research article renewal and renewal-intensity functions with minimal repair · 2019. 7. 31. ·...

Research ArticleRenewal and Renewal-Intensity Functions with Minimal Repair

Saeed Maghsoodloo and Dilcu Helvaci

Department of Industrial and Systems Engineering Auburn University Auburn AL 36849 USA

Correspondence should be addressed to Saeed Maghsoodloo maghsoodengauburnedu

Received 27 October 2013 Revised 2 January 2014 Accepted 16 January 2014 Published 19 March 2014

Academic Editor Elio Chiodo

Copyright copy 2014 S Maghsoodloo and D Helvaci This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

The renewal and renewal-intensity functions with minimal repair are explored for the Normal Gamma Uniform and Weibullunderlying lifetime distributions The Normal Gamma and Uniform renewal and renewal-intensity functions are derived bythe convolution method Unlike these last three failure distributions the Weibull except at shape 120573 = 1 does not have a closed-form function for the n-fold convolution Since the Weibull is the most important failure distribution in reliability analyses theapproximate renewal and renewal-intensity functions of Weibull were obtained by the time-discretizing method using theMean-Value Theorem for Integrals A Matlab program outputs all reliability and renewal measures

1 Introduction

Renewal (RNL) functions give the expected number of fail-ures of a system (or a component) during a time interval andthis is used to determine the optimal preventive maintenanceschedule of a system [1] Renewal functions (RNFs) haveparticular importance in analysis of warranty [2ndash6] Theyhave wide variety of applications in decision making such asinventory theory [7] supply chain planning [8 9] continuoussampling plans [10 11] insurance application and sequentialanalysis [2 8 12 13]

Since RNFs play an important role in many applicationsit is important to obtain them analytically if possible Basedon analytical approach the RNF 119872(119905) is the inverse ofits Laplace transform 119872(119904) where Laplace transforms willbe defined later Blischke and Murthy [6] state that ldquotheadvantage of analytical method is that one can carry outparametric studies of the RNF that is the behavior of119872(119905) asa function of the parameters of the distributionrdquoHowever formost distribution functions obtaining the RNF analytically iscomplicated and even impossible [8]Therefore developmentof computational techniques and approximations for RNFshas attracted researchers [14]

One of thewell-known approximations given by Tacklind[15] is119872(119905) ≃ 1199051205831015840

1+120583101584022120583101584021minus1 where 1205831015840

1and 12058310158401are the first

and second raw moments which is generally known as theasymptotic approximation and is also cited in numerouspapers such as Smith [16] The asymptotic approximationhas a closed-form expression and thus it is easy to applyto optimization problems that involve a RNL process [8]However since the asymptotic approximation is not accuratefor small values of 119905 Parsa and Jin [8] propose betterapproximation by keeping the positive features of asymptoticapproximations such as simplicity closed-form expressionand independence from higher moments of an underlyingdistribution Jiang [17] proposes an approximation for theRNF with an increasing failure rate (IFR) which is also usefulin areas such as optimization where a RNF needs to beevaluated

There are series methods available in the literature toapproximate RNFs such as Smith and Leadbetter [18] whodeveloped a method to compute the RNF for the Weibullby using power series expansion of 119905120573 where 120573 is the shapeparameter of theWeibull On the other hand instead of usingpower series expansion Lomnicki [19] proposes anothermethod by using the infinite series of appropriate Poissonianfunctions of 119905120573 There are also many other approximationsavailable such as Xie [20] Smeitink and Dekker [21] Baxteret al [22] Garg and Kalagnanam [23] and From [24] Thereare usually three criteria model simplicity applicability

Hindawi Publishing CorporationJournal of Quality and Reliability EngineeringVolume 2014 Article ID 857437 10 pageshttpdxdoiorg1011552014857437

2 Journal of Quality and Reliability Engineering

and accuracy to evaluate the value of RNF approximation[17] Increasing the complexity may lead to more accurateapproximation but may make the process complicated anddifficult to implement in practice [25]

Furthermore in the literature lower and upper boundson RNFs have been discussed which can be used to obtainupper and lower bounds on warranty costs such as Blischkeand Murthy [6] Marshall [26] provides lower and upperlinear bounds on the RNF of an ordinary RNLprocess Ayhanet al [27] provide tight lower and upper bounds for the RNFwhich are based on Riemann-Stieljes integration There arealso many other studies conducted about bounds on RNFsuch as Barlow [28] Leadbetter [29] Ozbaykal [30] Xie [31]Ran et al [32] and Politis and Koutras [33]

Finally simulation can be considered as an alternateapproach to estimate the value of a RNF Brown et al [34] usethe Monte Carlo simulation to estimate the RNF for a RNLprocess with known interarrival time distribution

This paper proposes a convolution method to obtainthe Gamma and Normal renewal and renewal-intensityfunctions throughout we are assuming that repair time isnegligible (or minimal) relative to Time to Failure (TTF)As a result of this assumption the replacement time of afailed component in a system is minimal A further objectiveis to obtain the renewal and RNL-intensity functions of theuniform distribution by using 119899 = 2 to 119899 = 12 convolutionsand applying the normal approximation for convolutionsbeyond 12 Because the Weibull distribution except at shape120573 = 1 does not have a closed-form function for its 119899-foldconvolution the last objective is to approximate its renewaland RNL-intensity functions by discretizing time using theMean-Value Theorem for Integrals We will also highlight thedifferences between the renewal-intensity function 120588(119905) andthe hazard function ℎ(119905)

2 Preliminaries

Suppose that component (or system) failures occur at times119879119899(119899 = 1 2 3 4 ) measured from zero and assuming

that replacement (or restoration time) is negligible relativeto operational time then 119879

119899represents the operating time

(measured from zero) until the 119899th failure Because theprobability density function (pdf) of 119879

1(or 119883

1) may be

different from the intervening times 1198832= (1198792minus 1198791) 1198833=

(1198793minus1198792)1198834= (1198794minus1198793) we consider only the simpler case

of pdf of time to first failure 1198911(119905) being identical to those of

intervening times1198832 1198833 1198834 That is all119883

119894rsquos are assumed

iid (independently and identically distributed) Therefore119879119899= sum119899

119894=1119883119894= sum of the times to the first failure from

zero plus the intervening times of 2nd failure until the 119899thfailure If 119899 gt 60 then the central limit theorem (CLT) statesthat the distribution of 119879

119899approaches normality with mean

119899120583 where 120583 (the expected time between successive renewals)= 119864(119883

119894) 119894 = 1 2 3 4 and with variance of 119879

119899 denoted

by 119881(119879119899) equal to 1198991205902 However if the pdf of 119883

119894 119891(119909) with

119883 being the parent variable is highly skewed andor 119899 isnot sufficiently large then the exact pdf of 119879

119899= sum119899

119894=1119883119894is

given by the 119899-fold convolution of 119891(119909) with itself denoted

by 119891119879119899

(119905) = 119891(119899)(119905) Thus by renewal we mean either the

replacement of a failed component with a brand-new oneor the case when the failed component can be repaired in anegligibly short time to an almost brand-new condition Ithas been proven bymany authors both in stochastic processesand reliability engineering that the RNF for the duration [0 119905]is given by

119872(119905) = 119864 [119873 (119905)] =infin

sum119899=1

119865(119899)(119905) (1a)

119881 [119873 (119905)] =infin

sum119899=1

(2119899 minus 1) 119865(119899)(119905) minus [119872 (119905)]

2 (1b)

where the random variable 119873(119905) represents the number (orcount) of renewals that occur during the time interval [0 119905]and 119865

(119899)(119905) is the cumulative distribution function (cdf) of

the 119899-fold convolution of 119891(119905) with itself that is 119865(119899)(119905) =

int119905

0119891(119899)(120591)119889120591 where 119891

(119899)(119905) is the 119899-fold convolution density

of 119891(119905) It is also very well known that the RNI (renewalintensity) of a distribution function119865(119905) by definition is givenby 120588(119905) = 119889119872(119905)119889119905

Authors in stochastic processes refer to120588(119905) as the renewaldensity because 120588(119905) times Δ119905 describes the (unconditional)probability element of a renewal during the interval (119905 119905+Δ119905)[35] further explanation is forthcoming in Section 52 whilea few authors in reliability engineering such as Ebeling [36]and Leemis [37] refer to 120588(119905) as intensity function Because120588(119905) is never a pdf over the support set of 119891(119905) for all failuredistributions throughout this paper we will refer to it as therenewal intensity function (RNIF)

The simplest and most common renewal process isthe homogeneous Poisson process (HPP) where the inter-vening times are exponentially distributed at the constantinterrenewal (or failure) rate 120582 Because 120582 is a constantand intervening times are iid a Poisson process is alsoreferred to as a homogeneous renewal process It is alsovery well known that for a HPP the pdf of interarrivals119883119894rsquos is given by 119891(119905) = 120582119890minus120582119905 and that of the time to

119899th failure (or renewal) measured from zero is givenby 119891119879119899

(119905) = 119891(119899)(119905) = (120582Γ(119899))(120582119905)119899minus1119890minus120582119905 (the Gamma

density with shape 119899 and scale 120573 = 1120582) As a result theuse of (1a) for the interval [0 119905] leads to the RNF 119872(119905) =

119864[119873(119905)] = suminfin

119899=1119865(119899)(119905) = sum

infin

119899=1int119905

119909=0(120582Γ(119899))(120582119909)119899minus1119890minus120582119909119889119909 =

int119905

119909=0120582119890minus120582119909sum

infin

119899=1((120582119909)119899minus1(119899 minus 1))119889119909 = int

119905

119909=0120582119890minus120582119909

suminfin

119899=0((120582119909)119899119899) 119889119909 = int

119905

119909=0120582 119889119909 = 120582119905 a fact that has been

known for well more than a century Further the RNIFfor a HPP is also a constant and is given by 120588(119905) =119889119872(119905)119889119905 = 119889(120582119905)119889119905 = 120582 Further it is also widely knownthat for a HPP the 119881[119873(119905)] = 120582119905

3 The Renewal and RNL-Intensity Functionsfor a Normal Baseline Failure Distribution

Suppose that the time between failures 119883119894 119894 = 1 2 3 4

are NID (120583 1205902) that is normally and independently dis-tributed with mean time between failures (MTBF) = 120583 and

Journal of Quality and Reliability Engineering 3

process variance 1205902 Then statistical theory indicates thattime to the 119899th failure (measured from zero) is the 119899-foldconvolution of 119873(120583 1205902) with itself that is time-to-the-119899th-failure 119879

119899= TTF

119899sim 119873(119899120583 1198991205902) Hence in the case of

minimal repair from (1a) the RNF is given by

119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

Φ(119905 minus 119899120583

120590radic119899) (2)

where Φ universally stands for the cumulative distributionfunction (cdf) of the standardized normal deviate 119873(0 1)and 119865

(119899)(119905) gives the probability of at least 119899 renewals by

time 119905 Cui and Xie [38] gave the same exact expression forthe Normal RNF as in the above equation (2) which theyused as an approximation for theWeibull renewal with shape120573 ge 3 It should be kept in mind that the normal failurelaw is approximately applicable in reliability analyses only ifthe coefficient of variation CVT le 15 because the supportfor the normal density is (minusinfininfin) while TTF can never benegative (this assures that the size of left-tail below zero is lessthan 1 times 10minus10) If the CVT is not sufficiently small then thetruncated Normal can qualify as a failure distribution [24]discussed theRNF for the truncatedNormal Fromapracticalstandpoint the restriction on CVT can be relaxed to less than20

As an example suppose that a cutting toolrsquos TTF hasthe lifetime distribution 119873 (120583 = MTBF = 15 operatinghours 1205902 = 225 hours2) (extracted from Example 95 of [36page 225]) with minimal repair (or replacement time) whereCV119879= 010 Then (2) shows that the expected number of

renewals (or replacements) during 42 hours of use is given by119872(42) = sum

infin

119899=1Φ((42 minus 15119899)15radic119899) = 2124107 Further for

a Laplace-Gaussian nonhomogeneous process letting 119911119899=

(119905minus119899120583)120590radic119899 the RNIF by direct differentiation of (2) is givenby

120588 (119905) =119889

119889119905

infin

sum119899=1

Φ(119911119899) =infin

sum119899=1

119889

119889119911119899

Φ(119911119899)119889119911119899

119889119905

=infin

sum119899=1

120593 (119911119899)

1

120590radic119899=infin

sum119899=1

1

radic2120587119890minus1199112

1198992 1

120590radic119899

(3)

or

120588 (119905) =infin

sum119899=1

1

radic21205871198991205902119890minus[(119905minus119899120583)radic119899120590

2]2

2

(4)

where the symbol 120593 stands for the standard normal density119873(0 1) The value of RNIF at 42 hours for the 119873(15 225)baseline distribution from (4) is 120588 (at 119905 = 42 hours) =007883674 failureshour Note that the value of the hazardrate at 42 hours is given by ℎ(42) = 119891(42)119877(42) = 10358138failureshour where 119877(119905) is the reliability function at time 119905Because the Normal failure probability (Pr) law always hasan increasing failure rate (IFR) 120588(119905) lt ℎ(119905) Ross [39 pages426-427] proves that 119905120583 = the glb (greatest lower bound) le119872(119905) le 119905120583 + 050(CV2

119879minus 1) = the lub (least upper bound)

only if the hazard-function (HZF) ℎ(119905) is a decreasingfailure-rate (ie ldquothe item is improvingrdquo [37 page 165]) and

Ross further proves when ℎ(119905) is a decreasing failure rate(DFR) then 120588(119905) ge ℎ(119905) for all 119905 ge 0 and as a result 119877(119905) ge119890minus119872(119905) and wemay add that equalities can occur only at 119905 = 0

4 The Renewal Function for a GammaBaseline Failure Distribution

In order to obtain the RNF and RNIF of a Gamma non-homogeneous Poisson process (NHPP) we first resort toLaplace transformation because it is the common procedurein stochastic processes It has been proven (see [40 pages277ndash280]) in theory of stochastic processes that the Laplacetransforms (LTs) of 120588(119905) and119872(119905) are respectively given by

L 120588 (119905) = 120588 (119904) = intinfin

0

119890minus119904119905120588 (119905) 119889119905 =119891 (119904)

1 minus 119891 (119904) 119904 gt 0

(5a)

119872(119904)=L 119872 (119905)=intinfin

0

119890minus119904119905119872(119905) 119889119905=119891 (119904)

119904 [1 minus 119891 (119904)]=120588 (119904)

119904

(5b)

Further it is also widely known that the LT of the Gammadensity 119891(119905) = (120582Γ(120572))(120582119905)120572minus1119890minus120582119905 is given by 119891(119904) = 120582120572(120582+119904)120572 120588(119904) = L120588(119905) = 120582120572((120582 + 119904)120572 minus 120582120572) and the Gamma119872(119904) = L119872(119905) = 120582120572119904[(120582 + 119904)120572 minus 120582120572] However these lastclosed-form expressions for LTs are valid only if the shape 120572 isan exact positive integer because the integration by parts doesnot terminate for noninteger values of 120572 As a result whenthe shape parameter 120572 is not a positive integer there exists noclosed-form inverse-Laplace transform for the Gamma119872(119904)however when the underlying failure distribution is Erlang(ie Gamma with positive integer shape) then there exists aclosed-form inverse Laplace-transform for 120572 = 2 3 4 Infact for the specific Erlang density with shape 120572 equiv 2 and scale120573 = 1120582 it is very well known that119872(119904) = int

infin

0119890minus119904119905119872(119905)119889119905 =

12058221199042(119904 + 2120582) = minus14119904 + 12058221199042 + 14(119904 + 2120582) Upon inversionto the t-space we obtain the well-known119872(119905) = 119864[119873(119905)] =

Lminus1119872(119904) = Lminus1minus14119904 + 12058221199042 + 14(119904 + 2120582) = minus14 +(1205822)119905 + (14)119890minus2120582119905 Hence at 120572 equiv 2 the RNIF is givenby 120588(119905) = 119889119872(119905)119889119905 = 1205822 minus (1205822)119890minus2120582119905 which is quitedifferent from the corresponding Gamma (at 120572 equiv 2) IFRℎ(119905) = 120582(120582119905)119890minus120582119905sum

120572minus1

119896=0((120582119905)119896119896)119890minus120582119905 = 120582(120582119905)(1 + 120582119905) gt 120588(119905)

for 119905 gt 0We usedMatlabrsquos ilaplace function as an aid in orderto obtain the RNF and RNIF for the Erlang at shapes 120572 equiv 3and 4 which are respectively given below

119872(119905 120572 = 3) =Lminus1

1205823

119904 [(120582 + 119904)3 minus 1205823]

=120582119905

3+ 119890minus(31205822)119905 [cos(120582119905radic3

4)

+sin (120582119905radic34)

radic3] times 3minus1 minus

1

3


120588 (119905 120572 = 3) =Lminus1

1205823

[(120582 + 119904)3 minus 1205823]

=120582

31 minus 119890minus(31205822)119905 [cos(120582119905radic3

4)

+radic3 sin(120582119905radic34)]

119872 (119905 120572 = 4) =120582119905

4+119890minus2120582119905

8

+119890minus120582119905 [cos (120582119905) + sin (120582119905)]

4minus3

8

120588 (119905 120572 = 4) =Lminus1

1205824

(120582 + 119904)4 minus 1205824

=120582

4[1 minus 119890minus2120582119905 minus 2119890minus120582119905 sin (120582119905)]

(6)

The Gamma HZF at 120572 equiv 3 is given by ℎ(119905) =

[120582(120582119905)2Γ(3)](1 + 120582119905 + 120582211990522) = 120582(120582119905)2(2 + 2120582119905 + 12058221199052)which exceeds the above 120588(119905) at 120572 equiv 3 for all 119905 gt 0

We now use the cdf 119865(119899)(119905) = Pr(119879

119899le 119905) = Pr[119873(119905) ge 119899)]

see the excellent text by Ebeling [36 page 226 Equation 910]directly in order to obtain the Gamma RNF from (1a)

119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

int120582119905

0

119906119899120572minus1

Γ (119899120572)119890minus119906119889119906 =

infin

sum119899=1

119868 (120582119905 119899120572)

(7)

where 119868(120582119905 119899120572) = Matlabrsquos gammainc (120582119905 119899120572) =

(1Γ(119899120572)) int120582119905

0119906119899120572minus1119890minus119906119889119906 represents the incomplete-Gamma

function at point 120582119905 and shape 119899120572 In fact 119868(120582119905 119899120572) gives thecdf of the standard Gamma density at point 120582119905 and shape 119899120572

41 The Gamma Renewal Intensity Function Next wedirectly differentiate (7) in order to obtain the Gamma RNIFThat is

120588 (119905) =119889119872 (119905)

119889119905

=infin

sum119899=1

119891(119899)(119905)

=120597

120597119905

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120597

120597119905int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120582

Γ (119899120572)(120582119905)119899120572minus1119890minus120582119905

(8)

However the function under the summation in (8) is simplythe Gamma density with shape 119899120572 and scale 120573 = 1120582 Thenwe used our Matlab program to obtain the value of (8) at119905 = 12 years at shape 120572 = 15 and scale 120573 = 35 yearswhich yielded 120588(119905 = 12) = 0190348 renewalsyear Thevalue of the HZF at 12 years is ℎ(12) = 119891(12)119877(12) =0019361300765932 = 0252781 failuresyear In order tocheck the validity of 120588(119905 = 12) = 0190348 we resort to thelimiting form lim

119905rarrinfin120588(119905) = 1120583 = 1MTBF Because the

expected TTF of the Gamma density is 120583 = 120572 times 120573 then forthis example 120583 = MTBF = 15 times 35 = 525 years whichyields 120588 (12 years) cong 1525 = 0190476year

Because the Gamma density is an IFRmodel if and only ifthe shape 120572 gt 1 then 120588(119905) lt ℎ(119905) for 119905 gt 0 and all 120572 gt 1 Onlyat 120572 = 1 the Gamma baseline failure density reduces to theexponential with CFR the only case for which 120588(119905) equiv ℎ(119905) equiv120582 Note that since the well-known renewal-type equation forthe RNIF 120588(119905) is given by 120588(119905) = 119891(119905)+int119905

0120588(119905minus119909)119891(119909)119889119909 this

last equation clearly shows that 120588(0) = 119891(0) further ℎ(119905) =119891(119905)119877(119905) for certain yields ℎ(0) = 119891(0) and hence 120588(0) =119891(0) = ℎ(0) for all baseline failure distributions Moreover ifthe minimum life 120575 gt 0 then for certain 120588(120575) = 119891(120575) = ℎ(120575)

42 Examining Results of the Convolution Method Jin andGonigunta [25] propose the use of generalized exponen-tial functions to approximate the underlying Weibull andGamma distributions and solve RNFs using Laplace trans-forms Table 1 shows a comparison of the RNF from (7) withtheir results They refer to 119867(119905) as the actual RNF [obtainedfrom Xiersquos [20] numerical integration] and 119867

119886(119905) is their

approximated RNF Table 1 shows that their 119867(119905) becomesmore accurate as compared to119872(119905) from (7) for larger valuesof 119905

Further it is well known from statistical theory that theskewness of Gamma density is given by 120573

3= 2radic120572 and

its kurtosis is 1205734= 6120572 both of these clearly showing that

their limiting values in terms of shape 120572 is zero which arethose of the corresponding Laplace-Gaussian 119873(120572120582 1205721205822)We compared our results using our Matlab program forthe Gamma at 120572 = 70 120573 = 15 and 119905 = 5000 whichyielded119872(5000) = 4186155 (Matlab gives 15 decimal accu-racy) while the corresponding Normal yielded 119872(5000) cong4185793 expected renewals

5 The Renewal and RNI Functions When theUnderlying TTF Distribution Is Uniform

51 Renewal Function When TTF Is Uniform Suppose thatthe TTF of a component or system (such as a network) isuniformly distributed over the real interval [119886 119887weeks] then119891(119905) = 1119888 119886 ge 0 119887 gt 119886 119888 = 119887 minus 119886 gt 0 and the cdfis 119865(119905) = (119905 minus 119886)119888 119886 le 119905 lt 119887 weeks Further succeedingfailures have identical failure distributions as 119880(119886 119887) Froma practical standpoint the common value of minimum life119886 = 0 Then the fundamental renewal equation is given by119872(119905) = 119865

1(119905) + int

119905

0119872(119905 minus 120591)119891(120591)119889120591 [40 pages 277ndash280] Since

we are considering the simplercase of time to first failure


Table 1 Comparison of119867119886(119905) with119872(119905) for the Gamma renewal function

119905120573 = 15 120573 = 3 120573 = 5

119867(119905) 119867119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905)

0 0000 0000 0000 0000 0000 0000 0000 0000 00001 0517 0508 0517 0081 0065 0081 0004 0001 00042 1170 1167 1170 0340 0331 0340 0053 0042 00533 1834 1833 1834 0665 0664 0665 0186 0179 01864 2500 2500 2500 0999 1000 0999 0379 0383 03795 3167 3167 3167 1333 1333 1333 0592 0597 0592

Table 2 Normal approximation to the 12-fold convolution of 119880(0 1)

119885 05 1 15 2 25 3 35 4119865(12)

(119911) 0689422 0839273 0932553 0977724 0994421 0998993 0999879 0999991Normal approx 0691462 0841345 0933193 097725 099379 099865 0999767 0999968Rel error 0002960 0002469 0000686 minus000049 minus000063 minus000034 minus000011 minus23119864 minus 05

distribution being identical to those of succeeding times tofailure then

119872(119905) = 119865 (119905) + int119905

0

119872(119905 minus 119909) 119891 (119909) 119889119909 (9a)

whereas 119865(119905) represents the cdf of 119879 = TTF HoweverHildebrand [41] proves that 119872(119904) times 119891(119904) = Lint

119905

0119872(119905 minus

119909)119891(119909)119889119909 = Lint119905

0119872(119905 minus 119909)119889119865(119909) the integral inside

brackets representing the convolution of 119872(119905) with 119891(119905)Conversely we conclude that 119891(119904) times 119872(119904) = Lint

119905

0119865(119905 minus

119909)119889119872(119909) Upon inversion of this last LT we obtain thewidely known Lminus1119891(119904) times 119872(119904) = int

119905

0119865(119905 minus 119909)119889119872(119909) =

int119905

0119865(119905 minus 119909)120588(119909)119889(119909) Hence (9a) can also be represented as

119872(119905) = 119865 (119905) + int119905

0

119865 (119905 minus 119909) 119889119872 (119909)

= 119865 (119905) + int119905

0

119865 (119905 minus 119909) 120588 (119909) 119889119909

(9b)

The renewal equation of the type (9b) has been given bymanywell-known authors such as Xie [20]Murthy et al [42]Leemis [37] and other notables

In order to obtain the RNF for the Uniform density wesubstitute into (9a) for the specific Uniform 119880(0 119887) baselinedistribution for whichminimum life 119886 = 0 in order to obtain119872(119905) = 119905119887 + int

119905

0119872(119905 minus 119909)119889119909119887 letting 119905 minus 119909 = 120591 in this last

equation yields

119872(119905) =119905

119887+1

119887int0

119905

119872(120591) (minus119889120591) =119905

119887+1

119887int119905

0

119872(120591) 119889120591 (10)

The above equation (10) shows that the RNIF is given by120588(119905) = 119889119872(119905)119889119905 = 1119887+119872(119905)119887 Solving this last differential

equation and applying the boundary condition119872(119905 = 0) =

0 we obtain 119872(119905) = 119890119905119887 minus 1 where time must start atzero (or the last RNL) that is this last expression is validonly for 0 le 119905 le 119887 119887 gt 0 Note that Ross [39 Problem37 page 154] gives the same identical 119872(119905) only for thestandard 119880(0 1) underlying failure density whose base isequal to 1 (It is worth mentioning the fact that Ross [39page 7] uses the common notation in stochastic processesof 119865(119909) = 1 minus 119865(119909) for the reliability function at lifetime119909 the authors of this paper are indebted to Rossrsquos text thathas provided so much help and guidance in writing thispaper)

Next in order to obtain the RNF for the 119880(119886 119887) wetransform the origin from zero to minimum life = 119886 gt 0 byletting 120591 = 119886+(119887minus119886)119905119887 = 119886+119888119905119887 in the RNF119872(119905) = 119890119905119887minus1This yields119872(120591) = 119890(120591minus119886)119888 minus 1 and hence119872(119905) = 119890(119905minus119886)119888 minus 10 le 119886 le 119905 lt 119887 and 119888 = 119887 minus 119886 gt 0 is the Uniform-densitybase The corresponding RNIF is given by 120588(119905) = [119890(119905minus119886)119888]1198880 le 119886 le 119905 lt 119887

Because the Uniform renewal function119872(119905) = 119890(119905minus119886)119888 minus1is valid only for the interval [119886 119887] we will obtain the 119899-foldconvolution of the 119880(119886 119887)-density which in turn will enableus to obtain 119872(119905) for 119905 gt 119887 by making use of (1a) that usesthe infinite-sumof convolution cdfs119865

(119899)(119905) As stated byOlds

[43] the convolutions of Uniform density of equal bases 119888have been known since Laplace However there are otherarticles on this topic since 1950 The specific convolutions ofthe Uniform density with itself over the interval [minus12 12]were obtained by Maghsoodloo and Hool [44] only for 119899 =2-fold 3 4 5 6 and 8-folds There are other articles onthe Uniform convolutions such as Shiu [45] Killmann andCollani [46] and Kang et al [47] We used the procedure inMaghsoodloo and Hool [44] but re-developed each of the119899 = 2 through 119899 = 8 convolutions of 119880(119886 119887) by a geometricalmathematical statistics method (the (1983) article is available


from the authors although the 8th convolution has sometypos) Further we programmed this last geometric methodin Matlab in order to obtain the exact 9 through 12-fold

convolutions of the 119880(119886 119887) with itself For example our 7-fold convolution-density of 119880(119886 119887) with itself is given belowwhere 119888 = 119887 minus 119886 gt 0

119891(7)(119905) =

(7119886 minus 119905)6

(7201198887) 7119886 le 119905 le 6119886 + 119887

minus [6(7119886 minus 119905)6 + 42119888(7119886 minus 119905)5 + 1051198882(7119886 minus 119905)4 + 1401198883(7119886 minus 119905)3 6119886 + 119887 le 119905 le 5119886 + 2119887

+1051198884(7119886 minus 119905)2 +421198885 (7119886 minus 119905) + 71198886] times (7201198887)minus1

[15(7119886 minus 119905)6 + 210119888(7119886 minus 119905)5 + 11551198882(7119886 minus 119905)4 + 32201198883(7119886 minus 119905)3 5119886 + 2119887 le 119905 le 4119886 + 3119887

+49351198884(7119886 minus 119905)2 + 39901198885 (7119886 minus 119905) + 13371198886] times (7201198887)minus1




+ 968101198884(7119886 minus 119905)2 + 1680001198885 (7119886 minus 119905) + 1191821198886] times (7201198887)minus1



(7119887 minus 119905)6

(7201198887) 119886 + 6119887 le 119905 le 7119887

(11)

52 The Renewal-Intensity Function Approximation WhenTTF Is Uniform The Uniform RNIF is given by 120588(119905) =

[119890(119905minus119886)119888]119888 only for the interval 0 le 119886 le 119905 lt 119887 Bartholomew[35] describes 120588(119905) times Δ119905 as the (unconditional) probabilityelement of a renewal during the interval (119905 119905 + Δ119905) and inthe case of negligible repair-time 120588(119905) also represents theinstantaneous failure intensity function at 119905 However asdescribed by nearly every author in reliability engineeringthe HZF ℎ(119905) gives the conditional hazard-rate at time 119905 onlyamongst survivors of age 119905 that is ℎ(119905) times Δ119905 = Pr(119905 le 119879 le119905 + Δ119905)119877(119905) The hazard function for the 119880(119886 119887) baselinedistribution is given by ℎ(119905) = 1(119887 minus 119905) 119886 le 119905 lt 119887 119887 gt 0which is infinite at the end of life interval 119887 as expectedBecause the uniform HZF is an IFR then for the uniformdensity it can be proven using the infinite geometric seriesfor ℎ(119905) = (1119887)(1 minus 119905119887) and the Maclaurin series for 119890119905119887that 120588(119905) lt ℎ(119905) for all 119886 lt 119905 le 119887

In order to compute the RNIF 120588(119905) for 119905 gt 119887 weused two different approximate procedures First by directlydifferentiating (1a) as follows 120588(119905) = (119889119889119905)sum

infin

119899=1119865(119899)(119905) =

suminfin

119899=1119891(119899)(119905) where the exact 119891

(1)(119905) = 119891(119905) and uniform

convolutions 119891(119899)(119905) for 119899 = 2 3 12 can be calculated

by our Matlab program For 119899 gt 12 the program uses theordinate of 119873(119899120583 1198991205902) approximation where 120583 = (119886 + 119887)2

and 1205902 = (119887 minus 119886)212 And the second method which usesthe right hand and left hand derivatives will be explained inSection 63

53 The Uniform Approximation Results The method wepropose uses the exact 119899 = 2 through 12 convolutions 119865

(119899)(119905)

and then applies the Normal approximation for convolutions

beyond 12 The question now arises how accurate is thenormal approximation for 119899 = 13 14 15 We usedour 12-fold convolution of the standard Uniform 119880(0 1)to determine the accuracy Clearly the partial sum 119879

12=

sum12

119894=1119883119894 each 119883

119894sim 119880(0 1) and mutually independent has

a mean of 6 and variance 12(119887 minus 119886)212 = 1 where 119886 = 0and maximum life 119887 = 1 The summary in Table 2 shows thenormal approximation to119865

(12)(119905) for intervals of length 050minus

120590 Table 2 clearly shows that the worst relative error occurs at1205902 and that the normal approximation improves as119885movestoward the right tail The accuracy is within 2 decimals up toone-120590 and 3 decimals beyond 149120590 Therefore we concludethat the normal approximation to each of Uniform 119865

(119899)(119905)

119899 = 13 14 15 due to the CLT should not have a relativeerror at 119885 = 050 exceeding 0002960

It should also be noted that the normal approximationto the Uniform 119865

(119899)(119905) 119899 = 13 14 15 must be very

accurate from the standpoint of the first 4 moments Becausethe skewness of 119899-fold convolution of 119880(119886 119887) with itself isidentically zero which is identical to the Laplace-Gaussian119873(119899(119886 + 119887)2 119899(119887 minus 119886)212) and hence a perfect matchbetween the first 3 moments of 119891

119879119899

(119905) = 119891(119899)(119905) with those

of the corresponding Normal It can be proven (the proofis available on request) that the skewness of the partial sum119879119899= sum119899

119894=1119883119894119883119894rsquos being iid like119883 is given by

1205733(119879119899) =

1205833(119879119899)

[1205832(119879119899)]32

=1198991205833(119883)

(1198991205902)32

=1205833(119883)

(1205903radic119899)=1205733(119883)

radic119899

(12a)


Further the kurtosis of 119879119899is given by

1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)

where 120583119894(119894 = 2 3 4) are the universal notation for the

119894th central moments Equation (12b) clearly shows that thekurtosis of the uniform 119891

(119899)(119905) for 119899 = 13 14 and 15

respectively is 1205734= minus12013 = minus009231 minus12014 =

minus008571 and minus 12015 = minus008000 with the amount minus120being the kurtosis of a 119880(119886 119887) underlying failure densityThus an 119899 = 120 is needed in order for the kurtosisof 119891(119899)(119905) to be within 001 of Laplace-Gaussian 119873(119899(119886 +

119887)2 119899(119887 minus 119886)212) Fortunately summary in Table 1 clearlyindicates that the Normal approximation is superior at thetails where kurtosis may play a more important role than themiddle of 119891

(119899)(119905) density [48 page 86]

6 Approximating the Renewal Function forUnknown Convolutions

61 The Three-Parameter Weibull Renewal Function Unlikethe Gamma Normal and Uniform underlying failure densi-ties theWeibull baseline distribution (except when the shapeparameter 120573 = 1) does not have a closed-form expressionfor its 119899-fold cdf convolution 119865

(119899)(119905) and hence (1a) cannot

directly be used to obtain the renewal function 119872(119905) for all120573 gt 0 When minimum life = 0 and shape 120573 = 2 the Weibullspecifically is called the Rayleigh pdf we do have a closed-form function for the Rayleigh119872(119904) but it cannot be invertedto yield a closed-form expression for its119872(119905)

It must be highlighted that there have been severalarticles on approximating the RNF such as Deligonul [49]and also on numerical solutions of 119872(119905) by Tortorella [50]From [24] and other notables The online supplement byTortorella provides extensive references on renewal theoryand applications Note that Murthy et al [42] provide anextensive treatise onWeibull Models referring to theWeibullwith zero minimum life as the standard model Murthy et al[42] state at the bottom of their page 37 the confusionand misconception that had resulted from the plethora ofterminologies for intensity and hazard functions We will usethe time-discretizing method used by Elsayed [1] and otherssuch as Xie [20] with the aid of Matlab to obtain anotherapproximation for119872(119905)

62 Discretizing Time in order to Approximate the RenewalEquation Because 119872(119905) = 119865(119905) + int

119905

0119872(119905 minus 120591)119891(120591)119889120591 and

the underling failure distributions are herein specified thefirst term on the right-hand side (RHS) of 119872(119905) 119865(119905) canbe easily computed However the convolution integral onthe RHS int119905

0119872(119905 minus 120591)119891(120591)119889120591 except for rare cases cannot

in general be computed and has to be approximated Thediscretization method was first applied by Xie [20] where he

called his procedure ldquoTHE RS-METHODrdquo RS for Riemann-Stieltjes However Xie [20] used the renewal type (9b) in hisRS-METHOD while we are discretizing (9a)

The first step in the discretization is to divide the specifiedinterval (0 119905) into equal-length subintervals and only for thesake of illustration we consider the interval (0 119905 = 5 weeks)and divide it into 10 subintervals (0 050 week) (050 1) (45 5) Note that Xiersquos method does not require equal-lengthsubintervals Thus the length of each subinterval in thisexample is Δ119905 = 050weeks As a result int5

0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942

(119894minus1)2119872(5 minus 120591)119891(120591)119889120591 where the index 119894 = 1 pertains

to the subinterval (0 050) and 119894 = 10 pertains to the lastsubinterval (45 5) We now make use of the Mean-ValueTheorem for Integrals which states the following if a function119891(119909) is continuous over the real closed interval [119886 119887] then forcertain there exists a real number 119909

0such that int119887

119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909

0) being the ordinate of the

integrand at 1199090 Because both 119872(119905 minus 120591) and the density f(t)

are continuous then for example applying the aboveMean-ValueTheorem for Integrals to the 4th subinterval there existsfor certain a real number 120591

4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591

4le 2 As a result int5

0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591

10le 5 Clearly the exact values

of119872(5 minus 120591119894) 119894 = 1 10 cannot in general be determined

and because in this example 119891(120591119894)(12) = int

1198942

(119894minus1)2119891(120591)119889120591 =

Pr[(119894minus1)2 le TTF le 1198942] it follows that int50119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1119872(5minus120591

119894) int1198942

(119894minus1)2119891(120591)119889120591 As proposed by Elsayed [1] who

used the end of each subinterval and a conditional probabilityapproach to arrive at his equation (710) we will approximatethis last function in the same manner by119872(5 minus 050119894) whichresults in

119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)

The above equation (13) is similar to that of (710) of Elsayed[1] where his subintervals are of length Δ119905 = 1 It shouldbe highlighted that using the interval midpoints (instead ofendpoints) inorder to attain more accuracy leads to morecomputational complexity We first used the information119872(0) equiv 0 at 119894 = 10 to calculate the last term of (13)


Table 3 Time-discretizing approximation method results

Δ119905 119872(119905) Approximate119872(119905) Relative error Elapsed time (seconds)10 100000000000 99501662508319 minus049834 402386171425 100000000000 98760351886669 minus123965 201870761450 100000000000 97541150998572 minus245885 82738732100 100000000000 95162581964040 minus483742 65112829200 100000000000 90634623461009 minus936538 32320588250 100000000000 88479686771438 minus1152031 21066944

further at 119894 = 9 (13) yields [1 +119872(5 minus 45)] int(8+1)2

82119891(120591)119889120591 =

[1 + 119872(050)] int45

4119891(120591)119889120591 However 119872(050) represents the

expected number of renewals during an interval of lengthΔ119905 = 050 Assuming that Δ119905 is sufficiently small relative to119905 such that 119873(119905) is approximately Bernoulli then 119872(Δ119905 =050) cong 1 times 119865(050) + 0 times 119877(050) Hence at 119894 = 9 thevalue of the term before last in (13) reduces approximatelyto [1 + 119865(050)] times int45

4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942

(119894minus1)2119891(120591)119889120591 where 119872(050) has been

approximated Continuing in this manner we solved (13)backward-recursively in order to approximate 119872(119905) Thesmaller Δ119905 always leads to a better approximation of119872(119905)

63 Renewal-Intensity FunctionApproximation for theWeibullDistribution Next after approximating the Weibull RNFhowdowe use its approximate119872(119905) to obtain a fairly accuratevalue of Weibull RNIF 120588(119905) Because 120588(119905) equiv 119889119872(119905)119889119905 equivLimΔ119905rarr0

(119872(119905 + Δ119905) minus 119872(119905))Δ119905 then for sufficiently smallΔ119905 gt 0 the approximate 120588(119905) asymp (119872(119905 + Δ119905) minus 119872(119905))Δ119905which uses the right-hand derivative and 120588(119905) asymp (119872(119905) minus119872(119905 minus Δ119905))Δ119905 using the left-hand derivative Because theRNF is not linear but strictly increasing our Matlab programcomputes both the left- and right-hand expressions andapproximates 120588(119905) by averaging the two where 119905 and Δ119905 areinputted by the user It is recommended that the user inputs0 lt Δ119905 lt 001119905 such that the probability of 2 or more failuresduring an interval of length Δ119905 is almost zero Further ourprogram shows that 120588(119905) lt ℎ(119905) if and only if the shape 120573 gt 1

64 Time-Discretization Accuracy The accuracy of abovediscretization method was checked in two different ways

First we verified that at 120573 cong 3439541 the Weibull meanmedian and mode become almost identical at which theWeibull skewness is 120573

3= 004052595 and Weibull kurtosis

is 1205734= minus0288751 with these last 2 standardized-moments

being very close to those of Laplace-Gaussian of identicallyequal to zero Our Matlab program at 120573 = 3439541 120575 = 0120579 = 120572 = 2000 and Δ119905 = 50 yielded the Weibull119872(4000) cong1753638831 (with CPU-time = 2404183 sec) while thecorresponding Normal (with MTBF = 179784459964 and120590 = 5779338342) resulted in119872(4000) = 177439715 (CPUcong 486) Secondly we ran our program for the exponential(which is the Weibull with minimum-life 120575 = 0 120579 = 120572 = 1120582and 120573 = 1) at 120582 = 0001 and 119905 = 10000 for varying

values ofΔ119905 Table 3 depicts the comparisons against the exact119872(119905) = 120582119905 = 0001 times 10000 = 10 expected RNLs

The above time-discretization-method using the Mean-Value Theorem for Integrals can similarly be applied to anylifetime distribution and should give fairly accurate results forsufficiently small subintervalsHowever Table 3 clearly showsthat even at Δ119905 = 001119905 the computational time exceeds oneminute and the corresponding error may not be acceptablefurther it is not a closed-form approximation

7 Conclusions

This paper provided the exact RNF and RNIF for the NormalGamma and Uniform underlying failure densities We havedevised a Matlab program that outputs nearly all the renewaland reliability measures of a 3-parameter Weibull NormalGamma and Uniform We have quantified the differencesbetween 120588(119905) and ℎ(119905) for 119905 gt 0 except in the only case ofCFR for which 120588(119905) equiv ℎ(119905) equiv 120582 Further when ℎ(119905) is an IFRthen 120588(119905) lt ℎ(119905) so that 119877(119905) lt 119890minus119872(0119905) 119905 gt 0 for the fourbaseline failure distributions that we have studied Howeverthis is not quite consistent with 119877(119905) = 119890minus119872(0119905) given by someauthors in reliability engineering such as Ebeling [36] fora NHPP Further some authors such as Leemis [37] use thesame notation 120582(119905) for theWeibull RNIF and also use 120582(119905) forthe Weibull HZF (see his Example 65 page 165)

Similar works for other underlying failure densities arein the immediate future Work is also in progress that incor-porates nonnegligible repair-time requiring some knowledgeofconvolution of TTF distribution with that of time-to-restore (TTR) Such a stochastic process is referred to as analternating renewal process [51 page 350]

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] E A Elsayed Reliability Engineering John Wiley amp SonsHoboken NJ USA 2nd edition 2012

[2] E W Frees ldquoWarranty analysis and renewal function estima-tionrdquo Naval Research Logistics Quarterly vol 33 pp 361ndash3721986


[3] H-G Kim and B M Rao ldquoExpected warranty cost of two-attribute free-replacementwarranties based on a bivariate expo-nential distributionrdquoComputers and Industrial Engineering vol38 no 4 pp 425ndash434 2000

[4] E P C Kao and M S Smith ldquoComputational approximationsof renewal process relating to a warranty problem the case ofphase-type lifetimesrdquoEuropean Journal of Operational Researchvol 90 no 1 pp 156ndash170 1996

[5] E W Frees ldquoNonparametric renewal function estimationrdquoTheAnnals of Statistics vol 14 no 4 pp 1366ndash1378 1986

[6] W Blischke and D N P MurthyWarranty Cost Analysis CRCPress New York NY USA 1st edition 1994

[7] S Karlin and A J Fabens ldquoGeneralized renewal functions andstationary inventory modelsrdquo Journal of Mathematical Analysisand Applications vol 5 no 3 pp 461ndash487 1962

[8] H Parsa and M Jin ldquoAn improved approximation for therenewal function and its integral with an application in two-echelon inventory managementrdquo International Journal of Pro-duction Economics vol 146 no 1 pp 142ndash152 2013

[9] S Cetinkaya E Tekin and C-Y Lee ldquoA stochastic modelfor joint inventory and outbound shipment decisionsrdquo IIETransactions vol 40 no 3 pp 324ndash340 2008

[10] G L Yang ldquoApplication of renewal theory to continous sam-pling sampling plansrdquo Technometrics vol 25 no 1 pp 59ndash671983

[11] G L Yang ldquoApplication of renewal theory to continous sam-pling sampling plansrdquo Naval Research Logistics Qsuarterly vol32 no 1 pp 45ndash51 1983

[12] T L Lai and D Siegmund ldquoA nonlinear renewal theory withapplications to sequential analysis IrdquoTheAnnals of Statistics vol5 no 5 pp 946ndash954 1977

[13] T L Lai and D Siegmund ldquoA nonlinear renewal theory withapplications to sequential analysis IIrdquo The Annals of Statisticsvol 7 no 1 pp 60ndash76 1979

[14] M L Spearman ldquoA simple approximation for IFR weibullrenewal functionsrdquoMicroelectronics Reliability vol 29 no 1 pp73ndash80 1989

[15] S Tacklind ldquoFourieranalytische behandlung vom erneuerung-sproblemrdquo Scandinavian Actuarial Journal vol 1945 no 1-2 pp68ndash105 1945

[16] W L Smith ldquoAsymptotic renewal theoremsrdquo Proceedings of theRoyal Society of Edinburgh A vol 64 no 01 pp 9ndash48 1954

[17] R Jiang ldquoA simple approximation for the renewal function withan increasing failure raterdquo Reliability Engineering and SystemSafety vol 95 no 9 pp 963ndash969 2010

[18] W L Smith andM R Leadbetter ldquoOn the renewal function forthe Weibull distributionrdquo Technometrics vol 5 no 3 pp 393ndash396 1963

[19] Z A Lomnicki ldquoA note on the Weibull renewal processrdquoBiometrika vol 53 no 3-4 pp 375ndash381 1966

[20] M Xie ldquoOn the solution of renewal-type integral equationsrdquoCommunications in Statistics vol 18 no 1 pp 281ndash293 1989

[21] E Smeitink and R Dekker ldquoSimple approximation to therenewal functionrdquo IEEE Transactions on Reliability vol 39 no1 pp 71ndash75 1990

[22] L A Baxter E M Scheuer D J McConalogue and WR Blischke ldquoOn the Tabulation of the Renewal FunctionrdquoTechnometrics vol 24 no 2 pp 151ndash156 1982

[23] A Garg and J R Kalagnanam ldquoApproximations for the renewalfunctionrdquo IEEE Transactions on Reliability vol 47 no 1 pp 66ndash72 1998

[24] S G From ldquoSome new approximations for the renewal func-tionrdquo Communications in Statistics B vol 30 no 1 pp 113ndash1282001

[25] T Jin and L Gonigunta ldquoWeibull and gamma renewal approx-imation using generalized exponential functionsrdquo Communica-tions in Statistics vol 38 no 1 pp 154ndash171 2008

[26] K T Marshall ldquoLinear bounds on the renewal functionrdquo SIAMJournal on Applied Mathematics vol 24 no 2 pp 245ndash2501973

[27] H Ayhan J Limon-Robles and M A Wortman ldquoAn approachfor computing tight numerical bounds on renewal functionsrdquoIEEE Transactions on Reliability vol 48 no 2 pp 182ndash188 1999

[28] R E Barlow ldquoBounds on integrals with applications to reliabil-ity problemsrdquoThe Annals of Mathematical Statistics vol 36 no2 pp 565ndash574 1965

[29] M R Leadbetter ldquoBounds on the error in the linear approxi-mation to the renewal functionrdquo Biometrika vol 51 no 3-4 pp355ndash364 1964

[30] T Ozbaykal ldquoBounds and approximations for the renewalfunctionrdquoUSNaval Postgraduate SchoolMonterey Calif USA1971 httparchiveorgdetailsboundsapproximat00ozba

[31] M Xie ldquoSome results on renewal equationsrdquo Communicationsin Statistics vol 18 no 3 pp 1159ndash1171 1989

[32] L Ran L Cui and M Xie ldquoSome analytical and numericalbounds on the renewal functionrdquo Communications in Statisticsvol 35 no 10 pp 1815ndash1827 2006

[33] K Politis andM V Koutras ldquoSome new bounds for the renewalfunctionrdquo Probability in the Engineering and InformationalSciences vol 20 no 2 pp 231ndash250 2006

[34] M Brown H Solomon andM A Stephens ldquoMonte Carlo sim-ulation of the renewal functionrdquo Journal of Applied Probabilityvol 18 no 2 pp 426ndash434 1981

[35] D J Bartholomew ldquoAn approximate solution of the integralequation of renewal theoryrdquo Journal of the Royal StatisticalSociety B vol 25 no 2 pp 432ndash441 1963

[36] C E EbelingAn Introduction To Reliability andMaintainabilityEngineeringWaveland Press LongGrove Ill USA 2nd edition2010

[37] L M Leemis Reliability Probabilistic Models and StatisticalMethods 2nd edition 2009

[38] L Cui and M Xie ldquoSome normal approximations for renewalfunction of large Weibull shape parameterrdquo Communications inStatistics B vol 32 no 1 pp 1ndash16 2003

[39] S M Ross Stochastic Processes John Wiley amp Sons New YorkNY USA 2nd edition 1996

[40] U N Bhat Elements of applied stochastic processes John Wileyamp Sons New York NY USA 2nd edition 1984

[41] F Hildebrand Advanced Calculus for Applications PrenticeHall New York NY USA 1962

[42] D N P Murthy M Xie and R Jiang Weibull Models JohnWiley amp Sons Hoboken NJ USA 2004

[43] E GOlds ldquoA note on the convolution of uniformdistributionsrdquoTheAnnals ofMathematical Statistics vol 23 no 2 pp 282ndash2851952 1023072236455

[44] S Maghsoodloo and J H Hool ldquoOn fourth moment limitsrequired to achieve 01 accuracy of normal approximation forsymetrical linear combinationsrdquo Journal of Alabama Academyof Science vol 54 no 1 pp 1ndash13 1983

[45] E S W Shiu ldquoConvolution of uniform distributions and ruinprobabilityrdquo Scandinavian Actuarial Journal no 3-4 pp 191ndash197 19871987


[46] F Killmann and V E Collani ldquoA note on the convolution ofthe uniform and related distributions and their use in qualitycontrolrdquo Economic Quality Control vol 16 no 1 pp 17ndash41 2001

[47] J S Kang S L Kim Y H Kim and Y S Jang ldquoGeneralizedconvolution of uniform distributionsrdquo Journal of Applied Math-ematics amp Informatics vol 28 no 5-6 pp 1573ndash1581 2010

[48] M G Kendall and A StuartThe Advanced Theory of Statisticsvol 1 Charles Griffin and Company London UK 1963

[49] Z Seyda Deligonul ldquoAn approximate solution of the integralequation of renewal theoryrdquo Journal of Applied Probability vol22 no 4 pp 926ndash931 1985

[50] M Tortorella ldquoNumerical solutions of renewal-type integralequationsrdquo INFORMS Journal on Computing vol 17 no 1 pp66ndash74 2005

[51] D R Cox and H D Miller The Theory of Stochastic ProcessesJohn Wiley amp Sons New York NY USA 1968

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Active and Passive Electronic Components

Control Scienceand Engineering

Journal of



RotatingMachinery


Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design



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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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DistributedSensor Networks



and accuracy to evaluate the value of RNF approximation[17] Increasing the complexity may lead to more accurateapproximation but may make the process complicated anddifficult to implement in practice [25]

Furthermore in the literature lower and upper boundson RNFs have been discussed which can be used to obtainupper and lower bounds on warranty costs such as Blischkeand Murthy [6] Marshall [26] provides lower and upperlinear bounds on the RNF of an ordinary RNLprocess Ayhanet al [27] provide tight lower and upper bounds for the RNFwhich are based on Riemann-Stieljes integration There arealso many other studies conducted about bounds on RNFsuch as Barlow [28] Leadbetter [29] Ozbaykal [30] Xie [31]Ran et al [32] and Politis and Koutras [33]

Finally simulation can be considered as an alternateapproach to estimate the value of a RNF Brown et al [34] usethe Monte Carlo simulation to estimate the RNF for a RNLprocess with known interarrival time distribution

This paper proposes a convolution method to obtainthe Gamma and Normal renewal and renewal-intensityfunctions throughout we are assuming that repair time isnegligible (or minimal) relative to Time to Failure (TTF)As a result of this assumption the replacement time of afailed component in a system is minimal A further objectiveis to obtain the renewal and RNL-intensity functions of theuniform distribution by using 119899 = 2 to 119899 = 12 convolutionsand applying the normal approximation for convolutionsbeyond 12 Because the Weibull distribution except at shape120573 = 1 does not have a closed-form function for its 119899-foldconvolution the last objective is to approximate its renewaland RNL-intensity functions by discretizing time using theMean-Value Theorem for Integrals We will also highlight thedifferences between the renewal-intensity function 120588(119905) andthe hazard function ℎ(119905)

2 Preliminaries

Suppose that component (or system) failures occur at times119879119899(119899 = 1 2 3 4 ) measured from zero and assuming

that replacement (or restoration time) is negligible relativeto operational time then 119879

119899represents the operating time

(measured from zero) until the 119899th failure Because theprobability density function (pdf) of 119879

1(or 119883

1) may be

different from the intervening times 1198832= (1198792minus 1198791) 1198833=

(1198793minus1198792)1198834= (1198794minus1198793) we consider only the simpler case

of pdf of time to first failure 1198911(119905) being identical to those of

intervening times1198832 1198833 1198834 That is all119883

119894rsquos are assumed

iid (independently and identically distributed) Therefore119879119899= sum119899

119894=1119883119894= sum of the times to the first failure from

zero plus the intervening times of 2nd failure until the 119899thfailure If 119899 gt 60 then the central limit theorem (CLT) statesthat the distribution of 119879

119899approaches normality with mean

119899120583 where 120583 (the expected time between successive renewals)= 119864(119883

119894) 119894 = 1 2 3 4 and with variance of 119879

119899 denoted

by 119881(119879119899) equal to 1198991205902 However if the pdf of 119883

119894 119891(119909) with

119883 being the parent variable is highly skewed andor 119899 isnot sufficiently large then the exact pdf of 119879

119899= sum119899

119894=1119883119894is

given by the 119899-fold convolution of 119891(119909) with itself denoted

by 119891119879119899

(119905) = 119891(119899)(119905) Thus by renewal we mean either the

replacement of a failed component with a brand-new oneor the case when the failed component can be repaired in anegligibly short time to an almost brand-new condition Ithas been proven bymany authors both in stochastic processesand reliability engineering that the RNF for the duration [0 119905]is given by

119872(119905) = 119864 [119873 (119905)] =infin

sum119899=1

119865(119899)(119905) (1a)

119881 [119873 (119905)] =infin

sum119899=1

(2119899 minus 1) 119865(119899)(119905) minus [119872 (119905)]

2 (1b)

where the random variable 119873(119905) represents the number (orcount) of renewals that occur during the time interval [0 119905]and 119865

(119899)(119905) is the cumulative distribution function (cdf) of

the 119899-fold convolution of 119891(119905) with itself that is 119865(119899)(119905) =

int119905

0119891(119899)(120591)119889120591 where 119891

(119899)(119905) is the 119899-fold convolution density

of 119891(119905) It is also very well known that the RNI (renewalintensity) of a distribution function119865(119905) by definition is givenby 120588(119905) = 119889119872(119905)119889119905

Authors in stochastic processes refer to120588(119905) as the renewaldensity because 120588(119905) times Δ119905 describes the (unconditional)probability element of a renewal during the interval (119905 119905+Δ119905)[35] further explanation is forthcoming in Section 52 whilea few authors in reliability engineering such as Ebeling [36]and Leemis [37] refer to 120588(119905) as intensity function Because120588(119905) is never a pdf over the support set of 119891(119905) for all failuredistributions throughout this paper we will refer to it as therenewal intensity function (RNIF)

The simplest and most common renewal process isthe homogeneous Poisson process (HPP) where the inter-vening times are exponentially distributed at the constantinterrenewal (or failure) rate 120582 Because 120582 is a constantand intervening times are iid a Poisson process is alsoreferred to as a homogeneous renewal process It is alsovery well known that for a HPP the pdf of interarrivals119883119894rsquos is given by 119891(119905) = 120582119890minus120582119905 and that of the time to

119899th failure (or renewal) measured from zero is givenby 119891119879119899

(119905) = 119891(119899)(119905) = (120582Γ(119899))(120582119905)119899minus1119890minus120582119905 (the Gamma

density with shape 119899 and scale 120573 = 1120582) As a result theuse of (1a) for the interval [0 119905] leads to the RNF 119872(119905) =

119864[119873(119905)] = suminfin

119899=1119865(119899)(119905) = sum

infin

119899=1int119905

119909=0(120582Γ(119899))(120582119909)119899minus1119890minus120582119909119889119909 =

int119905

119909=0120582119890minus120582119909sum

infin

119899=1((120582119909)119899minus1(119899 minus 1))119889119909 = int

119905

119909=0120582119890minus120582119909

suminfin

119899=0((120582119909)119899119899) 119889119909 = int

119905

119909=0120582 119889119909 = 120582119905 a fact that has been

known for well more than a century Further the RNIFfor a HPP is also a constant and is given by 120588(119905) =119889119872(119905)119889119905 = 119889(120582119905)119889119905 = 120582 Further it is also widely knownthat for a HPP the 119881[119873(119905)] = 120582119905

3 The Renewal and RNL-Intensity Functionsfor a Normal Baseline Failure Distribution

Suppose that the time between failures 119883119894 119894 = 1 2 3 4

are NID (120583 1205902) that is normally and independently dis-tributed with mean time between failures (MTBF) = 120583 and



119899= TTF



119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

Φ(119905 minus 119899120583

120590radic119899) (2)






infin




120588 (119905) =119889

119889119905

infin

sum119899=1

Φ(119911119899) =infin

sum119899=1

119889

119889119911119899

Φ(119911119899)119889119911119899

119889119905

=infin

sum119899=1

120593 (119911119899)

1


sum119899=1

1

radic2120587119890minus1199112

1198992 1

120590radic119899

(3)

or

120588 (119905) =infin

sum119899=1

1


2]2

2

(4)







L 120588 (119905) = 120588 (119904) = intinfin

0

119890minus119904119905120588 (119905) 119889119905 =119891 (119904)

1 minus 119891 (119904) 119904 gt 0

(5a)

119872(119904)=L 119872 (119905)=intinfin

0

119890minus119904119905119872(119905) 119889119905=119891 (119904)

119904 [1 minus 119891 (119904)]=120588 (119904)

119904

(5b)


infin

0119890minus119904119905119872(119905)119889119905 =



120572minus1

119896=0((120582119905)119896119896)119890minus120582119905 = 120582(120582119905)(1 + 120582119905) gt 120588(119905)


119872(119905 120572 = 3) =Lminus1

1205823

119904 [(120582 + 119904)3 minus 1205823]

=120582119905

3+ 119890minus(31205822)119905 [cos(120582119905radic3

4)

+sin (120582119905radic34)


1

3


120588 (119905 120572 = 3) =Lminus1

1205823

[(120582 + 119904)3 minus 1205823]

=120582


4)

+radic3 sin(120582119905radic34)]

119872 (119905 120572 = 4) =120582119905

4+119890minus2120582119905

8

+119890minus120582119905 [cos (120582119905) + sin (120582119905)]

4minus3

8

120588 (119905 120572 = 4) =Lminus1

1205824

(120582 + 119904)4 minus 1205824

=120582


(6)




119899le 119905) = Pr[119873(119905) ge 119899)]


119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

int120582119905

0

119906119899120572minus1

Γ (119899120572)119890minus119906119889119906 =

infin

sum119899=1

119868 (120582119905 119899120572)

(7)


(1Γ(119899120572)) int120582119905




120588 (119905) =119889119872 (119905)

119889119905

=infin

sum119899=1

119891(119899)(119905)

=120597

120597119905

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120597

120597119905int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120582

Γ (119899120572)(120582119905)119899120572minus1119890minus120582119905

(8)





0120588(119905minus119909)119891(119909)119889119909 this



119886(119905) is their



3= 2radic120572 and





1(119905) + int

119905





119905120573 = 15 120573 = 3 120573 = 5

119867(119905) 119867119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905)

0 0000 0000 0000 0000 0000 0000 0000 0000 00001 0517 0508 0517 0081 0065 0081 0004 0001 00042 1170 1167 1170 0340 0331 0340 0053 0042 00533 1834 1833 1834 0665 0664 0665 0186 0179 01864 2500 2500 2500 0999 1000 0999 0379 0383 03795 3167 3167 3167 1333 1333 1333 0592 0597 0592


119885 05 1 15 2 25 3 35 4119865(12)



119872(119905) = 119865 (119905) + int119905

0

119872(119905 minus 119909) 119891 (119909) 119889119909 (9a)


119905

0119872(119905 minus

119909)119891(119909)119889119909 = Lint119905



119905

0119865(119905 minus


119905

0119865(119905 minus 119909)119889119872(119909) =

int119905


119872(119905) = 119865 (119905) + int119905

0

119865 (119905 minus 119909) 119889119872 (119909)

= 119865 (119905) + int119905

0

119865 (119905 minus 119909) 120588 (119909) 119889119909

(9b)



119905


equation yields

119872(119905) =119905

119887+1

119887int0

119905

119872(120591) (minus119889120591) =119905

119887+1

119887int119905

0

119872(120591) 119889120591 (10)











119891(7)(119905) =

(7119886 minus 119905)6

(7201198887) 7119886 le 119905 le 6119886 + 119887











(7119887 minus 119905)6

(7201198887) 119886 + 6119887 le 119905 le 7119887

(11)




infin

119899=1119865(119899)(119905) =

suminfin

119899=1119891(119899)(119905) where the exact 119891

(1)(119905) = 119891(119905) and uniform





(119899)(119905)



12=

sum12

119894=1119883119894 each 119883





(119899)(119905)



(119899)(119905) 119899 = 13 14 15 must be very


119879119899

(119905) = 119891(119899)(119905) with those



1205733(119879119899) =

1205833(119879119899)

[1205832(119879119899)]32

=1198991205833(119883)

(1198991205902)32

=1205833(119883)

(1205903radic119899)=1205733(119883)

radic119899

(12a)



1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)



(119899)(119905) for 119899 = 13 14 and 15




(119899)(119905) density [48 page 86]







119905

0119872(119905 minus 120591)119891(120591)119889120591 and






0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942




119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909




4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591


0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591




1198942

(119894minus1)2119891(120591)119889120591 =


sum10

119894=1119872(5minus120591

119894) int1198942



119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)






82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











119899= TTF



119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

Φ(119905 minus 119899120583

120590radic119899) (2)






infin




120588 (119905) =119889

119889119905

infin

sum119899=1

Φ(119911119899) =infin

sum119899=1

119889

119889119911119899

Φ(119911119899)119889119911119899

119889119905

=infin

sum119899=1

120593 (119911119899)

1


sum119899=1

1

radic2120587119890minus1199112

1198992 1

120590radic119899

(3)

or

120588 (119905) =infin

sum119899=1

1


2]2

2

(4)







L 120588 (119905) = 120588 (119904) = intinfin

0

119890minus119904119905120588 (119905) 119889119905 =119891 (119904)

1 minus 119891 (119904) 119904 gt 0

(5a)

119872(119904)=L 119872 (119905)=intinfin

0

119890minus119904119905119872(119905) 119889119905=119891 (119904)

119904 [1 minus 119891 (119904)]=120588 (119904)

119904

(5b)


infin

0119890minus119904119905119872(119905)119889119905 =



120572minus1

119896=0((120582119905)119896119896)119890minus120582119905 = 120582(120582119905)(1 + 120582119905) gt 120588(119905)


119872(119905 120572 = 3) =Lminus1

1205823

119904 [(120582 + 119904)3 minus 1205823]

=120582119905

3+ 119890minus(31205822)119905 [cos(120582119905radic3

4)

+sin (120582119905radic34)


1

3


120588 (119905 120572 = 3) =Lminus1

1205823

[(120582 + 119904)3 minus 1205823]

=120582


4)

+radic3 sin(120582119905radic34)]

119872 (119905 120572 = 4) =120582119905

4+119890minus2120582119905

8

+119890minus120582119905 [cos (120582119905) + sin (120582119905)]

4minus3

8

120588 (119905 120572 = 4) =Lminus1

1205824

(120582 + 119904)4 minus 1205824

=120582


(6)




119899le 119905) = Pr[119873(119905) ge 119899)]


119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

int120582119905

0

119906119899120572minus1

Γ (119899120572)119890minus119906119889119906 =

infin

sum119899=1

119868 (120582119905 119899120572)

(7)


(1Γ(119899120572)) int120582119905




120588 (119905) =119889119872 (119905)

119889119905

=infin

sum119899=1

119891(119899)(119905)

=120597

120597119905

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120597

120597119905int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120582

Γ (119899120572)(120582119905)119899120572minus1119890minus120582119905

(8)





0120588(119905minus119909)119891(119909)119889119909 this



119886(119905) is their



3= 2radic120572 and





1(119905) + int

119905





119905120573 = 15 120573 = 3 120573 = 5

119867(119905) 119867119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905)

0 0000 0000 0000 0000 0000 0000 0000 0000 00001 0517 0508 0517 0081 0065 0081 0004 0001 00042 1170 1167 1170 0340 0331 0340 0053 0042 00533 1834 1833 1834 0665 0664 0665 0186 0179 01864 2500 2500 2500 0999 1000 0999 0379 0383 03795 3167 3167 3167 1333 1333 1333 0592 0597 0592


119885 05 1 15 2 25 3 35 4119865(12)



119872(119905) = 119865 (119905) + int119905

0

119872(119905 minus 119909) 119891 (119909) 119889119909 (9a)


119905

0119872(119905 minus

119909)119891(119909)119889119909 = Lint119905



119905

0119865(119905 minus


119905

0119865(119905 minus 119909)119889119872(119909) =

int119905


119872(119905) = 119865 (119905) + int119905

0

119865 (119905 minus 119909) 119889119872 (119909)

= 119865 (119905) + int119905

0

119865 (119905 minus 119909) 120588 (119909) 119889119909

(9b)



119905


equation yields

119872(119905) =119905

119887+1

119887int0

119905

119872(120591) (minus119889120591) =119905

119887+1

119887int119905

0

119872(120591) 119889120591 (10)











119891(7)(119905) =

(7119886 minus 119905)6

(7201198887) 7119886 le 119905 le 6119886 + 119887











(7119887 minus 119905)6

(7201198887) 119886 + 6119887 le 119905 le 7119887

(11)




infin

119899=1119865(119899)(119905) =

suminfin

119899=1119891(119899)(119905) where the exact 119891

(1)(119905) = 119891(119905) and uniform





(119899)(119905)



12=

sum12

119894=1119883119894 each 119883





(119899)(119905)



(119899)(119905) 119899 = 13 14 15 must be very


119879119899

(119905) = 119891(119899)(119905) with those



1205733(119879119899) =

1205833(119879119899)

[1205832(119879119899)]32

=1198991205833(119883)

(1198991205902)32

=1205833(119883)

(1205903radic119899)=1205733(119883)

radic119899

(12a)



1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)



(119899)(119905) for 119899 = 13 14 and 15




(119899)(119905) density [48 page 86]







119905

0119872(119905 minus 120591)119891(120591)119889120591 and






0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942




119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909




4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591


0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591




1198942

(119894minus1)2119891(120591)119889120591 =


sum10

119894=1119872(5minus120591

119894) int1198942



119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)






82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










120588 (119905 120572 = 3) =Lminus1

1205823

[(120582 + 119904)3 minus 1205823]

=120582


4)

+radic3 sin(120582119905radic34)]

119872 (119905 120572 = 4) =120582119905

4+119890minus2120582119905

8

+119890minus120582119905 [cos (120582119905) + sin (120582119905)]

4minus3

8

120588 (119905 120572 = 4) =Lminus1

1205824

(120582 + 119904)4 minus 1205824

=120582


(6)




119899le 119905) = Pr[119873(119905) ge 119899)]


119872(119905) =infin

sum119899=1

119865(119899)(119905) =

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

int120582119905

0

119906119899120572minus1

Γ (119899120572)119890minus119906119889119906 =

infin

sum119899=1

119868 (120582119905 119899120572)

(7)


(1Γ(119899120572)) int120582119905




120588 (119905) =119889119872 (119905)

119889119905

=infin

sum119899=1

119891(119899)(119905)

=120597

120597119905

infin

sum119899=1

int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120597

120597119905int119905

0

120582

Γ (119899120572)(120582119909)119899120572minus1119890minus120582119909119889119909

=infin

sum119899=1

120582

Γ (119899120572)(120582119905)119899120572minus1119890minus120582119905

(8)





0120588(119905minus119909)119891(119909)119889119909 this



119886(119905) is their



3= 2radic120572 and





1(119905) + int

119905





119905120573 = 15 120573 = 3 120573 = 5

119867(119905) 119867119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905)

0 0000 0000 0000 0000 0000 0000 0000 0000 00001 0517 0508 0517 0081 0065 0081 0004 0001 00042 1170 1167 1170 0340 0331 0340 0053 0042 00533 1834 1833 1834 0665 0664 0665 0186 0179 01864 2500 2500 2500 0999 1000 0999 0379 0383 03795 3167 3167 3167 1333 1333 1333 0592 0597 0592


119885 05 1 15 2 25 3 35 4119865(12)



119872(119905) = 119865 (119905) + int119905

0

119872(119905 minus 119909) 119891 (119909) 119889119909 (9a)


119905

0119872(119905 minus

119909)119891(119909)119889119909 = Lint119905



119905

0119865(119905 minus


119905

0119865(119905 minus 119909)119889119872(119909) =

int119905


119872(119905) = 119865 (119905) + int119905

0

119865 (119905 minus 119909) 119889119872 (119909)

= 119865 (119905) + int119905

0

119865 (119905 minus 119909) 120588 (119909) 119889119909

(9b)



119905


equation yields

119872(119905) =119905

119887+1

119887int0

119905

119872(120591) (minus119889120591) =119905

119887+1

119887int119905

0

119872(120591) 119889120591 (10)











119891(7)(119905) =

(7119886 minus 119905)6

(7201198887) 7119886 le 119905 le 6119886 + 119887











(7119887 minus 119905)6

(7201198887) 119886 + 6119887 le 119905 le 7119887

(11)




infin

119899=1119865(119899)(119905) =

suminfin

119899=1119891(119899)(119905) where the exact 119891

(1)(119905) = 119891(119905) and uniform





(119899)(119905)



12=

sum12

119894=1119883119894 each 119883





(119899)(119905)



(119899)(119905) 119899 = 13 14 15 must be very


119879119899

(119905) = 119891(119899)(119905) with those



1205733(119879119899) =

1205833(119879119899)

[1205832(119879119899)]32

=1198991205833(119883)

(1198991205902)32

=1205833(119883)

(1205903radic119899)=1205733(119883)

radic119899

(12a)



1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)



(119899)(119905) for 119899 = 13 14 and 15




(119899)(119905) density [48 page 86]







119905

0119872(119905 minus 120591)119891(120591)119889120591 and






0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942




119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909




4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591


0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591




1198942

(119894minus1)2119891(120591)119889120591 =


sum10

119894=1119872(5minus120591

119894) int1198942



119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)






82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











119905120573 = 15 120573 = 3 120573 = 5

119867(119905) 119867119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905) 119867(119905) 119867

119886(119905) 119872(119905)

0 0000 0000 0000 0000 0000 0000 0000 0000 00001 0517 0508 0517 0081 0065 0081 0004 0001 00042 1170 1167 1170 0340 0331 0340 0053 0042 00533 1834 1833 1834 0665 0664 0665 0186 0179 01864 2500 2500 2500 0999 1000 0999 0379 0383 03795 3167 3167 3167 1333 1333 1333 0592 0597 0592


119885 05 1 15 2 25 3 35 4119865(12)



119872(119905) = 119865 (119905) + int119905

0

119872(119905 minus 119909) 119891 (119909) 119889119909 (9a)


119905

0119872(119905 minus

119909)119891(119909)119889119909 = Lint119905



119905

0119865(119905 minus


119905

0119865(119905 minus 119909)119889119872(119909) =

int119905


119872(119905) = 119865 (119905) + int119905

0

119865 (119905 minus 119909) 119889119872 (119909)

= 119865 (119905) + int119905

0

119865 (119905 minus 119909) 120588 (119909) 119889119909

(9b)



119905


equation yields

119872(119905) =119905

119887+1

119887int0

119905

119872(120591) (minus119889120591) =119905

119887+1

119887int119905

0

119872(120591) 119889120591 (10)











119891(7)(119905) =

(7119886 minus 119905)6

(7201198887) 7119886 le 119905 le 6119886 + 119887











(7119887 minus 119905)6

(7201198887) 119886 + 6119887 le 119905 le 7119887

(11)




infin

119899=1119865(119899)(119905) =

suminfin

119899=1119891(119899)(119905) where the exact 119891

(1)(119905) = 119891(119905) and uniform





(119899)(119905)



12=

sum12

119894=1119883119894 each 119883





(119899)(119905)



(119899)(119905) 119899 = 13 14 15 must be very


119879119899

(119905) = 119891(119899)(119905) with those



1205733(119879119899) =

1205833(119879119899)

[1205832(119879119899)]32

=1198991205833(119883)

(1198991205902)32

=1205833(119883)

(1205903radic119899)=1205733(119883)

radic119899

(12a)



1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)



(119899)(119905) for 119899 = 13 14 and 15




(119899)(119905) density [48 page 86]







119905

0119872(119905 minus 120591)119891(120591)119889120591 and






0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942




119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909




4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591


0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591




1198942

(119894minus1)2119891(120591)119889120591 =


sum10

119894=1119872(5minus120591

119894) int1198942



119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)






82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation












119891(7)(119905) =

(7119886 minus 119905)6

(7201198887) 7119886 le 119905 le 6119886 + 119887











(7119887 minus 119905)6

(7201198887) 119886 + 6119887 le 119905 le 7119887

(11)




infin

119899=1119865(119899)(119905) =

suminfin

119899=1119891(119899)(119905) where the exact 119891

(1)(119905) = 119891(119905) and uniform





(119899)(119905)



12=

sum12

119894=1119883119894 each 119883





(119899)(119905)



(119899)(119905) 119899 = 13 14 15 must be very


119879119899

(119905) = 119891(119899)(119905) with those



1205733(119879119899) =

1205833(119879119899)

[1205832(119879119899)]32

=1198991205833(119883)

(1198991205902)32

=1205833(119883)

(1205903radic119899)=1205733(119883)

radic119899

(12a)



1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)



(119899)(119905) for 119899 = 13 14 and 15




(119899)(119905) density [48 page 86]







119905

0119872(119905 minus 120591)119891(120591)119889120591 and






0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942




119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909




4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591


0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591




1198942

(119894minus1)2119891(120591)119889120591 =


sum10

119894=1119872(5minus120591

119894) int1198942



119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)






82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











1205734(119879119899) =

1205724(119883)

119899+3 (119899 minus 1)

119899minus 3

997888rarr 1205734(119879119899) =

[1205724(119883) minus 3]

119899=1205734(119883)

119899

(12b)



(119899)(119905) for 119899 = 13 14 and 15




(119899)(119905) density [48 page 86]







119905

0119872(119905 minus 120591)119891(120591)119889120591 and






0119872(5minus120591)119891(120591)119889120591 equiv

sum10

119894=1int1198942




119886119891(119909)119889119909 equiv

119891(1199090) times (119887 minus 119886) 119886 le 119909

0le 119887 119891(119909




4such that int2

32119872(5 minus 120591)119891(120591)119889120591 equiv

119872(5 minus 1205914)119891(1205914)(2 minus 32) 32 le 120591


0119872(5 minus

120591)119891(120591)119889120591 equiv sum10

119894=1119872(5 minus 120591

119894)119891(120591119894)(12) where 0 le 120591

1le 050

05 le 1205912le 1 45 le 120591




1198942

(119894minus1)2119891(120591)119889120591 =


sum10

119894=1119872(5minus120591

119894) int1198942



119872(5) equiv 119865 (5) + int5

0

119872(5 minus 120591) 119891 (120591) 119889120591

cong 119865 (5) +10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

int1198942

(119894minus1)2

119891 (120591) 119889120591

+10

sum119894=1

119872(5 minus 050119894) int1198942

(119894minus1)2

119891 (120591) 119889120591

=10

sum119894=1

[1 +119872 (5 minus 050119894)] times int1198942

(119894minus1)2

119891 (120591) 119889120591

(13)






82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation













82119891(120591)119889120591 =

[1 + 119872(050)] int45



4119891(120591)119889120591 At 119894 = 8 the value of (13) is

given by [1 + 119872(1)] int4

35119891(120591)119889120591 where 119872(1) = sum

2

119894=1[1 +

119872(1 minus 050119894)] times int1198942












7 Conclusions





References
























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation






























































RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation









research article renewal and renewal-intensity functions with minimal repair · 2019. 7. 31. ·...

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