Probability distributions

Although there are a large number of probability or statistical distributions in published literature, this section presents just four such distributions considered useful for performing various types of reliability, maintainability, and safety analysis studies concerned with engineering systems [14-16].

Exponential distribution

This is one of the simplest continuous random variable distributions frequently used in the industrial sector, particularly for performing reliability studies. Its probability density function is expressed by [9,17]

f(t) = ae “, for t > 0, a > 0

(2.36)

where

f(t) is the probability density function.

t is the time (i.e., a continuous random variable).

a is the distribution parameter.

By inserting Equation (2.36) into Equation (2.21), we get the equation for the cumulative distribution function:

F(f) = l-e"“'

(2.37)

With the aid of Equations (2.26) and (2.36), we obtain the following equation for the distribution expected value (i.e., mean value):

E(f) = -

  • (2.38)
  • 7 a

Example 2.9

Assume that the mean time to failure of engineering system is 1000 hours. Calculate the failure probability of the engineering system during a 500-hour mission by using Equations (2.37) and (2.38).

By inserting the specified data value into Equation (2.38), we obtain

a = —i— = 0.001 failures per hour 1000 r

By substituting the calculated and the given data values into Equation (2.37), we get

F(500) = e‘(0tx,1)(500)

= 0.6065

Thus, the failure probability of the engineering system during the 500-hour mission is 0.6065.

Rayleigh distribution

This continuous random variable probability distribution is named after its founder, John Rayleigh (1842-1919) [1], and its probability density function is defined by

for 0 > 0, t > 0

(2.39)

where

0 is the distribution parameter.

By substituting Equation (2.39) into Equation (2.21), we obtain the following equation for the cumulative distribution function:

(2.40)

By inserting Equation (2.39) into Equation (2.26), we get the following equation for the distribution expected value:

E(t) = 0f

  • 2/
  • (2.41)

where

f(.) is the gamma function and is defined by

00

r(n) = forn>0

  • (2.42)
  • 0

Weibull distribution

This continuous random variable distribution was developed by Waloddi Weibull, a Swedish professor in mechanical engineering in the early 1950s and its probability density function is defined [18]

/(1) = ^^ fort>0,b>0,e>0 (2A3)

0

where

b and 0 are the distribution shape and scale parameters, respectively.

By inserting Equation (2.43) into Equation (2.21), we obtain the following equation for the cumulative distribution function:

F(t) = l-e’(f/0)i’ (2.44)

By substituting Equation (2.43) into Equation (2.26), we obtain the following equation for the distribution expected value:

E(t) = 0r|l + ^| (2.45)

It is to be noted that for b = 1 and b = 2, the exponential and Rayleigh distributions are the special cases of this distribution, respectively.

Bathtub hazard rate curve distribution

This is another continuous random variable distribution, and it can represent bathtub-shaped, increasing and decreasing hazard rates. This distribution was developed in 1981 [19], and in the published literature by other authors around the world, it is generally referred to as the Dhillon distribution/law/model [20-39].

The probability density function of the distribution is expressed by [19]

/(t) = 170(et)!,'1e I > for t>0,6>0,b>0 (2.46)

where

b and 9 are the distribution shape and scale parameters, respectively.

By substituting Equation (2.46) into Equation (2.21), we obtain the following equation for cumulative distribution function:

-W-il

F(f) = l-el 1 (2.47)

It is to be noted that for b = 0.5, this probability distribution gives the bathtub-shaped hazard rate curve, and for b = 1, it gives the extreme value probability distribution. In other words, the extreme value probability distribution is the special case of this probability distribution at b = 1.

Solving first-order differential equations with Laplace transforms

Usually, Laplace transforms are used to find solutions to first-order linear differential equations in reliability, maintainability, and safety analysis-related studies of engineering systems. The example presented below demonstrates the finding of solutions to a set of linear first-order differential equations, describing a engineering system in regard to reliability and safety, using Laplace transforms.

Example 2.10

Assume that an engineering system can be in any of the three states: operating normally, failed safely, or failed unsafely. The following three first-order linear differential equations describe the engineering system under consideration:

^iû+(xs+x„)po(t)=o

(2.48)

^-W) = 0 at

(2.49)

(2.50)

where

Xs is the engineering system constant safe failure rate, is the engineering system constant unsafe failure rate.

Pi (t) is the probability that the engineering system is in state i at time, for i = 0 (operating normally), i = 1 (failed safely), and i = 2 (failed unsafely).

At time t = 0, Po (0) = 1, Pi (0) = 0, and P2 (0) = 0.

Solve differential Equations (2.48)—(2.50) by using Laplace transforms.

Using Table 2.1, differential Equations (2.48)-(2.50), and the given initial conditions, we obtain

sP0(s)-l + (Xs+X„)Po(s) = O

(2.51)

sP!(s)-XsPo(s)=0

(2.52)

sft(s)-X„Pu(s) = 0

(2.53)

By solving Equations (2.51)—(2.53), we get

Po(s)=, ? . '

(s + A.s + X„)

Pl(s)= z x

A(s)= z

+ XM J

  • (2.54)
  • (2.55)
  • (2.56)

By taking the inverse Laplace transforms of Equations (2.54)-(2.56), we obtain

P0(t) = e‘<Xs+x“)'

(2.57)

  • (2.58)
  • (2.59)

Thus, Equations (2.57)-(2.59) are the solutions to differential Equations (2.48)-(2.50).

 
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