A pharmacokinetic and pharmacodynamic analysis of drug forgiveness

Abstract

Nonadherence to medication is a major public health problem. To combat nonadherence, some clinicians have suggested using “forgiving” drugs, which maintain efficacy in spite of delayed or missed doses. What pharmacokinetic (PK) and pharmacodynamic (PD) factors make a drug forgiving? In this paper, we address this question by analyzing a linear PK/PD model for a patient with imperfect adherence. We assume that the drug effect is far from maximal and consider direct effect, effect compartment (biophase), and indirect response PD models. We prove that the average drug effect relative to the clinically desired effect is simply the fraction of prescribed doses actually taken by the patient. Hence, under these assumptions, drug forgiveness cannot be defined in terms of the average effect. We argue that forgiveness should instead be understood in terms of effect fluctuations. We prove that the rates of PK absorption, PK elimination, and PD elimination are exactly equivalent for determining effect fluctuations. We prove all the aforementioned results for any pattern of nonadherence, including late doses, missed doses, drug holidays, extra doses, etc. To obtain quantitative estimates of effect fluctuations, we consider a simple statistical pattern of nonadherence and analytically calculate the coefficient of variation of effect. We further show how effect fluctuations can be reduced by taking an extra “make up” dose following a missed dose if any one of the aforementioned PK/PD rates is sufficiently slow. We illustrate some of our results for a nonlinear indirect response model of metformin.

Appendix

In this appendix, we present the mathematical proofs and derivations of some results in the main text.

Proof of formula for E(t) in (21)

To verify (21), note first that (21) and the definitions in (22) imply that $E(0)=0$, as desired. Next, the definition of E(t) in (18) and the ODE for $c_{\text {e}}$ in (6) imply that E(t) satisfies

$$\begin{aligned} \frac{\text {d}}{\text {d}t}E =\frac{E_{\max }}{\text {EC}_{\text {50}}}k_{ \text {p}\text {e}} {c_{ \text {p}}}-k_{\text {e}0} E. \end{aligned}$$

(49)

Differentiating (21) with respect to t and using the value of $c_{ \text {p}}$ in (4) shows that (21) indeed satisfies (49), which verifies (21).□

Proof that $\langle E \rangle =\mu \langle E^{\mathrm{perf}} \rangle$

To prove $\langle E \rangle =\mu \langle E^{\mathrm{perf}} \rangle$, we let $k>0$ and consider the large T behavior of

$$\begin{aligned} F(T,k):=\frac{1}{T}\int _{0}^{T}\sum _{n:t_{n}\le t}e^{-k(t-t_{n})}f_{n}\,\text {d}t. \end{aligned}$$

Interchanging the sum and integral yields

$$\begin{aligned} F(T,k)&=\frac{1}{T}\sum _{n:t_{n}\le T}\int _{t_{n}}^{T}e^{-k(t-t_{n})}f_{n}\,\text {d}t\nonumber \\&=\frac{1}{k}\frac{1}{T}\sum _{n:t_{n}\le T}f_{n} -\frac{1}{k}\frac{1}{T}\sum _{n:t_{n}\le T}e^{-k(T-t_{n})}f_{n}. \end{aligned}$$

(50)

By (30), the first term in (50) has the large T limit,

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{k}\frac{1}{T}\sum _{n:t_{n}\le T}f_{n} =\mu \frac{1}{k\tau }. \end{aligned}$$

(51)

We claim that the second term in (50) vanishes as $T\rightarrow \infty$. To see this, let $\varepsilon \in (0,1)$ and observe that

$$\begin{aligned}&\frac{1}{T}\sum _{n:t_{n}\le T}e^{-k(T-t_{n})}f_{n}\\&\quad =\frac{1}{T}\sum _{n:t_{n}\le (1-\varepsilon )T}e^{-k(T-t_{n})}f_{n} +\frac{1}{T}\sum _{n:(1-\varepsilon )T<t_{n}\le T}e^{-k(T-t_{n})}f_{n}\\&\quad \le e^{-\varepsilon kT}\frac{1}{T}\sum _{n:t_{n}\le (1-\varepsilon )T}f_{n} +\frac{1}{T}\sum _{n:(1-\varepsilon )T<t_{n}\le T}f_{n}. \end{aligned}$$

Since (30) implies that

$$\begin{aligned} \lim _{T\rightarrow \infty }e^{-\varepsilon kT}\frac{1}{T}\sum _{n:t_{n}\le (1-\varepsilon )T}f_{n} =0, \end{aligned}$$

it follows that

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\sum _{n:t_{n}\le T}e^{-k(T-t_{n})}f_{n} \le \lim _{T\rightarrow \infty }\frac{1}{T}\sum _{n:(1-\varepsilon )T<t_{n}\le T}f_{n}. \end{aligned}$$

(52)

To estimate the upper bound in (52), let $\varepsilon _{2}>0$ and observe that (30) implies that if T is sufficiently large, then

$$\begin{aligned} \frac{1}{T}\sum _{n:t_{n}\le T(1-\varepsilon )}f_{n} +\frac{1}{T}\sum _{n:T(1-\varepsilon )<t_{n}\le T}f_{n} \le \mu /\tau +\varepsilon _{2}, \end{aligned}$$

(53)

and

$$\begin{aligned} \mu /\tau -\varepsilon _{2} \le \frac{1}{T(1-\varepsilon )}\sum _{n:t_{n}\le T(1-\varepsilon )}f_{n}. \end{aligned}$$

(54)

Combining (53)–(54) yields

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\sum _{n:T(1-\varepsilon )<t_{n}\le T}f_{n}&\le \mu /\tau +\varepsilon _{2} -(1-\varepsilon )(\mu /\tau -\varepsilon _{2})\\&=2\varepsilon _{2}+\varepsilon \mu /\tau -\varepsilon \varepsilon _{2}. \end{aligned}$$

Since $\varepsilon \in (0,1)$ and $\varepsilon _{2}>0$ are arbitrary, we conclude from (52) that

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\sum _{n:t_{n}\le T}e^{-k(T-t_{n})}f_{n}=0, \end{aligned}$$

and therefore (50) and (51) imply

$$\begin{aligned} \lim _{T\rightarrow \infty }F(T,k) =\mu \frac{1}{k\tau }. \end{aligned}$$

(55)

Finally, using the expression for E(t) in (25), the values of $f_{n}$ and $t_{n}$ in (26) for perfect adherence, and the result in (55), we conclude that

$$\begin{aligned} \langle E \rangle =\mu \frac{E_{\max }}{\text {EC}_{\text {50}}}\frac{DF}{V_{}}\Big ( \frac{b_{ \text {p}0}}{k_{ \text {p}0}\tau } +\frac{b_{ \text {a} \text {p}}}{k_{ \text {a} \text {p}}\tau } +\frac{b_{\text {e}0}}{k_{\text {e}0}\tau } \Big ) =\mu \langle E^{\mathrm{perf}} \rangle . \end{aligned}$$

□

Derivation of $\text {CV}(E^{\text {single}})$

Using the model of nonadherence introduced in “Coefficient of variation estimate”, applying (33) gives

$$\begin{aligned}&\psi (t) :=\frac{E(t)}{\mu \langle E^{\mathrm{perf}} \rangle } =\frac{1}{\mu }\sum _{n=0}^{{\lfloor }{t/\tau }{\rfloor }}\Big (a_{ \text {p}0}e^{-k_{ \text {p}0}(t-n\tau )}\\&\quad +\,a_{ \text {a} \text {p}}e^{-k_{ \text {a} \text {p}}(t-n\tau )}+a_{\text {e}0}e^{-k_{\text {e}0}(t-n\tau )}\Big )f_{n}, \end{aligned}$$

where the coefficients $a_{ \text {p}0}$, $a_{ \text {a} \text {p}}$, and $a_{\text {e}0}$ are defined in (34), ${\lfloor }{t/\tau }{\rfloor }$ denotes the largest integer less than or equal to $t/\tau$, and $\{f_{n}\}_{n\in \mathbb {Z}}$ are as in (37). Note that we now take the sequence $\{f_{n}\}_{n\in \mathbb {Z}}$ to be bi-infinite (i.e. $n\in \mathbb {Z}$), which is convenient for the analysis below.

For $t\in [0,\tau ]$, define

$$\begin{aligned} \Psi (t) :=\frac{1}{\mu }(a_{ \text {p}0}e^{-k_{ \text {p}0}t}A_{ \text {p}0} +a_{ \text {a} \text {p}}e^{-k_{ \text {a} \text {p}}t}A_{ \text {a} \text {p}} +a_{\text {e}0}e^{-k_{\text {e}0}t}A_{\text {e}0}), \end{aligned}$$

(56)

where

$$\begin{aligned} A_{ij} :=\sum _{n=0}^{\infty }(e^{-k_{ij}\tau })^{n}f_{-n},\quad \text {for }i,j\in \{ \text {a}, \text {p},0\}. \end{aligned}$$

(57)

Note that $A_{ij}$ converges almost surely by the Weierstrass M-test since $|f_{n}|$ is bounded by a deterministic constant for all $n\in \mathbb {Z}$ (namely, $|f_{n}|\le 1$ in this case). Random variables of the form in (57) are often referred to as random pullback attractors [42,43,44,45,46].

It then follows from (32) and Theorem 1 in [20] that

$$\begin{aligned} \begin{aligned} \text {CV}(E^{\text {single}})&=\sqrt{\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}\Big (\psi (t)-1\Big )^{2}\,\text {d}t}\\&=\sqrt{\frac{1}{\tau }\int _{0}^{\tau }\mathbb {E}\Big [\big (\Psi (t)-1\big )^{2}\Big ]\,\text {d}t}, \end{aligned} \end{aligned}$$

(58)

where $\mathbb {E}$ denotes mathematical expectation. Using the definition of $\Psi (t)$ in (56), equation (58) implies that $\text {CV}(E^{\text {single}})$ requires calculating expectations of the form

$$\begin{aligned} \mathbb {E}[A_{ij}] \quad \text {and}\quad \mathbb {E}[A_{ij}A_{lm}],\quad \text {for }i,j,l,m\in \{ \text {a}, \text {p},0\}. \end{aligned}$$

(59)

These expectations are obtained immediately from Corollary 3 in [20]. In particular, we have that

$$\begin{aligned} \mathbb {E}[A_{ij}]&=\frac{p}{1-e^{-k_{ij}\tau }},\\ \mathbb {E}[A_{ij}A_{lm}]&=\frac{p}{1-e^{-k_{ij}\tau }e^{-k_{lm}\tau }}\\&\quad +\,p^{2}\Big (\frac{1}{(1-e^{-k_{ij}\tau })(1-e^{-k_{lm}\tau })}\\&\quad -\frac{1}{1-e^{-k_{ij}\tau }e^{-k_{lm}\tau }}\Big ). \end{aligned}$$

Plugging these formulas into (58) and performing the integration yields for the formula for $\text {CV}(E^{\text {single}})$ in (39).

Derivation of $\mu ^{\text {double}}$

For the double dose protocol in ““Make up” doses reduce variation for slow PK or PD”, it is immediate that $\{f_{n}\}_{n\in \mathbb {Z}}$ is an irreducible, time-homogeneous Markov chain with the following transition probabilities,

$$\begin{aligned} \mathbb {P}(f_{n+1}=0\,|\,f_{n}=i)&=1-p\quad \text {if }i\in \{0,1,2\},\\ \mathbb {P}(f_{n+1}=1\,|\,f_{n}=i)&={\left\{ \begin{array}{ll} 0 &{} \text {if }i=0,\\ p &{} \text {if }i=1,\\ p &{} \text {if }i=2, \end{array}\right. }\\ \mathbb {P}(f_{n+1}=2\,|\,f_{n}=i)&={\left\{ \begin{array}{ll} p &{} \text {if }i=0,\\ 0 &{} \text {if }i=1,\\ 0 &{} \text {if }i=2. \end{array}\right. } \end{aligned}$$

Using standard Markov chain results (for example, see section 1.7 in [47]), a quick calculation shows that the stationary distribution of $\{f_{n}\}_{n\in \mathbb {Z}}$ is

$$\begin{aligned} \mathbb {P}(f_{n}=i) ={\left\{ \begin{array}{ll} 1-p &{} \text {if }i=0,\\ p^{2} &{} \text {if }i=1,\\ p(1-p) &{} \text {if }i=2. \end{array}\right. } \end{aligned}$$

Using (30), Birkhoff’s ergodic theorem (for example, see section 7.2 in [48]) thus implies that the long-term fraction of doses taken for the double dose protocol is

$$\begin{aligned} \mu ^{\text {double}} =\lim _{N\rightarrow \infty }\frac{1}{N}\sum _{n=0}^{N-1}f_{n} =\sum _{i=0}^{2}i\mathbb {P}(f_{n}=i) =p+p(1-p). \end{aligned}$$

Derivation of $\text {CV}(E^{\text {double}})$

The derivation of the formula for $\text {CV}(E^{\text {double}})$ in (47) is identical to the derivation presented in “Derivation of $\text {CV}(E^{\text {single}})$” above except the formulas for the expectations in (59) are given by the formulas for the double dose protocol in Corollary 3 in [20].

Maximum plasma concentration for double dose protocol

Fig. 8

Contour plot of the factor $\varphi$ in (61) which bounds the maximum increase in plasma concentration caused by following the double dose protocol (see (60))

Theorem 5 in [20] yields the following upper bound for the maximum possible plasma compartment concentration obtained by following the double dose protocol,

$$\begin{aligned} c_{ \text {p}}^{\text {double}}(t) \le \max _{s\ge 0}c_{ \text {p}}^{\mathrm{perf}}(s) +\varphi \langle c_{ \text {p}}^{\mathrm{perf}} \rangle , \end{aligned}$$

(60)

where $\varphi$ is the following dimensionless factor,

$$\begin{aligned} \varphi :=\tau (k_{ \text {a} \text {p}})^{\frac{k_{ \text {p}0}}{k_{ \text {p}0}-k_{ \text {a} \text {p}}}}(k_{ \text {p}0})^{\frac{k_{ \text {a} \text {p}}}{k_{ \text {a} \text {p}}-k_{ \text {p}0}}}, \end{aligned}$$

(61)

and $\langle c_{ \text {p}}^{\mathrm{perf}} \rangle =\frac{DF}{V}\frac{1}{\tau }\frac{1}{k_{ \text {p}0}}$ is the long-term average plasma concentration for perfect adherence. The upper bound in (60) is valid for any time $t\ge 0$ and any sequence of remembering or forgetting $\{\xi _{n}\}_{n\ge 0}$ for the adherence model in ““Make up” doses reduce variation for slow PK or PD” above. The first term in the righthand side of (60) is the maximum plasma concentration obtained by a patient with perfect adherence. The second term in the righthand side of (60) bounds the maximum possible increase in plasma concentration caused by the double dose protocol.

In Fig. 8, we plot the factor $\varphi$ in (61). This plot shows that $\varphi \ll 1$ if $k_{ \text {a} \text {p}}\tau \ll 1$ and/or $k_{ \text {p}0}\tau \ll 1$. This means that if the PK absorption rate is slow compared to the dosing interval ($k_{ \text {a} \text {p}}\tau \ll 1$) and/or the PK elimination rate is slow compared to the dosing interval ($k_{ \text {p}0}\tau \ll 1$), then the double dose protocol can at most cause the plasma concentration to rise only slightly above the concentration for perfect adherence.