Hi Phill,

I understand the questioning of numbers – how trustful are they? That is something that I think it would be unnecessary to say that I and everybody properly qualified, who really feel responsible, always do when facing a new case.

quote:

To me this says, if the supply time cannot be reduced significantly then you must hold one unit. Analysis time - 30 seconds.

Compared to my 6 hours work, how do you think I could feel? Miserable? On the contrary, I am firmly convinced that I followed the right way, assuming that data are correct and the situation cannot be changed significantly. I even took advantage of this case as I am now better prepared to solve similar cases “in a clink of an eye”, that is less than the 30 seconds of yours… Furthermore, I don’t know why you didn’t take the opportunity to show Jaz and all of us your superior skills (or intuition?) .

Why not 2 or even 3 units? What guarantee do you have that a second failure won’t follow just a few hours after the restart of the tower? It all depends on the consequences whose magnitude has to be assessed and on the delivery time. If you cannot do anything about it, the number of spares “on the shelf” has to be increased. Math provides you solid grounds on which to base your conclusions and leave gut feelings apart.

In short, I would like to say that I agree with your methods but, despite, there are many occasions when maths will provide a complementary or more appropriate and definitive solution. We are agreed on this, aren’t we? If so, it is useless to discuss this point any further.
Regards,

Rui

This message has been edited. Last edited by: <Rui Assis>,

I have not read all the post's but to add to your question, if you have a local supplier who might carry supplies for your equipment, form a relationship with that supplier. Come up with kits which contain all the required spares for that PM. Keep the kit under one number and when it comes time to get that on site for the PM you have it under one number.

The other suggestion is to Benchmark every machine in your plant. Use historical trends to view failure rates. Also, ask your self "if this machine goes down what would happen to my business?" If the answer is "were dead!" then you need to make sure you have the spares for that machine. I would suggest you draft a BOM and put a supplier, part number and lead time after each item on that BOM. If some parts are long lead times you will need to have those on hand. If your historical trends show failure rates are high on certain parts, make sure those are on the shelf. The rest would fall into order as needed.

Hope that helps

Thanks G for your contribution.

The problem is that things cannot be reduced to black and white colours, as you know. Different grades of gray have to be considered, that is, apart from failure rates, you have to account for different levels of severity: negligible, marginal, critical, catastrophic (dead according to you) or others in between. I agree with your approach which is wise and recommendable but I wouldn’t neglect a quantitative analysis in a few cases where the balance between failure costs and holding costs doesn’t seem to favour any specific solution beforehand.

Rui

Rui,

Very well put response. I do believe there are different shades of gray not only in the magnitude of failure consequences, but also in the method used for deciding to hold spares. Quantitative analysis is necessary at times - as I believe you have mentioned in your post. This idea is what I felt when I made my last post, but you have stated so clearly here.

For example the work you performed on the cooling water tower definitely aids in building a business case to say, "I want to invest 40,000 dollars in a critical spare part." I have to compete for that money against others in the company that may want to use the capital for other purposes that may have equally compelling cases.

Jaz,

Thanks for being appreciative of my effort. I hope you’ll succeed in your struggle for the 40,000 dollars to be invested in a "good cause"…

Regards,

Rui

Rui,

If I have hurt your feelings I apologise, that was never my intension. I was merely trying to show that not all problems require an in depth quantitative analysis for solution.

So in the spirit of providing an alternative in the debate here is some more input.

I don't recall saying that I don't trust the numbers. In fact just the opposite - I think that the right set of numbers will answer the question. But there is the problem, what is the right set of numbers and more importantly what assumptions have been made around those numbers? In my opinion, most errors made in a calculation relate to the assumptions. I think that you would agree with this given your comment about different shades of gray. Is that right?

As engineers we often fall back on calculation methods. It is what we are taught at Uni and it is what we trust in. But as you have said in this forum there are very many shades of gray and things aren't black and white. Yet the 'calculation' approach is black and white - at least for the set of data that is being reviewed.

So, in order to manage the shades of gray we must limit the analysis to some 'logical' set of numbers - otherwise one must review every circumstance. In any problem solving scenario (whether it is inventory or anything else) there are always key factors that make the difference - you don't need to review every number just the ones that make the difference or in other words would change the decision.

As an aside this is why the inventory approach I advocate was developed using the scientific approach of hypothesis falsification. In this way you need only understand what makes a condition true or untrue. This approach has been the basis of all scientific development for 200 years.

Now back to the cooling towers.
In the case of the cooling towers the numbers that we can trust are the clearly defined values such as cost, lead time, repair time, cost of downtime. These are easily verifiable. Any value relating to failure rates is an assumption. The only certainty is that with 8 towers and no maintenace failure is likley at some point. So my logic was, as follows:
Failure is expensive in terms of downtime
The parts cannot be supplied in a short time frame
The probablity of failure is high
Therefore you need to hold the spare.

Now for the crucial part - question the constraints. Can lead time be reduced? Can maintenace be performed? and so on. It is only by applying this final step that a truly optimised solution can be found.

Don't take my word for it - go to google and type in Double Loop Learning and read all you can about this appraoch. It was developed by a Harvard Professor in the mid 1970's. (Alternatively you can read an article I wrote Double Loop Your Thinking which is on my website www.InitiateAction.com/articles.htm)
Finally, it seems that maybe the real issue was justifying the spare to management rather than identifying the need for a spare. Justifying the purchase to management is a completely different problem which ironically must involve calculation if you are to win the arguement!

Rui, keep up the good work. I think that your demonstration of the technique has helped inform some decisions and was not at all wasted.

Cheers,

Hi Phill,

quote:

Finally, it seems that maybe the real issue was justifying the spare to management rather than identifying the need for a spare. Justifying the purchase to management is a completely different problem which ironically must involve calculation if you are to win the argument!

Thanks for pointing me out this side of the issue. Funny that it may seem, I was never aware of it. In reality, my studies and advices, in what consultancy is concerned, have always been addressed to management people. That is why – now I understand – so many practitioners here seem to neglect the importance of quantitative methods tying in money and technical issues (the financial dimension). Money related issues; this has always been the nature of cases treated by me. This explains why I am not so prone towards qualitative methods of problem solving despite I know how they work.

I always prefer the numbers side of the picture. Statistics and simulation are such terrific tools to assist you in assessing reliability problems...

Thanks for taking the time and also for the links to the articles of yours that I have read already and honestly recommend.

I will also follow your sugestion and visit with interest "Double Loop Learning" from Harvard.

Regards,

Rui

Let me challenge you one more time with the case below. I am very curious to know how you would address it.

Suppose you have in your company 3 similar pumps A, B and C, whose gaskets XYZ (2 per pump) fail according to a 2-parameters Weibull distribution. The shape parameter is the same for all the pumps and equals 2.6. But the scale parameter varies:

Pump A = 900 running hours
Pump B = 1,200 running hours
Pump C = 1,600 running hours

Mean working time of the pumps are:

Pump A = 16 hours/day
Pump B = 14 hours/day
Pump C = 13 hours/day

Preventive replacement takes place periodically:

Pump A = 500 running hours
Pump B = 700 running hours
Pump C = 900 running hours

Pumps are needed 365 days per year. Lead time from the vendor is 1.5 months (just let it be this long - please, don´t advise to change the vendor... ).

If you want to guarantee a level of service (availability) of 98%, would you hold any gaskets in stock? In case so, how many?

Rui

This message has been edited. Last edited by: <Rui Assis>,

My wild guess (which could be wrong because cannot use quantitative method but pls give your answer after others give their replies)

Assuming all 3 pumps are always needed to run simultanously and use the same gaskets, how about set the min/max levels of gasket at 1/3 sets?

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

Thanks Josh for giving your guess. I´ll wait for other replies.

Regards

I will start with some assumptions:
(1) There are "significant" consequences to any one of these pumps failing,
(2) That the cost to stock the gasket sets is an undertaking that management would certainly be “concerned” with.

I performed my analysis with these assumptions in mind.

My answer is: I would set the re-order point at a minimum of 3 for 98% Availability, but because there are 2 gasket sets per pump I would then stock 4 (order in multiples of 2).

When I worked in mobile equipment maintenance we stocked a minimum 3 pistons for a 4-cylinder diesel engine – and I wouldn’t be making this comment if this minimum re-order point didn’t harm our availability.

At this time I will keep my thoughts on how I answered this problem to myself.

Jaz,

I am very curious to know how you reached those figures. We will compare our methods side by side, if you agree, after we get a few more replies.

Regards

Replies from more peers, specialy from Phill and Daryl who already participated in this topic, would surely be wellcome by us all.

Hi everyone,

First let me say that I studied Weibull distribution as part of my graduate studies in 1990/91. However, I have not used or referred to it once in the ensuing 15 years. This means that I can barely spell Weibull let alone apply the analysis!

Rui, I am sure that, within the assumptions of the problem your anaylsis will give the right result.

Here is my go at this. I am assuming that the gaskets are supplied and used as a set. I am also assuming that our maintenance system is able to order the parts for preventative replacement such that they arrive the week before they are needed (and that this is a cost effective way to order).

So, if I start with two in stock I never fall below 1 in stock for PM and sometimes have as many as 3 sets. If the PM stock is 'reserved' then I am holding only 1 set to manage unexpected failure but often have up to 3 sets available. Without further insight into the history of failure then I would hold about this level of stock. (By the way, if you start with just 1 there is a period of 14 weeks when there is zero and this is why I rejected that position.)

If the gaskets are cheap I would hold a couple of extra sets. If they are expensive then I would hire Rui to work out the number for me!

Jaz has already mentioned the need to understand the cost of the gasket and to my way of thinking this is a 'must know' piece of information. The reason? If the gaskets are cheap I wouldn't spend time solving the problem I would just stock plenty. I can explain later why this ought not be an issue.

My other observation is why 98%? Does this mean that 2 failures in a 100 have no stock available? If so the plant is down for 45 days each time. The gaskets would need to be very expesnive for this to be tolerable.

Anyway, there is my response. Rui, we all wait for you to reveal the grand result!

Thanks Phill for your viewpoints on this issue. I assume that parts needed for PM are ordered such that they arrive a couple of days before they are needed. Parts to keep in stock are aimed solely for CM tasks. Note that despite you perform PM, there is still a certain chance that any gasket fails before the PM. People are tempted to think that no stock would be needed if PM is adopted. That is why I put this question.

I didn´t give the gasket unit price but it is certainly low. If it were high, a compromise between holding costs and opportunity costs would have to be met (this would lead us to a similar case as the Jaz´colling tower). Instead, I gave the desired level of service (98%) whose meaning is that there is a 2% chance that one day, when you need one or more parts, you go to the warehouse and are told that a stockout occurred. You will have to wait until the day parts are received from the supplier but not 45 days at all, as you said, as parts must be already on the way if your maintenance system performs properly. Anyway, even if gaskets are cheap you still have to decide how much the reorder point will be.

Next weekend “I will reveal the grand result!”, using your own words, and I will explain how I did it. I hope that people here appreciate it and find it useful.

PS: Jaz has the right answer already...

Rui

I hope that you all appreciate the case that I attach (an Excel file and a PDF document) and find it usefull.

Regards,

Rui

This message has been edited. Last edited by: <Rui Assis>,

Gaskets_of_3_pumps.xls (124 Kb, 21 downloads)

It seems that I have to post twice in order to attach two documents.

Anyway, here is the PDF document.

Regards,

Rui

This message has been edited. Last edited by: <Rui Assis>,

Gaskets_of_3_pumps.pdf (23 Kb, 17 downloads)

I attach one more Excel file on the Weibull distribution that might interest you. It returns in the last 3 columns:

- The expected mean life of a part (column I) that fails until a certain point in time (mission in column C) if it is operating since it was new;
- The conditional probability of failure of a part (column J) if it is operating after an idle period (cell N3), that is, the part already aged and is embarking on a new mission (column C);
- The expected mean life of a part (column K) that fails until a certain point in time (mission in column C) if it is operating after an idle period (cell N3) and is embarking on a new mission (column C).

Notice that if you enter a number in cell N13 such that cell F25 returns a value approximately equal to 1, you get cell I25 = cell H27 = MTTF.

Notice also that if cell N3 is made equal to 0, you get column J = column F and column K = column I.

Regards,

Rui

This message has been edited. Last edited by: <Rui Assis>,

Weibull_distribution.xls (74 Kb, 14 downloads)

Rui,

You certainly have worked hard to answer the question. Thanks for all that work.

It seems that the quantatative approach has predicted a much greater number of gaskets than the rest of us. Here is the comparison.

Josh - 1 - 3 sets
Jaz - 4 gaskets (I read this as 2 sets)
Phill - 1- 3 sets
Rui - minimum of 4 sets held - not sure of the maximum.

The large number predicted by the quantative analysis surprised me and I think I know why - the gaskets are more unreliable in this example than would be expected (or tolerated) in real life!

Rui, I may need your help here. In the spreadsheet it seems that the probability of failure before the next PM for each pump is approx 20%. So, if pump A has its gaskets replaced every 3 weeks (500 hours) then we can expect a failure to occur, on average, every fifth cycle (20%) - is that correct? This means that despite our PM we experince a failure, on average, every 15 weeks. That is more than 3 times per year. To me this says that there is something wrong with the PM approach and/or pump design.

What does everyone else think - is that failure rate acceptable?

Rui, can you confirm that my logic is correct? I know that you were only constructing an example so may not have realized the implication of the numbers.

Cheers

Also "brute force" calculation, no weibull either.

Assumptions: the normal replenishment for PM is handled by the stores, thus I only need to give minimum stock (Wished it was that easy in real life )

minimum = 2 sets, read as 4 each.

Motivation: 3 pumps in operation (client wants to operate pump his way)
1 set can be screwed up, so you need another quick.

In our case vendor is not near, (lead time minimum = 2 months) unless you are willing to pay premium rates.

Pump gaskets are not found in every hardware store, especially not API ones.
Vendor wants to sell, it easier to buy 6 than 1