Find Annual Costs Using
A typical preventative maintenance strategy is to remove functioning
components from service before they fail.
This is done on a planed replacement strategy at an optimum replacement
interval. The purpose of a planned
replacement strategy is to avoid the high cost of a failure in service. A planned replacement strategy is motivated
1) a wear out failure mode—this means a Weibull shape factor, b>1 (infant mortality b<1 and chance failures b=1 will not result in an optimum replacement interval),
2) the cost of unplanned failure which is much greater than the small cost of a planned replacement.
This is a two-out-of-two requirement for a planned replacement. Failure to meet both requirements results in a run-to-failure strategy (this does not imply running the failure into the ground!!). Please note, if the planned replacement stops a sold out process, the cost for a planned replacement may approximately equal the cost of an unplanned failure. Thus it is very important to identify the real costs incurred.
One method for considering alternatives such as optimum replacements or run to failure strategies is to computer the life cycle cost (LCC). LCC analysis requires expenditures for each year of the project for calculating net present value (NPV). This includes costs of component failures, cost consequences of the failure, and costs of replacements all put into the context of money issues over a project time span. In many cases, the calculated NPV will be negative and thus selection of the least negative NPV from a series of alternatives results in the most economical decision.
The examples below described the life of a single component viewed through
the eyes of many occurrences of this life through the use of
A Weibull database is helpful for selecting:
• shape factor, beta, (b tells about failure modes) and
• characteristic life eta (h tells about durability).
The Weibull statistics tell how components fail. Failure costs follow component failures. You need yearly failure costs to produce NPV details.
Sometimes is it more cost effective to replace equipment before it fails rather than running to failure. Of course, sometimes planned replacement strategies are very cost ineffective for achieving the lowest long term cost of ownership. You need cost details and failure details to help make wise cost decisions when considering the cost of money for financial decisions during a project life time.
You can download an Excel worksheet for computing NPV for making LCC calculations. Also an Excel replacement Monte Carlo simulation is also available as a download for producing the graphs shown below. The simulation uses Weibull failure data, failure/replacement costs, and replacement intervals to find LCC for each annual period of the project for calculating a financial term called net present value (NPV).
Failures in early years of a project are highly priced in NPV terms. Failures in later years are lower priced because of discounted values. All the following examples use a discount rate = 12%, project time span of 20 years, and a 38% tax provision for calculating the NPVs. Out-of-pocket expenditures refer to the first 20 years which is a typical project life. Please note if you work for the Government, your tax rate is zero.
Timed replacements are described by Glasser’s paper (Glasser, Gerald J.,“Planned Replacement: Some Theory and its Application”, Journal of Quality Technology, Vol. 1, No. 2, April 1969, Pages 110-119). This article is available from ASQ. If you’re a member of ASQ, download the article as a PDF file directly from the ASQ website at no cost.) Dr. Glasser makes these points in his article:
1. Preventive maintenance refers to planned replacement of pieces of equipment that may fail in operation unless a replacement is made in time. The timed replacement of functioning components means you have thrown away unused and unconsumed component life!
2. You never know precisely when a part will fail. This lack of precision involves considerable uncertainty. If the unused life is expensive compared to the cost of an unplanned failure, then the throw-away of unused life may be very expensive. Throw-away of unused expensive life can be a disadvantage.
For finding the optimum replacement interval, the risk
of failure must increase with time.
• Increasing failure rates describes a wear out failure mode with increasing
failure rates (this is a Weibull b>1). Timed replacements of unfailed parts may be a cost advantage.
• Constant failure rates described chance failures (this is a Weibull b=1). Timed replacements are
a disadvantage. An old part is a good as a new part.
• Decreasing failure rates describe infant mortality (this is a Weibull b<1). Timed replacements
are a disadvantage. An older part (with it’s lower failure rate) is better
than a new part (with it’s higher failure rate).
Glasser says you can only get two shapes to the replacement cost curve:
a) A run to failure curve where cost per unit of time continuously decline, or
b) A declining cost curve with time which reaches a minimum cost and then cost begin to increase with time beyond the minimum.
4. Replacements can not be made too frequently or operating costs outweigh gains from prevention of failures.
Glasser’s optimum replacement equation is driven by
Weibull distributions for life:
Rw(t) is the cost per unit of time
Cp is the planned replacement cost in $’s
Cup is the unplanned replacement cost which is greater than CP$ replacement cost
is the Weibull reliability equation
1- is the Weibull unreliability equation or cumulative distribution function (CDF)
is the mean time to failure up to time t
Glasser’s optimum replacement equation is . The numerator has units of currency. The denominator has units of time. Thus Rw(t) has units of currency/time.
Here is the way the data looks for one specific
Figure 1 (a) gives the equation.
Figure 1 (b) gives the cost elements and the numerator values.
Figure 1 (c) has values for the denominator plus the long term mean time to failure for the first failure (MTTF) give by h*G(1+1/b) where G represents the gamma function. [The gamma function can be evaluated in Excel as h*(exp(gammaln(1+1/b))).] Of course the many replacements will result in the mean time between failures (MTBF).
Figure 1 (c) represents the results of Rw(t) which is given in $/year for this case.
7. The cost ration Cup/Cp is very important!! Cup/Cp ~1 motivates no optimum replacements. Cup/Cp ~10+ motivates significantly important optimum replacements. Small cost ratios of Cup/Cp ~2 to 4 may motivate optimum replacements as a technical solution but the technical solution may be a costly business decision which could better be handled as a fix-when-broken solution rather than by timing replacements. Watch out for the engineering trap of having the correct technical decision but the wrong business decision!!!!
An obvious optimum replacement interval in Figure 1 (d) clearly occurs at 8.1 years where the shape of the cost curve in reaches a minimum cost value. At 8.1 years, 41.2% of the population will fail (and 58.8% will survive this age) which says you will need to make a failure replacement before reaching the optimum replacement interval. This optimum replacement is a technical solution. The technical solution does not involve the time value of money. The technical solution may not be the best business solution involving NPV. In the business world, it’s the money that’s important not the technical answer!!!
In Figure 1 (d), costs to the left of the minimum at 8.1 years are known as the high cost of early replacements. Costs to the right of the minimum in Figure 1 (d) are known as the high cost of delayed maintenance. Costs beyond the characteristic life h become non-valid as Glasser’s equation does not take into account the second round of replacements—so take the cost values at times larger than h with a grain of salt.
Planned replacement intervals impose significantly more administrative record keeping for aging “time clocks” needed for each replacement. Furthermore when a failure occurs before the time replacement interval a repair must be undertaken, and a new time clock must be reset to take into account a new age to failure interval or a new replacement interval. Also note that optimum replacement does not mean no other failures will occur before the replacement, but it does mean the failures have already been priced out. To find the percentage of the population that will fail before the optimum replacement curve go to the Weibull cumulative distribution curve and either calculate the value or read it from the Weibull graph. The percent of items that will fail are calculated by: where F(t) is the fraction that will fail by time t.
Do conditions exist for violating the business decision (based on NPV) and performing earlier timed replacements?—the answer is yes. Regulatory restriction may require compliance which out trumps economics. When common sense dictates, such as for safety issues, (which also implies the price of failures has not been priced correctly). Other similar examples exist.
The figures shown below were obtained from the Monte Carlo simulation. Except as noted, the planned replacement cost is $5,000 and the unplanned cost is $10,000. For the NPV, the cost of money is 12%, the tax rate is 38%, and the project life is 20 years. Where do you get the optimum replacement values? Either you solve Glasser’s equation as shown in Figure 1(a) or you use the optimum replacement equations in WinSMITH Weibull.
Figure 2 for timed replacements at 7.8 years where 34.2% of the population will have failed (and 65.8% will survive the 8.1 years) produces a NPV = -$3,271 with a 20 year out-of-pocket expenditure of $15,284. The sum consists of $7,021 for replacements and $8,263 for repairs. Remember the timed replacement strategy causes a large expenditure early in the project life where NPV costs are high.
Compare these replacement details in Figure 2 to the December ’04 Problem Of The Month in which Figure 1 (use your browser back button to return to this page) with a fix it when it breaks strategy had a NPV of -$2,992 with a 20 year out-of-pocket expenditure of $17,698. This says the technical solution of an optimum replacement is not a business winner. The no-replacement strategy produces the most favorable NPV (i.e., select the least negative NPV as the most desirable conditions) by ~12%.
Now for the case in Figure 1, we’re into the confusing situation for most
1) A timed replacement cost less out-of-pocket over 20 years than the fix when broken strategy.
2) Timed replacement have larger expenditures early in the project life than the fix when broken strategy.
3) The timed replacement strategy is a bad business decision as the NPV is inferior.
4) In business, it’s all about the money!! It’s not about elegant technical solutions!
Engineers must think like MBA’s while acting like engineers to solve problems in the best interest of the business. Money is time and time is money.
Figure 2 shows the rhythms of renewal (renewals are the spikes in Figure 2). You can also see the slow increase in repairs with time which plummets after occurrences of renewals. Cost details are shown in Table 1 which is derived from the simulation.
The simulation also produces failures requiring repairs and replacement
details as shown in Table 2. Notice how
the replacements leap upward with the planned replacement cycles as outline by
the double bordered boxes.
Notice from Table 2 that a full set of replacements does not occur until year 16. This is because failures occurred before/after the replacement intervals which restart the time to replacement (that means you should expect record keeping problems with timed replacements).
Table 2 also shows 0.8263 failures have occurred during the 20 year project interval which makes the system MTBF = 20years/0.8263failures = 24.2 years/failure.
You quickly see that timed replacements carry much overhead for keeping up with events although you get gains in reliability (as measured by the MTBF). However, you’ve got to ask the business question--is the value received worth the extra cost paid—without mitigating factors not described in this example, timed replacements do not pass the business test criteria for this case.
Figure 3 shows a different Weibull shape factor, b = 1.8 with h = 10 years, with it’s appropriate optimum replacement interval of 12 years (where 75% of the population will have failed or 25% will survive the 12 year interval—remember the 12 year interval is beyond the eta value and thus beyond the accuracy limit for Glasser’s validity interval) for a planned replacement cost of $5,000 and unplanned cost of $10,000. The replacement strategy produces a NPV = -$4,219 for the 20 year project life based on an out-of-pocket expenditure of $19,185. Compare this to the fix-when-broken strategy of the December ’04 Figure 2 which had an NPV = -$3,680 for the 20 year project life based on an out-of-pocket expenditure of $19,157. Once again we have a fix-when-broken strategy that results in a better business strategy than the technical answer for optimum replacement.
Table 3 shows the annual costs derived from the simulation.
Table 4 shows the annual failures and replacements.
The bottom line for Figure 3 is simple. The out-of-pocket expenditures are nearly the same for the two strategies of fix-when-broken or planned replacement. However the planned replacements come early in the project life which contributes to an undesirable NPV = -$4,219. The clear business winner is the fix-when-broken strategy with it’s NPV = -$3,680.
Figure 4 shows a different Weibull shape factor, b = 1 with h = 10 years. No optimum replacement interval exists for this beta value. An arbitrary replacement interval of 9 years (where 59.3% of the population will have failed or 40.7% will survive the 9 year interval) for a planned replacement cost of $5,000 and unplanned cost of $10,000.
Figure 4’s arbitrary replacement strategy produces a NPV = -$6,012 for the 20 year project life based on an out-of-pocket expenditure of $25,270. Compare this to the fix-when-broken strategy of the December ’04 Figure 3 which had an NPV = -$4,635 for the 20 year project life based on an out-of-pocket expenditure of $20,035. Once again we have a fix-when-broken strategy that results in a better business strategy than the arbitrary replacement interval where no optimum replacement interval exists. The clear business winner is a fix-when-broken strategy.
Table 5 shows the annual costs derived from the simulation with an arbitrary replacement interval of 9 years.
Table 6 shows the annual failures and replacements.
Figure 5 shows a different Weibull shape factor, b = 0.7 with h = 10 years. No optimum replacement interval exists for this beta value. An arbitrary replacement interval of 9 years (where 60.5% of the population will have failed or 39.5% will survive the 9 year interval) for a planned replacement cost of $5,000 and unplanned cost of $10,000.
Figure 5’s arbitrary replacement strategy produces a NPV = -$7,276 for the 20 year project life based on an out-of-pocket expenditure of $29,139. Compare this to the fix-when-broken strategy of the December ’04 Figure 4 which had an NPV = -$5,374 for the 20 year project life based on an out-of-pocket expenditure of $20,386. Once again we have a fix-when-broken strategy that results in a better business strategy than the arbitrary replacement interval where no optimum replacement interval exists. The clear business winner is a fix-when-broken strategy.
Table 7 shows the annual costs derived from the simulation with an arbitrary replacement interval of 9 years.
Table 8 shows the annual failures and replacements.
In summary, the results of replacement strategies need to be based on NPV decisions rather than simply technical decisions. Often fix-when-broken strategies are simplest to administrate and economically the correct financial decision to make in face of the technical decision for optimum replacement. Arbitrary replacements, in disregard of the lack of technical motivation, often carry a large financial burden. In short, engineers must think like MBA while acting as engineers to achieve the best financial results.
Table 9 shows the out-of-pocket costs for 20 years of continuous service and net present values for several different conditions. In each case the durability, eta, is constant.
Table 9 shows two cases for beta=3.5 years, eta=10 years. One case is for planned renewal, and the
other case is run-to-failure (to do this, simply set the planned replacement
period to 50 years which will put the even outside of the time span for the NPV
spreadsheet). Consider what happens to
the MTBF from the Excel
Monte Carlo simulation which is shown in Figure 6.
• For unplanned replacements, the MTBF approaches the Weibull MTBF, see Figure 6, panel a).
• For optimum replacements, the MTBF exceeds the Weibull MTBF, see Figure 6, panel b).
• For earlier than planned replacements, the MTBF exceeds the Weibull MTBF, see Figure 6, panel c).
Compare the NPV results in Table 9 with the decision to run the components to failure (to set the simulation for run to failure, simply change the planned replacement period to say 50 years which will be outside of the NPV time span).
Figure 6 shows visual results of cost vs time and MTTB vs time. The graph on the left hand side shows annual costs for out-of-pocket expenditures. The graph on the right hand side shows MTBF vs time.
Figure 6: Summary Of MTBF Without and With Renewal At
Different Intervals Given:
a) No renewal, fix when broken with NPV=($3,483), Out-of-pocket expenditures = $17,698
b) Renew at 7.8 year interval with NPV=($3,271), Out-of-pocket expenditures = $15,284
c) Renew early at a 5 year interval with NPV=($4,429), Out-of-pocket expenditures = $21,711
Calculating the MTBF for the early years of Figure 6 is straight forward up to the point when failures start to kick in around the mean life where replacement components are required.
Table 10: Hand Calculated MTBF Up To When
Replacements Are Required,
In Table 10, the Weibull CDF= 1- e-(t/h)^b . For
the first year with one component, it tells the number of failures expected if
one part is exposed to service. The MTBF
for the first year is elapsed time/failures or 1/0.000316178 = 3163
years/failure for year one. For year two
and beyond, take the current year CDF and subtract the previous year’s CDF to
get an elapsed interval which for year two is 1/(0.0035713-0.000316178) =
1/0.003255122 = 307 years/failure for year two.
Of course this simple computation shows similar results to the
You can find other LCC resources on this website:
Return to the list of problems by clicking here.
Refer to the caveats on the Problem Of The Month Page about the limitations of the solution above. Maybe you have a better idea on how to solve the problem. Maybe you find where I've screwed-up the solution and you can point out my errors as you check my calculations. E-mail your comments, criticism, and corrections to: Paul Barringer by clicking here. Return to the top of this problem.
You can download a PDF copy of this Problem Of The Month by clicking here. By the way, the tables are readable in the PDF file.