## Calculation of time above MIC [🇷 for BE/BA]

Dear Ace,

❝ Prats et al. came up with a time of 102.5 hours and 94.9% of total time in SS.

I overlooked this hint. The end of the last dosage interval therefore is 102.5/.949 = 108 hours, which is strange. I would set it to the time of the last dose plus one dosage interval (tau). If tau=12h, then the end should be set to 120h.

❝ But anyway, with your reply, I started over and finally it works!

OK, checked it again, this time with the last time point 108 hours (later samples dropped):
time <- c(0.5,1,3,5,8,12,24,36,48,60,72,84,96,108) conc <- c(0.64,0.7,1.39,1.35,0.67,1.48,0.32,2.08,0.87,0.93,1.09,           0.59,1.01,0.47) dat  <- data.frame(time, conc) f <- function(dat, th) {   under <- 0     for (i in which(dat$conc < th)) { if (!is.na(dat$conc[i-1])) {         y <- dat$conc[c(i-1,i)] x <- dat$time[c(i-1,i)]         slope <- coef(lm(y~x))         under <- under + x-x-(th-y)/slope       }       if (!is.na(dat$conc[i+1])) { y <- dat$conc[c(i,i+1)]         x <- dat$time[c(i,i+1)] slope <- coef(lm(y~x)) under <- under + (th-y)/slope } } return(under) } th <- 0.517 last <- 108 occupancy <- last - f(dat, th) coverage <- 100*occupancy/last cat(" End of last dosing interval:",last,"\n", "Occupancy time:",occupancy,"\n", "Coverage:",coverage,"%\n") Now I’m getting 103.6 hours (Coverage 95.9%) in agreement with Excel and even a manual calculation. I have no idea how the reference’s results (102.5 hours/ 94.9%) were obtained (last 120 h yields in 114.5 h / 95.4%). The R code right now needs a little cosmetics for t=0/c=0 (returns 'error in if (!is.na(dat$conc[i - 1])) { : Argument has lenght 0') - although the result is still correct.

BTW, I looked the reference up; I’m afraid it will not be very helpful:

Skelly JP. Issues and controversies involving controlled-release drug product studies. Pharmacy International. Nov. 1986: 280–6.

Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes  Ing. Helmut Schütz 