Problem Of The Month

February 2001—Spare Equipment


Standard concerns about spare equipment are:

1.     When should I spare equipment? 

2.     How do I decide if I need a spare? 

3.     Why should I add spare equipment?

4.     How do I justify spare equipment?

Simple questions. 


Accurate and straightforward answers for the question are also easy:

1.     Add spare equipment for humanitarian reasons involving life/limb and when it’s the least cost alternative over the long term.

2.     The datum for sparing decision is the cost for NOT sparing

3.     Alternatives for adding spares must be described in terms of net present value (NPV).

4.     Add spare equipment when it’s cost effective, or standards/codes require it.

5.     Justify spare equipment for achieving the lowest long term cost of ownership which requires a NPV statement

You frequently make sparing decisions one piece of equipment at a time after you have consider the failures that will occur from a simple series configuration and their long term cost effects.  The spare equipment problem is simply a money problem.


Seldom can you plan for highly repeatable times for failures to occur—failures will occur when the physics of the failure dictate and not when you dictate the failure should occur!  See the January 2001 problem of the month for a Monte Carlo simulation, which illustrate the problem in a graphical manner.


Sparing decisions for complicated systems often requires building a block diagram using failure data from your records, failure details from your vendors or consultants, and cost data for equipment along with the gross margin losses for the business when failures occur plus other failure cost details.  This requires you to consider grades of equipment, grades of installation/use of the equipment, and maintenance strategies for preserving the inherent reliability of the equipment. 


Start considerations for sparing with a simple reliability block diagram.  Use no-cost/low-cost software such as RAPTOR to mechanize the process-- particularly for the first cuts at finding the magnitude of the problem.  Save the more complicated reliability modeling software packages such as SPAR for making the final complicated decisions after the rough cuts are completed with RAPTOR--SPAR will help with careful tailoring of the problem.


Consider a very simple spare equipment problem using exponential failure distributions and exponential repair distributions (yes, log-normal repair distribution are more representative of real life repair times and Weibull distributions better represent ages to failure with different modes of failure).

1.     We use one piece of equipment, which must function continuously for one year and routine maintenance occurs without stopping the equipment.

2.     If the equipment fails, we loose production at the rate of $1000/hr for gross margin loss from a plant that is sold out.

3.     The equipment usually requires 8 hours for repairs and the repair cost is $1800 per failure incident.

4.     The equipment has a mean time to failure of 4000 hours and cost $50,000 for the capital purchase and installation of spare equipment.

5.     To spare or not to spare—that is the question.  (First we start with arithmetic and then we consider NPV calculations.)

Find the datum, which is the cost of doing nothing, with the single piece of equipment for which the capital is sunk cost.


How many times a year will the equipment fail?

·       A one year mission time contains (24hrs/day)*(365days/year) = 8760 hours/year

·       The number of failures to expect on the average is (8760hrs/yr)/(4000 hrs/failure) = 2.19 failures/year

·       The time lost from outages is (8hours/failure)*(2.19failures/year) = 17.52 hours/year

·       Lost gross margin is ($1000/hr)*(17.52hour/year) = $17,520/year

·       Maintenance cost for the outage is ($1800/failure)*(2.19failures/year) = $3942/year

·       Total failure cost is $17,520/yr + $3,942/yr = $21,462/year

This is the datum for making sparing decisions.  The cost of doing nothing has a cost of $21,462/year for capital that is already sunk.


For ultra simplicity, assume a spare piece of equipment will prevent all failures (yes, we could calculate the real numbers) from occurring in the near future.

·       The capital cost of the spare equipment is $50,000 installed.

·       All failures of the system are cancelled by installation of the spare and we avoid $21,462/year of extra cost.

·       Would you spend $50,000 to save $21,462/year of losses?—this is a payback of $50,000/($21,462/year) = 2.3 years,

·       For small capital expenditures, the rule-of-thumb is 2-year paybacks are great and 3-year paybacks are marginally acceptable—therefore spend the money for the spare if cash is available.

Now consider the time value of money for this situation.

·       The discount rate is 12%

·       Tax provisions are 38% of profit before tax

·       The project life is 10 years

·       The solo equipment requires not capital equipment and incurs costs of $21,462/year for each of the ten years

·       The net present value for the do nothing case is ($75,184) using Excel spreadsheets from the Life Cycle Cost training session.

·       The dual equipment requires an expenditure of $50,000 in year zero and incurs no losses for each of ten years for the ultra simple case

·       The net present value for the spare is ($39,265).

·       Therefore select the spared equipment as having the LEAST NEGATIVE NPV on a one-to-one comparison.

·       As an alternate, merge the two cases together so the spreadsheet shows a $50,000 capital cost in year zero plus $21,462 savings/year and the NPV = $35,919 with a internal rate of return = 14.1%.  (Notice the delta between the earlier NPV’s is ?—such a strange coincidence!).  This is the data you need to sell your project.  By the way, if you want an explanation for IRR, check out the definition in your Excel help file.

·       Layout your Excel spreadsheet so your accounts can validate each detail (avoid use of the really neat Excel functions so everyone understands what’s happening to build trust and rapport)

Sparing equipment is all about financial decisions and alternatives using failure rates for equipment and reliability calculations for how systems work. The reason you use these ultra-simple examples is to get a quick go/not-go look at the problem so you don’t waste time working on a problem that is not a winner.  You can always flesh-out the ultra-simple problem with more details to obtain an accurate number for financial results.


By the way, if the plant for the solo equipment is NOT sold out, gross margin is not lost with each failure and the decisions are far different than shown above.  For the case of no gross margin losses, you cannot afford a spare equipment to protect against losses.  This is the reason equipment in the first quartile plants (the big guys are always sold out and need spare equipment to keep their cost low) follow a different strategy than equipment decisions in fourth quartile plants (the inferior guys are rarely sold out and cannot afford spare equipment).


Refer to the caveats on the Problem Of The Month Page about the limitations of the following solution. 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 top of page by clicking here.

Technical tools are only interesting toys for engineers until results are converted into a business solution involving money and time. Complete your analysis with a bottom line which converts $'s and time so you have answers that will interest your management team!

Last revised 1/26/2001
© Barringer & Associates, Inc. 2001

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