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R10252/exercises/Dependability/reliability-availability-solutions.tex
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\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}
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