source: trunk/docs/proposed/lossmodel.lyx

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\begin_body

\begin_layout Title
Tahoe Distributed Filesharing System Loss Model
\end_layout

\begin_layout Author
Shawn Willden
\end_layout

\begin_layout Date
07/22/2009
\end_layout

\begin_layout Address
South Weber, Utah
\end_layout

\begin_layout Email
shawn@willden.org
\end_layout

\begin_layout Abstract
The abstract goes here
\end_layout

\begin_layout Section
Problem Statement
\end_layout

\begin_layout Standard
The allmydata Tahoe distributed file system uses Reed-Solomon erasure coding
 to split files into 
\begin_inset Formula $N$
\end_inset

 shares which are delivered to randomly-selected peers in a distributed
 network.
 The file can later be reassembled from any 
\begin_inset Formula $k\leq N$
\end_inset

 of the shares, if they are available.
\end_layout

\begin_layout Standard
Over time shares are lost for a variety of reasons.
 Storage servers may crash, be destroyed or simply be removed from the network.
 To mitigate such losses, Tahoe network clients employ a repair agent which
 scans the peers once per time period 
\begin_inset Formula $A$
\end_inset

 and determines how many of the shares remain.
 If less than 
\begin_inset Formula $L$
\end_inset

 (
\begin_inset Formula $k\leq L\leq N$
\end_inset

) shares remain, then the repairer reconstructs the file shares and redistribute
s the missing ones, bringing the availability back up to full.
\end_layout

\begin_layout Standard
The question we're trying to answer is "What is the probability that we'll
 be able to reassemble the file at some later time 
\begin_inset Formula $T$
\end_inset

?".
 We'd also like to be able to determine what values we should choose for
 
\begin_inset Formula $k$
\end_inset

, 
\begin_inset Formula $N$
\end_inset

, 
\begin_inset Formula $A$
\end_inset

, and 
\begin_inset Formula $L$
\end_inset

 in order to ensure 
\begin_inset Formula $Pr[loss]\leq r$
\end_inset

 for some threshold probability 
\begin_inset Formula $r$
\end_inset

.
 This is an optimization problem because although we could obtain very low
 
\begin_inset Formula $Pr[loss]$
\end_inset

 by selecting conservative parameters, these choices have costs.
 The peer storage and bandwidth consumed by the share distribution process
 are approximately 
\begin_inset Formula $\nicefrac{N}{k}$
\end_inset

 times the size of the original file, so we would like to minimize 
\begin_inset Formula $\nicefrac{N}{k}$
\end_inset

, consistent with 
\begin_inset Formula $Pr[loss]\leq r$
\end_inset

.
 Likewise, a frequent and aggressive repair process keeps the number of
 shares available close to 
\begin_inset Formula $N,$
\end_inset

 but at a cost in bandwidth and processing time as the repair agent downloads
 
\begin_inset Formula $k$
\end_inset

 shares, reconstructs the file and uploads new shares to replace those that
 are lost.
\end_layout

\begin_layout Section
Reliability
\end_layout

\begin_layout Standard
The probability that the file becomes unrecoverable is dependent upon the
 probability that the peers to whom we send shares are able to return those
 copies on demand.
 Shares that are corrupted are detected and discarded, so there is no need
 to distinguish between corruption and loss.
\end_layout

\begin_layout Standard
Many factors affect share availability.
 Availability can be temporarily interrupted by peer unavailability due
 to network outages, power failures or administrative shutdown, among other
 reasons.
 Availability can be permanently lost due to failure or corruption of storage
 media, catastrophic damage to the peer system, administrative error, withdrawal
 from the network, malicious corruption, etc.
\end_layout

\begin_layout Standard
The existence of intermittent failure modes motivates the introduction of
 a distinction between 
\noun on
availability
\noun default
 and 
\noun on
reliability
\noun default
.
 Reliability is the probability that a share is retrievable assuming intermitten
t failures can be waited out, so reliability considers only permanent failures.
 Availability considers all failures, and is focused on the probability
 of retrieval within some defined time frame.
\end_layout

\begin_layout Standard
Another consideration is that some failures affect multiple shares.
 If multiple shares of a file are stored on a single hard drive, for example,
 failure of that drive may lose them all.
 Catastrophic damage to a data center may destroy all shares on all peers
 in that data center.
\end_layout

\begin_layout Standard
While the types of failures that may occur are quite consistent across peers,
 their probabilities differ dramatically.
 A professionally-administered server with redundant storage, power and
 Internet located in a carefully-monitored data center with automatic fire
 suppression systems is much less likely to become either temporarily or
 permanently unavailable than the typical virus and malware-ridden home
 computer on a single cable modem connection.
 A variety of situations in between exist as well, such as the case of the
 author's home file server, which is administered by an IT professional
 and uses RAID level 6 redundant storage, but runs on old, cobbled-together
 equipment, and has a consumer-grade Internet connection.
\end_layout

\begin_layout Standard
To begin with, let's use a simple definition of reliability:
\end_layout

\begin_layout Definition

\noun on
Reliability
\noun default
 is the probability 
\begin_inset Formula $p_{i}$
\end_inset

 that a share 
\begin_inset Formula $s_{i}$
\end_inset

 will survive to (be retrievable at) time 
\begin_inset Formula $T=A$
\end_inset

, ignoring intermittent failures.
 That is, the probability that the share will be retrievable at the end
 of the current repair cycle, and therefore usable by the repairer to regenerate
 any lost shares.
\end_layout

\begin_layout Standard
Reliability 
\begin_inset Formula $p_{i}$
\end_inset

 is clearly dependent on 
\begin_inset Formula $A$
\end_inset

.
 Short repair cycles offer less time for shares to 
\begin_inset Quotes eld
\end_inset

decay
\begin_inset Quotes erd
\end_inset

 into unavailability.
\end_layout

\begin_layout Subsection
Peer Reliability
\end_layout

\begin_layout Standard
Since peer reliability is the basis for any computations we may do on share
 and file reliability, we must have a way to estimate it.
 Reliability modeling of hardware, software and human performance are each
 complex topics, the subject of much ongoing research.
 In particular, the reliability of one of the key components of any peer
 from our perspective -- the hard drive where file shares are stored --
 is the subject of much current debate.
\end_layout

\begin_layout Standard
A common assumption about hardware failure is that it follows the 
\begin_inset Quotes eld
\end_inset

bathtub curve
\begin_inset Quotes erd
\end_inset

, with frequent failures during the first few months, a constant failure
 rate for a few years and then a rising failure rate as the hardware wears
 out.
 This curve is often flattened by burn-in stress testing, and by periodic
 replacement that assures that in-service components never reach 
\begin_inset Quotes eld
\end_inset

old age
\begin_inset Quotes erd
\end_inset

.
\end_layout

\begin_layout Standard
In any case, we're generally going to ignore all of that complexity and
 focus on the bottom of the bathtub, assuming constant failure rates.
 This is a particularly reasonable assumption as long as we're focused on
 failures during a particular, relatively short interval 
\begin_inset Formula $A$
\end_inset

.
 Towards the end of this paper, as we examine failures over many repair
 intervals, the assumption becomes more tenuous, and we note some of the
 issues.
\end_layout

\begin_layout Subsubsection
Estimate Adaptation
\end_layout

\begin_layout Standard
Even assuming constant failure rates, however, it will be rare that the
 duration of 
\begin_inset Formula $A$
\end_inset

 coincides with the available failure rate data, particularly since we want
 to view 
\begin_inset Formula $A$
\end_inset

 as a tunable parameter.
 It's necessary to be able adapt failure rates baselined against any given
 duration to the selected value of 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Standard
Another issue is that failure rates of hardware, etc., are necessarily continuous
 in nature, while the per-interval failure/survival rates that are of interest
 for file reliability calculations are discrete -- a peer either survives
 or fails during the interval.
 The continuous nature of failure rates means that the common and obvious
 methods for estimating failure rates result in values that follow continuous,
 not discrete distributions.
 The difference is minor for small failure probabilities, and converges
 to zero as the number of intervals goes to infinity, but is important enough
 in some cases to be worth correcting for.
\end_layout

\begin_layout Standard
Continuous failure rates are described in terms of mean time to failure,
 and under the assumption that failure rates are constant, are exponentially
 distributed.
 Under these assumptions, the probability that a machine fails at time 
\begin_inset Formula $t$
\end_inset

, is 
\begin_inset Formula \[
f\left(t\right)=\lambda e^{-\lambda t}\]

\end_inset

where 
\begin_inset Formula $\lambda$
\end_inset

 represents the per unit-time failure rate.
 The probability that a machine fails at or before time 
\begin_inset Formula $A$
\end_inset

 is therefore
\begin_inset Formula \begin{align}
F\left(t\right) & =\int_{0}^{A}f\left(x\right)dx\nonumber \\
 & =\int_{0}^{A}\lambda e^{-\lambda x}dx\nonumber \\
 & =1-e^{-\lambda A}\label{eq:failure-time}\end{align}

\end_inset


\end_layout

\begin_layout Standard
Note that 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $\lambda$
\end_inset

 in 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:failure-time"

\end_inset

 must be expressed in consistent time units.
 If they're different, unit conversions should be applied in the normal
 way.
 For example, if the estimate for 
\begin_inset Formula $\lambda$
\end_inset

 is 750 failures per million hours, and 
\begin_inset Formula $A$
\end_inset

 is one month, then either 
\begin_inset Formula $A$
\end_inset

 should be represented as 
\begin_inset Formula $30\cdot24/1000000=.00072$
\end_inset

, or 
\begin_inset Formula $\lambda$
\end_inset

 should be converted to failures per month.
 Or both may be converted to hours.
\end_layout

\begin_layout Subsubsection
Acquiring Peer Reliability Estimates
\end_layout

\begin_layout Standard
Need to write this.
\end_layout

\begin_layout Subsection
Uniform Reliability
\begin_inset CommandInset label
LatexCommand label
name "sub:Fixed-Reliability"

\end_inset


\end_layout

\begin_layout Standard
In the simplest case, the peers holding the file shares all have the same
 reliability 
\begin_inset Formula $p$
\end_inset

, and are all independent from one another.
 Let 
\begin_inset Formula $K$
\end_inset

 be a random variable that represents the number of shares that survive
 
\begin_inset Formula $A$
\end_inset

.
 Each share's survival can be viewed as an independent Bernoulli trial with
 a success probability of 
\begin_inset Formula $p$
\end_inset

, which means that 
\begin_inset Formula $K$
\end_inset

 follows the binomial distribution with parameters 
\begin_inset Formula $N$
\end_inset

 and 
\begin_inset Formula $p$
\end_inset

.
 That is, 
\begin_inset Formula $K\sim B(N,p)$
\end_inset

.
\end_layout

\begin_layout Theorem
Binomial Distribution Theorem
\end_layout

\begin_layout Theorem
Consider 
\begin_inset Formula $n$
\end_inset

 independent Bernoulli trials
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
A Bernoulli trial is simply a test of some sort that results in one of two
 outcomes, one of which is designated success and the other failure.
 The classic example of a Bernoulli trial is a coin toss.
\end_layout

\end_inset

 that succeed with probability 
\begin_inset Formula $p$
\end_inset

, and let 
\begin_inset Formula $K$
\end_inset

 be a random variable that represents the number, 
\begin_inset Formula $m$
\end_inset

, of successes, 
\begin_inset Formula $0\le m\le n$
\end_inset

.
 We say that 
\begin_inset Formula $K$
\end_inset

 follows the Binomial Distribution with parameters n and p, denoted 
\begin_inset Formula $K\sim B(n,p)$
\end_inset

.
 The probability mass function (PMF) of K is a function that gives the probabili
ty that 
\begin_inset Formula $K$
\end_inset

 takes a particular value 
\begin_inset Formula $m$
\end_inset

 (the probability that there are exactly 
\begin_inset Formula $m$
\end_inset

 successful trials, and therefore 
\begin_inset Formula $n-m$
\end_inset

 failures).
 The PMF of K is
\begin_inset Formula \begin{equation}
Pr[K=m]=f(m;n,p)=\binom{n}{m}p^{m}(1-p)^{n-m}\label{eq:binomial-pmf}\end{equation}

\end_inset


\end_layout

\begin_layout Proof
Consider the specific case of exactly 
\begin_inset Formula $m$
\end_inset

 successes followed by 
\begin_inset Formula $n-m$
\end_inset

 failures, because each success has probability 
\begin_inset Formula $p$
\end_inset

, each failure has probability 
\begin_inset Formula $1-p$
\end_inset

, and the trials are independent, the probability of this exact case occurring
 is 
\begin_inset Formula $p^{m}\left(1-p\right)^{\left(n-m\right)}$
\end_inset

, the product of the probabilities of the outcome of each trial.
\end_layout

\begin_layout Proof
Now consider any reordering of these 
\begin_inset Formula $m$
\end_inset

 successes and 
\begin_inset Formula $n$
\end_inset

 failures.
 Any such reordering occurs with the same probability 
\begin_inset Formula $p^{m}\left(1-p\right)^{\left(n-m\right)}$
\end_inset

, but with the terms of the product reordered.
 Since multiplication is commutative, each such reordering has the same
 probability.
 There are n-choose-m such orderings, and each ordering is an independent
 event, meaning we can sum the probabilities of the individual orderings,
 so the probability that any ordering of 
\begin_inset Formula $m$
\end_inset

 successes and 
\begin_inset Formula $n-m$
\end_inset

 failures occurs is given by
\begin_inset Formula \[
\binom{n}{m}p^{m}\left(1-p\right)^{\left(n-m\right)}\]

\end_inset

which is the right-hand-side of equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:binomial-pmf"

\end_inset

.
\end_layout

\begin_layout Standard
A file survives if at least 
\begin_inset Formula $k$
\end_inset

 of the 
\begin_inset Formula $N$
\end_inset

 shares survive.
 Equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:binomial-pmf"

\end_inset

 gives the probability that exactly 
\begin_inset Formula $i$
\end_inset

 shares survive, for any 
\begin_inset Formula $1\leq i\leq n$
\end_inset

, so the probability that fewer than 
\begin_inset Formula $k$
\end_inset

 survive is the sum of the probabilities that 
\begin_inset Formula $0,1,2,\ldots,k-1$
\end_inset

 shares survive.
 That is:
\end_layout

\begin_layout Standard
\begin_inset Formula \begin{equation}
Pr[file\, lost]=\sum_{i=0}^{k-1}\binom{n}{i}p^{i}(1-p)^{n-i}\label{eq:simple-failure}\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Independent Reliability
\begin_inset CommandInset label
LatexCommand label
name "sub:Independent-Reliability"

\end_inset


\end_layout

\begin_layout Standard
Equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:simple-failure"

\end_inset

 assumes that all shares have the same probability of survival, but as explained
 above, this is not necessarily true.
 A more accurate model allows each share 
\begin_inset Formula $s_{i}$
\end_inset

 an independent probability of survival 
\begin_inset Formula $p_{i}$
\end_inset

.
 Each share's survival can still be treated as an independent Bernoulli
 trial, but with success probability 
\begin_inset Formula $p_{i}$
\end_inset

.
 Under this assumption, 
\begin_inset Formula $K$
\end_inset

 follows a generalized binomial distribution with parameters 
\begin_inset Formula $N$
\end_inset

 and 
\begin_inset Formula $p_{1},p_{2},\dots,p_{N}$
\end_inset

.
\end_layout

\begin_layout Standard
The PMF for this generalized 
\begin_inset Formula $K$
\end_inset

 does not have a simple closed-form representation.
 However, the PMFs for random variables representing individual share survival
 do.
 Let 
\begin_inset Formula $K_{i}$
\end_inset

 be a random variable such that:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
K_{i}=\begin{cases}
1 & \textnormal{if }s_{i}\textnormal{ survives}\\
0 & \textnormal{if }s_{i}\textnormal{ fails}\end{cases}\]

\end_inset


\end_layout

\begin_layout Standard
The PMF for 
\begin_inset Formula $K_{i}$
\end_inset

 is very simple: 
\begin_inset Formula \[
Pr[K_{i}=j]=\begin{cases}
p_{i} & j=1\\
1-p_{i} & j=0\end{cases}\]

\end_inset

 which can also be expressed as
\begin_inset Formula \[
Pr[K_{i}=j]=f\left(j\right)=\left(1-p_{i}\right)\left(1-j\right)+p_{i}\left(j\right)\]

\end_inset


\end_layout

\begin_layout Standard
Note that since each 
\begin_inset Formula $K_{i}$
\end_inset

 represents the count of shares 
\begin_inset Formula $s_{i}$
\end_inset

 that survives (either 0 or 1), if we add up all of the individual survivor
 counts, we get the group survivor count.
 That is:
\begin_inset Formula \[
\sum_{i=1}^{N}K_{i}=K\]

\end_inset

Effectively, we have separated 
\begin_inset Formula $K$
\end_inset

 into the series of Bernoulli trials that make it up.
\end_layout

\begin_layout Theorem
Discrete Convolution Theorem
\end_layout

\begin_layout Theorem
Let 
\begin_inset Formula $X$
\end_inset

 and 
\begin_inset Formula $Y$
\end_inset

 be discrete random variables with probability mass functions given by 
\begin_inset Formula $Pr\left[X=x\right]=f(x)$
\end_inset

 and 
\begin_inset Formula $Pr\left[Y=y\right]=g(y).$
\end_inset

 Let 
\begin_inset Formula $Z$
\end_inset

 be the discrete random random variable obtained by summing 
\begin_inset Formula $X$
\end_inset

 and 
\begin_inset Formula $Y$
\end_inset

.
\end_layout

\begin_layout Theorem
The probability mass function of 
\begin_inset Formula $Z$
\end_inset

 is given by
\begin_inset Formula \[
Pr[Z=z]=h(z)=\left(f\star g\right)(z)\]

\end_inset

where 
\begin_inset Formula $\star$
\end_inset

 denotes the discrete convolution operation:
\begin_inset Formula \[
\left(f\star g\right)\left(n\right)=\sum_{m=-\infty}^{\infty}f\left(m\right)g\left(m-n\right)\]

\end_inset


\end_layout

\begin_layout Proof
The proof is beyond the scope of this paper.
\end_layout

\begin_layout Standard
If we denote the PMF of 
\begin_inset Formula $K$
\end_inset

 with 
\begin_inset Formula $f$
\end_inset

 and the PMF of 
\begin_inset Formula $K_{i}$
\end_inset

 with 
\begin_inset Formula $g_{i}$
\end_inset

 (more formally, 
\begin_inset Formula $Pr[K=x]=f(x)$
\end_inset

 and 
\begin_inset Formula $Pr[K_{i}=x]=g_{i}(x)$
\end_inset

) then since 
\begin_inset Formula $K=\sum_{i=1}^{N}K_{i}$
\end_inset

, according to the discrete convolution theorem 
\begin_inset Formula $f=g_{1}\star g_{2}\star g_{3}\star\ldots\star g_{N}$
\end_inset

.
 Since convolution is associative, this can also be written as 
\begin_inset Formula $ $
\end_inset


\begin_inset Formula \begin{equation}
f=(\ldots((g_{1}\star g_{2})\star g_{3})\star\ldots)\star g_{N})\label{eq:convolution}\end{equation}

\end_inset

Therefore, 
\begin_inset Formula $f$
\end_inset

 can be computed as a sequence of convolution operations on the simple PMFs
 of the random variables 
\begin_inset Formula $K_{i}$
\end_inset

.
 In fact, for large 
\begin_inset Formula $N$
\end_inset

, equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:convolution"

\end_inset

 turns out to be a more effective means of computing the PMF of 
\begin_inset Formula $K$
\end_inset

 than the binomial theorem.
 even in the case of shares with identical survival probability.
 The reason it's better is because the calculation of 
\begin_inset Formula $\binom{n}{m}$
\end_inset

 in equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:binomial-pmf"

\end_inset

 produces very large values that overflow unless arbitrary precision numeric
 representations are used.
\end_layout

\begin_layout Standard
Note also that it is not necessary to have very simple PMFs like those of
 the 
\begin_inset Formula $K_{i}$
\end_inset

.
 Any share or set of shares that has a known PMF can be combined with any
 other set with a known PMF by convolution, as long as the two share sets
 are independent.
 The reverse holds as well; given a group with an empirically-derived PMF,
 in it's theoretically possible to solve for an individual PMF, and thereby
 determine 
\begin_inset Formula $p_{i}$
\end_inset

 even when per-share data is unavailable.
\end_layout

\begin_layout Subsection
Multiple Failure Modes
\begin_inset CommandInset label
LatexCommand label
name "sub:Multiple-Failure-Modes"

\end_inset


\end_layout

\begin_layout Standard
In modeling share survival probabilities, it's useful to be able to analyze
 separately each of the various failure modes.
 For example, if reliable statistics for disk failure can be obtained, then
 a probability mass function for that form of failure can be generated.
 Similarly, statistics on other hardware failures, administrative errors,
 network losses, etc., can all be estimated independently.
 If those estimates can then be combined into a single PMF for a share,
 then we can use it to predict failures for that share.
\end_layout

\begin_layout Standard
Combining independent failure modes for a single share is straightforward.
 If 
\begin_inset Formula $p_{i,j}$
\end_inset

 is the probability of survival of the 
\begin_inset Formula $j$
\end_inset

th failure mode of share 
\begin_inset Formula $i$
\end_inset

, 
\begin_inset Formula $1\leq j\leq m$
\end_inset

, then 
\begin_inset Formula \[
Pr[K_{i}=k]=f_{i}(k)=\begin{cases}
\prod_{j=1}^{m}p_{i,j} & k=1\\
1-\prod_{j=1}^{m}p_{i,j} & k=0\end{cases}\]

\end_inset

is the survival PMF.
\end_layout

\begin_layout Subsection
Multi-share failures
\begin_inset CommandInset label
LatexCommand label
name "sub:Multi-share-failures"

\end_inset


\end_layout

\begin_layout Standard
If there are failure modes that affect multiple computers, we can also construct
 the PMF that predicts their survival.
 The key observation is that the PMF has non-zero probabilities only for
 
\begin_inset Formula $0$
\end_inset

 survivors and 
\begin_inset Formula $n$
\end_inset

 survivors, where 
\begin_inset Formula $n$
\end_inset

 is the number of shares in the set.
 If 
\begin_inset Formula $p$
\end_inset

 is the probability of survival, the PMF of 
\begin_inset Formula $K$
\end_inset

, a random variable representing the number of survivors is
\begin_inset Formula \[
Pr[K=k]=f(k)=\begin{cases}
p & k=n\\
0 & 0<i<n\\
1-p & k=0\end{cases}\]

\end_inset


\end_layout

\begin_layout Standard
Group failures due to multiple independent causes can be combined as in
 section 
\begin_inset CommandInset ref
LatexCommand ref
reference "sub:Multiple-Failure-Modes"

\end_inset

, as long as they apply to the whole group.
\end_layout

\begin_layout Example
Putting the Pieces Together
\end_layout

\begin_layout Standard
Sections 
\begin_inset CommandInset ref
LatexCommand ref
reference "sub:Fixed-Reliability"

\end_inset

 through 
\begin_inset CommandInset ref
LatexCommand ref
reference "sub:Multi-share-failures"

\end_inset

 provide ways of calculating the survival probability mass functions for
 a variety of share failure structures and modes.
 As an example of how these pieces can be used, consider a network with
 the following peers:
\end_layout

\begin_layout Itemize
Four servers located in a data center in Nebraska.
 The machines have multiply-redundant Internet connections, with a failure
 probability of 0.0001.
 They store their shares on RAID arrays with failure probability of 0.0002.
 The administrative staff makes data-destroying errors with probability
 0.003.
\end_layout

\begin_layout Itemize
Four servers located in a data center on the island of Hawaii.
 These servers have identical failure probabilities as the servers in Nebraska,
 except that the data center is near the edge of the crater on Mount Kilauea
 (nobody said examples had to be realistic).
 There is a 0.04 chance that the volcano will erupt and bury the data center
 in molten lava, destroying it entirely.
\end_layout

\begin_layout Itemize
Four PCs located in random homes, connected to the Internet via assorted
 cable modems and DSL.
 Their network connections fail with probability 0.009.
 Their disks fail with probability 0.001.
 Their users destroy data with probability 0.05.
\end_layout

\begin_layout Standard
If one share is placed on each of these 12 computers, what's the probability
 mass function of share survival? To more compactly describe PMFs, we'll
 denote them as probability vectors of the form 
\begin_inset Formula $\left[\alpha_{o},\alpha_{1},\alpha_{2},\ldots\alpha_{n}\right]$
\end_inset

 where 
\begin_inset Formula $\alpha_{i}$
\end_inset

 is the probability that exactly 
\begin_inset Formula $i$
\end_inset

 shares survive.
\end_layout

\begin_layout Standard
The servers in the two data centers have individual failure probabilities
 of RAID failure (.0002) and administrative error (.003) giving an individual
 survival probability of 
\begin_inset Formula \[
(1-.0002)\cdot(1-.003)=.9998\cdot.997=.9968\]

\end_inset


\end_layout

\begin_layout Standard
Using 
\begin_inset Formula $p=.9968,n=4$
\end_inset

 in equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:binomial-pmf"

\end_inset

 gives the survival PMF 
\begin_inset Formula \[
\left[1.049\times10^{-10},1.307\times10^{-7},6.105\times10^{-5},0.01271,0.9872\right]\]

\end_inset

which applies to each group of four servers.
 However, each data center also has a .0001 chance of data connection loss,
 which affects all four servers at once, and Hawaii has the additional .04
 probability of severe lava burn.
 If the network fails at a location, all the machines go offline together.
 The probability that 0 machines survive is the probability that they all
 fail for individual reasons (
\begin_inset Formula $1.049\cdot10^{-10}$
\end_inset

) plus the probability they all fail because of a network outage (
\begin_inset Formula $.0001$
\end_inset

) less the probability they fail for both reasons:
\begin_inset Formula \[
\left(1.049\times10^{-10}\right)+\left(0.0001\right)-\left[\left(1.049\times10^{-10}\right)\cdot\left(0.0001\right)\right]\approxeq0.0001\]

\end_inset


\end_layout

\begin_layout Standard
That's the 
\begin_inset Formula $i=0$
\end_inset

 element of the combined PMF.
 The combined probability of survival of 
\begin_inset Formula $0<i\leq4$
\end_inset

 servers is simpler: it's the probability they survive individual failure,
 from the individual failure PMF above, times the probability they survive
 network failure (.9999).
 So the combined survival PMF, which we'll denote as 
\begin_inset Formula $n(i)$
\end_inset

 of the Nebraska servers is
\begin_inset Formula \[
n(i)=\left[0.0001,1.306\times10^{-7},6.104\times10^{-5},0.01268,0.9872\right]\]

\end_inset

which has the interesting property that complete failure is 1000 times more
 likely than survival of one server.
 This is because the probability of a network outage is so much greater
 than simultaneous
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
Of course, the failures need not be truly simultaneous, they just have happen
 in the same interval between repair runs.
\end_layout

\end_inset

 independent failure of three servers.
\end_layout

\begin_layout Standard
We apply the same process for the Hawaii servers, but with group survival
 probability of 
\begin_inset Formula $(1-.0001)(1-.04)=.9799$
\end_inset

 gives the survival PMF 
\begin_inset Formula \[
h(i)=\left[0.0201,1.280\times10^{-7},5.982\times10^{-5},0.01242,0.9674\right]\]

\end_inset


\end_layout

\begin_layout Standard
Applying the convolution operator to 
\begin_inset Formula $n(i)$
\end_inset

 and 
\begin_inset Formula $h(i)$
\end_inset

, the survival PMF of all eight servers is:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\left(n\star h\right)\left(i\right)=\begin{cases}
2.010\times10^{-6} & i=0\\
2.639\times10^{-9} & i=1\\
1.233\times10^{-6} & i=2\\
2.560\times10^{-4} & i=3\\
0.01994 & i=4\\
1.769\times10^{-6} & i=5\\
2.756\times10^{-4} & i=6\\
0.02452 & i=7\\
0.9559 & i=8\end{cases}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset VSpace defskip
\end_inset


\end_layout

\begin_layout Standard
Note that losing four shares (
\begin_inset Formula $i=4$
\end_inset

) is 10,000 times more likely than losing three (
\begin_inset Formula $i=5$
\end_inset

).
 This is because both data centers have a whole-center failure mode, and
 the Hawaii center's lava burn probability is so high.
 Similarly, the probability of losing all of them is 1000 times higher than
 the probability of losing all but one.
\end_layout

\begin_layout Standard
For the home PCs, their individual probability of survival is
\begin_inset Formula \[
(1-.009)\cdot(1-.001)\cdot(1-.05)=.991\cdot.999\cdot.95=.9405\]

\end_inset


\end_layout

\begin_layout Standard
We can then apply equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:binomial-pmf"

\end_inset

 with 
\begin_inset Formula $N=4$
\end_inset

 and 
\begin_inset Formula $p=.9405$
\end_inset

 to compute the PMF 
\begin_inset Formula $g(i),0\leq i\leq4$
\end_inset

 for the PCs and finally compute 
\begin_inset Formula $f(i)=\left(g\star\left(n\star h\right)\right)\left(i\right)$
\end_inset

, the PMF of the whole share set.
 Summing the values of 
\begin_inset Formula $f(i)$
\end_inset

 for 
\begin_inset Formula $0\leq i\leq k-1$
\end_inset

 gives the probability that less than 
\begin_inset Formula $k$
\end_inset

 shares survive and the file is unrecoverable.
 For this example, those sums are shown in table 
\begin_inset CommandInset ref
LatexCommand vref
reference "tab:Example-PMF"

\end_inset

.
\begin_inset Float table
wide false
sideways false
status collapsed

\begin_layout Plain Layout
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="13" columns="4">
<features>
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $k$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $Pr[K=k]$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $Pr[file\, loss]=Pr[K<k]$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $N/k$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $1.60\times10^{-9}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $2.53\times10^{-11}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
12
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
2
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $3.80\times10^{-8}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $1.63\times10^{-9}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
6
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
3
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $4.04\times10^{-7}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $3.70\times10^{-8}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
4
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
4
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $2.06\times10^{-6}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $4.44\times10^{-7}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
3
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
5
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $2.10\times10^{-5}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $2.50\times10^{-6}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
2.4
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
6
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.000428$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $2.35\times10^{-5}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
2
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
7
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.00417$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.000452$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.7
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
8
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.0157$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.00462$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.5
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
9
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.00127$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.0203$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.3
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
10
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<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.0230$
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\end_layout

\end_inset
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<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.0216$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.2
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
11
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<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
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\begin_layout Plain Layout
\begin_inset Formula $0.208$
\end_inset


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<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
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\end_inset


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\end_inset
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<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.1
\end_layout

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</cell>
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<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
12
\end_layout

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</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.747$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0.253$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption

\begin_layout Plain Layout
\align left
\begin_inset CommandInset label
LatexCommand label
name "tab:Example-PMF"

\end_inset

Example PMF
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Standard
The table demonstrates the importance of the selection of 
\begin_inset Formula $k$
\end_inset

, and the tradeoff against file size expansion.
 Note that the survival of exactly 9 servers is significantly less likely
 than the survival of 8 or 10 servers.
 This is, again, an artifact of the group failure modes.
 Because of this, there is no reason to choose 
\begin_inset Formula $k=9$
\end_inset

 over 
\begin_inset Formula $k=10$
\end_inset

.
 Normally, reducing the number of shares needed for reassembly improve the
 file's chances of survival, but in this case it provides a minuscule gain
 in reliability at the cost of a 10% increase in bandwidth and storage consumed.
\end_layout

\begin_layout Subsection
Share Duplication
\end_layout

\begin_layout Standard
Before moving on to consider issues other than single-interval file loss,
 let's analyze one more possibility, that of 
\begin_inset Quotes eld
\end_inset

cheap
\begin_inset Quotes erd
\end_inset

 file repair via share duplication.
\end_layout

\begin_layout Standard
Initially, files are split using erasure coding, which creates 
\begin_inset Formula $N$
\end_inset

 unique shares, any 
\begin_inset Formula $k$
\end_inset

 of which can be used to to reconstruct the file.
 When shares are lost, proper repair downloads some 
\begin_inset Formula $k$
\end_inset

 shares, reconstructs the original file and then uses the erasure coding
 algorithm to reconstruct the lost shares, then redeploys them to peers
 in the network.
 This is a somewhat expensive process.
\end_layout

\begin_layout Standard
A cheaper repair option is simply to direct some peer that has share 
\begin_inset Formula $s_{i}$
\end_inset

 to send a copy to another peer, thus increasing by one the number of shares
 in the network.
 This is not as good as actually replacing the lost share, though.
 Suppose that more shares were lost, leaving only 
\begin_inset Formula $k$
\end_inset

 shares remaining.
 If two of those shares are identical, because one was duplicated in this
 fashion, then only 
\begin_inset Formula $k-1$
\end_inset

 shares truly remain, and the file can no longer be reconstructed.
\end_layout

\begin_layout Standard
However, such cheap repair is not completely pointless; it does increase
 file survivability.
 But by how much?
\end_layout

\begin_layout Standard
Effectively, share duplication simply increases the probability that 
\begin_inset Formula $s_{i}$
\end_inset

 will survive, by providing two locations from which to retrieve it.
 We can view the two copies of the single share as one, but with a higher
 probability of survival than would be provided by either of the two peers.
 In particular, if 
\begin_inset Formula $p_{1}$
\end_inset

 and 
\begin_inset Formula $p_{2}$
\end_inset

 are the probabilities that the two peers will survive, respectively, then
\begin_inset Formula \[
Pr[s_{i}\, survives]=p_{1}+p_{2}-p_{1}p_{2}\]

\end_inset


\end_layout

\begin_layout Standard
More generally, if a single share is deployed on 
\begin_inset Formula $n$
\end_inset

 peers, each with a PMF 
\begin_inset Formula $f_{i}(j),0\leq j\leq1,1\leq i\leq n$
\end_inset

, the share survival count is a random variable 
\begin_inset Formula $K$
\end_inset

 and the probability of share loss is 
\begin_inset Formula \[
Pr[K=0]=(f_{1}\star f_{2}\star\ldots\star f_{n})(0)\]

\end_inset


\end_layout

\begin_layout Standard
From that, we can construct a share PMF in the obvious way, which can then
 be convolved with the other share PMFs to produce the share set PMF.
\end_layout

\begin_layout Example
Suppose a file has 
\begin_inset Formula $N=10,k=3$
\end_inset

 and that all servers have survival probability 
\begin_inset Formula $p=.9$
\end_inset

.
 Given a full complement of shares, 
\begin_inset Formula $Pr[\textrm{file\, loss}]=3.74\times10^{-7}$
\end_inset

.
 Suppose that four shares are lost, which increases 
\begin_inset Formula $Pr[\textrm{file\, loss}]$
\end_inset

 to 
\begin_inset Formula $.00127$
\end_inset

, a value 
\begin_inset Formula $3400$
\end_inset

 times greater.
 Rather than doing a proper reconstruction, we could direct four peers still
 holding shares to send a copy of their share to new peer, which changes
 the composition of the shares from one of six, unique 
\begin_inset Quotes eld
\end_inset

standard
\begin_inset Quotes erd
\end_inset

 shares, to one of two standard shares, each with survival probability 
\begin_inset Formula $.9$
\end_inset

 and four 
\begin_inset Quotes eld
\end_inset

doubled
\begin_inset Quotes erd
\end_inset

 shares, each with survival probability 
\begin_inset Formula $2p-p^{2}\approxeq.99$
\end_inset

.
\end_layout

\begin_layout Example
Combining the two single-peer share PMFs with the four double-share PMFs
 gives a new file survival probability of 
\begin_inset Formula $6.64\times10^{-6}$
\end_inset

.
 Not as good as a full repair, but still quite respectable.
 Also, if storage were not a concern, all six shares could be duplicated,
 for a 
\begin_inset Formula $Pr[file\, loss]=1.48\times10^{-7}$
\end_inset

, which is actually three time better than the nominal case.
\end_layout

\begin_layout Example
The reason such cheap repairs may be attractive in many cases is that distribute
d bandwidth is cheaper than bandwidth through a single peer.
 This is particularly true if that single peer has a very slow connection,
 which is common for home computers -- especially in the outbound direction.
\end_layout

\begin_layout Section
Long-Term Reliability
\end_layout

\begin_layout Standard
Thus far, we've focused entirely on the probability that a file survives
 the interval 
\begin_inset Formula $A$
\end_inset

 between repair times.
 The probability that a file survives long-term, though, is also important.
 As long as the probability of failure during a repair period is non-zero,
 a given file will eventually be lost.
 We want to know the probability of surviving for time 
\begin_inset Formula $T$
\end_inset

, and how the parameters 
\begin_inset Formula $A$
\end_inset

 (time between repairs) and 
\begin_inset Formula $L$
\end_inset

 (allowed share low watermark) affect survival time.
\end_layout

\begin_layout Standard
To model file survival time, let 
\begin_inset Formula $T$
\end_inset

 be a random variable denoting the time at which a given file becomes unrecovera
ble, and 
\begin_inset Formula $R(t)=Pr[T>t]$
\end_inset

 be a function that gives the probability that the file survives to time
 
\begin_inset Formula $t$
\end_inset

.
 
\begin_inset Formula $R(t)$
\end_inset

 is the cumulative distribution function of 
\begin_inset Formula $T$
\end_inset

.
\end_layout

\begin_layout Standard
Most survival functions are continuous, but 
\begin_inset Formula $R(t)$
\end_inset

 is inherently discrete and stochastic.
 The time steps are the repair intervals, each of length 
\begin_inset Formula $A$
\end_inset

, so 
\begin_inset Formula $T$
\end_inset

-values are multiples of 
\begin_inset Formula $A$
\end_inset

.
 During each interval, the file's shares degrade according to the probability
 mass function of 
\begin_inset Formula $K$
\end_inset

.
\end_layout

\begin_layout Subsection
Aggressive Repair
\end_layout

\begin_layout Standard
Let's first consider the case of an aggressive repairer.
 Every interval, this repairer checks the file for share losses and restores
 them.
 Thus, at the beginning of each interval, the file always has 
\begin_inset Formula $N$
\end_inset

 shares, distributed on servers with various individual and group failure
 probabilities, which will survive or fail per the output of random variable
 
\begin_inset Formula $K$
\end_inset

.
\end_layout

\begin_layout Standard
For any interval, then, the probability that the file will survive is 
\begin_inset Formula $f\left(k\right)=Pr[K\geq k]$
\end_inset

.
 Since each interval success or failure is independent, and assuming the
 share reliabilities remain constant over time, 
\begin_inset Formula \begin{equation}
R\left(t\right)=f(k)^{t}\end{equation}

\end_inset


\end_layout

\begin_layout Standard
This simple survival function makes it simple to select parameters 
\begin_inset Formula $N$
\end_inset

 and 
\begin_inset Formula $K$
\end_inset

 such that 
\begin_inset Formula $R(t)\geq r$
\end_inset

, where 
\begin_inset Formula $r$
\end_inset

 is a user-specified parameter indicating the desired probability of survival
 to time 
\begin_inset Formula $t$
\end_inset

.
 Specifically, we can solve for 
\begin_inset Formula $f\left(k\right)$
\end_inset

 in 
\begin_inset Formula $r\leq f\left(k\right)^{t}$
\end_inset

, giving:
\begin_inset Formula \begin{equation}
f\left(k\right)\geq\sqrt[t]{r}\end{equation}

\end_inset


\end_layout

\begin_layout Standard
So, given a PMF 
\begin_inset Formula $f\left(k\right)$
\end_inset

, to assure the survival of a file to time 
\begin_inset Formula $t$
\end_inset

 with probability at least 
\begin_inset Formula $r$
\end_inset

, choose 
\begin_inset Formula $k$
\end_inset

 such that 
\begin_inset Formula $f\left(k\right)\geq\sqrt[t]{r}$
\end_inset

.
 For example, if 
\begin_inset Formula $A$
\end_inset

 is one month, and 
\begin_inset Formula $r=1-\nicefrac{1}{10^{6}}$
\end_inset

 and 
\begin_inset Formula $t=120$
\end_inset

, or 10 years, we calculate 
\begin_inset Formula $f\left(k\right)\geq\sqrt[120]{.999999}\approx0.999999992$
\end_inset

.
 Per the PMF of table 
\begin_inset CommandInset ref
LatexCommand ref
reference "tab:Example-PMF"

\end_inset

, this means 
\begin_inset Formula $k=2$
\end_inset

, achieves the goal, at the cost of a six-fold expansion in stored file
 size.
 If the lesser goal of no more than 
\begin_inset Formula $\nicefrac{1}{1000}$
\end_inset

 probability of loss is taken, then since 
\begin_inset Formula $\sqrt[120]{.9999}=.999992$
\end_inset

, 
\begin_inset Formula $k=5$
\end_inset

 achieves the goal with an expansion factor of 
\begin_inset Formula $2.4$
\end_inset

.
\end_layout

\begin_layout Subsection
Repair Cost
\end_layout

\begin_layout Standard
The simplicity and predictability of aggressive repair is attractive, but
 there is a downside: Repairs cost processing power and bandwidth.
 The processing power is proportional to the size of the file, since the
 whole file must be reconstructed and then re-processed using the Reed-Solomon
 algorithm, while the bandwidth cost is proportional to the number of missing
 shares that must be replaced, 
\begin_inset Formula $N-K$
\end_inset

.
\end_layout

\begin_layout Standard
Let 
\begin_inset Formula $c\left(s,d,k\right)$
\end_inset

 be a cost function that combines the processing cost of regenerating a
 file of size 
\begin_inset Formula $s$
\end_inset

 and the bandwidth cost of downloading a file of size 
\begin_inset Formula $s$
\end_inset

 and uploading 
\begin_inset Formula $d$
\end_inset

 shares each of size 
\begin_inset Formula $\nicefrac{s}{k}$
\end_inset

.
 Also, let 
\begin_inset Formula $D$
\end_inset

 denote the random variable 
\begin_inset Formula $N-K$
\end_inset

, which is the number of shares that must be redistributed to bring the
 file share set back up to 
\begin_inset Formula $N$
\end_inset

 after degrading during an interval.
 The probability mass function of 
\begin_inset Formula $D$
\end_inset

 is
\begin_inset Formula \[
Pr[D=d]=f(d)=\begin{cases}
Pr\left[K=N\right]+Pr[K<k] & d=0\\
Pr\left[K=N-d\right] & 0<d\leq N-k\\
0 & N-k<d\leq N\end{cases}\]

\end_inset


\end_layout

\begin_layout Standard
The expected cost of repairs in a given interval, then, is simply 
\begin_inset Formula $c\left(s,E\left[D\right],k\right)$
\end_inset

 where E is the expected value function -- in this case:
\begin_inset Formula \begin{align*}
E[D] & =\sum_{d=0}^{N}d\cdot Pr\left[D=d\right]\\
 & =0\cdot Pr\left[D=0\right]+\sum_{d=1}^{N-k}\left\{ d\cdot Pr\left[K=N-d\right]\right\} +\sum_{d=N-k+1}^{N}\left\{ d\cdot0\right\} \\
 & =\sum_{d=1}^{N-k}d\cdot Pr\left[K=N-d\right]\end{align*}

\end_inset


\end_layout

\begin_layout Standard
Since each interval starts with a full complement of shares, the expected
 repair cost for each interval is the same, and the cost for file that survives
 for 
\begin_inset Formula $t$
\end_inset

 intervals is 
\begin_inset Formula $t\cdot c\left(s,E\left[D\right]\right)$
\end_inset

.
 To calculate the lifetime repair cost, we just take the limit over all
 intervals as 
\begin_inset Formula $t\rightarrow\infty$
\end_inset

, discounting each cost by the probability that the file has already failed.
 So, the lifetime expected repair cost is
\begin_inset Formula \begin{align*}
\sum_{t=1}^{\infty}R\left(t-1\right)c\left(s,E\left[D\right],k\right) & =c\left(s,E\left[D\right],k\right)\sum_{t=1}^{\infty}R\left(t-1\right)\\
 & =c\left(s,E\left[D\right],k\right)\sum_{t=1}^{\infty}f\left(k\right)^{t-1}\\
 & =c\left(s,E\left[D\right],k\right)\cdot\frac{1}{1-f\left(k\right)}\\
 & =\frac{c\left(s,E\left[D\right],k\right)}{1-f\left(k\right)}\end{align*}

\end_inset


\end_layout

\begin_layout Standard
It is also necessary to discount future cost, since CPU and bandwidth are
 both going to get cheaper over time.
 To accommodate this, we throw in an addition per-period discount rate 
\begin_inset Formula $r$
\end_inset

.
 In accordance with common discount rate usage, the discount multiplier
 at time 
\begin_inset Formula $t$
\end_inset

 is 
\begin_inset Formula $\left(1-r\right)^{t}$
\end_inset

.
 This gives:
\begin_inset Formula \begin{align*}
\sum_{t=1}^{\infty}\left(1-r\right){}^{t}R\left(t-1\right)c\left(s,E\left[D\right],k\right) & =c\left(s,E\left[D\right],k\right)\sum_{t=1}^{\infty}\left(1-r\right)^{t}f\left(k\right)^{t-1}\\
 & =c\left(s,E\left[D\right],k\right)\sum_{t=1}^{\infty}\left(1-r\right)^{t}f\left(k\right)^{t-1}\\
 & =c\left(s,E\left[D\right],k\right)\left(1-r\right)\sum_{t=1}^{\infty}\left(1-r\right)^{t-1}f\left(k\right)^{t-1}\\
 & =\frac{c\left(s,E\left[D\right],k\right)\left(1-r\right)}{1-\left(1-r\right)f\left(k\right)}\end{align*}

\end_inset

If 
\begin_inset Formula $r=0$
\end_inset

 this collapses to the previous result, as one would expect.
\end_layout

\begin_layout Subsection
Non-aggressive Repair
\end_layout

\begin_layout Standard
Need to write this.
\end_layout

\begin_layout Section
Time-Sensitive Retrieval
\end_layout

\begin_layout Standard
The above work has almost entirely ignored the distinction between availability
 and reliability.
\end_layout

\begin_layout Standard
Need to write this.
\end_layout

\end_body
\end_document