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

Dear Ace,

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 (

OK, checked it again, this time with the last time point 108 hours (later samples dropped):

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

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

❝ 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))[2]

under <- under + x[2]-x[1]-(th-y[1])/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))[2]

under <- under + (th-y[1])/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.

—

Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮

Science Quotes

*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

### Complete thread:

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