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Q: MTBF for a system with several subsystems with individual MTBFs ( Answered 4 out of 5 stars,   7 Comments )
Question  
Subject: MTBF for a system with several subsystems with individual MTBFs
Category: Computers
Asked by: ravil-ga
List Price: $10.00
Posted: 19 Aug 2004 16:23 PDT
Expires: 18 Sep 2004 16:23 PDT
Question ID: 390140
I need to find out the MTBF of an appliance that we manufacture.  The
appliance consists of several subsystems - motherboard, powersupply,
fans, LCD etc.  If I know the MTBF of each of these subsystems, how do
I go about calculating the MTBF for the appliance.  I need links to
tutorials that describe this scenario in a clear and concise manner.
Answer  
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
Answered By: maniac-ga on 21 Aug 2004 10:16 PDT
Rated:4 out of 5 stars
 
Hello Dhrm77,

A complete answer will take into account both:
 - serial failures
 - parallel failures (or redundant components)
There are different formulas for these two basic scenarios which can
be combined to produce an accurate system level MTBF result.

Serial Failures:

Let's do the serial failure case first. In this case, the failure of
any one component will cause failure of the system. This may sometimes
be illustrated as:
  A -> B -> C -> D
to show the serial nature of the failure analysis. The comment by
jeannot52 is basically correct for this case (but does not address the
case with redundant components).

However I find it much easier to describe (and compute) by introducing
failures per million (lambda). For the remaining part of the answer:

  MTBF = mean time between failures (hours per failure)
  Lambda = failures per million hours
  F = failure rate or probability of failure in one hour
  R = reliability rate (probability of working in one hour)

You can convert between MTBF and Lambda with the following equations:
  Lambda = 1,000,000 / MTBF
  MTBF = 1,000,000 / Lambda
and assuming a constant failure rate (not necessarily true)
  F = Lambda / 1,000,000 or 1 / MTBF
  R = 1-F
The reason I talk in terms of Lambda (failures per million hours) is
you can still explain the comparisions - less failures per million is
better, and the arithmetic is simpler (until you *have* to compute
MTBF).

So if you have four components, A, B, C, and D each with MTBF of
20,000, 10,000, 15,000, and 30,000 hours respectively, using this
method, the MTBF of the system is calculated as:
  Lambda A = 1,000,000 / 20,000 = 50.0
  Lambda B = 1,000,000 / 10,000 = 100.0
  Lambda C = 1,000,000 / 15,000 = 66.67
  Lambda D = 1,000,000 / 30,000 = 33.33
  Lambda (composite system) = 50+100+66.67+33.33 = 250
  MTBF (composite system) = 1,000,000 / 250 = 4,000 hours
This is a relatively simple example. A system 10's or hundreds of
components can be calculated in the same way.

Redundant Components:

If you have two components in parallel (e.g., dual power supplies)
where a failure of both components is required to fail the system, the
MTBF of the system is MUCH less than either component. I will do a
simple example using both serial and parallel failures.

Assume A and B both have MTBF of 100 hours or Lambda = 10,000. The
failure rate F for A and B would then be 0.01 for each. For
comparison, the serial solution has Lambda = 20,000 failures per
million or MTBF = 50 hours.

For the redundant case, the probability (F) that both items are failed
at the same time is:
  F = FA * FB
  F = 0.01 * 0.01
  F = 0.0001
Solving for lambda gets
  Lambda = 100
or
  MTBF = 10,000 hours
So there is a substantial improvement in reliability when using
redundant components. Note that if you have serial components before /
after the redundant components, you still need to handle those in
series with the redundant components.

A composite system:

If you have both serial / parallel components, break up the system
into pieces and do the lambda calculations as serial or parallel. I
usually end up with several serial items to add at the end and then
compute the overall system MTBF value.

A few other sources or books include:

http://www.softwareresearch.net/site/teaching/SS2003/PDFdocs.EmbC/16_fault_tolerance.pdf
Part of a training class. Has the formulas and illustrations about 1/2
the way down the file. Also includes some historical information and
explanation of the different causes of failure (e.g., mechanical) and
why failure rates can vary over time. Note this uses R for most of the
calculations.

http://www.alericonetworks.com/support/analysis/harddisk/
An explanation of hard disk reliability and how different ways of
organizing the disks can affect overall reliability as well as how
reliability varies with time.

http://www.intellectuk.org/publications/relc.asp
A reliability guide from an UK industry group - price listed at 45
pounds. See the table of contents / introduction link for a good
summary of what is covered.

http://www.fetchbook.info/search_1587130173/tab_reviews.html
A book describing networks - but includes sections on computing
reliability with the various methods. Also addresses several issues
related to high availability. You can apparently download the "SHARC
Spreadsheet" from a link at the end of
  http://safariexamples.informit.com/1587130173/index.htm
for more general calculations.

http://www.enre.umd.edu/rmp.htm
Links to several reliability analysis programs with brief
descriptions. See also the links near the bottom for other resources
on this university site.

http://www.bmtrcl.com/reliability_calculator.html
A simple reliability calculator - converting to / from failures per
million hours (or failures per hour) to MTBF in hours or years.

Search phrases included:
  lambda failure per million hours
  RMA reliability formula
  "reliability calculator" mtbf
  reliability tutorial

If you have a lot of these reliability calculations to perform - I
strongly suggest getting a tool. Perhaps start with the SHARC
spreadsheet and then move on to a more comprehensive tool later.

Please use a clarification request if you need further details on
performing the reliability calculations or if some part of the answer
is unclear.

  --Maniac
ravil-ga rated this answer:4 out of 5 stars

Comments  
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: dhrm77-ga on 19 Aug 2004 20:34 PDT
 
Assuming that the design is well within all the recommended operating
conditions for each component, the over-all MTBF is the lowest MTBF of
all components.
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: msblack-ga on 20 Aug 2004 00:10 PDT
 
That doesn't sound correct.  Probabilities multiply.

For example, if a part has a 10% failure rate within one year, two
identical parts have a failure rate (either one or both failing) of
19%.
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: msblack-ga on 20 Aug 2004 07:36 PDT
 
Taken another way:

MTBF is Mean Time Between Failure.  That translates into the part has
a 50% chance of surving that length of time.  If A has an MTBF of 1
year and B has an MTBF of 1 year, either part has a 50% chance of
working after one year.  Taken together, there's only a 25% chance
that both parts will be working after one year.
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: ddh-ga on 20 Aug 2004 12:38 PDT
 
Assuming you have 3 components A,B,C. Assume failure rate within a
period (let's say 1 year) of component A,B,C are X%, Y%,Z%
respectively.
So, for component A not to fail during the one year is (100-X)%.
Similarly, for Componet B not to fail during the one year is (100-Y)%
Also, for Componet C not to fail during the one year is (100-Z)%
Therefore for all the 3 components not to fail is (100-X)/100 x
(100-Y)/100 x(100-Z)/100 = ((100-X)/100 x (100-Y)/100
x(100-Z)/100)x100%
Also for at least 1 component to fail within the period is 
(100 - ((100-X)/100 x (100-Y)/100 x(100-Z)))/100)%

You know the percentage it will fail during the fixed period (1 year),
so you can use this to work out MTBF. (I don't know how to work out
the MTBF from the failure rate)

This can also be ued for 4, 5, 6 and many more componenets.
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: ravil-ga on 20 Aug 2004 14:25 PDT
 
All I need is to provide this kind of information for our appliance:

MTBF at 40 degree C is xxx,xxx hours

How have others got this information knowing the MTBFs of the subsystems.
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: jeannot52-ga on 20 Aug 2004 21:39 PDT
 
The general answer to your question is that the reciprocal of the MTBF
of an entire system is equal to the sum of the reciprocal of all its
subsystems :

ie:
1/MTBF(total)=Sum[1/MTBF(subcomponent1) + 1/MTBF(subcomponent2) + etc....).
You can, thus, easily derive the system MTBF after adding a few
fractions and taking the inverse of the resulting fraction.

The trick is to make sure that the each of the subsystems' MTBF is
known (especially, if only an estimate is provided, you must check
that the assumed values are using the same hypotheses, ie working
condidtions such as temp, pressure, duty cycle, aso...).

Best of luck and best regards.

jeannot52
Subject: Re: MTBF for a system with several subsystems with individual MTBFs
From: dhrm77-ga on 24 Aug 2004 15:43 PDT
 
Ooops... 
I was just thinking about it again.. and yes I made a mistake.
I was thinking in term of the weakest link...and didn't realize that
we were talking about averages.
yes, probabilities do multiply.. 
The answer isn't really simple, however.
Let's say we have 2 systems, if both systems have a MTBF of 5 years.
that means 1/2 of each system will have failed after 5 years. so we
have 1/4 of the systems working. the problem is what exactly does the
failure curve look like ? is it a straight line ? is it a curve where
1/2 of the remaining systems break down after another 5 years ?
If the failure rate follows an arithmetic curve, then the over-all
MTBF would be 0.707 x 5 years or about 3.5 years.
It could be designed in such a way that 99% of the systems will work
correctly for 4 years, then within the next 2 or 3 years 80% of those
system will experience some failure... A typical example is a car.
They work fine for the first few years.. then you keep having to
replace some parts..
That makes the over-all MTBF very hard to calculate.

Furthermore, as "maniac" mentionned above, there are scenarii with
serial failures. I had one of those recently where a hole drilled in
the wrong place, went through a power supply, without simingly
affecting it. The cracked PC board allowed moisture to start corroding
copper traces. when the trace was gone, the power supply didn't
regulate anymore, which activated the protection cicuitry, which
started to draw power from nearby power supply, over-burdening and
burning up the over-current protection, and starting a fire.
But it really doesn't matter.. once a component has failed.. whether
or not some other components fails too in the process is sometimes
irrelevant, except if "backup" sub-systems will compensate for the
failing component...

As I read above, any attempt to calculate an MTBF assumes that the
known failure rates of each component is linear or arithmetic with
time... which is not necessarily the case.

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