diff --git a/exercises/Dependability/reliability-availability-solutions.pdf b/exercises/Dependability/reliability-availability-solutions.pdf
new file mode 100644
index 0000000..5e30f17
Binary files /dev/null and b/exercises/Dependability/reliability-availability-solutions.pdf differ
diff --git a/exercises/Dependability/reliability-availability-solutions.tex b/exercises/Dependability/reliability-availability-solutions.tex
new file mode 100644
index 0000000..36be0e4
--- /dev/null
+++ b/exercises/Dependability/reliability-availability-solutions.tex
@@ -0,0 +1,409 @@
+\documentclass{article}
+
+\usepackage[utf8]{inputenc}
+\usepackage{graphicx}
+\usepackage{geometry}
+\usepackage[hidelinks]{hyperref}
+%% \usepackage{enumitem}
+
+\newcommand{\Q}{\textbf{Q:}}
+\newcommand{\A}{\textbf{A:}}
+
+\title{\vspace{-.5cm}Industrial Automation\\ \vspace{1cm}Reliability and Availability Exercises\\ \vspace{1cm}Solutions}
+%% \author{Matej Pavlovic}
+\date{}
+
+\begin{document}
+
+\maketitle
+
+In our running example, we consider a chain of Christmas lights consisting of many light bulbs.
+
+\begin{enumerate}
+
+\item Let us first consider a single light bulb.
+\begin{enumerate}
+\item
+  \Q\ Under what conditions can we assume that a single light bulb's
+  failure rate is constant?
+
+  \A\ We can assume that a light bulb's failure rate is constant if
+  the life bulb is in the middle phase (also called ``useful life'')
+  of its life cycle.
+\item
+  \Q\ From now on, let us assume that the failure rate of a single
+  light bulb is indeed constant. Moreover, assume that a light bulb is
+  expected to burn out after 1000 hours. What is its failure rate?
+
+  \A\ The time after which the light bulb is expected to burn out is
+  nothing else than it's mean time to failure (MTTF). Note that we
+  also know that under the assumption of a constant failure rate
+  $\lambda$, MTTF $= 1/\lambda$. Thus, $\lambda = 1/1000$ $h^{-1}$
+\item
+  \Q\ What is the reliability of a single light bulb?
+
+  \A\ Unlike the mean time to failure, reliability is a function of
+  time. It expresses the probability of a system not having failed
+  until a given time. For a constant failure rate $\lambda$, the
+  reliability $R_B(t) = e^{-\lambda t} = e^{-t/1000}$. Note that the we
+  consider hours as the units of $t$, as we express $\lambda$ using
+  the unit $h^{-1}$.
+  
+\end{enumerate}
+
+\item We consider a chain of Christmas lights consisting of 50
+  serially connected light bulbs. This means that if any light bulb
+  burns out, the electric circuit is interrupted and all light bulbs
+  go off.
+\begin{enumerate}
+\item
+  \Q\ What is the reliability of the whole chain?
+
+  \A\ As all components must work in order for the whole system to
+  work, the reliability of the system is the product of the
+  reliabilities of all its components. For our light chain, it is the
+  product of reliabilities of each light bulb. Let $R_i(t)$ be the
+  reliability of light bulb $i$, for $1 \leq i \leq 50$. As all
+  light bulbs to have the same reliability $R_i(t) = R_B(t)$, the
+  reliability of the whole (serially connected) chain is
+  \begin{equation}
+  R_S(t) = R_1(t)\cdot R_2(t)\cdots R_{50} = e^{-t/1000}\cdot e^{-t/1000} \cdots e^{-t/1000} = e^{-50t/1000} = e^{-t/20}
+  \end{equation}
+\item
+  \Q\ How long can we expect the light chain to work?
+
+  \A\ The reliability of the whole chain being $e^{-t/20}$, we see
+  that the (constant) failure rate $\lambda = 1/20$ $h^{-1}$. The time
+  we expect the chain to work is the MTTF $ = 1/\lambda = 20$ hours.
+
+\end{enumerate}
+
+\item We now assume that the light bulbs are connected in
+  parallel. Thus, if a light bulb burns out, the others may continue
+  operating. We consider a light chain to be working as long as not
+  more than 10 light bulbs are burned out.
+\begin{enumerate}
+\item
+  \Q\ What is the reliability of the whole chain?
+
+  \A\ Let $R_B$ the reliability of a single light bulb. For simplicity
+  of notation, we omit the time argument ``(t)'', but remember that
+  the reliability is still a function of time!
+
+  The probability that exactly $i$ out of 50 light bulbs fail is
+  \begin{equation}
+    {50 \choose i} \cdot (1-R_B)^i \cdot R_B^{50-i}
+  \end{equation}
+  The part $(1-R_B)^i$ signifies $i$ particular bulbs failing while
+  $(R_B)^{50-i}$ stands for the remaining bulbs not failing. The
+  binomial coefficient represents the number of ways we can choose $i$
+  out of 50 light bulbs.
+  
+  If at most 10 light bulbs are allowed to fail, we need to calculate
+  the probability that exactly zero of them fail, or exactly one
+  fails, or exactly two fail ... or exactly 10 fail. Summing all of
+  these probabilities gives us the resulting system reliability of he
+  parallel version of the light chain:
+
+\begin{equation}
+R_P(t) = \sum_{i=0}^{10}{50 \choose i} \cdot (1-R_B)^i \cdot R_B^{50-i}
+       = \sum_{i=0}^{10}{50 \choose i} \cdot (1-e^{-t/1000})^i \cdot (e^{-t/1000})^{50-i}
+\end{equation}
+
+\item
+  \Q\ How long can we expect the light chain to work? (No need to
+  calculate the value, just give a mathematical expression describing
+  it.)
+
+  \A\ Again, we were asking for the mean time to failure. In general,
+  the MTTF is calculated as the integral of the reliability function
+  over time from 0 until infinity. Thus, in our case, the MTTF is
+
+  \begin{equation}
+    \int_0^{\infty} \sum_{i=0}^{10}{50 \choose i} \cdot (1-e^{-t/1000})^i \cdot (e^{-t/1000})^{50-i}dt
+  \end{equation}
+
+  This evaluates to more than 245 hours (we did not ask you to
+  calculate this), which is a substantial improvement over the serial
+  version.
+
+\end{enumerate}
+
+\item We assume that during the Christmas period, the light chain is
+  used every evening for 4 hours during the whole month of
+  December. Answer the following questions for both the serial and the
+  parallel variant of the light chain.
+\begin{enumerate}
+
+\item
+  \Q\ What is the probability that the light chain will remain
+  functional during the whole Christmas period? (Feel free to serve
+  yourself with mathematical tools like WolframAlpha to compute the
+  resulting numbers.)
+  
+  \A\ In our definition, the Christmas period lasts $31 * 4 = 124$
+  hours. The probability that a system (in our case the light chain)
+  will remain functional for a certain time $t$ is expressed by its
+  reliability. We only need to substitute $t$ for $124$ hours.
+
+  \begin{equation}
+    R_S(124) = e^{-124/20} \approx 0.2\%
+  \end{equation}
+  \begin{equation}
+    R_P(124) = \sum_{i=0}^{10}{50 \choose i} \cdot (1-e^{-0.124})^i \cdot (e^{-0.124})^{50-i} \approx 97.3\%
+  \end{equation}
+
+  Note that the serially connected light chain has extremely slim
+  chance of surviving a Christmas period, while the vast majority of
+  light chain survive (by our definition of ``survive'').
+
+\item
+  \Q\ What is the probability that the light chain survives the
+  Christmas periods of two consecutive years?
+
+  \A\ This means nothing else than doubling the number of hours the
+  light chain needs to remain functional, i.e. computing $R_S(248)$
+  and $R_P(248)$.
+
+  \begin{equation}
+    R_S(248) = e^{-248/20} \approx 0.0004\%
+  \end{equation}
+  \begin{equation}
+    R_P(248) = \sum_{i=0}^{10}{50 \choose i} \cdot (1-e^{-0.248})^i \cdot (e^{-0.248})^{50-i} \approx 44.72\%
+  \end{equation}
+
+  The serially connected light chain's survival chances are almost
+  zero, while almost half of the parallel chains survives two
+  consecutive Christmas periods.
+
+\item
+  \Q\ If, after the first year's Christmas period, 5 light bulbs burn
+  out, what is the probability of the light chain surviving the next
+  Christmas period? (Only relevant for the parallel case.)
+
+  \A\ This case is equivalent to considering a light chain consisting
+  of 45 lights out of which only 5 are allowed to burn out, and
+  evaluating its reliability (let us call it $R_P'$) for one Christmas
+  period (of 124 hours).
+
+  \begin{equation}
+    R_P'(124) = \sum_{i=0}^{5}{45 \choose i} \cdot (1-e^{-0.124})^i \cdot (e^{-0.124})^{45-i} \approx 57\%
+  \end{equation}
+
+\item
+  \Q\ Assume that the package with the light chain contains 2
+  replacement light bulbs. How do the answers to the 3 above questions
+  change?
+
+  \A\ Technically speaking, if ``surviving'' a Christmas means to
+  continue functioning without any intervention, the answers do not
+  change! The reliability of a system is a well defined term and does
+  not take into account whether there are replacement parts available
+  or not.
+
+  For replacement light bulbs to be useful, we need to make additional
+  assumptions on repairing a (partially) broken light chain. If we
+  assume that on the parallel version of the chain two light bulbs can
+  be replaced before the whole chain fails, then we can model it as a
+  system of 50 light bulbs out of which 12 (instead of 10) can fail.
+
+  For the serial version, we even need to redefine what ``surviving''
+  means. Otherwise, whenever a single light bulb fails, the chain is
+  considered as having failed, regardless of the presence of 2
+  replacement light bulbs in the box. We can say, for example, that
+  the serial version of the light chain fails if a light bulb goes of
+  AND there is no more replacement light bulbs (the immediate use of
+  which we assume for the first two failures). In this case, the
+  probability of surviving a Christmas period is close to the case of
+  a parallel version of the chain, where only 2 light bulbs are
+  allowed to fail. It is however, not exactly the same probability
+  (can you give an intuition why?).
+
+\item
+  \Q\ What are the answers to the previous 4 questions if we use LED
+  light bulbs with an expected life time of 5000 hours?
+
+  \A\ Here we use the very same formulas as above, but with different
+  parameters, namely with the parameter to the exponential function
+  being 5 times smaller (note that the MTTF of a single light bulb
+  appears in the denominator). At the root of this change is the
+  reliability of a single light bulb that changes from $e^{-t/1000}$
+  to $e^{-t/5000}$.
+
+  \begin{equation}
+    R_S(124) = e^{-124/100} \approx 28.94\%
+  \end{equation}
+  \begin{equation}
+    R_P(124) = \sum_{i=0}^{10}{50 \choose i} \cdot (1-e^{-0.124/5})^i \cdot (e^{-0.124/5})^{50-i} \approx 99.999997\%
+  \end{equation}
+
+  \begin{equation}
+    R_S(248) = e^{-248/100} \approx 8.37\%
+  \end{equation}
+  \begin{equation}
+    R_P(248) = \sum_{i=0}^{10}{50 \choose i} \cdot (1-e^{-0.248/5})^i \cdot (e^{-0.248/5})^{50-i} \approx 99.998\%
+  \end{equation}
+
+  \begin{equation}
+    R_P'(124) = \sum_{i=0}^{5}{45 \choose i} \cdot (1-e^{-0.124/5})^i \cdot (e^{-0.124/5})^{45-i} \approx 99.92\%
+  \end{equation}
+
+  We see that using better components substantially helps improving
+  the reliability of the chain in both the serial and the parallel
+  case. While the serial version remains very bad nonetheless, the
+  parallel version becomes extremely reliable (for a Christmas light
+  chain - for other systems, e.g. safety-critical ones, even such high
+  reliability might be insufficient in practice).
+
+\end{enumerate}
+
+\item Now consider a case of a light chain that is supposed to be on
+  permanently. If a light bulb goes off, it takes 1 hour to be noticed
+  and replaced on expectation. For simplicity, we assume that repairs
+  happen independently and can even be carried out in parallel.
+\begin{enumerate}
+\item
+  \Q What is the (stationary) availability of the serial variant of
+  the light chain?
+
+  \A The availability of a system expresses what fraction of the time
+  the system is in a working state. Thus, it is the time the system is
+  working divided by the sum of the time it is working and the time it
+  is being repaired, expressed mathematically, the availability is
+  \begin{equation}
+    \frac{MTTF}{MTTF + MTTR} = \frac{20}{20+1} \approx 95.24\%
+  \end{equation}
+  We substituted 20 for the MTTF, as we computed it in a previous
+  exercise, and 1 for MTTR, as given in this question. We can say
+  that, on average, the light chain works for 20 hours, then for one
+  hour it is being repaired, then it works for 20 hours again, and so
+  on...
+
+\item
+  \Q\ Draw a Markov chain modeling the reliability of both the serial
+  and the parallel variant of the light chain. For simplicity, assume
+  that in the parallel case, the light chain goes off automatically
+  when more than 10 light bulbs burn out.
+  
+  \A\ For the serial case, the Markov chain only consists of two
+  states, as there are only two possibilities for the light chain:
+  either all light bulbs work or the whole chain is off after the
+  first light bulb fails. The transition from the working state to the
+  failed state happens with the light chain's failure rate $(\lambda)
+  = 1/20$ $h^{-1}$, which we have already before. The repair rate, which
+  is, analogous to the failure rate, the inverse of the MTTR, is
+  $1$ $h^{-1}$
+
+
+  \begin{figure}[h]
+    \center
+    \includegraphics[height=2.5cm]{markov-serial}
+  \end{figure}
+
+  The parallel case is slightly more complex than the serial one, as
+  we need to distinguish more states of the light chain, depending on
+  how many light bulbs have failed. Also, the failure rates for each
+  depend on the number of working light bulbs. A single light bulb
+  having a failure rate of $1/1000$ $h^{-1}$, the combined failure
+  rate of $n$ light bulbs is $n\cdot 1/1000$ $h^{-1}$ (recall how we
+  computed the reliability in question 2.(a)). The analogous is true
+  for the repair rates.
+
+  \begin{figure}[h]
+    \center
+    \includegraphics[height=2.5cm]{markov-parallel}
+  \end{figure}
+
+\item
+  \Q\ Set up the differential equations describing the states of the
+  Markov chains.
+
+  \A\ For each state $p_i$, we have one equation constructed using the all
+  the incoming and outgoing edges of that state, of the form
+  \begin{equation}
+    \frac{dp_i}{dt} = \sum \lambda_kp_k(t) - \sum \lambda_ip_i(t)
+  \end{equation}
+  where $k$ goes through all the neighboring states from which there
+  is an edge to $p_i$, while $\lambda_i$ goes through all the outgoing
+  edges of $p_i$.
+
+  Thus, the equations for the serial version are
+
+  \begin{equation}
+    \frac{dp_0(t)}{dt} = 1 \cdot p_1(t) - \frac{50}{1000} \cdot p_0(t)
+  \end{equation}
+  \begin{equation}
+    \frac{dp_1(t)}{dt} = \frac{50}{1000} \cdot p_0(t) - 1 \cdot p_1(t)
+  \end{equation}
+
+  and the equations for the parallel version are
+
+  \begin{equation}
+    \frac{dp_0(t)}{dt} = 1 \cdot p_1(t) - \frac{50}{1000} \cdot p_0(t)
+  \end{equation}
+  \begin{equation}
+    \frac{dp_1(t)}{dt} = \frac{50}{1000} \cdot p_0(t) + 2 \cdot p_2(t) - \frac{49}{1000} \cdot p_1(t) - 1 \cdot p_1(t)
+  \end{equation}
+  \begin{equation}
+    \frac{dp_2(t)}{dt} = \frac{49}{1000} \cdot p_1(t) + 3 \cdot p_3(t) - \frac{48}{1000} \cdot p_2(t) - 2 \cdot p_2(t)
+  \end{equation}
+  \begin{equation}
+    \frac{dp_3(t)}{dt} = \frac{48}{1000} \cdot p_2(t) + 4 \cdot p_4(t) - \frac{47}{1000} \cdot p_3(t) - 3 \cdot p_3(t)
+  \end{equation}
+  \begin{equation}
+    \cdots
+  \end{equation}
+  \begin{equation}
+    \frac{dp_{10}(t)}{dt} = \frac{41}{1000} \cdot p_9(t) + 11 \cdot p_{11}(t) - \frac{40}{1000} \cdot p_{10}(t) - 10 \cdot p_{10}(t)
+  \end{equation}
+  \begin{equation}
+    \frac{dp_{11}(t)}{dt} = \frac{40}{1000} \cdot p_{10}(t) - 11 \cdot p_{11}(t)
+  \end{equation}
+
+\item
+  \Q\ What are the initial conditions of the differential equations
+  for a new light chain (i.e. when none of the light bulbs burned out
+  yet)?
+
+  \A\ When the light chain is new, we assume all the light bulbs to be
+  working. Thus, for both the serial and parallel version of the
+  chain, at time $t = 0$, $p_0(0) = 1$ and $p_i(0) = 0$ for all $i >
+  0$.
+\item
+  \Q\ How would you calculate the mean time to failure (MTTF) of the
+  light chains? (Describe the steps, without performing the actual calculation.)
+
+  \A\ As described in class, to compute the MTTF of a system modeled
+  by a Markov chain, we need to
+  \begin{enumerate}
+    \item Set up the differential equations.
+    \item Identify the terminal (absorbing) states, i.e. the states
+      with no outgoing edges.
+    \item Solve the linear equation system using the Laplace transform.
+    \item Compute the MTTF as the sum of the integrals over the
+      functions corresponding to all non-terminal states.
+  \end{enumerate}
+\item
+  \Q\ How would you calculate the availability of the light chains?
+  (Describe the steps, without performing the actual calculation.)
+
+  \A\ To compute the availability of a system, we need to:
+  \begin{enumerate}
+    \item Set up the differential equations according to the Markov chain.
+    \item Identify all states in which the system is considered up
+      (working) and down (failed).
+    \item Remove one state equation (in principle an arbitrary one,
+      but for numerical reasons, one corresponding to an unlikely
+      state should be removed.)
+    \item Add $1 = \sum p$ as the first equation.
+    \item Solve the linear equation system, yielding the fraction of
+      time each state is visited.
+    \item Compute the \underline{un}availability as the sum of all the
+      down states.
+  \end{enumerate}
+\end{enumerate}
+
+\end{enumerate}
+
+
+\end{document}
diff --git a/exercises/Dependability/reliability-availability.pdf b/exercises/Dependability/reliability-availability.pdf
new file mode 100644
index 0000000..4230a42
Binary files /dev/null and b/exercises/Dependability/reliability-availability.pdf differ
diff --git a/exercises/Dependability/reliability-availability.tex b/exercises/Dependability/reliability-availability.tex
new file mode 100644
index 0000000..c7deb2e
--- /dev/null
+++ b/exercises/Dependability/reliability-availability.tex
@@ -0,0 +1,97 @@
+\documentclass{article}
+
+\usepackage[utf8]{inputenc}
+\usepackage{graphicx}
+\usepackage{geometry}
+\usepackage[hidelinks]{hyperref}
+
+\title{\vspace{-.5cm}Industrial Automation\\ \vspace{1cm}Reliability and Availability Exercises}
+%% \author{Matej Pavlovic}
+\date{}
+
+\begin{document}
+
+\maketitle
+
+In our running example, we consider a chain of Christmas lights consisting of many light bulbs.
+
+\begin{enumerate}
+
+\item Let us first consider a single light bulb.
+\begin{enumerate}
+\item
+  Under what conditions can we assume that a single light bulb's
+  failure rate is constant?
+\item From now on, let us assume that the failure rate of a single
+  light bulb is indeed constant. Moreover, assume that a light bulb is
+  expected to burn out after 1000 hours. What is its failure rate?
+\item What is the reliability of a single light bulb?
+\end{enumerate}
+
+\item We consider a chain of Christmas lights consisting of 50
+  serially connected light bulbs. This means that if any light bulb
+  burns out, the electric circuit is interrupted and all light bulbs
+  go off.
+\begin{enumerate}
+\item What is the reliability of the whole chain?
+\item How long can we expect the light chain to work?
+\end{enumerate}
+
+\item We now assume that the light bulbs are connected in
+  parallel. Thus, if a light bulb burns out, the others may continue
+  operating. We consider a light chain to be working as long as not
+  more than 10 light bulbs are burned out.
+\begin{enumerate}
+\item What is the reliability of the whole chain?
+\item How long can we expect the light chain to work? (No need to
+  calculate the value, just give a mathematical expression describing
+  it.)
+\end{enumerate}
+
+\item We assume that during the Christmas period, the light chain is
+  used every evening for 4 hours during the whole month of
+  December. Answer the following questions for both the serial and the
+  parallel variant of the light chain.
+\begin{enumerate}
+\item What is the probability that the light chain will remain
+  functional during the whole Christmas period? (Feel free to serve
+  yourself with mathematical tools like WolframAlpha to compute the
+  resulting numbers.)
+\item What is the probability that the light chain survives the
+  Christmas periods of two consecutive years?
+\item If, after the first year's Christmas period, 5 light bulbs burn
+  out, what is the probability of the light chain surviving the next
+  Christmas period? (Only relevant for the parallel case.)
+\item Assume that the package with the light chain contains 2
+  replacement light bulbs. How do the answers to the 3 above questions
+  change?
+\item What are the answers to the previous 4 questions if we use LED
+  light bulbs with an expected life time of 5000 hours?
+\end{enumerate}
+
+\item Now consider a case of a light chain that is supposed to be on
+  permanently. If a light bulb goes off, it takes 1 hour to be noticed
+  and replaced on expectation. For simplicity, we assume that repairs
+  happen independently and can even be carried out in parallel.
+\begin{enumerate}
+\item What is the (stationary) availability of the serial variant of the light
+  chain?
+\item Draw a Markov chain modeling the reliability of both the serial
+  and the parallel variant of the light chain. For simplicity, assume
+  that in the parallel case, the light chain goes off automatically
+  when more than 10 light bulbs burn out.
+\item Set up the differential equations describing the states of the
+  Markov chains.
+\item What are the initial conditions of the differential equations
+  for a new light chain (i.e. when none of the light bulbs burned out
+  yet)?
+\item How would you calculate the mean time to failure (MTTF) of the
+  light chains? (Describe the steps, without performing the actual calculation.)
+\item How would you calculate the availability of the light chains?
+  (Describe the steps, without performing the actual calculation.)
+\end{enumerate}
+
+\end{enumerate}
+
+
+\end{document}
diff --git a/exercises/Exercise 1.docx b/exercises/Exercise 1.docx
new file mode 100644
index 0000000..7437279
Binary files /dev/null and b/exercises/Exercise 1.docx differ
diff --git a/exercises/Exercise 1.pages b/exercises/Exercise 1.pages
new file mode 100755
index 0000000..a0a2169
Binary files /dev/null and b/exercises/Exercise 1.pages differ
diff --git a/exercises/Exercise 1.pdf b/exercises/Exercise 1.pdf
new file mode 100644
index 0000000..4eade21
Binary files /dev/null and b/exercises/Exercise 1.pdf differ
diff --git a/exercises/Exercise 2.docx b/exercises/Exercise 2.docx
new file mode 100644
index 0000000..301b690
Binary files /dev/null and b/exercises/Exercise 2.docx differ
diff --git a/exercises/Exercise 2.pages b/exercises/Exercise 2.pages
new file mode 100755
index 0000000..5de1c29
Binary files /dev/null and b/exercises/Exercise 2.pages differ
diff --git a/exercises/Exercise 2.pdf b/exercises/Exercise 2.pdf
new file mode 100644
index 0000000..56a3c2a
Binary files /dev/null and b/exercises/Exercise 2.pdf differ
diff --git a/exercises/Exercise 3.docx b/exercises/Exercise 3.docx
new file mode 100644
index 0000000..d2a1dd8
Binary files /dev/null and b/exercises/Exercise 3.docx differ
diff --git a/exercises/Exercise 3.pages b/exercises/Exercise 3.pages
new file mode 100755
index 0000000..5fb8b82
Binary files /dev/null and b/exercises/Exercise 3.pages differ
diff --git a/exercises/Exercise 3.pdf b/exercises/Exercise 3.pdf
new file mode 100644
index 0000000..4637490
Binary files /dev/null and b/exercises/Exercise 3.pdf differ
diff --git a/exercises/Solution 1.docx b/exercises/Solution 1.docx
new file mode 100644
index 0000000..bc47c65
Binary files /dev/null and b/exercises/Solution 1.docx differ
diff --git a/exercises/Solution 1.pages b/exercises/Solution 1.pages
new file mode 100755
index 0000000..24fa28a
Binary files /dev/null and b/exercises/Solution 1.pages differ
diff --git a/exercises/Solution 1.pdf b/exercises/Solution 1.pdf
new file mode 100644
index 0000000..9a67949
Binary files /dev/null and b/exercises/Solution 1.pdf differ
diff --git a/exercises/Solution 2.docx b/exercises/Solution 2.docx
new file mode 100644
index 0000000..56c3db5
Binary files /dev/null and b/exercises/Solution 2.docx differ
diff --git a/exercises/Solution 2.pages b/exercises/Solution 2.pages
new file mode 100755
index 0000000..f3beb2c
Binary files /dev/null and b/exercises/Solution 2.pages differ
diff --git a/exercises/Solution 2.pdf b/exercises/Solution 2.pdf
new file mode 100644
index 0000000..ad060bf
Binary files /dev/null and b/exercises/Solution 2.pdf differ
diff --git a/exercises/Solution 3.docx b/exercises/Solution 3.docx
new file mode 100644
index 0000000..e54d0ad
Binary files /dev/null and b/exercises/Solution 3.docx differ
diff --git a/exercises/Solution 3.pages b/exercises/Solution 3.pages
new file mode 100755
index 0000000..bf8eb28
Binary files /dev/null and b/exercises/Solution 3.pages differ
diff --git a/exercises/Solution 3.pdf b/exercises/Solution 3.pdf
new file mode 100644
index 0000000..6fd8e9a
Binary files /dev/null and b/exercises/Solution 3.pdf differ
diff --git a/homework/Groups.pdf b/homework/Groups.pdf
new file mode 100644
index 0000000..0e3eba2
Binary files /dev/null and b/homework/Groups.pdf differ
diff --git a/homework/Groups.xlsx b/homework/Groups.xlsx
new file mode 100644
index 0000000..a4312e0
Binary files /dev/null and b/homework/Groups.xlsx differ
diff --git a/homework/Homework_2020_(GS).docx b/homework/Homework_2020_(GS).docx
new file mode 100644
index 0000000..7856f99
Binary files /dev/null and b/homework/Homework_2020_(GS).docx differ
diff --git a/homework/Homework_2020_(GS).pdf b/homework/Homework_2020_(GS).pdf
new file mode 100644
index 0000000..2967474
Binary files /dev/null and b/homework/Homework_2020_(GS).pdf differ
diff --git a/homework/Project Timeline and Evaluation.pages b/homework/Project Timeline and Evaluation.pages
new file mode 100644
index 0000000..cce3cf8
Binary files /dev/null and b/homework/Project Timeline and Evaluation.pages differ
diff --git a/homework/Project Timeline and Evaluation.pdf b/homework/Project Timeline and Evaluation.pdf
new file mode 100644
index 0000000..9a56a2b
Binary files /dev/null and b/homework/Project Timeline and Evaluation.pdf differ
diff --git a/homework/evaluation.txt b/homework/evaluation.txt
new file mode 100644
index 0000000..086b2c8
--- /dev/null
+++ b/homework/evaluation.txt
@@ -0,0 +1,82 @@
+Project:			1-6
+Name of evaluator:		First name, last name
+Format:				Is the report complete? If not, what is missing?
+
+Use EPFL grading (1-6) for each section.
+
+1. Overall architecture (Weight 2)
+*************************************************
+- Is it easy to read and understandable? 
+- Explanation and justification of assumptions
+- Are the most important elements of the system described?
+- Mapping of functionality onto hardware and software components?
+- Description of interaction between components (information flow)?
+- Partitioning into levels (are the different parts on the right level of the pyramid?)?
+- Constraints / Regulations?
+- Source of information listed?
+
+
+1.a Control system  (Weight 2)
+*************************************************
+- What are the requirements? (What needs to be monitored, regulated and controlled? How does this translate into requirements?)
+- Is it easy to read and understandable?
+- Explanation and justification of assumptions
+- Which parts are continuous, which parts discrete? Why?
+- Selection of PLCs/IO parts (suitable? explained? How many?)
+- Source of information listed
+
+1.b Communication architecture (Weight 2)
+*************************************************
+- What are the requirements?  (How much information, how frequent, how far? How many devices? What kind of latencies are acceptable? How does this translate into requirements?)
+- Is it easy to read and understandable?
+- Explanation and justification of assumptions
+- Assignment of real-time / non -real time requirements
+- Which parts are cyclic / event driven?
+- Does protocol selection match requirements?
+- Communication equipment (suitable? explained?)
+- Synchronization
+- Source of information listed
+
+1.c SCADA (Weight 2)
+*************************************************
+- What are the requirements?
+- Is it easy to read and understandable?
+- Explanation and justification of assumptions
+- Selection of HW (suitable? explained? How many?)
+- Source of information listed
+
+2. Supervision Tango part (Weight 2.5)
+*************************************************
+- List of signals to monitor and control complete and explained?
+- List of alarms and priorities complete and justified?
+- Architecture choices are described and justified
+	- how many device servers are needed?
+	- where do the device servers run?
+	- what are the device servers connected to? through which protocol(s)?
+- Are the HMI principles adhered to in the synoptic views?
+- Does the device server work? I.e., can we run it?
+
+3. Manufacturing Execution System (Weight 1)
+*************************************************
+- Are the purposes identified correctly?
+- Is the proposed performance measure understandable and relevant?
+- Are the dependencies and implications on the other layers identified and described?
+
+4. FMEA / FTA (Weight 2.5)
+*************************************************
+- Is it easy to read and understandable?
+- Explanation and justification of choice of method
+- Explanation and justification of assumptions
+- Source of information
+- Completeness (are the most important elements of the analysis described?)
+
+
+5. Presentation (Weight 2)
+*************************************************
+- Is it easy to follow?
+- Do the slides support the presentation?
+
+
+5. Choice
+**********
+- If you as a customer had to pick between this and the other project bid, which one would you choose and why?
\ No newline at end of file
diff --git a/slides/IA_00_Overview.key b/slides/IA_00_Overview.key
new file mode 100644
index 0000000..55c85df
Binary files /dev/null and b/slides/IA_00_Overview.key differ
diff --git a/slides/IA_00_Overview.pdf b/slides/IA_00_Overview.pdf
new file mode 100644
index 0000000..3f6c6c1
Binary files /dev/null and b/slides/IA_00_Overview.pdf differ
diff --git a/slides/IA_01_Introduction.key b/slides/IA_01_Introduction.key
new file mode 100755
index 0000000..68058b8
Binary files /dev/null and b/slides/IA_01_Introduction.key differ
diff --git a/slides/IA_01_Introduction.pdf b/slides/IA_01_Introduction.pdf
new file mode 100644
index 0000000..11e6eec
Binary files /dev/null and b/slides/IA_01_Introduction.pdf differ
diff --git a/slides/IA_02_Control_and_Field_Devices.pdf b/slides/IA_02_Control_and_Field_Devices.pdf
new file mode 100644
index 0000000..2daa0da
Binary files /dev/null and b/slides/IA_02_Control_and_Field_Devices.pdf differ
diff --git a/slides/IA_02_Control_and_Field_Devices.pptx b/slides/IA_02_Control_and_Field_Devices.pptx
new file mode 100644
index 0000000..ef6b870
Binary files /dev/null and b/slides/IA_02_Control_and_Field_Devices.pptx differ
diff --git a/slides/IA_02_PLC_Programming.key b/slides/IA_02_PLC_Programming.key
new file mode 100644
index 0000000..bb26d00
Binary files /dev/null and b/slides/IA_02_PLC_Programming.key differ
diff --git a/slides/IA_02_PLC_Programming.pdf b/slides/IA_02_PLC_Programming.pdf
new file mode 100644
index 0000000..6bd3df8
Binary files /dev/null and b/slides/IA_02_PLC_Programming.pdf differ
diff --git a/slides/IA_03_Communication_Networks.key b/slides/IA_03_Communication_Networks.key
new file mode 100755
index 0000000..a80575a
Binary files /dev/null and b/slides/IA_03_Communication_Networks.key differ
diff --git a/slides/IA_03_Communication_Networks.pdf b/slides/IA_03_Communication_Networks.pdf
new file mode 100644
index 0000000..bbb67a9
Binary files /dev/null and b/slides/IA_03_Communication_Networks.pdf differ
diff --git a/slides/IA_03_OSI.key b/slides/IA_03_OSI.key
new file mode 100755
index 0000000..7cfe59f
Binary files /dev/null and b/slides/IA_03_OSI.key differ
diff --git a/slides/IA_03_OSI.pdf b/slides/IA_03_OSI.pdf
new file mode 100644
index 0000000..589f489
Binary files /dev/null and b/slides/IA_03_OSI.pdf differ
diff --git a/slides/IA_04_Application_Layer_Protocols_MES.key b/slides/IA_04_Application_Layer_Protocols_MES.key
new file mode 100755
index 0000000..d2821bf
Binary files /dev/null and b/slides/IA_04_Application_Layer_Protocols_MES.key differ
diff --git a/slides/IA_04_Application_Layer_Protocols_MES.pdf b/slides/IA_04_Application_Layer_Protocols_MES.pdf
new file mode 100644
index 0000000..153ef29
Binary files /dev/null and b/slides/IA_04_Application_Layer_Protocols_MES.pdf differ
diff --git a/slides/IA_05_SCADA.pdf b/slides/IA_05_SCADA.pdf
new file mode 100644
index 0000000..0c45a68
Binary files /dev/null and b/slides/IA_05_SCADA.pdf differ
diff --git a/slides/IA_05_SCADA.pptx b/slides/IA_05_SCADA.pptx
new file mode 100644
index 0000000..38b8f19
Binary files /dev/null and b/slides/IA_05_SCADA.pptx differ
diff --git a/slides/IA_ISA95.pdf b/slides/IA_ISA95.pdf
new file mode 100644
index 0000000..6f82a48
Binary files /dev/null and b/slides/IA_ISA95.pdf differ
diff --git a/slides/IA_ISA95.pptx b/slides/IA_ISA95.pptx
new file mode 100644
index 0000000..a986e84
Binary files /dev/null and b/slides/IA_ISA95.pptx differ