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Module 2/exo 5: mise à jour et version anglaise

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#+TITLE: Analysis of the risk of failure of the O-rings on the Challenger shuttle
#+AUTHOR: Arnaud Legrand
#+LANGUAGE: en
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/readtheorg.css"/>
#+HTML_HEAD: <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
#+HTML_HEAD: <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/lib/js/jquery.stickytableheaders.js"></script>
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
#+LATEX_HEADER: \usepackage[utf8]{inputenc}
#+LATEX_HEADER: \usepackage[T1]{fontenc}
#+LATEX_HEADER: \usepackage[a4paper,margin=.8in]{geometry}
#+LATEX_HEADER: \usepackage[french]{babel}
# #+PROPERTY: header-args :session :exports both
On January 27, 1986, the day before the takeoff of the shuttle /Challenger/, had
a three-hour teleconference was held between
Morton Thiokol (the manufacturer of one of the engines) and NASA. The
discussion focused on the consequences of the
temperature at take-off of 31°F (just below
0°C) for the success of the flight and in particular on the performance of the
O-rings used in the engines. Indeed, no test
had been performed at this temperature.
The following study takes up some of the analyses carried out that
night with the objective of assessing the potential influence of
the temperature and pressure to which the O-rings are subjected
on their probability of malfunction. Our starting point is
the results of the experiments carried out by NASA engineers
during the six years preceding the launch of the shuttle
Challenger.
* Loading the data
We start by loading this data:
#+begin_src R :results output :session *R* :exports both
data = read.csv("shuttle.csv",header=T)
data
#+end_src
#+RESULTS:
#+begin_example
Date Count Temperature Pressure Malfunction
1 4/12/81 6 66 50 0
2 11/12/81 6 70 50 1
3 3/22/82 6 69 50 0
4 11/11/82 6 68 50 0
5 4/04/83 6 67 50 0
6 6/18/82 6 72 50 0
7 8/30/83 6 73 100 0
8 11/28/83 6 70 100 0
9 2/03/84 6 57 200 1
10 4/06/84 6 63 200 1
11 8/30/84 6 70 200 1
12 10/05/84 6 78 200 0
13 11/08/84 6 67 200 0
14 1/24/85 6 53 200 2
15 4/12/85 6 67 200 0
16 4/29/85 6 75 200 0
17 6/17/85 6 70 200 0
18 7/2903/85 6 81 200 0
19 8/27/85 6 76 200 0
20 10/03/85 6 79 200 0
21 10/30/85 6 75 200 2
22 11/26/85 6 76 200 0
23 1/12/86 6 58 200 1
#+end_example
The data set shows us the date of each test, the number of O-rings
(there are 6 on the main launcher), the
temperature (in Fahrenheit) and pressure (in psi), and finally the
number of identified malfunctions.
* Graphical inspection
Flights without incidents do not provide any information
on the influence of temperature or pressure on malfunction.
We thus focus on the experiments in which at least one O-ring was defective.
#+begin_src R :results output :session *R* :exports both
data = data[data$Malfunction>0,]
data
#+end_src
#+RESULTS:
: Date Count Temperature Pressure Malfunction
: 2 11/12/81 6 70 50 1
: 9 2/03/84 6 57 200 1
: 10 4/06/84 6 63 200 1
: 11 8/30/84 6 70 200 1
: 14 1/24/85 6 53 200 2
: 21 10/30/85 6 75 200 2
: 23 1/12/86 6 58 200 1
We have a high temperature variability but
the pressure is almost always 200, which should
simplify the analysis.
How does the frequency of failure vary with temperature?
#+begin_src R :results output graphics :file "freq_temp.png" :exports both :width 600 :height 400 :session *R*
plot(data=data, Malfunction/Count ~ Temperature, ylim=c(0,1))
#+end_src
#+RESULTS:
[[file:freq_temp.png]]
At first glance, the dependence does not look very important, but let's try to
estimate the impact of temperature $t$ on the probability of O-ring malfunction.
* Estimation of the temperature influence
Suppose that each of the six O-rings is damaged with the same
probability and independently of the others and that this probability
depends only on the temperature. If $p(t)$ is this probability, the
number $D$ of malfunctioning O-rings during a flight at
temperature $t$ follows a binomial law with parameters $n=6$ and
$p=p(t)$. To link $p(t)$ to $t$, we will therefore perform a
logistic regression.
#+begin_src R :results output :session *R* :exports both
logistic_reg = glm(data=data, Malfunction/Count ~ Temperature, weights=Count,
family=binomial(link='logit'))
summary(logistic_reg)
#+end_src
#+RESULTS:
#+begin_example
Call:
glm(formula = Malfunction/Count ~ Temperature, family = binomial(link = "logit"),
data = data, weights = Count)
Deviance Residuals:
2 9 10 11 14 21 23
-0.3015 -0.2836 -0.2919 -0.3015 0.6891 0.6560 -0.2850
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.389528 3.195752 -0.435 0.664
Temperature 0.001416 0.049773 0.028 0.977
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 1.3347 on 6 degrees of freedom
Residual deviance: 1.3339 on 5 degrees of freedom
AIC: 18.894
Number of Fisher Scoring iterations: 4
#+end_example
The most likely estimator of the temperature parameter is 0.001416
and the standard error of this estimator is 0.049, in other words we
cannot distinguish any particular impact and we must take our
estimates with caution.
* Estimation of the probability of O-ring malfunction
The expected temperature on the take-off day is 31°F. Let's try to
estimate the probability of O-ring malfunction at
this temperature from the model we just built:
#+begin_src R :results output graphics :file "proba_estimate.png" :exports both :width 600 :height 400 :session *R*
# shuttle=shuttle[shuttle$r!=0,]
tempv = seq(from=30, to=90, by = .5)
rmv <- predict(logistic_reg,list(Temperature=tempv),type="response")
plot(tempv,rmv,type="l",ylim=c(0,1))
points(data=data, Malfunction/Count ~ Temperature)
#+end_src
#+RESULTS:
[[file:proba_estimate.png]]
As expected from the initial data, the
temperature has no significant impact on the probability of failure of the
O-rings. It will be about 0.2, as in the tests
where we had a failure of at least one joint. Let's get back to the initial dataset to estimate the probability of failure:
#+begin_src R :results output :session *R* :exports both
data_full = read.csv("shuttle.csv",header=T)
sum(data_full$Malfunction)/sum(data_full$Count)
#+end_src
#+RESULTS:
: [1] 0.06521739
This probability is thus about $p=0.065$. Knowing that there is
a primary and a secondary O-ring on each of the three parts of the
launcher, the probability of failure of both joints of a launcher
is $p^2 \approx 0.00425$. The probability of failure of any one of the
launchers is $1-(1-p^2)^3 \approx 1.2%$. That would really be
bad luck.... Everything is under control, so the takeoff can happen
tomorrow as planned.
But the next day, the Challenger shuttle exploded and took away
with her the seven crew members. The public was shocked and in
the subsequent investigation, the reliability of the
O-rings was questioned. Beyond the internal communication problems
of NASA, which have a lot to do with this fiasco, the previous analysis
includes (at least) a small problem.... Can you find it?
You are free to modify this analysis and to look at this dataset
from all angles in order to to explain what's wrong.
module2/exo5/exo5_en.Rmd 0 → 100644
View file @ a1512d41
---
title: "Analysis of the risk of failure of the O-rings on the Challenger shuttle"
author: "Arnaud Legrand"
date: "28 juin 2018"
output: html_document
---
On January 27, 1986, the day before the takeoff of the shuttle _Challenger_, had
a three-hour teleconference was held between
Morton Thiokol (the manufacturer of one of the engines) and NASA. The
discussion focused on the consequences of the
temperature at take-off of 31°F (just below
0°C) for the success of the flight and in particular on the performance of the
O-rings used in the engines. Indeed, no test
had been performed at this temperature.
The following study takes up some of the analyses carried out that
night with the objective of assessing the potential influence of
the temperature and pressure to which the O-rings are subjected
on their probability of malfunction. Our starting point is
the results of the experiments carried out by NASA engineers
during the six years preceding the launch of the shuttle
Challenger.
# Loading the data
We start by loading this data:
```{r}
data = read.csv("shuttle.csv",header=T)
data
```
The data set shows us the date of each test, the number of O-rings
(there are 6 on the main launcher), the
temperature (in Fahrenheit) and pressure (in psi), and finally the
number of identified malfunctions.
# Graphical inspection
Flights without incidents do not provide any information
on the influence of temperature or pressure on malfunction.
We thus focus on the experiments in which at least one O-ring was defective.
```{r}
data = data[data$Malfunction>0,]
data
```
We have a high temperature variability but
the pressure is almost always 200, which should
simplify the analysis.
How does the frequency of failure vary with temperature?
```{r}
plot(data=data, Malfunction/Count ~ Temperature, ylim=c(0,1))
```
At first glance, the dependence does not look very important, but let's try to
estimate the impact of temperature $t$ on the probability of O-ring malfunction.
# Estimation of the temperature influence
Suppose that each of the six O-rings is damaged with the same
probability and independently of the others and that this probability
depends only on the temperature. If $p(t)$ is this probability, the
number $D$ of malfunctioning O-rings during a flight at
temperature $t$ follows a binomial law with parameters $n=6$ and
$p=p(t)$. To link $p(t)$ to $t$, we will therefore perform a
logistic regression.
```{r}
logistic_reg = glm(data=data, Malfunction/Count ~ Temperature, weights=Count,
family=binomial(link='logit'))
summary(logistic_reg)
```
The most likely estimator of the temperature parameter is 0.001416
and the standard error of this estimator is 0.049, in other words we
cannot distinguish any particular impact and we must take our
estimates with caution.
# Estimation of the probability of O-ring malfunction
The expected temperature on the take-off day is 31°F. Let's try to
estimate the probability of O-ring malfunction at
this temperature from the model we just built:
```{r}
# shuttle=shuttle[shuttle$r!=0,]
tempv = seq(from=30, to=90, by = .5)
rmv <- predict(logistic_reg,list(Temperature=tempv),type="response")
plot(tempv,rmv,type="l",ylim=c(0,1))
points(data=data, Malfunction/Count ~ Temperature)
```
As expected from the initial data, the
temperature has no significant impact on the probability of failure of the
O-rings. It will be about 0.2, as in the tests
where we had a failure of at least one joint. Let's get back to the initial dataset to estimate the probability of failure:
```{r}
data_full = read.csv("shuttle.csv",header=T)
sum(data_full$Malfunction)/sum(data_full$Count)
```
This probability is thus about $p=0.065$. Knowing that there is
a primary and a secondary O-ring on each of the three parts of the
launcher, the probability of failure of both joints of a launcher
is $p^2 \approx 0.00425$. The probability of failure of any one of the
launchers is $1-(1-p^2)^3 \approx 1.2%$. That would really be
bad luck.... Everything is under control, so the takeoff can happen
tomorrow as planned.
But the next day, the Challenger shuttle exploded and took away
with her the seven crew members. The public was shocked and in
the subsequent investigation, the reliability of the
O-rings was questioned. Beyond the internal communication problems
of NASA, which have a lot to do with this fiasco, the previous analysis
includes (at least) a small problem.... Can you find it?
You are free to modify this analysis and to look at this dataset
from all angles in order to to explain what's wrong.
module2/exo5/exo5_en.ipynb 0 → 100644
View file @ a1512d41
This diff is collapsed.
module2/exo5/exo5.Rmd → module2/exo5/exo5_fr.Rmd
View file @ a1512d41
......@@ -5,7 +5,7 @@ date: "28 juin 2018"
output: html_document
---
Le 27 Janvier 1986, veille du décollage de la navette /Challenger/, eu
Le 27 Janvier 1986, veille du décollage de la navette _Challenger_, eu
lieu une télé-conférence de trois heures entre les ingénieurs de la
Morton Thiokol (constructeur d'un des moteurs) et de la NASA. La
discussion portait principalement sur les conséquences de la
......@@ -93,12 +93,25 @@ plot(tempv,rmv,type="l",ylim=c(0,1))
points(data=data, Malfunction/Count ~ Temperature)
```
La probabilité d'échec des joints toriques est donc d'environ 0.2
(comme dans les essais précédents) et comme on pouvait s'attendre au
vu des données initiales, la température n'a pas d'impact notable. La
probabilité que tous les joints toriques dysfonctionnent est de
$0.2^6 \approx 6.4\times10^{-5}$. Tout est sous contrôle, le décollage
peut donc avoir lieu demain comme prévu.
Comme on pouvait s'attendre au vu des données initiales, la
température n'a pas d'impact notable sur la probabilité d'échec des
joints toriques. Elle sera d'environ 0.2, comme dans les essais
précédents où nous il y a eu défaillance d'au moins un joint. Revenons
à l'ensemble des données initiales pour estimer la probabilité de
défaillance d'un joint:
```{r}
data_full = read.csv("shuttle.csv",header=T)
sum(data_full$Malfunction)/sum(data_full$Count)
```
Cette probabilité est donc d'environ $p=0.065$, sachant qu'il existe
un joint primaire un joint secondaire sur chacune des trois parties du
lançeur, la probabilité de défaillance des deux joints d'un lançeur
est de $p^2 \approx 0.00425$. La probabilité de défaillance d'un des
lançeur est donc de $1-(1-p^2)^3 \approx 1.2%$. Ça serait vraiment
pas de chance... Tout est sous contrôle, le décollage peut donc avoir
lieu demain comme prévu.
Seulement, le lendemain, la navette Challenger explosera et emportera
avec elle ses sept membres d'équipages. L'opinion publique est
......@@ -109,4 +122,3 @@ fiasco, l'analyse précédente comporte (au moins) un petit
problème... Saurez-vous le trouver ? Vous êtes libre de modifier cette
analyse et de regarder ce jeu de données sous tous les angles afin
d'expliquer ce qui ne va pas.
module2/exo5/exo5.ipynb → module2/exo5/exo5_fr.ipynb
View file @ a1512d41
......@@ -709,7 +709,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.5rc1"
"version": "3.6.1"
}
},
"nbformat": 4,
......
module2/exo5/exo5_python-en.org → module2/exo5/exo5_python_en.org
View file @ a1512d41
#+TITLE: Analysis of the risk of failure of the O-rings on the Challenger shuttle
#+AUTHOR: Arnaud Legrand
#+LANGUAGE: fr
#+LANGUAGE: en
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/readtheorg.css"/>
......@@ -202,7 +202,7 @@ This probability is thus about $p=0.065$. Knowing that there is
a primary and a secondary O-ring on each of the three parts of the
launcher, the probability of failure of both joints of a launcher
is $p^2 \approx 0.00425$. The probability of failure of any one of the
launchers is $1-(1-p^2)^3 \approximately 1.2%$. That would really be
launchers is $1-(1-p^2)^3 \approx 1.2%$. That would really be
bad luck.... Everything is under control, so the takeoff can happen
tomorrow as planned.
......
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