Math "Stumper"  What is a 1% incremental improvement in AVAILABILITY/RELIABILITY WORTH?
I have a challenging situation to determine what a 1% improvement would mean from a value perspective and I need help from an EE Pro who is very mathematically oriented.
Here is the Scenario: We have a piece of complex equipment that is used in remote locations. We have a 95% reliability/availability rating on this Asset. What we are considering is what would be the incremental value in going to a 96% or a 97% reliability level? For purposes of solving this, consider each field asset costs $1M and requires $30K / yr. for maintenance. The intangible value of being "more competitive" in our market is certainly an intangible and hard to measure. But for actual value recognition (and add any assumptions you would like that can explain how to approach this mathematical problem), what formula should be used or explanation of considerations should be put forth.
Your help and assistance is appreciated. Thank you in advance,
Thibault St john Cholmondeleyffeatherstonehaugh the 2ndCommented:
I think you need a calculated definition and costing for the item at 95%.
I'm presuming that 95% reliability means that for 5 out of every 100 times it is used that it requires a callout to fix it. 1 in 20 times. How much does this callout cost?
If you change to the 96% option you will save by the cost of one callout every 100 times it is used. Does this represent a saving in the overall cost?
Are you likely to generate more revenue because of the improved reliability? If so add this to your 96% savings.
I enjoy mathematics, but statistics worry me, they can be twisted any way you like to present a 'best' option.
From an engineering perspective I would ask why there is that difference. What can you do to bring the reliability to better that 96%? Expecting four fails in one hundred isn't great, imagine it's you trying to start your car in the morning.
There are many other considerations. You say that it is now available 95% of the time. But is it used? If nobody uses it, a 1% increase in availability is worthless.
How much will the1% availability cost? If it is $10 Million it is worthless.
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Bright01Author Commented:
Thibault St john Cholmondeleyffeatherstonehaugh the 2nd and aburr,
Thank you guys for working on this..... both of you have great comments about this.
This Asset has to do with Availability and has to do with a Asset that has to be available as much as possible (i.e. 95%+) in the event IT IS NEEDED. BUT, to aburr's point, it may never have to be used. So the value may be "intangible" but "required".
What about the trade off issues around increasing cost and diminishing value? In other words, it may cost more to improve availability as you get closer to 100%, but there may be less value in going up 1% as you also get closer (Law of diminishing returns?)? Thoughts?
And Thanks!
B.
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1. Is it safe to say that these rates do not include burnin time?
For some electronic equipment, once constructed, the failure rate may be relatively high to the published specs. In this case the company does extensive QA testing to verify the equipment is reliable. The company pays its fixing costs up front before shipping to the consumer. Some companies may produce the same product but do very little QA thereby reducing the price to the consumer. They pay the cost of fixing when the product is returned. https://en.wikipedia.org/wiki/Bathtub_curve
2. What model are these percentages based on?
Some products, for example, automobiles, have an availability rate that decreases over time. Eventually, it is not worth fixing. Other products behave as if the availability rate is constant over time. That is, if the product has not failed in 4 years, then the availability is still the same as when delivered to the consumer.
The bathtub effect noted earlier assumes an expected lifespan where the failure rate increases significantly, and whose failure rate is constant between the early high failure rate and end of lifetime.
For the case where we assume that the failure rate is constant (and hence the availability rate is constant), then IIRC, we are dealing with a Poisson distribution. "Can be used to evaluate the probability of an isolated event occurring a specific number of times in a given time interval, e.g. # of faults, # of lightning strokes time interval" http://www.engr.usask.ca/classes/EE/445/notes/notes67d.pdf
In practice, the mean time between failures (MTBF, 1/Î») is often reported instead of the failure rate. This is valid and useful if the failure rate may be assumed constant â€“ often used for complex units / systems, electronics â€“ and is a general agreement in some reliability standards (Military and Aerospace). It does in this case only relate to the flat region of the bathtub curve, also called the "useful life period".
The link gives a formula for computing downtime assuming 24 x 7 and 8 x 7 hour usage.
Assuming 24 x 7 usage, Let Pa = Availability Probability; T = Total Time in use in a year; D = downtime in a year
Pa = (T  D)/T = 1  D/T
Solving for D:
D = (1  Pa) * T
For 24 hours day, T = 24 * 365 = 8760 hoursD = (1  Pa) * 8760For 95% availability: D = .05 * 8760 = 438For 96% availability: D = .04 * 8760 = 350.4For 97% availability: D = .03 * 8760 = 262.8For 98% availability: D = .02 * 8760 = 175.2For 99% availability: D = .01 * 8760 = 87.60For 99.9% availability: D = .001 * 8760 = 8.76
Thibault St john Cholmondeleyffeatherstonehaugh the 2ndCommented:
The importance of the reliability/availability should also influence your decision. If it's a signal on a golf course that players can use to tell following players that they are clear of the hole it may be important, but not as important as an electronic defibrillator in a busy shopping centre or at an airport.
The 1% is a bit of a distraction. You are really looking at the change from 96 to 97 which is close to 1%, but you seem to be comparing it with a 1% increase all over the scale. An increase of 1% from 1 will double the effectiveness of your item which is vastly different from the slight 1% increase from 95%.
A fixed costing for each state should give you a better idea than trying to calculate a percentage difference
"This Asset has to do with Availability and has to do with a Asset that has to be available as much as possible (i.e. 95%+) in the event IT IS NEEDED. BUT, to aburr's point, it may never have to be used. So the value may be "intangible" but "required"."

If it is "required" do it and stop calculating.
If it is truly "intangible" you may have to go with the "gut feeling" of the most experienced person involved. That's why he is paid the big bucks (He is paid big bucks, right?).


"What about the trade off issues around increasing cost and diminishing value? In other words, it may cost more to improve availability as you get closer to 100%, {right} but there may be less value in going up 1% as you also get closer (Law of diminishing returns?) {again, right} ? Thoughts?"
Therefore think twice about spending the money, and then think again.

There is a value to "good will". In absence of redundant systems, if your customer loses money or information when your system goes down, then your "good will" value probably goes down too (unless you have a very understanding, and tolerant customer). In almost all of my systems, there was a specified requirement as to what the reliability had to be. In one case, it was 59's.
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Bright01Author Commented:
You guys have given me some "great" insight on this issue. Yes.... it's all about maximum "availability" irrigardless of actual "use". And the consequences of a failure is "catastrophic". So you can imagine how hard it is to actually determine how much to invest in order to improve reliability.
Take a look at this worksheet and let me know if this is correct in figuring out seconds of exposure per year.....
If " the consequences of a failure is "catastrophic" by all means stop calculating and get another machine. No matter what the calculation, debt is better than bankruptcy.
@aburr,
I second the motion, except that we went with triple redundancy to achieve 59's.
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Bright01Author Commented:
Hey guys.... take a quick look at my spreadsheet. Is the math right in looking at a reliability rate in the 95% range and overlaid by what 5/9s will achieve?
I'm presuming that 95% reliability means that for 5 out of every 100 times it is used that it requires a callout to fix it. 1 in 20 times. How much does this callout cost?
If you change to the 96% option you will save by the cost of one callout every 100 times it is used. Does this represent a saving in the overall cost?
Are you likely to generate more revenue because of the improved reliability? If so add this to your 96% savings.
I enjoy mathematics, but statistics worry me, they can be twisted any way you like to present a 'best' option.
From an engineering perspective I would ask why there is that difference. What can you do to bring the reliability to better that 96%? Expecting four fails in one hundred isn't great, imagine it's you trying to start your car in the morning.