Weibull Distribution

Weibull distribution is very flexible in fitting empirical data, such as strength or time to failure. Several methods for the determination of parameters are described, including di‐ rect fitting using solvers available in universal programs. Also finding of parameters of exponential distribution is described. The use of Weibull distribution is illustrated on ex‐ amples.

A special position in reliability assessment pertains to Weibull distribution, which offers great flexibility in fitting empirical data.The distribution function (Fig. 1a) is with parameters a, b, and t 0 .The scale parameter a is related to the values of t and ensures that the distribution is independent of the units of t (e.g.minutes or hours).The constant b is shape parameter.Depending on its value, Weibull distribution can approximate various, even very different shapes (Fig. 5 in Chapter 2).It is suitable for the characterization of time to failure as well as strength or load; therefore, it became popular in reliability assessment.The constant t 0 is the threshold value that corresponds to the minimum possible value and characterizes the position of the distribution on the t-axis.(t is the usual symbol for time; for other quantities, other symbols may be used.)

Determination of parameters in a two-parameter distribution
The strength or time to failure cannot attain negative values, so that the threshold parameter is often assumed zero, t 0 = 0.The distribution function (1) will thus have only two parameters: ( ) ( ) Parameters a and b can be found easily, as the transformed data can be fitted by a straight line.Double logarithmic transformation and rearrangement change Equation (2) to which corresponds to the equation of straight line (Fig. 1b) where The method of linearization was very popular in the past, and it is still often used for the determination of parameters from the operation data via a special diagram, called Weibull paper (Fig. 1b).For its construction, the individual measured values t j and the corresponding values F j of the empirical distribution function are needed.The t j values are obtained by rankordering the n data from operation (e.g. times to failure) from the minimal value (j = 1) to maximal (j = n).The corresponding values of distribution function are calculated as ( ) j is the rank number and n is the total number of measured values.The explanation of formula (5), common for order statistics, is simple.If we have, say, 100 values and order them from the minimal to maximal, then the probability F that t will be smaller or equal to the lowest of 100 values, t 1 , is 1:100.The probability of t ≤ t 2 is 2/100, etc.; generally, F j = j/n.In Equation ( 5), 1 was added to the denominator because of mathematical correctness; the probability F that t will be smaller or equal t n must be smaller than 1, simply because if more measurements would be done, values higher than t n could appear.Also other formulas exist for the calculation of empirical F j values [e.g.F j = (j -½)/n], but none can be recommended unequivocally, especially when considering the fact that bigger errors in the determination of distribution parameters can arise due to the small amount of data than due to the formula used for F j .
The regression constants A, B can be obtained by fitting the empirical data by a straight line (using Weibull paper or a program for curve fitting, such as "Insert Trendline" in Excel).Then, the constants in the distribution function (2) are obtained from A and B by inverse transformation: ( ) Plotting the empirical data into the coordinate system X = ln{ln[1/(1 -F)]}, Y = ln t, enables a good visual check.In the ideal case, if Equation ( 2) is valid, the data lie on a straight line.

Determination of parameters in a three-parameter distribution
A two-parameter distribution is not always suitable.Sometimes, the transformed data do not lie on a straight line, or it is obvious that the distribution should have a threshold value t 0 higher than zero.In such case, the use of a two-parameter distribution as a base for dimensioning could lead to uneconomical design, and a three-parameter function (1) would be better.
The parameters in this distribution can be found by the procedure for a two-parameter function if t in Equation ( 2) is replaced by the expression t -t 0 ; the constant t 0 must be chosen in advance.For various t 0 values, the shape of empirical distribution varies.The best t 0 value is such for which the transformed data best resemble a straight line.However, a more straightforward procedure exists.

Direct determination of parameters
The constants a, b, and t 0 can also be obtained in a simpler way without any transformation.The solution of Equation (1) for t gives the formula for quantiles: This equation and the least-squares method are used in search for such values of a, b, and t 0 , which minimize the sum of squared differences between the measured and the calculated values of t, If a suitable solver is available for such minimization (one is present also in Excel), it is then sufficient to prepare one series of measured data, t j,meas , and another series of the t j,calc values, calculated via Equation (7) for the same values of Fj using the parameters a, b, and t 0 .Solver's command to minimize the expression (8) by changing a, b, and t 0 will do the job.An example is shown at the end of this chapter.
Remark: Formula (7) is also suitable for the determination of a "minimum guaranteed value"(e.g.strength or time to failure) for acceptably low probability F.
In addition to flexibility, Weibull distribution has one more advantage.The shape parameter b in Equation ( 1) or ( 2) is related to the character of failures.This is well visible at the bathtub curve (Fig. 1 in Chapter 4).The values b < 1 are typical of decreasing failure rate λ and may thus indicate the period of early failures.On the contrary, b >1 corresponds to increasing failure rate λ and is typical of the period of aging or wear out.The value b = 1 corresponds to the constant failure rate λ = const, with failures from many various reasons (see Chapter 4).The exponent b thus can inform generally about the possible kind of failures and about the period in the life of an object even if the amount of data is not large.However, caution is necessary.
If the data from a long period are fitted by Weibull distribution, failures from various reasons and stages can be mixed, and the relation of b to the kind of failures is not unambiguous.
Remark: Weibull distribution was proposed in 1939 by the Swedish engineer Waloddi Weibull, who studied the strength of materials, life endurance of ball bearings, and fatigue life of mechanical components and other quantities.Later, it appeared that this very useful distribution belongs to the family of extreme value distributions [1, 2].More on Weibull distribution and its applications can be found, for example, in [3 -5].

Exponential distribution
Let us now look at a special and very important case.With the shape parameter b = 1, Weibull distribution simplifies to exponential distribution The probability density and distribution function are depicted in Fig. 5.The parameters a and t 0 can be determined similarly as described above.If t 0 = 0, the remaining parameter a is usually calculated from the mean time to failure, as it will be shown in Chapter 20.Typical of exponential distribution is that the standard deviation has the same or similar value as the mean.
The determination of parameters and use of Weibull and exponential distribution will be demonstrated in the following examples.

Example 1
The strength (S) of a new alloy was measured on seven specimens, with the following results: 203, 223, 248, 265, 290, 313, and 342 MPa.Solve the following three problems: A. Determine the parameters of Weibull distribution for this alloy using: a. Two-parameter distribution and linearized data; b. Two-parameter distribution, applying Solver on the original data without transformation); and c.Three-parameter distribution, applying Solver on the nontransformed data.
B. Calculate (for each case) the probability that the strength will be lower than 120 MPa.
C. Calculate (for each distribution) the "minimum guaranteed" strength such that the probability of the actual strength being lower equals: 0.05 -0.01 -0.001.

Solution.
Task A. Determination of distribution parameters a. Linearized two-parameter Weibull distribution.The strength values, ordered from minimum to maximum, are given in Table 1 together with the values of distribution function, calculated as F j = j/(n + 1), with n = 7; see also Fig. 2. The distribution function ] was transformed to linear form; see Equation ( 4) and the following formulas.The transformed values are in the columns X j and Y j .Note: The values of distribution function are fixed (deterministic), as they correspond to the number of measured values, whereas the strengths exhibit random variations.Therefore, F is the independent variable and t is the dependent variable.The transformed values were fitted by linear function (4); see columns X j and Y j in Table 2.The regression constants were A = 5.673642 and B = 0.194844.The inverse transformation has given a = exp A = 291.0928.13231 ].The corresponding calculated values S j are in column S j,c,lin2 and depicted by a curve in Fig. 2.
b. Two-parameter distribution, application of Solver on untransformed data.In this case, the strength values t j,calc were calculated for the individual F j values using Equation ( 7), with t 0 = 0. (See column t j,calc,2p .)Now, a quantity for characterization of the quality of the fit was defined: sum of the squared differences of the measured and calculated strengths, ∑(t meas -t calc ) 2 .The Solver then changes the constants a and b of the distribution function (7) automatically until the sum of squared differences attains a minimum.In the investigated example, the "optimum" constants were a = 291.7807and b = 4.964505, near to the results of the linearized problem.The calculated values are in column S j,c,sol2 .
c. Three-parameter distribution, application of Solver on untransformed data.The difference from the previous case is the full form of distribution function (7).Also here the sum of the squared differences of the measured and calculated strength, ∑(t meas -t calc ) 2 , was minimized.The resultant constants were a = 154.9796,b = 2.4156, and t 0 = 133.7975.The calculated values are in column S j,c,sol2 .The calculated distribution function is plotted by the thick curve in Fig. 2. Also the distribution function of a two-parameter distribution function is shown (thin curve).The curves for cases (a) and (b) were very close to each other.

Task B. Determination of probability S ≤ 120 MPa
The probabilities are as follows: a. 0.010533, (b) 0.012069, and (c) 0; the minimum possible value is t 0 = 133.8MPa.

Task C. Determination of guaranteed strength
Concise Reliability for Engineers The results are in the following table.Note the big difference between the two-and three-parameter distributions for very low failure probabilities (cf.also Fig. 2).According to the three-parameter model, the minimum (threshold) strength is 133.8MPa.Calculate also the standard deviation, so that you can assess whether exponential distribution may be used for the time to failure.

Figure 2 .
Figure 2. Measured values of strength (S) and approximate distribution functions (F) for various approximations in Example 1. Thick curve -case c, three-parameter function; thin curves -cases a, b, two-parameter curves.

Example 2
Eight components (n = 8) were tested until failure.The failures occurred at the following times t j : 65, 75, 90, 120, 250, 510, 520, and 760 h.Calculate the mean time to failure and failure rate.

Table 1 .
Measured values S(F j ) and those calculated using three methods.