1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
|
context("Complex test case from Schaefer et al. (2007) Piacenza paper")
test_that("Complex test case from Schaefer (2007) can be reproduced (10% tolerance)", {
skip_on_cran()
schaefer07_complex_model <- mkinmod(
parent = list(type = "SFO", to = c("A1", "B1", "C1"), sink = FALSE),
A1 = list(type = "SFO", to = "A2"),
B1 = list(type = "SFO"),
C1 = list(type = "SFO"),
A2 = list(type = "SFO"), use_of_ff = "max", quiet = TRUE)
schaefer07_long <- mkin_wide_to_long(schaefer07_complex_case, time = "time")
fit.default <- mkinfit(schaefer07_complex_model, schaefer07_long, quiet = TRUE)
s <- summary(fit.default)
r <- schaefer07_complex_results
with(as.list(fit.default$bparms.optim), {
r$mkin <<- c(
k_parent,
s$distimes["parent", "DT50"],
s$ff["parent_A1"],
k_A1,
s$distimes["A1", "DT50"],
s$ff["parent_B1"],
k_B1,
s$distimes["B1", "DT50"],
s$ff["parent_C1"],
k_C1,
s$distimes["C1", "DT50"],
s$ff["A1_A2"],
k_A2,
s$distimes["A2", "DT50"])
}
)
r$means <- (r$KinGUI + r$ModelMaker)/2
r$mkin.deviation <- abs(round(100 * ((r$mkin - r$means)/r$means), digits=1))
expect_equal(r$mkin.deviation < 10, rep(TRUE, 14))
# In previous versions of mkinfit, if we used optimisation algorithm 'Marq'
# we got a local minimum with a sum of squared residuals of 273.3707
# When using 'Marq', we needed to give a good starting estimate e.g. for k_A2 in
# order to get the optimum with sum of squared residuals 240.5686
ssr <- sum(fit.default$data$residual^2)
expect_equal(round(ssr, 4), 240.5686)
})
|
