Optimizing Antimicrobial Treatment Schedules: Some Fundamental Analytical Results

Abstract

This work studies fundamental questions regarding the optimal design of antimicrobial treatment protocols, using pharmacodynamic and pharmacokinetic mathematical models. We consider the problem of designing an antimicrobial treatment schedule to achieve eradication of a microbial infection, while minimizing the area under the time-concentration curve (AUC), which is equivalent to minimizing the cumulative dosage. We first solve this problem under the assumption that an arbitrary antimicrobial concentration profile may be chosen, and prove that the ideal concentration profile consists of a constant concentration over a finite time duration, where explicit expressions for the optimal concentration and the time duration are given in terms of the pharmacodynamic parameters. Since antimicrobial concentration profiles are induced by a dosing schedule and the antimicrobial pharmacokinetics, the ‘ideal’ concentration profile is not strictly feasible. We therefore also investigate the possibility of achieving outcomes which are close to those provided by the ‘ideal’ concentration profile, using a bolus+continuous dosing schedule, which consists of a loading dose followed by infusion of the antimicrobial at a constant rate. We explicitly find the optimal bolus+continuous dosing schedule, and show that, for realistic parameter ranges, this schedule achieves results which are nearly as efficient as those attained by the ‘ideal’ concentration profile. The optimality results obtained here provide a baseline and reference point for comparison and evaluation of antimicrobial treatment plans.

Appendices

Appendix A Proof of Lemma 1

Using assumptions (I), (II) and the assumption $\alpha >1$, we have

$$\begin{aligned}{} & {} c<zMIC\;\;\Rightarrow \;\; f_r(c)<0,\;\;\;c>zMIC\;\;\Rightarrow \;\; f_r(c)>0,\\ {}{} & {} \lim _{c\rightarrow +\infty }f_r(c)=0.\end{aligned}$$

These facts imply that $f_r(c)$ has a global maximizer on $(0,\infty )$, which we denote by $c_{opt}$.

It remains to show that $c_{opt}$ is the unique critical point of $f_r(c)$. We have

$$\begin{aligned} f_r'(c)=\frac{k'(c)-\frac{k(c)}{c}}{c}+\frac{r}{c^2} =\frac{r-h(c)}{c^2} \end{aligned}$$

(A1)

where

$$\begin{aligned} h(c)=k(c)-ck'(c),\end{aligned}$$

(A2)

so any critical point of $f_r(c)$ satisfies $h(c)=r$. Note that

$$\begin{aligned} h'(c)=-ck''(c),\end{aligned}$$

(A3)

hence:

(a) if k is concave then $h'(c)>0$, so h(c) is increasing in $(0,\infty )$, hence the critical point $c_{opt}$ of $f_r(c)$ is unique.

(b) If k is sigmoidal, then $h'(c)>0$ for $c>c_{infl}$, hence h(c) is increasing in this range, so that $f_r(c)$ has at most one critical point in the interval $[c_{infl},\infty )$. To prove uniqueness it therefore suffices to show that $f_r(c)$ has no critical point in $[0,c_{infl}]$. But note that since k(c) is convex on $[0,c_{infl}]$, hence $k'(c)$ is increasing on this interval, we have

$$\begin{aligned} c\in (0,c_{infl}]\;\;\Rightarrow \;\;k'(c)=\frac{1}{c}\int _0^c k'(c)du>\frac{1}{c}\int _0^c k'(u)du=\frac{k(c)}{c},\end{aligned}$$

(A4)

hence, by (A1), $f_r'(c)>0$ for $c\in (0,c_{infl}]$. Note that this also shows that $c_{opt}>c_{infl}$.

Appendix B Proof of Theorem 2

(i) is trivially true, since if $\hat{T}>T_{opt}$ then $C_{opt}(t)$ leads to eradication before time $\hat{T}$, and, by Theorem 1 it achieves the lowest possible value of AUC.

(ii) If $\hat{T}\le T_{min}$ then for any $T\le \hat{T}$ we have

$$\begin{aligned}\ln (10)\cdot LR(T){} & {} = \int _0^T [k(C(t))-r]dt< T\cdot [k_{max}-r]\le T_{min}\cdot [k_{max}-r]\\{} & {} =\ln (10)\cdot LR_{target},\end{aligned}$$

so that the target reduction cannot be achieved.

(iii) Assume now that

$$\begin{aligned} T_{min}<\hat{T}<T_{opt}. \end{aligned}$$

(B1)

Note first that $\hat{T}>T_{min}$ implies that, using assumption (II),

$$\begin{aligned}\lim _{c\rightarrow \infty }\hat{T}\cdot [k(c)-r]=\hat{T}\cdot [k_{max}-r]>T_{min}\cdot [k_{max}-r]=\ln (10)\cdot LR_{target},\end{aligned}$$

hence by (I) the equation (26) has a unique solution $\hat{c}$.

(26) implies that

$$\begin{aligned}\frac{1}{\ln (10)}\int _0^{\hat{T}}[k(\hat{C}(t))-r]dt=\hat{T}\cdot [k(\hat{c})-r]=LR_{target},\end{aligned}$$

so that the profile $\hat{C}(t)$ achieves eradication at time $\hat{T}$.

To show that $\hat{C}(t)$, defined by (25) is the solution of Problem 2, we assume that C(t) is any concentration profile for which $LR(T)=LR_{target}$, where $T\le \hat{T}$, and we need to show that if $AUC[C(t)]\ge AUC[\hat{C}(t)]$, with equality iff $C(t)=\hat{C}(t)$ for a.e. t.

By (26),(B1), and (15), we have

$$\begin{aligned}k(\hat{c})-r=\frac{\ln (10)\cdot LR_{target}}{\hat{T}}> \frac{\ln (10)\cdot LR_{target}}{T_{opt}}=k(c_{opt})-r,\end{aligned}$$

which, by (I), implies

$$\begin{aligned} \hat{c}>c_{opt}.\end{aligned}$$

(B2)

We now define

$$\begin{aligned} \hat{r}=k(\hat{c})-\hat{c}k'(\hat{c}),\end{aligned}$$

(B3)

and claim that

$$\begin{aligned} \hat{r}>r. \end{aligned}$$

(B4)

To show this we recall that in the proof of Lemma 1 (Appendix A) it was shown that the function h(c) defined by (A2) is monotone increasing for all $c>0$ in the concave case, and for all $c>c_{infl}$ in the sigmoidal case, in which case we also have $c_{opt}>c_{infl}$. Therefore, since $h(c_{opt})=r$ and $h(\hat{c})=\hat{r}$, (B2) implies (B4).

By (B3) we have

$$\begin{aligned}k'(\hat{c})=\frac{k(\hat{c})-\hat{r}}{\hat{c}},\end{aligned}$$

which, using Lemma 1, applied with $\hat{r}$ replacing r, implies that $\hat{c}$ is the maximizer of $f_{\hat{r}}(c)$, so that

$$\begin{aligned}\frac{k(\hat{c})-\hat{r}}{\hat{c}}=\max _{c>0}\frac{k(c)-\hat{r}}{c}.\end{aligned}$$

We therefore have, for all $c>0$,

$$\begin{aligned} k(c)-\hat{r}\le c\cdot \frac{k(\hat{c})-\hat{r}}{\hat{c}}.\end{aligned}$$

(B5)

If C(t) is a concentration profile for which $LR(T)=LR_{target}$, where $T\le \hat{T}$, then we have, using (B4),(B5)

$$\begin{aligned} \ln (10)\cdot LR_{target}= & {} \int _0^T [k(C(t))-r]dt=\int _0^T [k(C(t))-\hat{r}]dt +T\cdot (\hat{r}-r)\nonumber \\\le & {} \int _0^T [k(C(t))-\hat{r}]dt +\hat{T}\cdot (\hat{r}-r)\nonumber \\\le & {} \frac{k(\hat{c})-\hat{r}}{\hat{c}} \int _0^T C(t)dt +\hat{T}\cdot (\hat{r}-r)\nonumber \\\le & {} \frac{k(\hat{c})-\hat{r}}{\hat{c}} \cdot AUC[C(t)] +\hat{T}\cdot (\hat{r}-r), \end{aligned}$$

(B6)

which, togther with (26), implies

$$\begin{aligned}AUC[C(t)]\ge & {} \left[ \ln (10)\cdot LR_{target}-\hat{T}\cdot (\hat{r}-r)\right] \cdot \frac{\hat{c}}{k(\hat{c})-\hat{r}}\nonumber \\ {}= & {} \left[ \hat{T}\cdot [k(\hat{c})-r]-\hat{T}\cdot (\hat{r}-r)\right] \cdot \frac{\hat{c}}{k(\hat{c})-\hat{r}}=\hat{T}\cdot \hat{c}=AUC[\hat{C}(t)], \end{aligned}$$

proving that $\hat{C}(t)$ indeed attains the minimal AUC among the relevant concentration profiles. To show uniqueness, note that the equality $AUC[C(t)]=AUC[\hat{C}(t)]$ can hold only if all inequalities in (B6) are in fact equalities, which in particular implies $T=\hat{T}$ and $k(C(t))=k(\hat{c})$, hence $C(t)=\hat{c}$ for a.e. $0\le t\le \hat{T}$, and also that $\int _{\hat{T}}^\infty C(t)dt=0$, so that $C(t)=0$ for a.e. $t> \hat{T}$, hence $C(t)=\hat{C}(t)$ a.e..

Appendix C Proof of Theorem 3

To begin the analysis leading to Theorem 3, we calculate the values $LR_{max}$ and AUC corresponding to a bolus+continuous dosing schedule.

Lemma 2

Consider a bolus+continuous schedule $d_{bc}(t)$ (see (36)), with $\bar{c}>zMIC$, and the induced antimicrobial concentration profile $C_{bc}(t)$ (see (37)). Then:

(i) The maximal log-reduction corresponding to this concentration profile is

$$\begin{aligned} LR_{max}=\frac{1}{\ln (10)}\left[ T_{bc}\cdot (k(\bar{c})-r)+k_e^{-1}\phi (\bar{c}) \right] , \end{aligned}$$

(C1)

where the function $\phi (c)$ is defined by:

$$\begin{aligned} \phi (c)=\int _{zMIC}^{c} \frac{k(u)-r}{u}du.\end{aligned}$$

(C2)

(ii) The AUC corresponding to this dosage schedule is

$$\begin{aligned} AUC=[k_e^{-1}+T_{bc}]\cdot \bar{c}.\end{aligned}$$

(C3)

Proof

(i) By (10), the log-reduction of the microbial load corresponding to (37), at time T, is

$$\begin{aligned}LR(T)=\frac{1}{\ln (10)}\cdot \int _0^T [k(C_{bc}(t))-r]dt,\end{aligned}$$

hence

$$\begin{aligned}T<T_{bc}\;\;\Rightarrow \;\; LR'(T)=k(C_{bc}(T))-r=k(\bar{c})-r>0,\end{aligned}$$

and

$$\begin{aligned}\lim _{T\rightarrow \infty }LR'(T)=\lim _{T\rightarrow \infty }(k(C_{bc}(T))-r)=\lim _{T\rightarrow \infty }(k(\bar{c}e^{-k_e (T-T_{bc})})-r)=-r<0.\end{aligned}$$

We thus have that LR(T) is an increasing function for $T<T_{bc}$, and a decreasing function for sufficiently large T, so that its maximum attained at some $T^*>T_{bc}$ satisfying $LR'(T^*)=0$, that is

$$\begin{aligned}k(C_{bc}(T^*))-r=0\;\;\Leftrightarrow & {} \;\; C_{bc}(T^*)=zMIC \;\;\Leftrightarrow \;\; \bar{c}e^{-k_e (T^*-T_{bc})}=zMIC \\\Leftrightarrow & {} \;\; T^*=T_{bc}+\frac{1}{k_e}\cdot \ln \left( \frac{\bar{c}}{zMIC} \right) . \end{aligned}$$

Using the change of variable $u=\bar{c}e^{-k_e\cdot (t-T_{bc})}$ in the integral below, we conclude that

$$\begin{aligned}\ln (10)\cdot LR_{max}= & {} \ln (10)\cdot \max _{T>0}LR(T)=\ln (10)\cdot LR\left( T^* \right) \\= & {} T_{bc}\cdot (k(\bar{c})-r)+\int _{T_{bc}}^{T^*} \left[ k\left( \bar{c}e^{-k_e\cdot (t-T_{bc})}\right) -r\right] dt \\= & {} T_{bc}\cdot (k(\bar{c})-r)+\frac{1}{k_e}\int _{\bar{c}e^{-k_e\cdot (T^*-T_{bc})}}^{\bar{c}} \frac{k(u)-r}{u}du \\= & {} T_{bc}\cdot (k(\bar{c})-r)+\frac{1}{k_e}\int _{zMIC}^{\bar{c}} \frac{k(u)-r}{u}du \\ {}= & {} T_{bc}\cdot (k(\bar{c})-r)+k_e^{-1}\phi (\bar{c}), \end{aligned}$$

where the function $\phi$ is defined by (C2).

(ii) Using (34), the AUC corresponding to the concentration profile generated by the dosing schedule (36) is given by

$$\begin{aligned}AUC=V^{-1}k_e^{-1}\int _0^{\infty }d(t)dt=V^{-1}k_e^{-1}\left[ V\bar{c}+k_e V\bar{c}T_{bc} \right] =[k_e^{-1}+T_{bc}]\cdot \bar{c}.\end{aligned}$$

$\square$

Proof of Theorem 3

By (C1), in order to achieve a given log-reduction $LR_{target}$ using a dosing schedule of the form (36), the parameters $\bar{c},T_{bc}$ defining this schedule must satisfy the constraint $LR_{max}[C(t)]=LR_{target}$, or

$$\begin{aligned} T_{bc}\cdot (k(\bar{c})-r)+k_e^{-1}\phi (\bar{c}) =\ln (10)\cdot LR_{target}.\end{aligned}$$

(C4)

We need to minimize the expression (C3) over $(\bar{c},T_{bc})$, under the constraints (C4) and

$$\begin{aligned} \bar{c}\ge zMIC,\;\;\; T_{bc}\ge 0. \end{aligned}$$

(C5)

The constraint (C4) can be written as

$$\begin{aligned} T_{bc} =\frac{\ln (10)\cdot LR_{target}- k_e^{-1}\phi (\bar{c})}{k(\bar{c})-r},\end{aligned}$$

(C6)

and the inequality constraints (C5) imply that $\bar{c}$ must satisfy

$$\begin{aligned} zMIC\le \bar{c}\le \phi ^{-1}(\ln (10)\cdot LR_{target}\cdot k_e)=c^*, \end{aligned}$$

(C7)

where $c^*$ is the solution of (44).

Substituting (C6) into (C3) we get

$$\begin{aligned} AUC=AUC(\bar{c})=\left[ k_e^{-1}+ \frac{\ln (10)\cdot LR_{target}-k_e^{-1} \phi (\bar{c})}{k(\bar{c})-r}\right] \cdot \bar{c},\end{aligned}$$

(C8)

which must be minimized over $\bar{c}$ satisfying (C7). Noting that the expression (C8) goes to $+\infty$ when $\bar{c}\rightarrow zMIC$, we see that the minimum is attained either at (a) an interior point of the interval (C7), or (b) at $\bar{c}=c^*$.

If $c_{opt}<c^*$, then since $c_{opt}$ is the maximizer of $f_r(c)$ given by (13), we have

$$\begin{aligned}{} & {} \int _{c_{opt}}^{c^*} \frac{k(u)-r}{u}du< (c^*-c_{opt})\cdot \frac{k(c_{opt})-r}{c_{opt}}\\\Leftrightarrow & {} \;\; \phi (c^*)- \phi (c_{opt})< \frac{c^*-c_{opt}}{c_{opt}}\cdot (k(c_{opt})-r)\\\Leftrightarrow & {} \;\;\left[ k_e^{-1}+ \frac{\ln (10)\cdot LR_{target}- k_e^{-1}\phi (c_{opt})}{k(c_{opt})-r}\right] \cdot c_{opt}< k_e^{-1}c^* \\\Leftrightarrow & {} AUC(c_{opt})<AUC(c^*), \end{aligned}$$

so that the minimum of $AUC(\bar{c})$ in the interval (C7) is attained at an interior point, at which $AUC'(\bar{c})$ must vanish, and using the fact that $\phi '(c)=\frac{k(c)-r}{c}$ we compute

$$\begin{aligned}AUC'(\bar{c})= & {} \;\;\frac{[\ln (10)\cdot LR_{target}- k_e^{-1}\phi (\bar{c})][k(\bar{c})-r-k'(\bar{c})\cdot \bar{c}]}{(k(\bar{c})-r)^2}=0\\ {}\Leftrightarrow & {} \;\;k(\bar{c})-r-k'(\bar{c})\cdot \bar{c}=0 \;\;\Leftrightarrow \;\;\bar{c}=c_{opt}.\end{aligned}$$

Thus, from (C6) we get (40), and from (C3) we get (43).

On the other hand, if $c_{opt}\ge c^*$, the above calculation shows that the derivative of $AUC(\bar{c})$ does not vanish in the interior of the interval (C7), so that the minimum is attained at $\bar{c}=c^*$, proving part (ii) of the theorem. $\square$