Reliabil ity: Statistical Statistical Aspec Aspects ts
Piet Watté Philips Industrial Applications, Global Technology Development Compact High Intensity Discharge Discharge lamps Ceramic Metal Halide Steenweg op Gierle, 417 B-2300 Turnhout, PO 57 tel: (+32) (0) 14 401 631 fax: (+32) (0) 14 407 647 e-mail:
[email protected]
1. Introduction
Waloddi Weibull, a Swedish scientist published many papers on rupture in solids, bearings, ...[1] The well-kn well-known own Weibull distribution distributi on in reliability is named after him. The following paragraphs paragraphs describe this distribution function and how it can be applied to rel iability data.
The weibull distribution function and derived functions How to use these functions, in the framework framework of reliabilit y studies Practical applications in the li ghting industry and two algorithms algorithms to make Weibull plots
2. The Weibull cumulative distribution function
Assume a 2 parameter Weibull distribution with the following shape, b, and scale parameter, a.
t b F t a b 1 exp a
Eq. 1
F is the cumulative probability for failure of a product, t is its it s lifetime expressed in years of operation 1.05 0.84 F ( t 1 .5) F ( t 3 5)
0.63
F ( t 2 1) 0.42 0.21 0
0
2
4
6
t
The figure below is the same, but in a logarithmic plot 1
F ( t 1 .5) F ( t 3 5)
0.1
F ( t 2 1)
0.01 0.01
0.1
1 t
10
Figure 1: Influence of the scale and shape parameter on the Weibull distribution function. Shape and scale factors were respectively 1 and 0.5 (red curve), 3 and 5 (blue curve) and 2 and 1 (brown curve). In practice this could be three Weibull curves related to a specific failure mode. 3. From a Weibull distribution to a Weibull distribution density function f(t,a,b)
The Weibull distribution density function is defined as: d f t a b F t a b d t b
1 b1 t a t b e
a
f t a b
a
R is coined as the survival function. It lists the fraction of products that still operate after a time t b
R t a b
1 F t a b
1 t a R t a b e
This brings us to the failure rate function. This function lists the failure rate or the speed with which the products are failing over the lifetime of a product. The failure rate function r(t): definition
r t a b
f t a b R t a b
b1
t b a r t a b a
Examples a 2
b 5
R 5 a b
1
0
a
0.5
20
1
15 R ( t a b) 0.5
r ( t a b) 10 5
0
0
1
2
0
3
0
1
t
2
3
t
Figure 2: The reliability (or survival) function is plotted at the left. It gives the probability that products with a life t are still operating. On the right the failure rate function is plotted. It depicts the rate with which the products are failing. As can be seen the failure rate increases progressively over the product life The failure rate function is discussed below in the framework of the "concept" of the bath tub curve. 4. Graphical representation of the Weibull distribution density function
2
f ( t 1 0.5) f ( t 2 5)
1.5
1
f ( t 1 1) 0.5
0
0
1
2
3
t
Figure 3: Weibull density functions for different values of a and b.
5. Construction of the bath tub curve out of the function r(t)
In reliability literature [8] the t erm 'bath tub' curve refers to the failure rate function. Below it is shown what is meant by this. In figure 4 the failure rate functions are computed for three different Weibull curves. We have chosen the shape and scale parameters in order to yield a bath tub curve. One can classify failures by means of the shape parameter of the Weibull curve:
low values of b point to the i nfant mortality phase (red curve) a value of b equal to 1 points to the constant failure rate phase (magenta curve) a value of b higher than 1 points to the wear out phase (blue curve)
3
r ( t 3 0.5) 2 r ( t 3 5) r ( t 3 1)
1
0
0
1
2 t
3
Figure 4: Construction of a so-called bath tub curve. The bath tub shape is achieved by making the envelope of the red curve t = 0.72, the magenta curve to t = 2 and the blue curve after t = 2
Most products or equipment are freed from infant mortalities at the end of the design phase and establish only wear out modes 6. Intrepretation of Weibull data (see eq. 3)
Two frequently asked questions in reliability engineering are: a. What is the probability that products, having a Weibull distribution density function as described above, operate more than 2 years without failure?
P T
a
=
R a
For a equal to 2 years R 2 a b
or 37% b. What fraction of the products that have survived 1 year of operation, will fail in year 2? P a
T
b
F 2 a b
0.368
T
a
=
F a R a
F b
F 1 a b 0.62 R 1 a b
Eq. 2
or 62%.
7. Average life and 50% rated life from the Weibull distribution
7.1. Weibull average life time The average life time is equal to (here evaluated with a setting of a = 2 and b = 5): 2 Γ
∞
0
t f t a b d t
1 5
5
Here we exploit a property of the gamma function. In general it can be shown that the Mean Time To Failure of a two-parameter Weibull function defined above is equal to: Γ
1 1 a b
Let us check this, as the MTTF (Mean Time To Failure), or the Mean is equal to the first moment of the Probability Density Function being the first moment of the PDF. ∞
M1 0
t f t a b d t
is the Weibull average life time is the gamma function A property of the gamma function is Γ x 1 that M1
1 b
1.836
Γ a 1
b
=
xΓ x
1.836
Thus the Weibull average life time, or the MTTF (mean time to failure) is equal to: MTTF
=
1 b
Γ a 1
b
Eq. 3
As can be verified very easily this i s not equal to the 50%
rate e 7.2 Weibull 50% rated lifetime The 50% rated life t50 % is defined as
t50% b 1 exp a
=
0.5
From which we can compute that 1 b
t50% a ln 2
Eq. 4
As you can observe it is not equal to the MTTF.
t50%
1.859
8. Standard deviation of the Weibull density function
The standard deviation of a pdf is per definiti on [2] σ
M2 M1
=
2
in which the second moment M2 is equal to ∞
M2
=
0
2
t f t a b d t
=
∞
0 Γ
2
t f t a b dt
1 2 a 2 3.549 b
3.549
Γ
1 2 a 2 b
or 2
2 Γ 1 1 a σ Γ 1 b b σ
0.421
9. Rule of mixture
Sometimes, when dealing with multiple failure modes, we use a rule of mixture in order to find the overall Weibull pdf (see eq. 6) Suppose that we have 3 failure modes:
Failure mode 1 appears for 20% in the failure data and has a Weibull distribution F(t,1,.5) Failure mode 2 appears for 35% in the failure data and has a Weibull distribution F(t,3,5) Failure mode 3 is the remaining failure mode and has a Weibull distribution F(t,2,1)
The corresponding weighing factors are q1 .2 q2 .35 The overall Weibull function can be calculated using this rule of mixture as: Foverall t q1 q2
1 q1 1 F t 1 .5 q2 1 F t 3 5 1 q1 q2 1 F t 2 1
Figure 5 below depicts the overall Weibull probability density functions, together with the one pertaining to constituent pdfs of t he individual failure modes
1.05 0.875 Foverall ( t q1 q2)
0.7
F ( t 1 .5) F ( t 3 5)
0.525
F ( t 2 1)
0.35 0.175 0
0
1.5
3
4.5
6
t
1
Foverall ( t q1 q2) F ( t 1 .5) F ( t 3 5)
0.1
F ( t 2 1)
0.01 0.1
1
10
t
Figure 5: Application of the rule of mixture for constructing a general Weibull curve, out of three separate failure modes.
9. Practical cases
These are typical examples from the lighting industry, where reliability tools are a useful help in order to select the most optimum light solution. 9.1. Case 1
A lamp installer has to chose between two different kind of lamp systems for an installation in a fashion shop lit by 50 l amps. Manufacturer A gives a system performance with a MTTF of 12000 h and a shape factor b of 4, whereas manufacturer B gives a system performance with a MTTF of 14500 h and a shape factor b of 2.3 Compare both manufacturers on the basis of their 10% failure point. The installer wants to have the largest 10% failure point. In this respect, what lamp system should he prefer? Solution
a) Manufacturer A holds the following MTTF 12000 b 4 a
MTTF 1 b
Γ 1
1.324 10
4
b
From equation 1, it is straightforward to derive that the x% failure point can be computed by means of the function FP 1 b
100 FP x a b a ln 100 x FP 10 13240 4
7.543 10
3
b) Manufacturer B it is equal to MTTF 14500 b 2.3 MTTF
a
1 b
Γ 1
1.637 10
4
b
FP 10 16370 2.3
6.154 10
3
So from the point of view of 10% failure point he should chose manufacturer A. Although the MTTF is lower, the scatter on the mean life is larger for B, yielding a lower 10% failure point. 9.2 Case 2
The shop owner of the fashion store debates with the installer about the time to replace its installed lamp park. He prefers to do a group replacement, but preferably later than 6000 h. He still doubts whether to do a group replacement after 5 or after 10 lamps of the 50 installed ones fail. What is the probability for group replacement for both criteria. Lamp manufacturer A offers him an improved concept: the mean life has not changed, but the shape factor is now 6. Calculate the probability that 1 to N lamps fail as a function of lifetime.
Solution
MTTF 13240 b 6
MTTF
a
1 b
Γ 1
1.427 104
b
With some calculus, we find that the probability t hat exactly M lamps fail out of N installed lamps is equal to [7]: P1 M N t
combin N M F t a b
M
1 F t a b NM
The probability that 1 or 2 or ... M lamps (out of the total of N) fail is equal to [7]: M
P2 M N t
i
combin N i F t a b
i
1 F t a b Ni
1
Figure 6 depicts the probability for group replacement as a function of the number of failed lamps
1.1
P2 ( 25 50 t) P2 ( 5 50 t)
0.825
0.55
P2 ( 10 50 t) 0.275
0 0
3
2.510
3
510
3
7.510
4
4
4
110 1.2510 1.510
t
Figure 6: Probability of failure of 5, 10 and 25 of the 50 installed lamps as a function of lifetime
The criterion for group replacement when 5 lamps fail is already met after 7800 h, the criterion for 10 failed lamps is met after 9000 h 9.3 Fitting of Weibull curves 9.3.1 Case of non censored data [7]
Data are uncensored if all the l amps of the population burnt to failure. Suppose that this was the case and that the lamp system failed at following lamp lifes: 1216, 5029,13125,15987 and 29301 h. Calculate its Weibull shape and scale parameter.
1216 5029 Life 13125 15987
the failure data has to be fed into the vector Life
29301 Note that by plotting ln(ln(1/(1-F)) vs ln(t) you achieve a linear plot with the shape parameter as the slope and the intercept on the y axis equal to -b x ln(a) or -shape times ln(scale). This follows from the definition of the Weibull function. This enables your to find these parameters by means of linear regression. The following function returns the shape and scale parameter of a Weibull fit to the data contained in the vector L. In the for loop in the third li ne it uses the formula for the estimator of the rank F, explained in Nelson [4]
Weib_uncens L
X ln L N length L for j 0 N 1 Y j
ln ln
1 1
j 1 0.3 N 0.4
A0 slope X Y A1 exp
intercept X Y A0
A3 corr X Y 1
A2 A1 ln 2
A0
A
The first element is the shape factor, the second element the scale factor (63% failure point), the third element is the 50% rated life, and the fourth element is the correlation coefficient of the fit
0.824 1.492 104 Weib_uncens Life 3 9.563 10 0.987
graph_Weib_uncens L
X ln L N length L for j 0 N 1 Y j
ln ln
1
j 1 0.3 N 0.4 GW augment X Y 1
Weib_uncens Life 0 i b i b ln rl x
b ln Weib_uncens Life 1
X ln Life
b x i
G graph_Weib_uncens Life 1
0 ) ) F 1 ( / 1 ( n l ( n l
0
rl G
1
1
G
2
3
7
8
9
10
11
0
G
ln(t) Figure 7: Weibull plot of the data in exercise 9.3.1 The algorithm with censored data is a little bit more complicated, it is also listed in [7].
9.3.2 Case of censored data
Suppose we have failure data from a product. Not all the individual products failed; the ones that continue to operate are called censored products. Although they have not operated to catastrophic failure, they can be taken up in the Weibull analysis. The failure data has to be fed into the matrix Life_cens. The first column of the matrix contains the lifetime of the products. The second column contains a c (censored) or a f (failed) product
105 217 255 598 641 Life_cens 812 922 1030 1059 1208
"c"
"c" "c" "c" "f" "f" "c" "c" "f" "c"
The function below calculates the Weibull shape, scale parameter and MTTF for the censored data in the matrix above
Weib_censor L
0 N rows L for i 0 N 1 Q j 0 i 1 if Li 1 j
Q j 1 Li 0 if Li 1 j
j 1
if Li 1
=
=
=
The first for loop extracts out of the life input matrix only the data with a "f" or failed censoring value.
"f" "f"
"f"
r 0 0 for j 1 rows Q r j
r j 1
1 r j 1 rows L 1 Q j 1 0 1 rows L
The second for look computes the adjusted ranks to the lamps that failed in the input matrix.
V ln Q 1
for k 0 rows Q
1
Wk ln ln
1
1
r k1 0.3 rows L 0.4
A0 slope V W A1 exp
intercept V W A0 1
A2 A1 ln 2
A0
A3 corr V W
The third for loop calculates the data of the Weibull plot using these adjusted ranks.
4.767 1.164 103 Weib_censor Life_cens 3 1.078 10 0.988
The first element is the shape factor, the second one the scale factor. The 50% rated life is the third element, and the last element is the correlation coefficient of the Weibull plot.
Using the earlier defined function FP, no matter what failure point can be calculated when knowing the Weibull shape and scale parameter. Here is a function to create the Weibull plot. graph_Weib_censor L
0 N rows L for i 0 N 1 Q j 0 i 1 if Li 1 j
Q j 1 Li 0 if Li 1 j
j 1
if Li 1
=
=
=
"f" "f"
"f"
r 0 0 for j 1 rows Q r j
r j 1
1 r j 1 rows L 1 Q j 1 0 1 rows L
V ln Q 1
for k 0 rows Q
1
Wk ln ln
1
1
r k1 0.3 rows L 0.4
GW augment V W
b Weib_censor Life_cens 0 i
b ln Weib_censor Life_cens 1
rl x
b x i
G graph_Weib_censor Life_cens
1
) ) t ( F 1 / 1 ( n l ( n l
0
1
G
0
rl G
1
2
6.6
6.8
7
0
G
ln(t) Figure 8: Weibull plot of the data in exercise 9.3.2 10. References
1. http://www.weibull.com. 2. http://en.wikipedia.org/wiki/Weibull_distribution. 3. S. Nahmias, 'Production and Operations analysis, IRWIN, (1993), Chapter 11. 4. W. Nelson, "Accelerated testing, statistical models, test plans and data analysis", Wiley, 2004. 5. W. Meeker, L. Escobar, 'Statistical methods for reliability data', Wiley, (1998). 6. P. Zinck, J.F. Gérard, H.D. Wagner, Engineering Fracture Mechanics, 69 (2002), 1049. 7. http://www.mathpages.com/home/kmath122/kmath122.htm. 8. K. and A. Bhote, "World class reliability, using MEOST to make it happen," Amacom, 2004.
