In Auburngrads/SMRD: Statistical Methods For Reliability Data
SMRD:::vinny()
library(SMRD)
In this echapter...
Get Planning Values by Specifying a Point and a Slope
Specify a slope value and a single point $t, F(t)$
plan.values1 <- get.plan.values("Weibull",
beta = 2,
prob = .1,
time = 100,
time.units = "Hours")
summary(plan.values1)
plot(plan.values1)
failure.probabilities(plan.values1)
Carry out life test simulation using specified planning values. The function below plots several individual test simulations (the number of simulations is equal to number.detail) along with a plot of several simulations depending on the value of the argument number.sim (default is 2000).
life.test.simulation(plan.values1,
n = 50,
censor.time = 120,
number.detail = 5,
quantile.mark = 0.2,
number.sim = 200)
Try a longer test
life.test.simulation(plan.values1,
n = 50,
censor.time = 300,
number.detail = 5,
number.sim = 200)
Other optional arguments and skip detail
life.test.simulation(plan.values1,
n = 50,
censor.time = 1000,
number.sim = 50,
quantile.mark = 0.2)
Now try lognormal distribution
plan.values2 <- get.plan.values("Lognormal",
sigma = 0.5,
prob = 0.1,
time = 100,
time.units = "Hours")
summary(plan.values2)
plot(plan.values2)
plot(plan.values2,
censor.time = 1000,
grids = F)
life.test.simulation(plan.values2,
n = 50,
censor.time = 1000,
quantile.mark = 0.1)
Get planning values by specifying two points
plan.values3 <- get.plan.values("Weibull",
prob = c(.2,.12),
time = c(1000,500),
time.units = "Hours")
plan.values4 <- get.plan.values("Weibull",
prob = c(.05,.15),
time = c(40000,100000),
time.units = "Hours")
summary(plan.values3)
plot(plan.values3)
life.test.simulation(plan.values3,
n = 50,
censor.time = 1000,
quantile.mark = 0.1)
#compare the simulated value with the large-sample approx below
asd.quant(plan.values3,
n = 50,
censor.time = 1000,
quantile.mark = 0.1)
#compare:
asd.quant(plan.values3,
n = 50,
censor.time = 1000,
quantile.mark = 0.1) * sqrt(50)
asd.quant(plan.values3,
n = 500,
censor.time = 1000,
quantile.mark = 0.1) * sqrt(500)
asd.quant(plan.values3,
n = 5000,
censor.time = 1000,
quantile.mark = 0.1) * sqrt(5000)
get a single variance factor like in Figures 10.5 and 10.6
# For the normal distribution
variance.factor(distribution = 'normal',
type = 'quantile',
quantile.of.interest = 0.02,
proportion.failing = 0.2)
# For the smallest extreme value distribution
variance.factor(distribution = 'sev',
type = 'quantile',
quantile.of.interest = 0.02,
proportion.failing = 0.2)
asym.test.plan.properties(plan.values3,
n = 50,
proportion.failing = 0.1)
asd.quant(plan.values3,
n = 50,
censor.time = 1000,
quantile.mark = c(0.1, 0.3, 0.5, 0.63))
Scaled fisher information matrix elements from the Escobar & Meeker algorithm
lsinf(seq(-1,1, by = 0.1),"right","sev")
lsinf(seq(-2,2, by = 0.2),"right","normal")
Generate information table like that shown in Table C.20
table.lines(seq(-1,1,by=.1),"sev")
table.lines(seq(-1,1,by=.1),"normal")
variance.factor("sev", type = 'quantile')
variance.factor("normal", type = 'quantile')
variance.factor("logistic", type = 'quantile')
variance.factor("sev", type = 'hazard')
variance.factor("normal", type = 'hazard')
variance.factor("logistic", type = 'hazard')
Number of test time units needed for minimum-sized demonstration
zero.failure.plan(xlim = c(1.51,3.99),
ylim = c(.1,29),
krange = c(1.5,3.83))
zero.failure.plan(betavec = c( 1., 2.),
quantile = 0.01,
conlev = 0.95,
xlim = c(1.51,10),
ylim = c(.1,199),
krange = c(1.5,10),
grid = T,
bw = FALSE)
Determine test length (multiple of spec)
zero.failure.k(beta = 2, quantile = 0.1, conlev = 0.99, n = 5)
zero.failure.k(beta = 1, quantile = 0.01, conlev = 0.95, n = 5)
zero.failure.k(beta = 2, quantile = 0.01, conlev = 0.95, n = 5)
Determine sample size for a 0-failure test
zero.failure.n(conlev = 0.95, quantile = 0.01, k = 14, beta = 1)
zero.failure.n(conlev = 0.95, quantile = 0.01, k = 3.369, beta = 2)
zero.failure.prsd(alpha.vec = c(0.05,0.1), quantile = 0.01, pfactor = 3)
Plan for a light bulb test
bulb.plan.values1 <- get.plan.values("normal",
sigma = 85,
prob = 0.5,
time = 1000,
time.units = "Hours")
summary(bulb.plan.values1)
plot(bulb.plan.values1)
life.test.simulation(bulb.plan.values1,
n = 50,
censor.time = 1000,
number.detail = 5,
quantile.mark = 0.5)
plot(plan.values3,
censor.time = 100,
quantile.of.interest = 0.1)
#here is an example using type 2 censoring
plot(plan.values3,
fraction.failing = 0.1,
quantile.of.interest = 0.1)
# In actual application, use number.sim = 10000 to get smoother curves
asym.sample.size(plan.values3,
censor.time = 500,
Rvalue = 1.5,
quantile.of.interest = 0.1)
asym.sample.size(plan.values3,
fraction.failing = 0.1,
Rvalue = 1.5,
quantile.of.interest = 0.1)
asym.sample.size(bulb.plan.values1,
fraction.failing = 0.1,
HalfWidth = 50,
quantile.of.interest = 0.1)
mstt1 <- multiple.simulate.type.two(n = c(20,30,40),
r = c(20,30,40),
distribution = "lognormal",
number.sim = 1000)
plot(mstt1, qprob = 0.9)
plot(mstt1, qprob = 0.8)
plot(mstt1, qprob = 0.5)
mstt2 <- multiple.simulate.type.two(n = c(20,30,40),
r = c(10,15,20),
distribution = "lognormal",
number.sim = 1000)
plot(mstt2, qprob = 0.9)
plot(mstt2, qprob = 0.8)
plot(mstt2, qprob = 0.5, grids = T)
mstt3 <- multiple.simulate.type.two(n = c(24,28,33),
r = c(24,28,33),
distribution = "normal",
number.sim = 1000)
plot(mstt3, qprob = 0.9)
The following are for testing only and actual simulations should use at least 2000 trials
SMRD:::plot.prob.cs.type2("lognormal",
k = 2,
n = c(5,10,20),
r = c(3,6,12),
number.sim = 100)
SMRD:::plot.prob.cs.type2("loglogistic",
k = 2,
n = c(5,10,20),
r = c(3,6,12),
number.sim = 100)
SMRD:::plot.prob.cs.type2("weibull",
k = 2,
n = c(5,10,20),
r = c(3,6,12),
number.sim = 100)
SMRD:::plot.prob.cs.type2("frechet",
k = 2,
n = c(5,10,20),
r = c(3,6,12),
number.sim = 100)
Auburngrads/SMRD documentation built on Sept. 14, 2020, 2:21 a.m.
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
SMRD:::vinny() library(SMRD)
In this echapter...
Get Planning Values by Specifying a Point and a Slope
Specify a slope value and a single point $t, F(t)$
plan.values1 <- get.plan.values("Weibull", beta = 2, prob = .1, time = 100, time.units = "Hours")
summary(plan.values1) plot(plan.values1) failure.probabilities(plan.values1)
Carry out life test simulation using specified planning values. The function below plots several individual test simulations (the number of simulations is equal to number.detail) along with a plot of several simulations depending on the value of the argument number.sim (default is 2000).
life.test.simulation(plan.values1, n = 50, censor.time = 120, number.detail = 5, quantile.mark = 0.2, number.sim = 200)
Try a longer test
life.test.simulation(plan.values1, n = 50, censor.time = 300, number.detail = 5, number.sim = 200)
Other optional arguments and skip detail
life.test.simulation(plan.values1, n = 50, censor.time = 1000, number.sim = 50, quantile.mark = 0.2)
Now try lognormal distribution
plan.values2 <- get.plan.values("Lognormal", sigma = 0.5, prob = 0.1, time = 100, time.units = "Hours") summary(plan.values2) plot(plan.values2) plot(plan.values2, censor.time = 1000, grids = F)
life.test.simulation(plan.values2, n = 50, censor.time = 1000, quantile.mark = 0.1)
Get planning values by specifying two points
plan.values3 <- get.plan.values("Weibull", prob = c(.2,.12), time = c(1000,500), time.units = "Hours") plan.values4 <- get.plan.values("Weibull", prob = c(.05,.15), time = c(40000,100000), time.units = "Hours") summary(plan.values3) plot(plan.values3)
life.test.simulation(plan.values3, n = 50, censor.time = 1000, quantile.mark = 0.1)
#compare the simulated value with the large-sample approx below asd.quant(plan.values3, n = 50, censor.time = 1000, quantile.mark = 0.1) #compare: asd.quant(plan.values3, n = 50, censor.time = 1000, quantile.mark = 0.1) * sqrt(50) asd.quant(plan.values3, n = 500, censor.time = 1000, quantile.mark = 0.1) * sqrt(500) asd.quant(plan.values3, n = 5000, censor.time = 1000, quantile.mark = 0.1) * sqrt(5000)
get a single variance factor like in Figures 10.5 and 10.6
# For the normal distribution variance.factor(distribution = 'normal', type = 'quantile', quantile.of.interest = 0.02, proportion.failing = 0.2) # For the smallest extreme value distribution variance.factor(distribution = 'sev', type = 'quantile', quantile.of.interest = 0.02, proportion.failing = 0.2)
asym.test.plan.properties(plan.values3, n = 50, proportion.failing = 0.1) asd.quant(plan.values3, n = 50, censor.time = 1000, quantile.mark = c(0.1, 0.3, 0.5, 0.63))
Scaled fisher information matrix elements from the Escobar & Meeker algorithm
lsinf(seq(-1,1, by = 0.1),"right","sev") lsinf(seq(-2,2, by = 0.2),"right","normal")
Generate information table like that shown in Table C.20
table.lines(seq(-1,1,by=.1),"sev") table.lines(seq(-1,1,by=.1),"normal")
variance.factor("sev", type = 'quantile') variance.factor("normal", type = 'quantile') variance.factor("logistic", type = 'quantile') variance.factor("sev", type = 'hazard') variance.factor("normal", type = 'hazard') variance.factor("logistic", type = 'hazard')
Number of test time units needed for minimum-sized demonstration
zero.failure.plan(xlim = c(1.51,3.99), ylim = c(.1,29), krange = c(1.5,3.83)) zero.failure.plan(betavec = c( 1., 2.), quantile = 0.01, conlev = 0.95, xlim = c(1.51,10), ylim = c(.1,199), krange = c(1.5,10), grid = T, bw = FALSE)
Determine test length (multiple of spec)
zero.failure.k(beta = 2, quantile = 0.1, conlev = 0.99, n = 5) zero.failure.k(beta = 1, quantile = 0.01, conlev = 0.95, n = 5) zero.failure.k(beta = 2, quantile = 0.01, conlev = 0.95, n = 5)
Determine sample size for a 0-failure test
zero.failure.n(conlev = 0.95, quantile = 0.01, k = 14, beta = 1) zero.failure.n(conlev = 0.95, quantile = 0.01, k = 3.369, beta = 2) zero.failure.prsd(alpha.vec = c(0.05,0.1), quantile = 0.01, pfactor = 3)
Plan for a light bulb test
bulb.plan.values1 <- get.plan.values("normal", sigma = 85, prob = 0.5, time = 1000, time.units = "Hours") summary(bulb.plan.values1) plot(bulb.plan.values1)
life.test.simulation(bulb.plan.values1, n = 50, censor.time = 1000, number.detail = 5, quantile.mark = 0.5)
plot(plan.values3, censor.time = 100, quantile.of.interest = 0.1) #here is an example using type 2 censoring plot(plan.values3, fraction.failing = 0.1, quantile.of.interest = 0.1) # In actual application, use number.sim = 10000 to get smoother curves asym.sample.size(plan.values3, censor.time = 500, Rvalue = 1.5, quantile.of.interest = 0.1) asym.sample.size(plan.values3, fraction.failing = 0.1, Rvalue = 1.5, quantile.of.interest = 0.1) asym.sample.size(bulb.plan.values1, fraction.failing = 0.1, HalfWidth = 50, quantile.of.interest = 0.1)
mstt1 <- multiple.simulate.type.two(n = c(20,30,40), r = c(20,30,40), distribution = "lognormal", number.sim = 1000) plot(mstt1, qprob = 0.9) plot(mstt1, qprob = 0.8) plot(mstt1, qprob = 0.5) mstt2 <- multiple.simulate.type.two(n = c(20,30,40), r = c(10,15,20), distribution = "lognormal", number.sim = 1000) plot(mstt2, qprob = 0.9) plot(mstt2, qprob = 0.8) plot(mstt2, qprob = 0.5, grids = T) mstt3 <- multiple.simulate.type.two(n = c(24,28,33), r = c(24,28,33), distribution = "normal", number.sim = 1000) plot(mstt3, qprob = 0.9)
The following are for testing only and actual simulations should use at least 2000 trials
SMRD:::plot.prob.cs.type2("lognormal", k = 2, n = c(5,10,20), r = c(3,6,12), number.sim = 100) SMRD:::plot.prob.cs.type2("loglogistic", k = 2, n = c(5,10,20), r = c(3,6,12), number.sim = 100) SMRD:::plot.prob.cs.type2("weibull", k = 2, n = c(5,10,20), r = c(3,6,12), number.sim = 100) SMRD:::plot.prob.cs.type2("frechet", k = 2, n = c(5,10,20), r = c(3,6,12), number.sim = 100)
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
Embedding an R snippet on your website
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.