Estimate Stress-Strength Reliability Model Using Rayleigh

Research ArticleEstimate Stress-Strength Reliability Model Using Rayleigh andHalf-Normal Distribution

Osama Abdulaziz Alamri1 M M Abd El-Raouf 2 Eman Ahmed Ismail3

Zahra Almaspoor 4 Basim S O Alsaedi 1 Saima Khan Khosa5 and M Yusuf 6

1Department of Statistics Faculty of Science University of Tabuk Tabuk 71491 Saudi Arabia2Basic and Applied Science Institute Arab Academy for Science Technology and Maritime Transport (AASTMT)Alexandia Egypt3College of International Transport and LogisticsArab Academy for Science Technology and Maritime Transport Smart Village Campus Cairo Egypt4Department of Statistics Yazd University PO Box 89175-741 Yazd Iran5Department of Statistics Bahauddin Zakariya University Multan Pakistan6Department of Mathematics Faculty of science Helwan University Cairo Egypt

Correspondence should be addressed to Zahra Almaspoor zalmaspoorstuyazdacir

Received 25 May 2021 Accepted 23 June 2021 Published 5 July 2021

Academic Editor Ahmed Mostafa Khalil

Copyright copy 2021 Osama Abdulaziz Alamri et al )is is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in anymedium provided the original work isproperly cited

In the field of life testing it is very important to study the reliability of any component under testing One of the most importantsubjects is the ldquostress-strength reliabilityrdquo term which always refers to the quantity P (XgtY) in any statistical literature Itresamples a system with random strength (X) that is subjected to a random strength (Y) such that a system fails in case the stressexceeds the strength In this study we consider stress-strength reliability where the strength (X) follows Rayleigh-half-normaldistribution and stress (Y1 Y2 Y3 and Y4) follows Rayleigh-half-normal distribution exponential distribution Rayleigh dis-tribution and half-normal distribution respectively)is effort comprises determining the general formulations of the reliabilitiesof a system Also the maximum likelihood estimation approach and method of moment (MOM) will be utilized to estimate theparameters Finally reliability has been attained utilizing various values of stress and strength parameters

1 Introduction

)e life of a component is described using the stress-strengthmodels in reliability theory that is including a randomstrength (X) which is subjected to a random stress (Y) )efailure of a component is occurred instantaneously when thestress level applied to it exceeds the level of the strength)us the component reliability is measured byR P(YltX) )is measurement has a variety of applica-tions most notably in the engineering industry such as thedegradation of rocket motors and structures the fatiguefailure of aircraft structures the ageing of concrete pressurevessels and static fatigue of ceramic components )ereforethe estimation of R P(YltX) has a great importance in the

practical applications )e literature demonstrates that re-liability estimation (R) has already been performed when thedistributions of (X) and (Y) are Weibull exponential or lognormal

Church and Harris [1] firstly introduced the term stress-strength Many authors have adopted various distributionstypes for stress and strength )e works of Church andHarris Surles and Padgett [2] Raqab and Kundu [3]Mokhlis [4] and Saraccediloglu et al [5] contain the discussion ofthe estimation problems of the stress-strength reliabilitymodel for different distributions Recently a review of allmethods and results on the stress-strength reliability havepresented by Kotz et al [6] Bayes estimators and reliabilityfunction and the parameters of the Consul Geeta and size-

HindawiComputational Intelligence and NeuroscienceVolume 2021 Article ID 7653581 10 pageshttpsdoiorg10115520217653581

biased Geeta distributions are obtained by Khan Adil andJan [7] Akman et al [8] studied the estimation of reliabilityusing a finite mixture of inverse Gaussian distributions )eestimation of R P(YltX) is studied by AI-Hussaini [9]based on a finite mixture of lognormal components Formore reading see [10ndash14]

2 Finite Mixture of Rayleigh and Half-Normal Distribution

)e Rayleigh-half-normal distribution is denoted asRHN(θ) by Abd El-Monsef and Abd El-Raouf [15] Amixture of Rayleigh and half-normal distribution with aparameter (1

2θ

radic) is used to represent this model

f(x θ) KfR

x middot 12θ

radic1113888 1113889 +(1 minus K)fHNx middot 1

2θ

radic1113888 1113889

K 2θxeminus θx2

1113874 1113875 +(1 minus K) 2

θπ

1113971

eminus θx2

⎛⎝ ⎞⎠

(1)

where K (1(1 +πθ

radic))

)us the Rayleigh-half normal distribution probabilitydensity function (pdf) is given by

f(x θ) 2θ(x + 1)e

minus θx2

1 +πθ

radic x θgt 0 (2)

)e corresponding cumulative distribution function isgiven by

F(x θ) 1 minus e

minus θx2+

πθ

radicerf(

θ

radicx)

1 +πθ

radic x θ gt 0 (3)

where erf(u) is the Gauss error function defined as

erf(u) 2π

radic 1113946u

0e

minus t2dt (4)

21 =e Survival Function and the Hazard Function )ereliability function or the survival function S(x) tests thechance of occurring of a breakdown of units beyond certaingiven point in time For monitoring a unit lifetime acrossthe support of its lifetime distribution generally theprobability that an item will work properly for a specifiedtime period with no failure is the survival function )edefinition of the survival function is represented as follows

S(x) 1 minus F(x) e

minus θx2+

πθ

radicerfc(

θ

radicx)

1 +πθ

radic (5)

where erfc(u) is the complementary error function and itsdefinition is

erfc(u) 1 minus erf(u) 2π

radic 1113946infin

ue

minus t2dt (6)

)e definition of the hazard rate function is the ratiobetween the density function and its survival function whichmeasures the tendency to die or to fail depending on thereached age and therefore it has a critical role in theclassification of the distributions of lifetime so the hazardrate function of the RHN distribution is given by

h(x) f(x)

S(x)

2θ(1 + x)

1 + eθx2

πθradic

erf(θ

radicx)

(7)

3 Stress-Strength Reliability Computations

In this section the reliability R P(YltX) was derivedwhere the random variables (X) and (Y) are the independentrandom variables where the strength X follows Rayleigh-half normal distribution and the stress (Y) takes differentcases (Rayleigh-half normal distribution exponential dis-tribution Rayleigh distribution and half-normaldistribution)

Let (X) and (Y) be two independent random variableswhere (X) represents ldquostrengthrdquo and (Y) represents ldquostressrdquoand (X) and (Y) follows a joint pdf f(x θ) thus thecomponent reliability is

R P(YltX) 1113946infin

minus infin1113946

x

minus infinf(x y)dydx (8)

In case that the random variables are statistically in-dependent then f(x y) f(x)g(y) so that

R 1113946infin

minus infin1113946

x

minus infinf(x)g(x)dydx (9)

where f(x) and g(y) are pdfrsquos of X and Y respectively

31=e Stress and the Strength Follows Rayleigh-Half-NormalDistribution As the strength X sim RHN(θ) andY1 sim RHN(θ1) they are independent random variables withpdf f (x) and g(y1) respectively

f(x) 2θ(x + 1)e

minus θx2

1 +πθ

radic 0lt θ middot x

g y1( 1113857 2θ1 y1 + 1( 1113857e

minus θ1y21

1 +πθ1

1113968 0lt θ1 middot y1

(10)

We derive the reliability R P(YltX) as follows

2 Computational Intelligence and Neuroscience

R1 P(YltX) 1113946infin

01113946

x

0f(x)g y1( 1113857dydx

1113946infin

01113946

x

0

2θ1 y1 + 1( 1113857eminus θ1y2

1

1 +πθ1

1113968⎛⎝ ⎞⎠2θ(x + 1)e

minus θx2

1 +πθ

radic⎛⎝ ⎞⎠dydx

(11)

And we get after the simplification

R1 1 +

πθ1

1113968+ 2

θθ1

1113968Tanminus 1

θθ11113969

1113874 1113875 +π

radicθ minus θ1( 1113857

θ + θ11113969

1113874 1113875 minus θ1 θ + θ1( 1113857( 11138571113874 1113875

1 +πθ

radic+

πθ1

1113968+ π

θθ1

1113968 (12)

32 =e Strength Follows RHN Distribution and the StressFollows Exponential Distribution In this case the proba-bility density function (pdf) for the stress Y2 that follows theexponential distribution is given by

g y2( 1113857 θ2eminus y1θ2 y2 θ2 gt 0 (13)

)en reliability function R2 for the independent randomvariables X and Y2

R2 1113946infin

01113946

x

0θ2e

minus y2θ2 2θ(x + 1)eminus θx2

1 +πθ


R2

π

radic

2(θ

radic+

π

radicθ)

2θ minus 2θ minus θ2( 1113857eθ224θ( )erfc

θ22

θ

radic1113888 11138891113888 1113889

(14)

where the strength follows RHN distribution

33 =e Strength Follows RHN Distribution and the StressFollows Rayleigh Distribution In this case the probabilitydensity function (pdf) for the stress Y3 that follows theRayleigh distribution is given by

g y3( 1113857 y3

θ22e

minus y232θ

23( ) y θ3 gt 0 (15)

)en reliability function R3 for the independent randomvariables X and Y3 is

R3 1113946infin

01113946

x

0

y3

θ23e

minus y232θ

23( )1113888 1113889

2θ(x + 1)eminus θx2

1 +πθ


2θ

(1 +πθ

radic)

1113946infin

0(x + 1)e

minus θx21 minus e

minus x22θ23( )1113874 1113875dx

R3 θ

1 +πθ

radic1θ

+

π

radic

θ

radic minus2

2θ + 1θ231113872 1113873minus

2π

radic

2θ + 1θ231113872 1113873

1113969⎛⎜⎜⎜⎝ ⎞⎟⎟⎟⎠

(16)


34 =e Strength Follows RHN Distribution and the StressFollows Half-Normal Distribution In this case the proba-bility density function (pdf) for the stress Y4 that followshalf-normal distribution is given by

g y4( 1113857

2

radic

θ4π

radic1113888 1113889eminus y2

42θ24( ) y θ4 gt 0 (17)


R4 1113946infin

01113946

x

0

2

radic

θ4π

radic1113888 1113889e

minusy24

2θ24⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠2θ(x + 1)e

minus θx2

1 +πθ


2θ

(1 +πθ

radic)

1113946infin

0Erf

x2

radicθ4

1113888 1113889(x + 1)eminus θx2

dx

R4 1

θ

radic(1 +

πθ

radic)

2θCotminus 1 θ42θ

radic1113872 1113873π

radic +1

θ4

2 + 1θθ241113872 1113873

1113969⎛⎜⎜⎜⎝ ⎞⎟⎟⎟⎠

(18)where the strength follows RHN distribution

Computational Intelligence and Neuroscience 3

4 Estimation of Stress-Strength Reliability

In the literature a discussion of the estimation R P(YltX)when random variables (X) and (Y) are following the specifieddistributions have been presented including engineeringstatistics quality control medicine reliability biostatisticsand psychology )is quantity for a limited number of casescould be calculated in a closed form (Nadarajah [16] andBarreto-Souza et al [17]) Several authors including Milanand Vesna [18] have considered the estimation of (R) forindependent variables and normally distributed (X) and (Y)Later a list of papers related to the estimation problem of (R)were reported by Greco and Venture [19] when (X) and (Y)are independent and follow a class of lifetime distributionscontaining Gamma distributions exponential generalizedexponential bivariate exponential Weibull distribution Burrtype t model and others

41 Method of Moment (MOM) Estimation of R )e esti-mation of reliability is very common in the statistical lit-erature Now to compute 1113954R we need to estimate theparameters θ and θi i 1 2 3 4 in four cases of stress

Since the strengths X follow RHN (θ)the stress havefour cases

(i) Y1 follows Rayleigh-half normal distribution withparameter θ1

(ii) Y2 follows exponential distribution with parameterθ2

(iii) Y3 follows Rayleigh distribution with parameter θ3(iv) Y4 follows half-normal distribution with parameter

θ4 then their population means are given by

x 2

θ

radic+

π

radic

2θ

radic(1 +

θπ

radic)

y1 2

θ1

1113968+

π

radic

2θ1

11139681 +

θ1π

1113968( 1113857

y2 1θ2

y3 θ3

π2

1113970

y4 θ4

2π

1113970

(19)

)e MErsquos of θ θ1 θ2 θ3 and θ4 denoted by 1113954θ 1113954θ1 1113954θ2 1113954θ3and 1113954θ4 respectively can be obtained by solving (x y1 y2 y3and y4) numerically

1113954θ 1113936

mi1 xi minus n( 1113857

2+ nπ 1113936

mi1 xi

2π 1113936mi1 xi( 1113857

2 +

1113936mi1 xi minus n( 1113857

4+ 2nπ 1113936

mi1 xi 1113936


21113969

2π 1113936mi1 xi( 1113857

2

1113954θ1 1113936

mj1 y1j

minus m1113874 11138752

+ mπ 1113936mj1 y1j

2π 1113936mj1 y1j

1113874 11138752 +

1113936mj1 y1j

minus n1113874 11138754

+ 2mπ 1113936mj1 y1j

1113936mj1 y1j

minus m1113874 11138752

1113970

2π 1113936mj1 y1j

1113874 11138752

1113954θ2 m

1113936mj1 y2j

1113954θ3

2π

11139701113936

mj1 y3j

m1113888 1113889

1113954θ4

π2

11139701113936

mj1 y4j

m1113888 1113889

(20)

)e ME of R denoted by 1113954R1 middot 1113954R2 middot 1113954R3 and 1113954R4 is obtainedby substitute 1113954θ with 1113954θ1 middot 1113954θ2 middot 1113954θ3 and 1113954θ4 in R1 middot R2 middot R3 and R4

42 =e Maximum Likelihood Estimators of R )e maxi-mum likelihood estimator (MLE) is the most popularmethod for reliability estimation R p(YltX) becauseof its generality and flexibility )is method can be used

if the joint distribution of the strength (X) and thestress (Y) is a known function with some unknownparameters

Suppose x1 middot x2 middot middot middot middot middot xn is a random sample from RHNdistribution with θ and y11 middot y12 middot middot middot middot middot y1m is a randomsample from RHN distribution with θ1 )en the likelihoodfunction is given by


L θ middot θ1 x middot y1( 1113857 2n+mθnθm1 minus (1 +

πθ

radic)n

minus 1 +

πθ11113969

1113874 1113875m

1113945

n

i1xi + 1( 1113857e

minus θx2i 1113945

m

j1y1j + 11113872 1113873e

minus θ1y21j (21)

And the log-likelihood function of the observed samplesis

ln L θ middot θ1( 1113857 (m + n)ln(2) + n ln(θ) + m ln θ1( 1113857 minus n ln(1 +πθ

radic) minus m ln 1 +

πθ11113969

1113874 1113875

minus θ1113944n

i1x2i minus θ1 1113944

m

j1y21j + 1113944

n

i1ln xi + 1( 1113857 + 1113944

m

j1ln y1j + 11113872 1113873

(22)

By solving the following equations the MLE of θ and θ2can be obtained

z ln L θ middot θ1( 1113857( 1113857

zθ

n

θminus

(nπ

radic)

(2θ

radic(1 +

πθ

radic))

minus 1113944n

(i1)

x2i 0

z ln L θ middot θ1( 1113857

zθ1

m

θ1minus

mπ

radic

2θ1

11139681 +

πθ1

1113968( 1113857

minus 1113944m

j1y21j 0

(23)

)e MLEs of θ and θ1 can be obtained respectively as

1113954θ 1

6πA2 B + 2A(A + nπ) +

A2

n2π2

+ 4A(A minus 4nπ)1113872 1113873

B⎛⎝ ⎞⎠

1113954θ1 1

6πC2 D + 2C(C + mπ) +

C2

m2π2

+ 4C(C minus 4mπ)1113872 1113873

D⎛⎝ ⎞⎠

(24)

where A 1113936ni0 x2

i middot C 1113936mj0 y2

1j

B 8A6

minus 48πnA5

+ 51π2n2A4

minus π3n3A3

+ 33

radicπ32

n3A7 minus 16A2 + 71πnA minus 2π2n2( )1113968

1113872 111387313

(25)

D (8C6 minus 48πmC5 + 51π2m2C4 minus π3m3C3 + 33

radic

π32m3C7(minus 16C2 + 71πmC minus 2π2m2)

1113968)13

)en the maximum likelihood estimator of R when thestrength X follows RHN(θ) distribution and stress Y followsRHN(θ1) distribution is given as

1113954R1 1 +

π1113954θ11113969

+ 21113954θ1113954θ1

1113969

Tanminus 11113954θ1113954θ11113872 1113873

1113969

1113874 1113875 +π

radic1113954θ minus 1113954θ11113872 1113873

1113954θ + 1113954θ1

1113969

1113874 1113875 minus 1113954θ1 1113954θ + 1113954θ11113872 11138731113872 11138731113874 1113875

1 +π1113954θ

1113968+

π1113954θ11113969

+ π1113954θ1113954θ1

1113969 (26)

Similarly we perform the same steps to find (MLE) inother cases we can obtain

(i) When the stress Y2 that follows the exponentialdistribution with parameter θ2 the MLE of R2 isgiven as

1113954R2

π

radic

21113954θ

1113968+

π

radic 1113954θ21113874 1113875

21113954θ minus 21113954θ minus 1113954θ21113872 1113873e1113954θ2

241113954θ1113872 1113873Erfc1113954θ2

21113954θ

11139681113888 11138891113888 1113889

(27)

where the strength X follows Rayleigh-half-normaldistribution with parameter θ

(ii) When the stress Y3 that follows Rayleigh distri-bution with parameter θ3 and the strength X followsRayleigh-half-normal distribution with parameterθ the MLE of R3 is given as

1113954R3 1113954θ

1 +π1113954θ

111396811113954θ

+

π

radic

1113954θ

1113968 minus2

21113954θ + 11113954θ231113874 1113875

minus

2π

radic

21113954θ + 11113954θ231113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(28)

(iii) When the stress Y4 that follows half-normal dis-tribution with parameter θ4 the M L E of R2 isgiven as


1113954R4 1

1113954θ

1113968(1 +

π1113954θ

1113968)

21113954θCotminus 1 1113954θ2

21113954θ1113969

1113888 1113889

π

radic +1

1113954θ4

2 + 11113954θ1113954θ241113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(29)

where the strength X follows Rayleigh-half-normal distri-bution with parameter θ

5 Numerical Evaluation

In different cases the system reliability R has evaluated forsome specific values of the parameters involved in the ex-pression of R

51 Case 1 Strength and Stress Follows RHN DistributionFrom Table 1 and Figures 1 and 2 it is noticed that with theincrease in the strength parameter values the reliabilityvalue decreases If the stress parameter increases then thevalue of reliability increases

52 Case 2 Strength Follows RHN Distribution and StressFollows Exponential Distribution From Table 2 and Fig-ures 3 and 4 it is observed that if the strength parameterincreases then the value of reliability increases If the stressparameter increases then the value of reliability increases

53 Case 3 Strength Follows RHN Distribution and StressFollows Rayleigh Distribution From Table 3 and Figures 5and 6 it is noticed that with increasing the value ofthe strength and stress parameter the reliability valuedecreases

54 Case 4 Strength Follows RHN Distribution and StressFollows Half-Normal Distribution From Table 4 and Fig-ures 7 and 8 it is noticed that with increasing the value ofthe strength and stress parameter the reliability valuedecreases

55 Simulation Study In this section some results arerepresented depending on Monte-Carlo simulation forcomparing the estimates of (R) performance using MLE andMOM estimators fundamentally for many sample sizes )efollowing sample sizes are considered (n m) (5 5) (1010) (20 20) (30 30) (40 40) (50 50) and (100 100) Fromeach sample the estimates are computed for the parametersusing MLE and method of moment estimation Once the

Table 1 Variation in R1 when strength and stress has RHN distribution

θθ1

01 02 03 04 05 06 07 08 09 1

01 0208 0359 0472 0558 0624 0677 0718 0752 0780 080302 0149 0267 0364 0442 0507 0561 0606 0645 0678 070603 0121 0222 0307 0379 0440 0493 0538 0577 0611 064104 0105 0194 0271 0337 0395 0445 0490 0529 0563 059405 0093 0174 0245 0307 0362 0410 0453 0491 0526 055606 0085 0159 0225 0284 0336 0382 0424 0461 0495 052607 0078 0148 0210 0265 0315 0360 0400 0437 0470 050008 0073 0138 0197 0250 0297 0341 0380 0416 0448 047809 0068 0130 0186 0237 0283 0325 0363 0398 0430 04591 0065 0124 0177 0226 0270 0311 0348 0382 0413 0443

0000010002000300040005000600070008000900

θ1 = 01θ1 = 05θ1 = 1

02 04 06 08 1 120Strength parameter

Figure 1 Variation in R1 for constant stress

θ = 01θ = 05

0000010002000300040005000600070008000900

Relia

bilit

y

02 04 06 08 1 120Stress parameter

Figure 2 Variation in R1 for constant strength


Table 2 Variation in R2 when strength has RHN distribution and stress has Exponential distribution

θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1


Table 3 Variation in R3 when strength has RHN distribution and stress has Rayleigh distribution

θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



Table 4 Variation in R4 when strength has RHN distribution and stress has half-normal distribution

θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




Table 5 Average MSE of the simulated estimates of R1

(θ middot θ1)(n m)

(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00196 00180 minus 00181 minus 00073 minus 00041 minus 00034 00005minus 00720 minus 00318 minus 00193 00138 minus 00084 00076 minus 00029

(1 05) minus 00594 minus 00476 minus 00132 minus 00120 minus 00101 minus 00043 00048minus 00790 minus 00747 minus 00643 minus 00662 minus 00617 minus 00658 minus 00555

(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370

(1 2) minus 00125 00115 00111 00113 minus 00073 00044 minus 0001500800 minus 00777 00687 00663 00656 00649 00638

(05 1) 00252 00146 00122 minus 00097 minus 00030 minus 00018 0001100743 00737 00686 00656 00654 00639 00627

(15 1) 00594 minus 00090 00076 minus 00064 00062 00019 00019minus 00807 minus 00413 minus 00402 minus 00364 minus 00352 minus 00164 00125

(2 1) 00153 00066 00066 00061 00048 00048 00045minus 00997 minus 00726 minus 00719 minus 00647 minus 00624 minus 00618 minus 00584

θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046

(15 1) 00147 00097 00024 00024 minus 00018 00014 minus 0000800221 minus 00124 minus 00090 00062 00027 00022 00014

(2 1) 00121 00064 00059 00030 00024 00018 minus 0000300478 minus 00272 00256 minus 00145 minus 00076 00050 00010


parameters are estimated the estimates of R1 is obtained)e average biases of R1 is reported in Table 5 and meansquared errors (MSEs) of R1 are in Table 6

)e first row includes the average bias of R1 using theMLE and second row includes the average bias of R1 usingthe MOM in each cell

)e first row includes the average MSE of R1 using theMLE and second row includes the average MSE of R1 usingthe MOM in each cell

6 Conclusions

)e proposed model in this paper the stress-strength re-liability has been studied for Rayleigh-half normal whenthe strength (X) follows Rayleigh-half normal distributionand the stress (Y) takes Rayleigh-half normal distributionexponential distribution Rayleigh distribution and half-normal distribution Based on the computations andgraphs (i) it has been noticed that when the stress pa-rameter is increased the reliability value lowers and whenthe strength parameter is increased the reliability valueincreases )e numerical assessment demonstrates thatincreasing the stress parameter decreases the dependabilityvalue in case (ii) whereas increasing the strength parameterincreases the reliability value In cases (iii) and (vi) in-creasing the stress parameter decreases the reliability valuewhereas increasing the strength parameter increases it Acomparison is carried out between two methods of reli-ability estimation R P(XgtY) when (Y) and (X) bothfollow Rayleigh-half normal distributions for various pa-rameters scale We provide MLE and MOM procedure forestimating the unknown parameters that are used for re-liability estimation (R) Based on the simulation findingswe can conclude that MLE outperforms MOM in terms ofaverage bias and average MSE for a variety of parameterchoices

Data Availability

All data used to support the findings of the study areavailable within the article

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] J D Church and B Harris ldquo)e estimation of reliability fromstress-strength relationshipsrdquo Technometrics vol 12 no 1pp 49ndash54 1970

[2] J G Surles and W J Padgett ldquoInference for reliability andstress-strength for scald Burr X distributionrdquo Lifetime DataAnalysis vol 7 no 2 pp 187ndash200 2001

[3] M Z Raqab and D Kundu ldquoComparison of different esti-mates of P(Y lt X) for a scaled Burr type X distributionrdquoCommunications in Statistics - Simulation and Computationvol 34 no 2 pp 465ndash483 2005

[4] N Mokhlis ldquoReliability of a stress-strength model with Burrtype III distributionsrdquo Communications in Statisticsmdash=eoryand Methods vol 34 no 7 pp 1643ndash1657 2005

[5] B Saraccediloglu I Kınacı and D Kundu ldquoOn estimation of (P(Ylt X) for exponential distribution under progressive type-IIcensoringrdquo Journal of Statistical Computation and Simulationvol 82 no 5 pp 729ndash744 2011

[6] S Kotz Y Lumelskii and M Pensky =e Stress-StrengthModel and its Generalizations =eory and ApplicationsWorld Scientific Publishing Singapore 2003

[7] H Khan Adil and T R Jan ldquoOn estimation of reliabilityfunction of Consul and Geeta distributionsrdquo InternationalJournal of Advanced Scientific and Technical Research vol 4no 4 pp 96ndash105 2014

[8] O Akman P Sansgiry and M C Minnotte ldquoOn the esti-mation of reliability based on mixture inverse Gaussiandistributionsrdquo Applied Statistical Science Vol IV NovaScience Publishers New York NY USA 1999

[9] E K AI-Hussaini ldquoBayesian prediction under a mixture oftwo exponential components model based on type 1 cen-soringrdquo Journal of Applied Statistical Science vol 8pp 173ndash185 1999

[10] M M M El-Din M M Amein H E El-Attar andE H Hafez ldquoSymmetric and asymmetric bayesian estimationfor lindley distribution based on progressive first failurecensored datardquo Mathematical Sciences Letters vol 6 no 3pp 255ndash260 2017 thorn

[11] J Zhao Z Ahmad E Mahmoudi E H Hafez andM M Mohie El-Din ldquoA new class of heavy-tailed distribu-tions modeling and simulating actuarial measuresrdquo Com-plexity vol 2021 Article ID 5580228 2021

[12] H M Almongy E Almetwally H M AljohaniA S Alghamdi and E H Hafez ldquoA new extended Rayleighdistribution with applications of COVID-19 datardquo Results inPhysics vol 23 no 60 thorn Article ID 104012 2021

[13] H M Almongy F Y Alshenawy E M Almetwally andD A Abdo ldquoApplying transformer insulation using Weibullextended distribution based on progressive censoringschemerdquo Axioms vol 10 no 2 p 100 2021

[14] M H Abu-Moussa A M Abd-Elfattah and E H HafezldquoEstimation of stress-strength parameter for Rayleigh distri-bution based on progressive type-II censoringrdquo InformationSciences Letters vol 10 no 1 p 12 2021

[15] M M E Abd El-Monsef and M M Abd El-Raouf ldquoRayleigh-half normal distribution for modeling tooth movement datardquoJournal of Biopharmaceutical Statistics vol 30 no 3pp 481ndash494 2020

[16] S Nadarajah ldquo)e exponentiated Gumbel distribution withclimate applicationrdquo Environmetrics vol 17 no 1 pp 13ndash232006

[17] W Barreto-Souza A L De Morais and G M Cordeiro ldquo)eWeibull-geometric distributionrdquo Journal of Statistical Com-putation and Simulation vol 81 no 5 pp 645ndash657 2011

[18] J Milan and R Vesna ldquoEstimation of px lt y for gammaexponential modelrdquo Yugoslav Journal of Operations Researchvol 24 no 2 pp 283ndash291 2014

[19] L Greco and L Ventura ldquoRobust inference for the stress-strength reliabilityrdquo Statistical Papers vol 52 no 4pp 773ndash788 2011


biased Geeta distributions are obtained by Khan Adil andJan [7] Akman et al [8] studied the estimation of reliabilityusing a finite mixture of inverse Gaussian distributions )eestimation of R P(YltX) is studied by AI-Hussaini [9]based on a finite mixture of lognormal components Formore reading see [10ndash14]

2 Finite Mixture of Rayleigh and Half-Normal Distribution

)e Rayleigh-half-normal distribution is denoted asRHN(θ) by Abd El-Monsef and Abd El-Raouf [15] Amixture of Rayleigh and half-normal distribution with aparameter (1

2θ

radic) is used to represent this model

f(x θ) KfR

x middot 12θ

radic1113888 1113889 +(1 minus K)fHNx middot 1

2θ

radic1113888 1113889

K 2θxeminus θx2

1113874 1113875 +(1 minus K) 2

θπ

1113971

eminus θx2

⎛⎝ ⎞⎠

(1)

where K (1(1 +πθ

radic))

)us the Rayleigh-half normal distribution probabilitydensity function (pdf) is given by

f(x θ) 2θ(x + 1)e

minus θx2

1 +πθ

radic x θgt 0 (2)

)e corresponding cumulative distribution function isgiven by

F(x θ) 1 minus e

minus θx2+

πθ

radicerf(

θ

radicx)

1 +πθ

radic x θ gt 0 (3)

where erf(u) is the Gauss error function defined as

erf(u) 2π

radic 1113946u

0e

minus t2dt (4)

21 =e Survival Function and the Hazard Function )ereliability function or the survival function S(x) tests thechance of occurring of a breakdown of units beyond certaingiven point in time For monitoring a unit lifetime acrossthe support of its lifetime distribution generally theprobability that an item will work properly for a specifiedtime period with no failure is the survival function )edefinition of the survival function is represented as follows

S(x) 1 minus F(x) e

minus θx2+

πθ

radicerfc(

θ

radicx)

1 +πθ

radic (5)

where erfc(u) is the complementary error function and itsdefinition is

erfc(u) 1 minus erf(u) 2π

radic 1113946infin

ue

minus t2dt (6)

)e definition of the hazard rate function is the ratiobetween the density function and its survival function whichmeasures the tendency to die or to fail depending on thereached age and therefore it has a critical role in theclassification of the distributions of lifetime so the hazardrate function of the RHN distribution is given by

h(x) f(x)

S(x)

2θ(1 + x)

1 + eθx2

πθradic

erf(θ

radicx)

(7)

3 Stress-Strength Reliability Computations

In this section the reliability R P(YltX) was derivedwhere the random variables (X) and (Y) are the independentrandom variables where the strength X follows Rayleigh-half normal distribution and the stress (Y) takes differentcases (Rayleigh-half normal distribution exponential dis-tribution Rayleigh distribution and half-normaldistribution)

Let (X) and (Y) be two independent random variableswhere (X) represents ldquostrengthrdquo and (Y) represents ldquostressrdquoand (X) and (Y) follows a joint pdf f(x θ) thus thecomponent reliability is

R P(YltX) 1113946infin

minus infin1113946

x

minus infinf(x y)dydx (8)

In case that the random variables are statistically in-dependent then f(x y) f(x)g(y) so that

R 1113946infin

minus infin1113946

x

minus infinf(x)g(x)dydx (9)

where f(x) and g(y) are pdfrsquos of X and Y respectively

31=e Stress and the Strength Follows Rayleigh-Half-NormalDistribution As the strength X sim RHN(θ) andY1 sim RHN(θ1) they are independent random variables withpdf f (x) and g(y1) respectively

f(x) 2θ(x + 1)e

minus θx2

1 +πθ

radic 0lt θ middot x

g y1( 1113857 2θ1 y1 + 1( 1113857e

minus θ1y21

1 +πθ1

1113968 0lt θ1 middot y1

(10)

We derive the reliability R P(YltX) as follows



01113946

x

0f(x)g y1( 1113857dydx

1113946infin

01113946

x

0

2θ1 y1 + 1( 1113857eminus θ1y2

1

1 +πθ1

1113968⎛⎝ ⎞⎠2θ(x + 1)e

minus θx2

1 +πθ


(11)


R1 1 +

πθ1

1113968+ 2

θθ1

1113968Tanminus 1

θθ11113969

1113874 1113875 +π


θ + θ11113969

1113874 1113875 minus θ1 θ + θ1( 1113857( 11138571113874 1113875

1 +πθ

radic+

πθ1

1113968+ π

θθ1

1113968 (12)




R2 1113946infin

01113946

x

0θ2e


1 +πθ


R2

π

radic

2(θ

radic+

π

radicθ)


θ22

θ

radic1113888 11138891113888 1113889

(14)



g y3( 1113857 y3

θ22e

minus y232θ

23( ) y θ3 gt 0 (15)


R3 1113946infin

01113946

x

0

y3

θ23e

minus y232θ

23( )1113888 1113889


1 +πθ


2θ

(1 +πθ

radic)

1113946infin

0(x + 1)e

minus θx21 minus e

minus x22θ23( )1113874 1113875dx

R3 θ

1 +πθ

radic1θ

+

π

radic

θ

radic minus2

2θ + 1θ231113872 1113873minus

2π

radic

2θ + 1θ231113872 1113873

1113969⎛⎜⎜⎜⎝ ⎞⎟⎟⎟⎠

(16)



g y4( 1113857

2

radic

θ4π


42θ24( ) y θ4 gt 0 (17)


R4 1113946infin

01113946

x

0

2

radic

θ4π

radic1113888 1113889e

minusy24

2θ24⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠2θ(x + 1)e

minus θx2

1 +πθ


2θ

(1 +πθ

radic)

1113946infin

0Erf

x2

radicθ4

1113888 1113889(x + 1)eminus θx2

dx

R4 1

θ

radic(1 +

πθ

radic)


radic1113872 1113873π

radic +1

θ4

2 + 1θθ241113872 1113873

1113969⎛⎜⎜⎜⎝ ⎞⎟⎟⎟⎠











x 2

θ

radic+

π

radic

2θ

radic(1 +

θπ

radic)

y1 2

θ1

1113968+

π

radic

2θ1

11139681 +

θ1π

1113968( 1113857

y2 1θ2

y3 θ3

π2

1113970

y4 θ4

2π

1113970

(19)


1113954θ 1113936


2+ nπ 1113936

mi1 xi

2π 1113936mi1 xi( 1113857

2 +

1113936mi1 xi minus n( 1113857

4+ 2nπ 1113936

mi1 xi 1113936


21113969

2π 1113936mi1 xi( 1113857

2

1113954θ1 1113936

mj1 y1j

minus m1113874 11138752

+ mπ 1113936mj1 y1j

2π 1113936mj1 y1j

1113874 11138752 +

1113936mj1 y1j

minus n1113874 11138754

+ 2mπ 1113936mj1 y1j

1113936mj1 y1j

minus m1113874 11138752

1113970

2π 1113936mj1 y1j

1113874 11138752

1113954θ2 m

1113936mj1 y2j

1113954θ3

2π

11139701113936

mj1 y3j

m1113888 1113889

1113954θ4

π2

11139701113936

mj1 y4j

m1113888 1113889

(20)







πθ

radic)n

minus 1 +

πθ11113969

1113874 1113875m

1113945

n

i1xi + 1( 1113857e

minus θx2i 1113945

m

j1y1j + 11113872 1113873e

minus θ1y21j (21)




πθ11113969

1113874 1113875

minus θ1113944n


m

j1y21j + 1113944

n

i1ln xi + 1( 1113857 + 1113944

m

j1ln y1j + 11113872 1113873

(22)


z ln L θ middot θ1( 1113857( 1113857

zθ

n

θminus

(nπ

radic)

(2θ

radic(1 +

πθ

radic))

minus 1113944n

(i1)

x2i 0


zθ1

m

θ1minus

mπ

radic

2θ1

11139681 +

πθ1

1113968( 1113857

minus 1113944m

j1y21j 0

(23)


1113954θ 1

6πA2 B + 2A(A + nπ) +

A2

n2π2

+ 4A(A minus 4nπ)1113872 1113873

B⎛⎝ ⎞⎠

1113954θ1 1

6πC2 D + 2C(C + mπ) +

C2

m2π2

+ 4C(C minus 4mπ)1113872 1113873

D⎛⎝ ⎞⎠

(24)



1j

B 8A6

minus 48πnA5

+ 51π2n2A4

minus π3n3A3

+ 33

radicπ32


1113872 111387313

(25)


radic


1113968)13


1113954R1 1 +

π1113954θ11113969

+ 21113954θ1113954θ1

1113969

Tanminus 11113954θ1113954θ11113872 1113873

1113969

1113874 1113875 +π

radic1113954θ minus 1113954θ11113872 1113873

1113954θ + 1113954θ1

1113969

1113874 1113875 minus 1113954θ1 1113954θ + 1113954θ11113872 11138731113872 11138731113874 1113875

1 +π1113954θ

1113968+

π1113954θ11113969

+ π1113954θ1113954θ1

1113969 (26)



1113954R2

π

radic

21113954θ

1113968+

π

radic 1113954θ21113874 1113875


241113954θ1113872 1113873Erfc1113954θ2

21113954θ

11139681113888 11138891113888 1113889

(27)



1113954R3 1113954θ

1 +π1113954θ

111396811113954θ

+

π

radic

1113954θ

1113968 minus2

21113954θ + 11113954θ231113874 1113875

minus

2π

radic

21113954θ + 11113954θ231113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(28)



1113954R4 1

1113954θ

1113968(1 +

π1113954θ

1113968)

21113954θCotminus 1 1113954θ2

21113954θ1113969

1113888 1113889

π

radic +1

1113954θ4

2 + 11113954θ1113954θ241113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(29)










θθ1

01 02 03 04 05 06 07 08 09 1

01 0208 0359 0472 0558 0624 0677 0718 0752 0780 080302 0149 0267 0364 0442 0507 0561 0606 0645 0678 070603 0121 0222 0307 0379 0440 0493 0538 0577 0611 064104 0105 0194 0271 0337 0395 0445 0490 0529 0563 059405 0093 0174 0245 0307 0362 0410 0453 0491 0526 055606 0085 0159 0225 0284 0336 0382 0424 0461 0495 052607 0078 0148 0210 0265 0315 0360 0400 0437 0470 050008 0073 0138 0197 0250 0297 0341 0380 0416 0448 047809 0068 0130 0186 0237 0283 0325 0363 0398 0430 04591 0065 0124 0177 0226 0270 0311 0348 0382 0413 0443

0000010002000300040005000600070008000900

θ1 = 01θ1 = 05θ1 = 1



θ = 01θ = 05

0000010002000300040005000600070008000900

Relia

bilit

y





θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1



θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References






















01113946

x

0f(x)g y1( 1113857dydx

1113946infin

01113946

x

0

2θ1 y1 + 1( 1113857eminus θ1y2

1

1 +πθ1

1113968⎛⎝ ⎞⎠2θ(x + 1)e

minus θx2

1 +πθ


(11)


R1 1 +

πθ1

1113968+ 2

θθ1

1113968Tanminus 1

θθ11113969

1113874 1113875 +π


θ + θ11113969

1113874 1113875 minus θ1 θ + θ1( 1113857( 11138571113874 1113875

1 +πθ

radic+

πθ1

1113968+ π

θθ1

1113968 (12)




R2 1113946infin

01113946

x

0θ2e


1 +πθ


R2

π

radic

2(θ

radic+

π

radicθ)


θ22

θ

radic1113888 11138891113888 1113889

(14)



g y3( 1113857 y3

θ22e

minus y232θ

23( ) y θ3 gt 0 (15)


R3 1113946infin

01113946

x

0

y3

θ23e

minus y232θ

23( )1113888 1113889


1 +πθ


2θ

(1 +πθ

radic)

1113946infin

0(x + 1)e

minus θx21 minus e

minus x22θ23( )1113874 1113875dx

R3 θ

1 +πθ

radic1θ

+

π

radic

θ

radic minus2

2θ + 1θ231113872 1113873minus

2π

radic

2θ + 1θ231113872 1113873

1113969⎛⎜⎜⎜⎝ ⎞⎟⎟⎟⎠

(16)



g y4( 1113857

2

radic

θ4π


42θ24( ) y θ4 gt 0 (17)


R4 1113946infin

01113946

x

0

2

radic

θ4π

radic1113888 1113889e

minusy24

2θ24⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠2θ(x + 1)e

minus θx2

1 +πθ


2θ

(1 +πθ

radic)

1113946infin

0Erf

x2

radicθ4

1113888 1113889(x + 1)eminus θx2

dx

R4 1

θ

radic(1 +

πθ

radic)


radic1113872 1113873π

radic +1

θ4

2 + 1θθ241113872 1113873

1113969⎛⎜⎜⎜⎝ ⎞⎟⎟⎟⎠











x 2

θ

radic+

π

radic

2θ

radic(1 +

θπ

radic)

y1 2

θ1

1113968+

π

radic

2θ1

11139681 +

θ1π

1113968( 1113857

y2 1θ2

y3 θ3

π2

1113970

y4 θ4

2π

1113970

(19)


1113954θ 1113936


2+ nπ 1113936

mi1 xi

2π 1113936mi1 xi( 1113857

2 +

1113936mi1 xi minus n( 1113857

4+ 2nπ 1113936

mi1 xi 1113936


21113969

2π 1113936mi1 xi( 1113857

2

1113954θ1 1113936

mj1 y1j

minus m1113874 11138752

+ mπ 1113936mj1 y1j

2π 1113936mj1 y1j

1113874 11138752 +

1113936mj1 y1j

minus n1113874 11138754

+ 2mπ 1113936mj1 y1j

1113936mj1 y1j

minus m1113874 11138752

1113970

2π 1113936mj1 y1j

1113874 11138752

1113954θ2 m

1113936mj1 y2j

1113954θ3

2π

11139701113936

mj1 y3j

m1113888 1113889

1113954θ4

π2

11139701113936

mj1 y4j

m1113888 1113889

(20)







πθ

radic)n

minus 1 +

πθ11113969

1113874 1113875m

1113945

n

i1xi + 1( 1113857e

minus θx2i 1113945

m

j1y1j + 11113872 1113873e

minus θ1y21j (21)




πθ11113969

1113874 1113875

minus θ1113944n


m

j1y21j + 1113944

n

i1ln xi + 1( 1113857 + 1113944

m

j1ln y1j + 11113872 1113873

(22)


z ln L θ middot θ1( 1113857( 1113857

zθ

n

θminus

(nπ

radic)

(2θ

radic(1 +

πθ

radic))

minus 1113944n

(i1)

x2i 0


zθ1

m

θ1minus

mπ

radic

2θ1

11139681 +

πθ1

1113968( 1113857

minus 1113944m

j1y21j 0

(23)


1113954θ 1

6πA2 B + 2A(A + nπ) +

A2

n2π2

+ 4A(A minus 4nπ)1113872 1113873

B⎛⎝ ⎞⎠

1113954θ1 1

6πC2 D + 2C(C + mπ) +

C2

m2π2

+ 4C(C minus 4mπ)1113872 1113873

D⎛⎝ ⎞⎠

(24)



1j

B 8A6

minus 48πnA5

+ 51π2n2A4

minus π3n3A3

+ 33

radicπ32


1113872 111387313

(25)


radic


1113968)13


1113954R1 1 +

π1113954θ11113969

+ 21113954θ1113954θ1

1113969

Tanminus 11113954θ1113954θ11113872 1113873

1113969

1113874 1113875 +π

radic1113954θ minus 1113954θ11113872 1113873

1113954θ + 1113954θ1

1113969

1113874 1113875 minus 1113954θ1 1113954θ + 1113954θ11113872 11138731113872 11138731113874 1113875

1 +π1113954θ

1113968+

π1113954θ11113969

+ π1113954θ1113954θ1

1113969 (26)



1113954R2

π

radic

21113954θ

1113968+

π

radic 1113954θ21113874 1113875


241113954θ1113872 1113873Erfc1113954θ2

21113954θ

11139681113888 11138891113888 1113889

(27)



1113954R3 1113954θ

1 +π1113954θ

111396811113954θ

+

π

radic

1113954θ

1113968 minus2

21113954θ + 11113954θ231113874 1113875

minus

2π

radic

21113954θ + 11113954θ231113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(28)



1113954R4 1

1113954θ

1113968(1 +

π1113954θ

1113968)

21113954θCotminus 1 1113954θ2

21113954θ1113969

1113888 1113889

π

radic +1

1113954θ4

2 + 11113954θ1113954θ241113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(29)










θθ1

01 02 03 04 05 06 07 08 09 1

01 0208 0359 0472 0558 0624 0677 0718 0752 0780 080302 0149 0267 0364 0442 0507 0561 0606 0645 0678 070603 0121 0222 0307 0379 0440 0493 0538 0577 0611 064104 0105 0194 0271 0337 0395 0445 0490 0529 0563 059405 0093 0174 0245 0307 0362 0410 0453 0491 0526 055606 0085 0159 0225 0284 0336 0382 0424 0461 0495 052607 0078 0148 0210 0265 0315 0360 0400 0437 0470 050008 0073 0138 0197 0250 0297 0341 0380 0416 0448 047809 0068 0130 0186 0237 0283 0325 0363 0398 0430 04591 0065 0124 0177 0226 0270 0311 0348 0382 0413 0443

0000010002000300040005000600070008000900

θ1 = 01θ1 = 05θ1 = 1



θ = 01θ = 05

0000010002000300040005000600070008000900

Relia

bilit

y





θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1



θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References





























x 2

θ

radic+

π

radic

2θ

radic(1 +

θπ

radic)

y1 2

θ1

1113968+

π

radic

2θ1

11139681 +

θ1π

1113968( 1113857

y2 1θ2

y3 θ3

π2

1113970

y4 θ4

2π

1113970

(19)


1113954θ 1113936


2+ nπ 1113936

mi1 xi

2π 1113936mi1 xi( 1113857

2 +

1113936mi1 xi minus n( 1113857

4+ 2nπ 1113936

mi1 xi 1113936


21113969

2π 1113936mi1 xi( 1113857

2

1113954θ1 1113936

mj1 y1j

minus m1113874 11138752

+ mπ 1113936mj1 y1j

2π 1113936mj1 y1j

1113874 11138752 +

1113936mj1 y1j

minus n1113874 11138754

+ 2mπ 1113936mj1 y1j

1113936mj1 y1j

minus m1113874 11138752

1113970

2π 1113936mj1 y1j

1113874 11138752

1113954θ2 m

1113936mj1 y2j

1113954θ3

2π

11139701113936

mj1 y3j

m1113888 1113889

1113954θ4

π2

11139701113936

mj1 y4j

m1113888 1113889

(20)







πθ

radic)n

minus 1 +

πθ11113969

1113874 1113875m

1113945

n

i1xi + 1( 1113857e

minus θx2i 1113945

m

j1y1j + 11113872 1113873e

minus θ1y21j (21)




πθ11113969

1113874 1113875

minus θ1113944n


m

j1y21j + 1113944

n

i1ln xi + 1( 1113857 + 1113944

m

j1ln y1j + 11113872 1113873

(22)


z ln L θ middot θ1( 1113857( 1113857

zθ

n

θminus

(nπ

radic)

(2θ

radic(1 +

πθ

radic))

minus 1113944n

(i1)

x2i 0


zθ1

m

θ1minus

mπ

radic

2θ1

11139681 +

πθ1

1113968( 1113857

minus 1113944m

j1y21j 0

(23)


1113954θ 1

6πA2 B + 2A(A + nπ) +

A2

n2π2

+ 4A(A minus 4nπ)1113872 1113873

B⎛⎝ ⎞⎠

1113954θ1 1

6πC2 D + 2C(C + mπ) +

C2

m2π2

+ 4C(C minus 4mπ)1113872 1113873

D⎛⎝ ⎞⎠

(24)



1j

B 8A6

minus 48πnA5

+ 51π2n2A4

minus π3n3A3

+ 33

radicπ32


1113872 111387313

(25)


radic


1113968)13


1113954R1 1 +

π1113954θ11113969

+ 21113954θ1113954θ1

1113969

Tanminus 11113954θ1113954θ11113872 1113873

1113969

1113874 1113875 +π

radic1113954θ minus 1113954θ11113872 1113873

1113954θ + 1113954θ1

1113969

1113874 1113875 minus 1113954θ1 1113954θ + 1113954θ11113872 11138731113872 11138731113874 1113875

1 +π1113954θ

1113968+

π1113954θ11113969

+ π1113954θ1113954θ1

1113969 (26)



1113954R2

π

radic

21113954θ

1113968+

π

radic 1113954θ21113874 1113875


241113954θ1113872 1113873Erfc1113954θ2

21113954θ

11139681113888 11138891113888 1113889

(27)



1113954R3 1113954θ

1 +π1113954θ

111396811113954θ

+

π

radic

1113954θ

1113968 minus2

21113954θ + 11113954θ231113874 1113875

minus

2π

radic

21113954θ + 11113954θ231113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(28)



1113954R4 1

1113954θ

1113968(1 +

π1113954θ

1113968)

21113954θCotminus 1 1113954θ2

21113954θ1113969

1113888 1113889

π

radic +1

1113954θ4

2 + 11113954θ1113954θ241113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(29)










θθ1

01 02 03 04 05 06 07 08 09 1

01 0208 0359 0472 0558 0624 0677 0718 0752 0780 080302 0149 0267 0364 0442 0507 0561 0606 0645 0678 070603 0121 0222 0307 0379 0440 0493 0538 0577 0611 064104 0105 0194 0271 0337 0395 0445 0490 0529 0563 059405 0093 0174 0245 0307 0362 0410 0453 0491 0526 055606 0085 0159 0225 0284 0336 0382 0424 0461 0495 052607 0078 0148 0210 0265 0315 0360 0400 0437 0470 050008 0073 0138 0197 0250 0297 0341 0380 0416 0448 047809 0068 0130 0186 0237 0283 0325 0363 0398 0430 04591 0065 0124 0177 0226 0270 0311 0348 0382 0413 0443

0000010002000300040005000600070008000900

θ1 = 01θ1 = 05θ1 = 1



θ = 01θ = 05

0000010002000300040005000600070008000900

Relia

bilit

y





θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1



θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References






















πθ

radic)n

minus 1 +

πθ11113969

1113874 1113875m

1113945

n

i1xi + 1( 1113857e

minus θx2i 1113945

m

j1y1j + 11113872 1113873e

minus θ1y21j (21)




πθ11113969

1113874 1113875

minus θ1113944n


m

j1y21j + 1113944

n

i1ln xi + 1( 1113857 + 1113944

m

j1ln y1j + 11113872 1113873

(22)


z ln L θ middot θ1( 1113857( 1113857

zθ

n

θminus

(nπ

radic)

(2θ

radic(1 +

πθ

radic))

minus 1113944n

(i1)

x2i 0


zθ1

m

θ1minus

mπ

radic

2θ1

11139681 +

πθ1

1113968( 1113857

minus 1113944m

j1y21j 0

(23)


1113954θ 1

6πA2 B + 2A(A + nπ) +

A2

n2π2

+ 4A(A minus 4nπ)1113872 1113873

B⎛⎝ ⎞⎠

1113954θ1 1

6πC2 D + 2C(C + mπ) +

C2

m2π2

+ 4C(C minus 4mπ)1113872 1113873

D⎛⎝ ⎞⎠

(24)



1j

B 8A6

minus 48πnA5

+ 51π2n2A4

minus π3n3A3

+ 33

radicπ32


1113872 111387313

(25)


radic


1113968)13


1113954R1 1 +

π1113954θ11113969

+ 21113954θ1113954θ1

1113969

Tanminus 11113954θ1113954θ11113872 1113873

1113969

1113874 1113875 +π

radic1113954θ minus 1113954θ11113872 1113873

1113954θ + 1113954θ1

1113969

1113874 1113875 minus 1113954θ1 1113954θ + 1113954θ11113872 11138731113872 11138731113874 1113875

1 +π1113954θ

1113968+

π1113954θ11113969

+ π1113954θ1113954θ1

1113969 (26)



1113954R2

π

radic

21113954θ

1113968+

π

radic 1113954θ21113874 1113875


241113954θ1113872 1113873Erfc1113954θ2

21113954θ

11139681113888 11138891113888 1113889

(27)



1113954R3 1113954θ

1 +π1113954θ

111396811113954θ

+

π

radic

1113954θ

1113968 minus2

21113954θ + 11113954θ231113874 1113875

minus

2π

radic

21113954θ + 11113954θ231113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(28)



1113954R4 1

1113954θ

1113968(1 +

π1113954θ

1113968)

21113954θCotminus 1 1113954θ2

21113954θ1113969

1113888 1113889

π

radic +1

1113954θ4

2 + 11113954θ1113954θ241113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(29)










θθ1

01 02 03 04 05 06 07 08 09 1

01 0208 0359 0472 0558 0624 0677 0718 0752 0780 080302 0149 0267 0364 0442 0507 0561 0606 0645 0678 070603 0121 0222 0307 0379 0440 0493 0538 0577 0611 064104 0105 0194 0271 0337 0395 0445 0490 0529 0563 059405 0093 0174 0245 0307 0362 0410 0453 0491 0526 055606 0085 0159 0225 0284 0336 0382 0424 0461 0495 052607 0078 0148 0210 0265 0315 0360 0400 0437 0470 050008 0073 0138 0197 0250 0297 0341 0380 0416 0448 047809 0068 0130 0186 0237 0283 0325 0363 0398 0430 04591 0065 0124 0177 0226 0270 0311 0348 0382 0413 0443

0000010002000300040005000600070008000900

θ1 = 01θ1 = 05θ1 = 1



θ = 01θ = 05

0000010002000300040005000600070008000900

Relia

bilit

y





θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1



θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References





















1113954R4 1

1113954θ

1113968(1 +

π1113954θ

1113968)

21113954θCotminus 1 1113954θ2

21113954θ1113969

1113888 1113889

π

radic +1

1113954θ4

2 + 11113954θ1113954θ241113874 1113875

1113970⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(29)










θθ1

01 02 03 04 05 06 07 08 09 1

01 0208 0359 0472 0558 0624 0677 0718 0752 0780 080302 0149 0267 0364 0442 0507 0561 0606 0645 0678 070603 0121 0222 0307 0379 0440 0493 0538 0577 0611 064104 0105 0194 0271 0337 0395 0445 0490 0529 0563 059405 0093 0174 0245 0307 0362 0410 0453 0491 0526 055606 0085 0159 0225 0284 0336 0382 0424 0461 0495 052607 0078 0148 0210 0265 0315 0360 0400 0437 0470 050008 0073 0138 0197 0250 0297 0341 0380 0416 0448 047809 0068 0130 0186 0237 0283 0325 0363 0398 0430 04591 0065 0124 0177 0226 0270 0311 0348 0382 0413 0443

0000010002000300040005000600070008000900

θ1 = 01θ1 = 05θ1 = 1



θ = 01θ = 05

0000010002000300040005000600070008000900

Relia

bilit

y





θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1



θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References






















θθ2

01 02 03 04 05 06 07 08 09 1

01 0500 0349 0274 0228 0197 0175 0157 0144 0133 012402 0651 0500 0412 0354 0312 0280 0256 0236 0219 020503 0726 0588 0500 0438 0392 0357 0328 0304 0285 026804 0772 0646 0562 0500 0453 0415 0384 0359 0337 031805 0803 0688 0608 0547 0500 0462 0430 0403 0380 036106 0825 0720 0643 0585 0538 0500 0468 0441 0417 039607 0843 0744 0672 0616 0570 0532 0500 0472 0448 042708 0856 0764 0696 0641 0597 0559 0528 0500 0476 045509 0867 0781 0715 0663 0620 0583 0552 0524 0500 04791 0876 0795 0732 0682 0639 0604 0573 0545 0521 0500

θ2 = 01θ2 = 05θ2 = 1

0000

0200

0400

0600

0800

1000

Relia

bilit

y



0000

0200

0400

0600

0800

1000

Relia

bilit

y


θ = 01θ = 05θ = 1



θθ3

01 02 03 04 05 06 07 08 09 1

01 0983 0963 0941 0917 0891 0864 0835 0806 0777 074702 0970 0936 0898 0858 0816 0773 0730 0687 0645 060403 0959 0913 0863 0810 0756 0702 0650 0600 0552 050804 0949 0892 0831 0768 0706 0645 0587 0533 0483 043805 0940 0874 0804 0732 0662 0596 0535 0479 0429 038506 0932 0857 0778 0700 0625 0555 0492 0436 0386 034307 0924 0841 0755 0671 0591 0519 0455 0399 0351 030908 0917 0826 0734 0644 0562 0488 0424 0368 0321 028109 0910 0812 0714 0620 0535 0460 0396 0342 0296 02571 0903 0799 0695 0598 0511 0435 0372 0318 0274 0238


θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References





















θ3 = 01θ3 = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y



θ = 01θ = 05

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y




θθ4

01 02 03 04 05 06 07 08 09 1

01 0989 0977 0964 0950 0934 0918 0901 0884 0866 084802 0981 0960 0937 0913 0888 0862 0835 0808 0781 075503 0974 0946 0915 0883 0850 0816 0783 0750 0718 068704 0968 0933 0896 0857 0817 0778 0740 0703 0668 063405 0963 0921 0878 0834 0789 0745 0704 0664 0627 059206 0957 0911 0862 0812 0764 0717 0672 0630 0592 055707 0952 0901 0847 0793 0741 0691 0644 0602 0562 052708 0948 0891 0833 0775 0720 0668 0620 0576 0537 050109 0943 0883 0820 0759 0701 0647 0598 0554 0514 04791 0939 0874 0808 0744 0683 0628 0578 0533 0494 0459

θ4 = 01θ4 = 05θ4 = 1

0000

0200

0400

0600

0800

1000

1200

Relia

bilit

y






(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References























(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)



(1 15) minus 00419 00262 00172 00146 minus 00063 00024 0001100678 00643 00557 00478 00423 00372 00370





θ = 01θ = 05θ = 1

0000020004000600080010001200

Relia

bilit

y





(5 5) (10 10) (20 20) (30 30) (40 40) (50 50) (100 100)

(1 1) 00051 00020 00020 00013 00010 00005 0000300159 00104 00090 00035 00024 00020 00013

(1 05) 00123 00095 00037 00033 00025 00018 0000900148 00112 00051 00049 00046 00044 00044

(1 15) 00077 00072 00044 00030 00022 00020 minus 0000100135 00091 00054 00051 00034 00021 00020

(1 2) 00055 00050 00048 00033 00022 00019 minus 0000300115 00086 00059 00059 00049 00049 00009

(05 1) 00090 00038 00038 00036 00031 00024 0001000092 00087 00070 00050 00050 00048 00046







6 Conclusions


Data Availability




References
























6 Conclusions


Data Availability




References





















Documents

Estimate Stress-Strength Reliability Model Using Rayleigh