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MTBF MADE EASY

MTBF MADE EASY. BUT FIRST - A SHORT REVIEW RELIABILITY DEFINED MTBF AND HOW IT RELATES TO LAMBDA AND FIT SOME SYSTEMS WILL HAVE HIGHER FAILURE RATES ESTIMATING THE FIRST FAILURE IMPACT ON WARRANTY COST KEEPING IT SIMPLE WITH PARALLEL RESISTORS. BUT FIRST A REVIEW IN A FAMILIAR AREA.

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MTBF MADE EASY

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  1. MTBF MADE EASY BUT FIRST - A SHORT REVIEW RELIABILITY DEFINED MTBF AND HOW IT RELATES TO LAMBDA AND FIT SOME SYSTEMS WILL HAVE HIGHER FAILURE RATES ESTIMATING THE FIRST FAILURE IMPACT ON WARRANTY COST KEEPING IT SIMPLE WITH PARALLEL RESISTORS

  2. BUT FIRST A REVIEW IN A FAMILIAR AREA

  3. e(t) = Ef – [Ef – Ei] -t/RC (1)e(t) is the instantaneous voltage at time t.Ef is the final voltage that the capacitor will eventually discharge to.Ei is the initial voltage that the capacitor is charged to. is the base (2.71)t is the instantaneous time.R is the resistor value in ohms.C is the capacitance value in farads.

  4. Reliability Defined: The probability that a product will operate satisfactorily for a required amount of time under stated conditions to perform the function for which it was designed.R(t) = -t (2)is the constant failure rate (in failures per million hours).Examples:An Acme power supply has an MTBF of five million hours at 30C.A CAT Scanner has an MTBF of 1,200 hours at 25C.A Sony VCR has an MTBF of a Bazillion hours in the home. So when it breaks you feel you got your money’s worth and go out and buy another Sony VCR.

  5.  = 1/MTBF (4)MTBF is Mean Time Between FailuresSo (2) becomes:R(t) = N-(t)/(MTBF) (5)Example: Let, t = 1,000 hoursand let MTBF = 1,000 hoursN = 100 new VCRs or TV sets, or ATE systemsR(t) = 100 * -(1000hrs)/(1000Hrs between failures) = 100 *2.71-1000/1000 = 100 * 2.71-1 = 100 * .37 R(t) = 37 Units “STILL WORKING WITHOUT A FAILURE”This also means that 63 units had failures. But how many failures were there?

  6. With say a 1000 hour MTBF there will be 1/1000 or .001 lambda per million failure rate. This means that every 1000 hours a system will have a failure. With 100 systems then there will be 100 failures in those 1000 hours. But these failures will show up in only 63 units; the other 37 units will exhibit no failures during this time period.This must mean that of the 63 units that had failures, some had more than one failure.Make note that there were 63 units that exhibited 100 total failures in one MTBF period, and there were 37 units that went through this period without a failure.So how many of the 63 units had 1, 2, 3, . . . . . . n failures?P(n) = [(n * tn)/n!] * -t(6)Where:P(n) is the percent of units exhibiting n failures.t is the time durationn is the number of failures in a single system, (e.g. 1, 2, 3, . . . . . n)Let’s learn how many units will have 1, then 2, then 3, etc., failures per unit in the group of 63 units that will exhibit these 100 failures.For zero failures: (Note: 0! Is defined as equaling 1.)

  7. P(0) = [(0.0010 * 10000) / 1] * 2.71-(.001 * 1000) = [ (1 * 1)] / 1 * 2.71-1 = 1 * .37P(0) = .37 or 37%So with 100 units there will be 37 units exhibiting zero failures in one MTBF time period.P(1) = [(0.0011 * 10001) / 1]* 2.71-1P(1) = 1 / 1 * .37P(1) = .37, or 37% will exhibit one failure.So with 100 units there will be 37 units exhibiting one failure in one MTBF time period.P(2) = [(0.0012 * 10002)] / 2 * .37P(2) = (1 / 2) * .37P(2) = 18%So with 100 units there will be 18 or 19 units exhibiting two failures in one MTBF time period.

  8. P(3) = 6 units exhibiting 3 failures in one MTBF.P(4) = 1 or 2 units exhibiting 4 failures in one MTBF.P(5) = maybe 1 unit exhibiting 5 failures in one MTBF.A more simple way of finding the percentage of failures encountered in some time period is:P(f) = t (7)Find how many will fail in one one-hundredth of an MTBF time period (or in 10 hours).P(f) = 0.001 * 10P(f) = 0.01 or 1%Using 100 units this means that 1 unit exhibits the very first failure in 10 hours. So the time to first failure is 10 hours!!!!!Which one out of the 100 units is a mystery, however. . . . . . . ?MOVING TO THE WARRANTY ISSUE:

  9. Let a system have an MTBF of 4,000 hours.An engineer makes a recommendation for a hardware change that costs $400.00 per unit to install, and this raises the system’s MTBF to 6,500 hours.Given: The average cost of a field failure is $ 1,800.00/failure.The system operates 16 hours/day, 6 days per week.This roughly equals 16hrs * 6days * 52weeks/yr = 5,000 hours/yearThe warranty period is one year.So the number of failures = t = (1/4000)(5000) = 125%/year, or 125 annual failures per 100 systems shipped. (Assume the 100 units were all shipped at the same time.)At $1,800.00/failure, then $1800 * 125 = $225,000, annual warranty cost.If the MTBF is increased to 6,500 hours the number of failures= t = 5000/6500 = 77% of the 100 units shipped, or 77 failures.At $1,800.00/failure then $1800 * 77 = $138,600 annual warranty cost.

  10. The net savings is: Original Warranty Cost = $ 225,000.New Warranty Cost = - 138,600. Improvement Cost = - 40,000.Savings = $ 46,400.The Return on the Investment is:$46,400/$40,000 = 1.16The good news is that as the MTBF increases, the number of failures that occur per system, decreases. The following table shows how many units had 1, 2, 3, etc. failures in both MTBF cases.Number of Failures If the MTBF = 4000 hours If the MTBF = 6500 hours 0 28 to 29 46 to 47 1 35 to 36 35 to 36 2 22 to 23 13 to 14 3 9 to 10 3 to 4 4 2 to 3 0 to 1Range of TotalNumber of Failures = 114 to 124 70 to 80

  11. The following will demonstrate that the math is analogous between resistors and the MTBF of assemblies operating in parallel, or as a system.With 2 parallel resistors:Rt = 1/(1/R1 + 1/R2)Rt = 1/[R2/(R1*R2) + R1/(R1*R2)]Rt = 1/[(R1 + R2)/(R1*R2)Rt = (R1*R2)/(R1 + R2)For identical parallel resistors:Rt = R/n (8)Where n is the number of resistors in parallel.For larger parallel arrays of mixed value resistors:Rt = 1/(1/R1 + 1/R2 + 1/R3 + . . . . . +1/Rn)

  12. Given the above, we can look at the MTBF of assemblies operating in parallel(as a system) to be:MTBF = 1/(1/MTBF1 + 1/MTBF2 + 1/MTBF3 + . . . . +1/MTBFn) (9)Or more clearly, MTBF = 1/ [(1/MTBF1)+ (1/MTBF2)+ (1/MTBF3)+ (1/MTBF4)+ (1/MTBF5)]Note: When a system consists of subassemblies, each subassembly must have an MTBF significantly greater than the desired MTBF of the end system. If the goal MTBF of the system is specified, the MTBF of each subassembly must be allocated so that the combination of the subassemblies meets the desired MTBF of the whole system.

  13. Example: A typical allocation might appear as follows:Required MTBF = 4,000 hoursThen, given the following subassembly MTBFs:Subassembly 1 MTBF = 20,000 hoursSubassembly 2 MTBF = 14,000 hoursSubassembly 3 MTBF = 33,000 hoursSubassembly 4 MTBF = 24,000 hoursSubassembly 5 MTBF = 18,000 hoursThe system MTBF is derived using the relationship:MTBF Total = 1/ [(1/MTBF1)+ (1/MTBF2)+ (1/MTBF3)+ (1/MTBF4)+ (1/MTBF5)]

  14. Or looking at Lambdas and conductances in resistors . . . . MTBF Total = 1/ [ 1 + 2 + 3 + 4 + 5 ]MTBF Total = 1/ 0.000041 + 0.000071 + 0.000030 + 0.000042 + 0.000056] = 1/0.000240orMTBF Total = 4,167hrs.The system, consisting of these five subassemblies, has a systemMTBF of 4,167 hours.This is an acceptable figure because this figure exceeds the specified 4,000 hours.

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