm
## language (EN)
## collate English_United States.1252
## ctype English_United States.1252
## tz Europe/Paris
## date 2020-06-15
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## - Packages -------------------------------------------------------------------
## package * version date lib source
## assertthat 0.2.1 2019-03-21 [1] CRAN (R 4.0.0)
## backports 1.1.7 2020-05-13 [1] CRAN (R 4.0.0)
## callr 3.4.3 2020-03-28 [1] CRAN (R 4.0.0)
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## devtools 2.3.0 2020-04-10 [1] CRAN (R 4.0.0)
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## ggplot2 * 3.3.1 2020-05-28 [1] CRAN (R 4.0.0)
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## knitr 1.28 2020-02-06 [1] CRAN (R 4.0.0)
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## magrittr 1.5 2014-11-22 [1] CRAN (R 4.0.0)
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## yaml 2.2.1 2020-02-01 [1] CRAN (R 4.0.0)
##
## [1] C:/Users/V330/Documents/R/win-library/4.0
## [2] C:/Program Files/R/R-4.0.0/libraryLoading and inspecting data
data <- read.csv("~/R/data.csv")
data## Date Count Temperature Malfunction Pressure
## 1 04/12/1981 6 66 0 50
## 2 11/12/1981 6 70 1 50
## 3 03/22/1982 6 69 0 50
## 4 11/11/1982 6 68 0 50
## 5 04/04/1983 6 67 0 50
## 6 06/18/1982 6 72 0 50
## 7 08/30/1983 6 73 0 100
## 8 11/28/1983 6 70 0 100
## 9 02/03/1984 6 57 1 200
## 10 04/06/1984 6 63 1 200
## 11 08/30/1984 6 70 1 200
## 12 10/05/1984 6 78 0 200
## 13 11/08/1984 6 67 0 200
## 14 01/24/1985 6 53 2 200
## 15 04/12/1985 6 67 0 200
## 16 04/29/1985 6 75 0 200
## 17 06/17/1985 6 70 0 200
## 18 7/2903/85 6 81 0 200
## 19 08/27/1985 6 76 0 200
## 20 10/03/1985 6 79 0 200
## 21 10/30/1985 6 75 2 200
## 22 11/26/1985 6 76 0 200
## 23 01/12/1986 6 58 1 200We know from our previous experience on this data set that filtering data is a really bad idea. We will therefore process it as such.
Let’s visually inspect how temperature affects malfunction:
plot(data=data, Malfunction/Count ~ Temperature, ylim=c(0,1))
Logistic regression
Let’s assume O-rings independently fail with the same probability which solely depends on temperature. A logistic regression should allow us to estimate the influence of temperature.
logistic_reg = glm(data=data, Malfunction/Count ~ Temperature, weights=Count, family=binomial(link='logit'))
summary(logistic_reg)##
## Call:
## glm(formula = Malfunction/Count ~ Temperature, family = binomial(link = "logit"),
## data = data, weights = Count)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.95227 -0.78299 -0.54117 -0.04379 2.65152
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.08498 3.05247 1.666 0.0957 .
## Temperature -0.11560 0.04702 -2.458 0.0140 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 24.230 on 22 degrees of freedom
## Residual deviance: 18.086 on 21 degrees of freedom
## AIC: 35.647
##
## Number of Fisher Scoring iterations: 5The maximum likelyhood estimator of the intercept and of Temperature are thus \(\hat{\alpha}=5.0849\) and \(\hat{\beta}=-0.1156\) and their standard errors are \(s_{\hat{\alpha}} = 3.052\) and \(s_{\hat{\beta}} = 0.04702\). The Residual deviance corresponds to the Goodness of fit \(G^2=18.086\) with 21 degrees of freedom. I have therefore managed to replicate the results of the Dalal et al. article.
Predicting failure probability
The temperature when launching the shuttle was 31°F. Let’s try to estimate the failure probability for such temperature using our model.:
# shuttle=shuttle[shuttle$r!=0,]
tempv = seq(from=30, to=90, by = .5)
rmv <- predict(logistic_reg,list(Temperature=tempv),type="response")
plot(tempv,rmv,type="l",ylim=c(0,1))
points(data=data, Malfunction/Count ~ Temperature)
This figure is very similar to the Figure 4 of Dalal et al. I have managed to replicate the Figure 4 of the Dalal et al. article.
Confidence on the prediction
Let’s try to plot confidence intervals with ggplot2.
ggplot(data, aes(y=Malfunction/Count, x=Temperature)) + geom_point(alpha=.2, size = 2, color="blue") +
geom_smooth(method = "glm", method.args = list(family = "binomial"), fullrange=T) +
xlim(30,90) + ylim(0,1) + theme_bw()## `geom_smooth()` using formula 'y ~ x'## Warning in eval(family$initialize): non-integer #successes in a binomial glm!
Mmmh, I have a warning from ggplot2 indicating “non-integer #successes in a binomial glm!”. This seems fishy. Furthermore, this confidence region seems huge… It seems strange to me that the uncertainty grows so large for higher temperatures. And compared to my previous call to glm, I haven’t indicated the weight which accounts for the fact that each ratio Malfunction/Count corresponds to Count observations (if someone knows how to do this…). There must be something wrong.
So let’s provide the “raw” data to ggplot2.
data_flat=data.frame()
for(i in 1:nrow(data)) {
temperature = data[i,"Temperature"];
malfunction = data[i,"Malfunction"];
d = data.frame(Temperature=temperature,Malfunction=rep(0,times = data[i,"Count"]))
if(malfunction>0) {
d[1:malfunction, "Malfunction"]=1;
}
data_flat=rbind(data_flat,d)
}
dim(data_flat)## [1] 138 2str(data_flat)## 'data.frame': 138 obs. of 2 variables:
## $ Temperature: int 66 66 66 66 66 66 70 70 70 70 ...
## $ Malfunction: num 0 0 0 0 0 0 1 0 0 0 ...Let’s check whether I obtain the same regression or not:
logistic_reg_flat = glm(data=data_flat, Malfunction ~ Temperature, family=binomial(link='logit'))
summary(logistic_reg)##
## Call:
## glm(formula = Malfunction/Count ~ Temperature, family = binomial(link = "logit"),
## data = data, weights = Count)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.95227 -0.78299 -0.54117 -0.04379 2.65152
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.08498 3.05247 1.666 0.0957 .
## Temperature -0.11560 0.04702 -2.458 0.0140 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 24.230 on 22 degrees of freedom
## Residual deviance: 18.086 on 21 degrees of freedom
## AIC: 35.647
##
## Number of Fisher Scoring iterations: 5Perfect. The estimates and the standard errors are the same although the Residual deviance is difference since the distance is now measured with respect to each 0/1 measurement and not to ratios. Let’s use plot the regression for data_flat along with the ratios (data).
ggplot(data=data_flat, aes(y=Malfunction, x=Temperature)) +
geom_smooth(method = "glm", method.args = list(family = "binomial"), fullrange=T) +
geom_point(data=data, aes(y=Malfunction/Count, x=Temperature),alpha=.2, size = 2, color="blue") +
geom_point(alpha=.5, size = .5) +
xlim(30,90) + ylim(0,1) + theme_bw()## `geom_smooth()` using formula 'y ~ x'
This confidence interval seems much more reasonable (in accordance with the data) than the previous one. Let’s check whether it corresponds to the prediction obtained when calling directly predict. Obtaining the prediction can be done directly or through the link function.
Here is the “direct” (response) version I used in my very first plot:
pred = predict(logistic_reg_flat,list(Temperature=30),type="response",se.fit = T)
pred## $fit
## 1
## 0.834373
##
## $se.fit
## 1
## 0.2293304
##
## $residual.scale
## [1] 1The estimated Failure probability for 30° is thus \(0.834\). However, the \(se.fit\) value seems pretty hard to use as I can obviously not simply add \(\pm 2 se.fit\) to \(fit\) to compute a confidence interval.
Here is the “link” version:
pred_link = predict(logistic_reg_flat,list(Temperature=30),type="link",se.fit = T)
pred_link## $fit
## 1
## 1.616942
##
## $se.fit
## [1] 1.659473
##
## $residual.scale
## [1] 1logistic_reg$family$linkinv(pred_link$fit)## 1
## 0.834373I recover \(0.834\) for the estimated Failure probability at 30°. But now, going through the linkinv function, we can use \(se.fit\):
critval = 1.96
logistic_reg$family$linkinv(c(pred_link$fit-critval*pred_link$se.fit,
pred_link$fit+critval*pred_link$se.fit))## 1 1
## 0.1630612 0.9923814The 95% confidence interval for our estimation is thus [0.163,0.992]. This is what ggplot2 just plotted me. This seems coherent.
I am now rather confident that I have managed to correctly compute and plot the uncertainty of my prediction. Let’s be honnest, it took me a while. My first attempts were plainly wrong (I didn’t know how to do this so I trusted ggplot2, which I was misusing) and did not use the correct statistical method. I also feel confident now because this has been somehow validated by other colleagues but it will be interesting that you collect other kind of plots values that you obtained, that differ and that you would probably have kept if you didn’t have a reference to compare to. Please provide us with as many versions as you can.