# NOT RUN {
# Estimated running time for all examples was calculated
# using a computer with 16 GB of RAM and one core of
# an Intel i7-6500U processor. Please take this into
# account when interpreting the run time given.
# Find the optimal testing configuration for
# non-informative two-stage hierarchical testing.
res1 <- OTC1(algorithm = "D2", p = 0.01, Se = 0.99, Sp = 0.99,
group.sz = 2:100, obj.fn = c("ET", "MAR", "GR1"),
weights = matrix(data = c(1,1), nrow = 1, ncol = 2))
summary(res1)
# Find the optimal testing configuration for
# informative three-stage hierarchical testing
res2 <- OTC1(algorithm = "ID3", p = 0.025,
Se = c(0.95, 0.95, 0.99), Sp = c(0.96, 0.96, 0.98),
group.sz = 3:15, obj.fn = c("ET", "MAR"), alpha = 2)
summary(res2)
# Find the optimal testing configuration for
# informative array testing without master pooling.
# This example takes approximately 30 seconds to run.
# }
# NOT RUN {
res3 <- OTC1(algorithm = "IA2", p = 0.05, alpha = 2,
Se = 0.90, Sp = 0.90, group.sz = 2:20,
obj.fn = "ET")
summary(res3)
# }
# NOT RUN {
# Find the optimal testing configuraiton for
# informative two-stage hierarchical testing.
Se <- matrix(data = c(rep(0.95, 2), rep(0.99, 2)),
nrow = 2, ncol = 2, byrow = FALSE)
Sp <- matrix(data = c(rep(0.96, 2), rep(0.98, 2)),
nrow = 2, ncol = 2, byrow = FALSE)
res4 <- OTC2(algorithm = "ID2",
alpha = c(18.25, 0.75, 0.75, 0.25),
Se = Se, Sp = Sp, group.sz = 12)
summary(res4)
# Find the optimal testing configuration for
# non-informative three-stage hierarchical testing.
# This example takes approximately 1 minute to run.
Se <- matrix(data = c(rep(0.95, 6)), nrow = 2, ncol = 3)
Sp <- matrix(data = c(rep(0.99, 6)), nrow = 2, ncol = 3)
# }
# NOT RUN {
res5 <- OTC2(algorithm = "D3",
p.vec = c(0.95, 0.0275, 0.0175, 0.005),
Se = Se, Sp = Sp, group.sz = 5:20)
summary(res5)
# }
# NOT RUN {
# Find the optimal testing configuration for
# non-informative array testing with master pooling.
# This example takes approximately 10 seconds to run.
# }
# NOT RUN {
res6 <- OTC2(algorithm = "A2M", p.vec = c(0.90, 0.04, 0.04, 0.02),
Se = rep(0.99, 2), Sp = rep(0.99, 2), group.sz = 2:20)
summary(res6)
# }
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