[ExI] physics question

Tue Aug 21 22:06:55 UTC 2018

On Tue, Aug 21, 2018 at 4:20 PM Mike Dougherty <msd001 at gmail.com> wrote:

> On Tue, Aug 21, 2018, 4:13 PM John Clark <johnkclark at gmail.com> wrote:
>
>> On Tue, Aug 21, 2018 at 2:54 PM, Mike Dougherty <msd001 at gmail.com> wrote:
>>
>> >
>>> What's the error rate for the biological substrate humans that nature
>>> gave us for a starting/reference point?
>>>
>>
>> In DNA replication one error in 10 billion base pairs.
>>
>>
> In the context of transhuman existence measured in gigayears, I was
> imagining the error rate in the connectome rather than genome. That's an
> interesting metric though.
>
>
>>

For a given data storage component, there are two characteristics of its
reliability:

1. *R* = the time to repair a failed memory component, and
2. *F* = the average time to time to failure of the memory component

In a redundant system composed of *N* such components, where *M* are
redundant/extra components beyond the *K* components needed to store the
data (N = K + M), the system reliability (in terms of mean time to
irrecoverable data loss) is given by the formula:

       F / (K * (N choose K)) * (F / R)^(N-K)

Increasing the number of extra components (M) e.g. increasing (N-K) serves
to increase the exponent in the formula.  The base of this exponent is the
ratio of (F / R), either decreasing repair time, or increasing the mean
time between component failures is equally important.

For a replication based system, K=1, while N=replica count.  For erasure
coded systems, K and N are arbitrary, so long as 1 <= K <= N.

To give some common examples for modern data storage systems, a hard drive
with a 4% Annual Failure Rate has F = 25 years.  Rebuilding a drive that
has failed requires replacement of the drive and reconstructing its lost
data. The time for this is based on the throughput of the drive, the time
to read K other drives, and reconstruct the data in relation to how large
the drive is.  For drives around 8 TB with 200 MB/s throughput, this is
minimally 11 hours, but more typically 3 - 5 days (since the other drives
are busy serving normal reads at the same time).

What this means is that a single drive system with no redundancy has a mean
time to data loss of:

*25 years / (1 * (1 choose 1)) * (25 years / 3 days)^(1 - 1) = 25 years / 1
* (25 years / 3 days)^0
<https://www.google.com/search?ei=KIh8W8GjOcrIjwTYsq_wCw&q=25+years+%2F+%281+*+%281+choose+1%29%29+*+%2825+years+%2F+3+days%29%5E%281+-+1%29&oq=25+years+%2F+%281+*+%281+choose+1%29%29+*+%2825+years+%2F+3+days%29%5E%281+-+1%29&gs_l=psy-ab.3...155202.157531.0.157763.4.4.0.0.0.0.105.231.2j1.3.0....0...1c.1.64.psy-ab..1.0.0....0.f373FrMpd7w>
= 25 years*

A system of two copies of drives, has a mean time to data loss of:

*25 years / (1 * (2 choose 1)) * (25 years / 3 days)^(2 - 1) = (25 years /
2) * (25 years / 3 days)^1
<https://www.google.com/search?q=25+years+%2F+(1+*+(2+choose+1))+*+(25+years+%2F+3+days)%5E(2+-+1)&oq=25+years+%2F+(1+*+(2+choose+1))+*+(25+years+%2F+3+days)%5E(2+-+1)>
= 38,046 years*

A system of 3 drives, with 2 redundancies (e.g. RAID 6), has a mean time to
data loss of:

*25 years / (3 * (5 choose 3)) * (25 years / 3 days)^(5 - 3) = (25 years /
30) * (25 years / 3 days)^2
<https://www.google.com/search?ei=yoh8W_T_Joy-jwTkuKGgDQ&q=25+years+%2F+%283+*+%285+choose+3%29%29+*+%2825+years+%2F+3+days%29%5E%285+-+3%29&oq=25+years+%2F+%283+*+%285+choose+3%29%29+*+%2825+years+%2F+3+days%29%5E%285+-+3%29&gs_l=psy-ab.3...90563.90563.0.91023.1.1.0.0.0.0.132.132.0j1.1.0....0...1c.2.64.psy-ab..0.0.0....0.kujbr_lk51A>
= 7,720,015 years*

Note, however, that a mean time to data loss (MTTDL) does not mean there
will be no failure for that period of time.  It is a statistical
measurement with a very specific meaning. If the MTTDL were 25 years, it
means that if you had 25 such systems, and ran all 25 for 1 year, you would
expect one of them to fail.  Failures are random and can happen at any
time.  To estimate the probability of a failure occuring over a given
period of time, you use the following formula:

Probability of Data Loss = 1 - e^(-time / MTTDL)

So the two-copy system above, with a MTTDL of 38,000 years would have:

Over 1 year: 1 - e^(-1 year/38,000 years)
<https://www.google.com/search?ei=Zop8W5erMKzbjwTs7qXQBg&q=1+-+e%5E%28-1+year%2F38%2C000+years%29+in+percent&oq=1+-+e%5E%28-1+year%2F38%2C000+years%29+in+percent&gs_l=psy-ab.3..33i160k1l3.61505.64041.0.64159.13.12.0.0.0.0.179.1071.3j5.8.0....0...1c.1.64.psy-ab..5.5.796...33i21k1.0.GrFAasEfoYk>
= 0.002631544% change of data loss

Over 10 years:  1 - e^(-1 year/38,000 years)
<https://www.google.com/search?ei=p4p8W43IMMnVjwSSvrGADA&q=1+-+e%5E%28-10+years%2F38%2C000+years%29+in+percent&oq=1+-+e%5E%28-10+years%2F38%2C000+years%29+in+percent&gs_l=psy-ab.3...42584.43805.0.44204.2.2.0.0.0.0.89.160.2.2.0....0...1c.1.64.psy-ab..0.0.0....0.xVyeNsZ3ESY>
= 0.0263123272%
change of data loss

Over 500,000 years:  1 - e^(-1 year/38,000 years) <http://1 - e^(-500000
years/38,000 years) in percent> = 99.999807% change of data loss

Extropians, if they are to have lifetimes measured in billions of years,
need to design substrates with MTTDL's many orders of magnitude greater
than a billion years, to have a high chance of surviving that long.  Even
then, there is still no guarantee that you won't get unlucky and suffer a
data loss event before MTTDL has elapsed, in fact, 63% of the time this
happens (a failure before MTTDL time has passed). This is given by (1 -
e^(-1/1)) ~= 63%.

Fortunately, it is not hard to make MTTDL obscenely high.  For example,
setting K = 100 and N = 200.  Using erasure codes, this only has a the
overhead of 2 copies, but results in a system that can tolerate 100
concurrent faults.  With hard drives with the parameters given above, you
gain about a 100X improvement in MTTDL for each increase of M by 1.  A
system with M = 100 based on components with similar failure and rebuild
times as hard drives could have a MTTDL on the order of 1 followed by 200
zeros.

Jason
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