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Posted
This is the place to post questions about Vee Narayan's Reliability Roadmap web workshop presentation " Determining Failure Finding Test Intervals".

We will post a recording of the Reliability Roadmap web workshop on Friday afternoon.

Note: You must be a registered member of MaintenanceForums.com to post a question or reply. Your privacy is assured - we do not rent-trade-sell our member list.
 
Posts: 776 | Location: Southwest Florida Gulf | Registered: 03 April 2004Reply With QuoteEdit or Delete MessageReport This Post
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Here are some articles and excerpts by Vee Narayan:

Detective Maintenance

Effective Maintenance Management Excerpt

Very down to earth information.

Enjoy

Terry O
 
Posts: 776 | Location: Southwest Florida Gulf | Registered: 03 April 2004Reply With QuoteEdit or Delete MessageReport This Post
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Vee - Great workshop today - thanks again.

You said you can change the availablity by changing the test frequency.

By that logic - you can get 99.99% availablity by posting a live sentry 24/7.

Is there an economic component to the equation to keep things sensible?

Terry O
 
Posts: 776 | Location: Southwest Florida Gulf | Registered: 03 April 2004Reply With QuoteEdit or Delete MessageReport This Post
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V.

Great job! Looking forward to meeting you in Las Vegas.

Howard


Howard W Penrose, Ph.D., CMRP
President, SUCCESS by DESIGN Reliability Services
Author: "Physical Asset Management for the Executive (Caution: Don't Read this on an Airplane)" and;
"Electrical Motor Diagnostics: 2nd Edition"
 
Posts: 844 | Location: Connecticut, Michigan and Illinois | Registered: 12 April 2005Reply With QuoteEdit or Delete MessageReport This Post
Vee
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Terrence,

The economics comes at the front end, in determining the 'required availability'

That is determined by two factors:
1. The demand rates i.e., how often we ask the item to work, so e.g., a Relief valve operating at 85% of its cold set pressure is likely to be asked to work more often than another operating at 50% of the set pressure. Similarly if the process pressure varialtions are very low (dead man's heartbeat), the demand is low. If these are frequent and large in amplitude, the demand on the Relief valve to operate are high.
2. The actual or potential consequences; can it injure or kill one man or a hundred? Will the leak be a few ounces of oil on a pump bed or a few hundred barrels in the Gulf of Mexico?

The higher these two are, the greater the risks. In addition, we have to factor in qualitative risks, relating to e.g. public opinions, which need not be based on fact or logic. There are a number of factors affecting this, explained excellently by many experts (Reason, Kletz, Lorenzo, Slovic and others). It is also discussed in brief on pages 118-122 of my book,

The higher the total risk (quantitative + qualitative), the less likely we are to tolerate the failure. The less our tolerance, the higher the availability we expect from the devices that protect us from such failures. That sets the acceptable figure for availability. We must accept that if we want higher availability, we should be prepared to pay for it.

In general, for failure affecting Safety or the Environment, we set the availability at 95% or higher, moving up to 99% or more for cases where multiple fatalities could occur, e.g. for safety features on aircraft. Production loss related failures can have lower availabilities, ranging from say 85% to 95% or so, depending on the $$$ values involved.

Have I answered your question?

Howard, thanks for your words of encouragement. I too look forward to meeting you at RCM 2006 and working with you. Regards.


Regards,
V.Narayan (Vee)
Lead Author, 100 Years of Maintenance: Practical Lessons from Three Lifetimes, Industrial Press.NY ISBN-13: 978-0831133238
Author, Effective Maintenance Management: Risk and Reliability Strategies for Optimizing Performance, 2004, Industrial Press NY ISBN-13: 978-0831131784
 
Posts: 770 | Location: Scotland, UK. | Registered: 16 May 2004Reply With QuoteEdit or Delete MessageReport This Post
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How to access this?
 
Posts: 2596 | Location: Borneo | Registered: 13 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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JOsh

We were hoping to have the recorded playback today however it looks kmore likely it will posted on Tuesday Feb 21.

Sorry for the delay.

Terry O
 
Posts: 776 | Location: Southwest Florida Gulf | Registered: 03 April 2004Reply With QuoteEdit or Delete MessageReport This Post
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Terry, I've received the link but no playback or pdf files being attached. How to participate in the web workshop?
 
Posts: 2596 | Location: Borneo | Registered: 13 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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Here is the link for Vee's Web Workshop

It is in Flash format

There is no cost to access it.

Terry O
 
Posts: 776 | Location: Southwest Florida Gulf | Registered: 03 April 2004Reply With QuoteEdit or Delete MessageReport This Post
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I'm trying to calculate the availability with the various test intervals & MTBF=5 years but couldn't get the same figures stated in the presentation?

Here is my calculation:
-T/MTBF = ln(2A-1)
T=2 months, MTBF=5 years x 12 months= 60 months

-2/60 = ln(2A-1)
e to power of -1/30 = 2A-1
2A-1 = 0.9672
Therefore, A= 1.972/2 = 0.9836 or 98.36%

In your presentation, A=87.5%

Where is my mistake?

This message has been edited. Last edited by: Josh,
 
Posts: 2596 | Location: Borneo | Registered: 13 February 2005Reply With QuoteEdit or Delete MessageReport This Post
Vee
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Josh,

Good question.

The illustration in the presentation is for a single bulb. But when doing these calcs. in an industrial situation, there are hundreds of bulbs or gas detectors or relief valves to consider which have not failed after 1/2 month! The formula applies for Mean Availability not for a single event.

So, on average, by the end of the year, most light bulbs would still be working, even though the one we chose had failed. What the equation says is that if you had tested 10000 bulbs, not just the one, 9836 would have been working at the end of the year, with a test interval of 2 months and immediate replacement of burnt out bulbs.

For the single bulb in the powerpoint slide, its availability was 87.5%. This is like the surgeon saying the probability of a successful operation is 99%. Does that mean that the patient who is now on the table will have a successful operation? He may be the one accounting for the 1% failure statistic. That slide is simply to illustrate that there is a relationship between MTBF, test interval and availability. It is not what we need to compute Mean availability. It tells us that there is a 98.36% probability that a single bulb will survive the year, if we adopt a policy to test every 2 months and replace any burnt out ones promptly.

I hope all this make sense to you. It is always easier to explain all this face-to-face.


Regards,
V.Narayan (Vee)
Lead Author, 100 Years of Maintenance: Practical Lessons from Three Lifetimes, Industrial Press.NY ISBN-13: 978-0831133238
Author, Effective Maintenance Management: Risk and Reliability Strategies for Optimizing Performance, 2004, Industrial Press NY ISBN-13: 978-0831131784
 
Posts: 770 | Location: Scotland, UK. | Registered: 16 May 2004Reply With QuoteEdit or Delete MessageReport This Post
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How did you get 87.5%?
 
Posts: 2596 | Location: Borneo | Registered: 13 February 2005Reply With QuoteEdit or Delete MessageReport This Post
Vee
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Josh,

In the example of the single bulbunder consideration, it works in all for 1/2 month before it failed, then for 10 months after the first 2-monthly test, since a new bulb was installed. The first bulb had a random failure after 1/2 month, while, on average, these bulbs last 5 years (MTBF). If instead of just one bulb, you looked at say 100 bulbs, statistically, with a 2-month test interval and immediate replacement of any failed bulb, you would expect to see 98.36% of the bulbs in a working state.

The reason for showing the single bulb case is to show how witha given time to failut=re (1/2m) and MTBF (5 years), the avialability is related to the test interval. If we extend this to a family of bulbs rather than just one bulb, we will get the probability of bulb survival at anypoint in time.

Oh, to address your question (1/2+10)/12 = 0.875; that is how we get teh 87.5%.


Regards,
V.Narayan (Vee)
Lead Author, 100 Years of Maintenance: Practical Lessons from Three Lifetimes, Industrial Press.NY ISBN-13: 978-0831133238
Author, Effective Maintenance Management: Risk and Reliability Strategies for Optimizing Performance, 2004, Industrial Press NY ISBN-13: 978-0831131784
 
Posts: 770 | Location: Scotland, UK. | Registered: 16 May 2004Reply With QuoteEdit or Delete MessageReport This Post
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Ok thanks Vee for the explanations on the meaning of the calcns. Does this mean we cannot get 100% of trip loops working & available all the time and a few percentage are bound to fail if the test interval equals the shutdown interval of every 3 years let's say?

Ah, ok you assume the one bulb failed after 1/2 months & remain undetected/hidden/unavailable until after the various test intervals. After each test, you found the hidden failure & replace to AGAN, right?

I thought you calculated the availability using the ln formula. Where can I see the derivation of the ln formula from first principles?

How do you define risk level and demand rate and translate them into the require mean availability? Is there a sample table to select the appropriate values to suit the various scenarios? I know from above, you said for safety A=95 to 99% & for prod loss, A=85 to 95%.

This message has been edited. Last edited by: Josh,
 
Posts: 2596 | Location: Borneo | Registered: 13 February 2005Reply With QuoteEdit or Delete MessageReport This Post
Vee
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Josh,


My answers are embedded into your note.

quote:
Does this mean we cannot get 100% of trip loops working & available all the time <<

Vee: The reliability of any system is 100% ONLY at the instant it is installed. One millisecond later it is less than 100%. By how much? That depends on the item's 'goodness' which you can measure by its MTBF. The higher the MTBF, the less the reliability falls.
For hidden functions, the availability on demand equals its reliability at that time. If you test an item, and it has already failed, you reset the clock for that particular item, by replacing it with a new item.

>>and a few percentage are bound to fail if the test interval equals the shutdown interval of every 3 years let's say?<<

Vee: It all depends on the MTBF values. If you wait for 3 years, you are pretty sure to have several failures. If the MTBF was say, 60 years, you should expect approx 5% of the items to have failed in 3 years. If the MTBF was 30 years, you can expect about 7% to have failed.

>>Ah, ok you assume the one bulb failed after 1/2 months & remain undetected/hidden/unavailable until after the various test intervals. After each test, you found the hidden failure & replace to AGAN, right? <<

Just think of it thus: In the example used earlier, if the same bulb failed randomly after not 0.5m, but say,
- 1m, then for the same MTBF and the same 2m test regime, the availability is now 91.7%
- 1 1/2m, the availability is 95.83%
- 1 3/4m, the availability is 97.9%

Within that test period of 2m, the bulb could fail randomly at any time. So, statistically, the mean availabilitry represesnts the probable average value, because, we DO NOT know when the failure takes place. Whether it is after 1/2m, 1m, 1.5m 1.75m or not at all is unknown, as it is a hidden function. So we look at larger samples, not just one item, and ESTIMATE the probable availability of the family of items. THAT is what the formula for Mean Availability tells you.

>>I thought you calculated the availability using the ln formula. Where can I see the derivation of the ln formula from first principles?


Vee: Pages 34-39 of my book; see also pages 61-64 of the original Nowlan & Heap book on RCM, which is available in the reliabiltyweb.com site.

This message has been edited. Last edited by: Vee,


Regards,
V.Narayan (Vee)
Lead Author, 100 Years of Maintenance: Practical Lessons from Three Lifetimes, Industrial Press.NY ISBN-13: 978-0831133238
Author, Effective Maintenance Management: Risk and Reliability Strategies for Optimizing Performance, 2004, Industrial Press NY ISBN-13: 978-0831131784
 
Posts: 770 | Location: Scotland, UK. | Registered: 16 May 2004Reply With QuoteEdit or Delete MessageReport This Post
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So we need the MTBF to be as high as possible. Does manufacturers provide MTBF or we have to rely on actual MTBF from the fields eg OREDA?

This message has been edited. Last edited by: Josh,
 
Posts: 2596 | Location: Borneo | Registered: 13 February 2005Reply With QuoteEdit or Delete MessageReport This Post
Vee
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Josh,

The 'design' reliability is often far higher than what we achieve in practice. But we often operate and maintain equipment in such a way that we lose a lot of the theoretical run lengths.

In realistic terms, the only MTBF that matters is what you actually achieve under your operating conditions, not what is 'possible' according to the vendor or what other people have achieved e.g., using generic data sources such as OREDA. Of course if you dont have your own data because you dont collect and analyze it, then generic data sources are a good enough starting point


Regards,
V.Narayan (Vee)
Lead Author, 100 Years of Maintenance: Practical Lessons from Three Lifetimes, Industrial Press.NY ISBN-13: 978-0831133238
Author, Effective Maintenance Management: Risk and Reliability Strategies for Optimizing Performance, 2004, Industrial Press NY ISBN-13: 978-0831131784
 
Posts: 770 | Location: Scotland, UK. | Registered: 16 May 2004Reply With QuoteEdit or Delete MessageReport This Post
Vee
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Josh,

I missed your question earlier:
quote:
Originally posted by Josh:
How do you define risk level and demand rate and translate them into the require mean availability? Is there a sample table to select the appropriate values to suit the various scenarios? I know from above, you said for safety A=95 to 99% & for prod loss, A=85 to 95%.


In an earlier post in this thread, Terrence asked a similar question and I have tried to explain the pribciple involved in my reply. On pages 163 to 172 of my book, you can see a number of practical examples of how to apply these principles.


Regards,
V.Narayan (Vee)
Lead Author, 100 Years of Maintenance: Practical Lessons from Three Lifetimes, Industrial Press.NY ISBN-13: 978-0831133238
Author, Effective Maintenance Management: Risk and Reliability Strategies for Optimizing Performance, 2004, Industrial Press NY ISBN-13: 978-0831131784
 
Posts: 770 | Location: Scotland, UK. | Registered: 16 May 2004Reply With QuoteEdit or Delete MessageReport This Post
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