Bioequivalence and Bioavailability Forum

❝ China's regulatory agency (NMPA/CDE) refused to approve multiple highly-variable products which conducted their bioequivalence study in a four-period crossover design. The reason for this acion was based on "The similarity of plasma-concentration curves of the same subject between TEST and REFERENCE is larger than the similarity between two repeated REFERENCE".

❝ I tried to calculate the f2 factor using the Cmax value of the reference product as a correction factor, and I found many different products had a similar pattern that "The similarity of plasma-concentration curves of the same subject between TEST and REFERENCE is larger than the similarity between two repeated REFERENCE".

❝ I believe this is due to the nature of highly-variable products and randomness, but I am not sure if this is correct. Does FDA or EMA have similar statments?

❝ By the way, why single $ doesn't work for inline math?

f2 <- function(R, T, formula) {
  if (!length(R) == length(T))
    stop ("R and T vectors must have the same length.")
  if (missing(formula)) formula <- 2 # default
  if (!formula %in% 1:2)
    stop ("formula must be 1 or 2.")
  ms <- mean((R - T)^2) # instead of the clumsy formula
  if (formula == 1) {
    x <- 50 * log10((1 + ms)^-0.5 * 100)
  } else {
    x <- 50 * log10(100 / sqrt(1 + ms))
  }
  return(x)
}

n       <- 20
mu_R    <- mu_T    <- 100
sigma_R <- sigma_T <- 20
# is the package "truncnorm" installed?
inst    <- installed.packages()["truncnorm", "Package"]
if (length(inst) == 1) library(truncnorm)  # if yes, attach it
set.seed(123456)                           # for reproducibility
if (length(inst) == 1) { # truncated normal distribution
  R <- sort(rtruncnorm(n = n, a = 0, b = Inf,
                       mean = mu_R, sd = sigma_R))
  T <- sort(rtruncnorm(n = n, a = 0, b = Inf,
                       mean = mu_T, sd = sigma_T))
} else {                 # workaround to avoid negative values
  R <- sort(abs(rnorm(n = n, mean = mu_R, sd = sigma_R)))
  T <- sort(abs(rnorm(n = n, mean = mu_T, sd = sigma_T)))
}
f2_1 <- f2(R, T, 1)
f2_2 <- f2(R, T, 2)
cat(n, "sorted random values of R and T",
    "\nmean (R) =", sprintf("%8.3f", mean(R)),
    "range:", paste(signif(signif(range(R), 5)), collapse = " – "),
    "\nmean (T) =", sprintf("%8.3f", mean(T)),
    "range:", paste(signif(signif(range(T), 5)), collapse = " – "),
    "\n  f2 (1) =", sprintf("%.4g", f2_1),
    "\n  f2 (2) =", sprintf("%.4g", f2_2),
    "\nAre the results by the two formulas identical?",
    isTRUE(all.equal(f2_1, f2_2)), "\n")

20 sorted random values of R and T
mean (R) =  110.477 range: 77.721 – 150.05
mean (T) =   97.769 range: 45.023 – 131.2
  f2 (1) = 42.37
  f2 (2) = 42.37
Are the results by the two formulas identical? TRUE

mu_R    <- 100
sigma_R <- 20
mu_T    <- 10
sigma_T <- 2
...
20 sorted random values of R and T
mean (R) =  110.477 range: 77.721 – 150.05
mean (T) =    9.777 range: 4.5023 – 13.12
  f2 (1) = -0.4946
  f2 (2) = -0.4946
Are the results by the two formulas identical? TRUE

❝ An example: if the average difference is 100%, then you'll have $$f_2=50\,\log_{10} \frac{100}{\sqrt{1+\frac{1}{n}\sum_{i=1}^{i=n}100^2}} = -0.001.$$

❝ With average difference of 110%, \(f_2=-2.07\).

f2_ms <- function(ms) {
  if (ms < 0) stop ("ms must not be negative.")
  50 * log10(100 / sqrt(1 + ms))
}
ms <- c(0, 99, 100, 110, 99^2, 100^2, 110^2)
for (j in seq_along(ms)) {
  cat(sprintf("Mean sq. diff. = %5.0f, f2 = %+8.3f\n",
              ms[j], f2_ms(ms[j])))
}

Mean sq. diff. =     0, f2 = +100.000
Mean sq. diff. =    99, f2 =  +50.000
Mean sq. diff. =   100, f2 =  +49.892
Mean sq. diff. =   110, f2 =  +48.867
Mean sq. diff. =  9801, f2 =   +0.217
Mean sq. diff. = 10000, f2 =   -0.001
Mean sq. diff. = 12100, f2 =   -2.071