Drug concentration
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From: Jeff Johnson on 7 Aug 2010 16:05 I was curious about the concentration levels that a drug as in the body after repeated intake. Supposedly the drug concentration in the body after one intake can be modeled as f(t,C) = C*2^(-t/H)*Heaviside(t) where C was the initial intake amount and H is the half life(this assumes the drug is completely absorbed immediately upon intake which may be a fair approximation). If we assume identical dosage at regular intervals then for 2 dosages we would have f(t,C) if t <= T, f(t - T, C + f(T - e)) if t > T when the second dose comes at time T. f(T - e) is simply the concentration right before the second dose is taken. But it turns out that the above function is simply the some of two independent functions, f(t,C) + f(t - T, C) This makes it easy to add the effects of an arbitrary number of dosages, g(t, C) = sum(f(t - T(k), C)) = sum(C*2^(-(t - T(k))/H)*Heaviside(t - T(k)) = C*2^(-t/H)*sum(2^(T(k)/H)*Heaviside(t - T(k))) Basically what happens is each time we take a dose our concentration spikes up and then dies down depending on the half life. In the real case the dosage times and concentrations are not exact and cause the function to "wobble" in time but should be effectively represented by the above function. It seems the maximum and minium concentration(rather the envelope of the above function) reach a constant limit in the long run. I would like to know what the maximum and mininum concentrations are (depending on C and H) along with the average concentration(which is probably just an average of the above two). It seems if one scales the dosage rate(T(k)) and the half life by the same amount then characteristics I am looking for are effectively the same. so C*2^(t/sH)*sum(2^(sk/sH)*Heaviside(t - s*k)) The maximum seems to follow a simple geometric series. For example, with C = 1, H = 1, 1 - 2^(-t) and 2 - 2^(-t) seem to be the envelopes which suggests the max concentration is 2. Anyone mind enlightening me? It seems it is a simple geometric series should have the same maximum and minimum concentrations for for all s for large enough time. What I am interested in is if I take a drug with a half life of H then how long will it take to significantly reduce the concentration(say by 99%) after I stop taking it. The above analysis says that it will simply act like the last dosage was a single dose of a larger amount(the maximum concentration I'm looking for). This says that 2^(-t/H) effectively governs the decay percentage which would say it would take 6.643856190*H time intervals(days, half days, or whatever the dosage rate was) to decrease to 1% of the maximum. Instead of percentages I would like an absolute relation between the maximum and individual(C) concentrations as that 1% may still be a significant amount of concentration.
From: Ray Koopman on 7 Aug 2010 17:13 On Aug 7, 1:05 pm, "Jeff Johnson" <Jeff_John...(a)Hotmail.com> wrote: > I was curious about the concentration levels that a drug as in the body > after repeated intake. > > Supposedly the drug concentration in the body after one intake can be > modeled as > > f(t,C) = C*2^(-t/H)*Heaviside(t) where C was the initial intake amount and H > is the half life(this assumes the drug is completely absorbed immediately > upon intake which may be a fair approximation). > > If we assume identical dosage at regular intervals then > > for 2 dosages we would have > > f(t,C) if t <= T, > f(t - T, C + f(T - e)) if t > T > > when the second dose comes at time T. f(T - e) is simply the concentration > right before the second dose is taken. > > But it turns out that the above function is simply the some of two > independent functions, > > f(t,C) + f(t - T, C) > > This makes it easy to add the effects of an arbitrary number of dosages, > > g(t, C) = sum(f(t - T(k), C)) > > = sum(C*2^(-(t - T(k))/H)*Heaviside(t - T(k)) = > C*2^(-t/H)*sum(2^(T(k)/H)*Heaviside(t - T(k))) > > Basically what happens is each time we take a dose our concentration spikes > up and then dies down depending on the half life. In the real case the > dosage times and concentrations are not exact and cause the function to > "wobble" in time but should be effectively represented by the above > function. > > It seems the maximum and minium concentration(rather the envelope of the > above function) reach a constant limit in the long run. I would like to know > what the maximum and mininum concentrations are (depending on C and H) along > with the average concentration(which is probably just an average of the > above two). > > It seems if one scales the dosage rate(T(k)) and the half life by the same > amount then characteristics I am looking for are effectively the same. > > so > > C*2^(t/sH)*sum(2^(sk/sH)*Heaviside(t - s*k)) > > The maximum seems to follow a simple geometric series. > > For example, with C = 1, H = 1, 1 - 2^(-t) and 2 - 2^(-t) seem to be the > envelopes which suggests the max concentration is 2. > > Anyone mind enlightening me? > > It seems it is a simple geometric series > should have the same maximum and minimum concentrations for for all s for > large enough time. > > What I am interested in is if I take a drug with a half life of H then how > long will it take to significantly reduce the concentration(say by 99%) > after I stop taking it. The above analysis says that it will simply act like > the last dosage was a single dose of a larger amount(the maximum > concentration I'm looking for). > > This says that 2^(-t/H) effectively governs the decay percentage which would > say it would take 6.643856190*H time intervals(days, half days, or whatever > the dosage rate was) to decrease to 1% of the maximum. > > Instead of percentages I would like an absolute relation between the maximum > and individual(C) concentrations as that 1% may still be a significant > amount of concentration. Try asking at http://groups.google.ca/group/medstats/
From: Jeff Johnson on 8 Aug 2010 15:34
It seems the max is simply C/(1 - 2^(-1/H)) with the positive envelope being C*[1 - 2^(-(t+1)/H)]/[1 - 2^(-1/H)] and the min is C(1/(1 - 2^(-1/H)) - 1) with negative envelope C*([1 - 2^(-(t+1)/H)]/[1 - 2^(-1/H)] - 1). |