Improving the performance of supply chains in clinical trials with delays: an optimization approach to determining the number of recruitment sites

Abstract

Clinical testing accounts for almost 70% of the R &D expenditure in the pharmaceutical industry and most clinical trials experience considerable delays that result in extra costs. One of the major sources of these delays is the patient recruitment process. It is estimated that about 48% of the recruitment sites of a clinical trial fail to meet their deadlines due to slow recruitment rates and 11% do not recruit a single patient. Thus, determining the correct number of recruitment sites is paramount for more efficient supply chains in clinical trials. The calculation of the number of recruitment sites in clinical trials is based on factors such as the target number of patients, their estimated recruitment rates, and the associated operational costs. However, these calculations typically ignore the time needed for sites to be operational. In this paper, we analyze the impact of incorporating these site activation delays in determining the optimal number of clinical sites. Our study shows that ignoring these delays leads to overestimating the number of sites needed, thus resulting in an excess of expenditure. Consequently, the insights herein explained help quantify the impact of making better decisions in designing more efficient operations in clinical trials, both in terms of time and money.

Appendices

Appendix A: Mathematical proofs of lemmas and theorems

Proof of Proposition 4.1

To prove the case for $j=1,2,\dots$, we can condition on the event of a site opening at time s. From Johnson et al. (2005), the number of arrivals that arrive to a site by time t given that the site opens at time s follows a negative binomial distribution with parameters $(\alpha ,\beta /(t-s+\beta ))$. Then, $P\{N_m(t)=j\}=\int _{0}^{\infty } {j+\alpha - 1\atopwithdelims ()\alpha -1}(\beta /(t-s+\beta ))^\alpha (1-\beta /(t-s+\beta ))^{j} f_s(s) ds=\int _{0}^{t} {j+\alpha - 1\atopwithdelims ()\alpha -1}(\beta /(t-s+\beta ))^\alpha (1-\beta /(t-s+\beta ))^{j} f_s(s) ds$, since patients will not arrive to a closed site.

For the case $j=0$, we condition again on the event that the site opens at time s, i.e., $P\{N_m(t)=0\}=\int _0^{t}P\{N_m(t)=0|s_m=s\}f_s(s)ds+\int _t^{\infty }P\{N_m(t)=0|s_m=s\}f_s(s)ds$. Since at time t there will not be any patients in a site that is closed, the second term becomes $\int _t^{\infty }P\{N_m(t)=0|s_m=s\}f_s(s)ds=\bar{F}_s(t)$. The first term can in turn be rewritten as $\int _0^t {\alpha -1\atopwithdelims ()\alpha -1}(\beta /(t-s+\beta ))^\alpha (1-\beta /(t-s+\beta ))^0f_s(s)ds=\int _0^t (\beta /(t-s+\beta ))^\alpha f_s(s)ds$, whence our result is derived. $\square$

Proof of Proposition 4.2

To prove (i), we calculate $\mathbb {E}[N_m(t)]$ by conditioning, i.e., $\mathbb {E}[N_m(t)] = \mathbb {E}[\mathbb {E}[N_m(t) | s_m]]=\int _0^{\infty }\mathbb {E}\left[ N_m(t)|s_m=s\right] f_s(s)ds =\int _0^{t}\mathbb {E}\left[ N_m(t)|s_m=s\right] f_s(s)ds+\int _t^{\infty }\mathbb {E}\left[ N_m(t)|s_m=s\right] f_s(s)ds$. The second term in the last equality is 0 since arrivals cannot occur before the site m has opened. Thus, we have that $\mathbb {E}[N_m(t)]=\mathbb {E}\left[ \mathbb {E}\left[ N_m(t)|s_m\le t\right] \right]$. In turn, by virtue of the Poisson–gamma model, the expected number of arrivals that occur between the activation time and t is $\mathbb {E}\left[ N_m(t)|s_m\le t\right] =\alpha (t-s_m)_+/\beta$ (Karlis and Xekalaki (2005)). Therefore, the expected number of arrivals at time t is

$$\begin{aligned} \mathbb {E}\left[ N_m(t)\right]&=\mathbb {E}\left[ \frac{\alpha }{\beta }(t-s_m)_+\right] = \frac{\alpha }{\beta } \int \limits _0^t (t-s) f_s(s) d s=\frac{\alpha }{\beta } \left( \int \limits _0^t t f_s(s) ds - \int \limits _0^t s f_s(s) d s \right) \nonumber \\&= \frac{\alpha }{\beta } \left( \int \limits _0^t t f_s(s) ds - F_s(t)\int \limits _0^t s \frac{f_s(s)}{F_s(t)} d s \right) = \frac{\alpha }{\beta }F_s(t) \left( t - \mathbb {E}[s_m | s_m \le t] \right) . \end{aligned}$$

(A1)

To prove (ii), we will calculate $Var[N_m(t)]$ by using the well-known result $Var[X]=\mathbb {E}\left[ Var[X|Y]\right] +Var\left[ \mathbb {E}[X|Y]\right]$, which, applied to this case, results in

$$\begin{aligned} Var[N_m(t)] =&\mathbb {E}[Var[N_m(t) | s_m\le t]] + Var[\mathbb {E}[N_m(t) | s_m \le t]]. \end{aligned}$$

(A2)

The second term in the right-hand side of (A2) can be expanded as

$$\begin{aligned} Var[\mathbb {E}[N_m(t) | s_m\le t]]=&Var\left[ \displaystyle \frac{\alpha }{\beta }(t-s_m)_+\right] =\mathbb {E}\left[ \left( \frac{\alpha }{\beta }(t-s_m)_+\right) ^2\right] -\mathbb {E}\left[ \frac{\alpha }{\beta }(t-s_m)_+\right] ^2\nonumber \\ =&\frac{\alpha ^2}{\beta ^2}\left( \int \limits _0^t (t-s)^2f_s(s)ds-\mu ^2\right) , \end{aligned}$$

(A3)

where $\mu =\mathbb {E}[N_m(t)]$, as shown in (A1). Now, per Karlis and Xekalaki (2005), the first term in the right-hand side of (A2) is

$$\begin{aligned} \mathbb {E}[Var[N_m(t) | s_m\le t]]=&\mathbb {E}\left[ \frac{\alpha }{\beta }(t-s)_+ + \frac{\alpha }{\beta ^2}(t-s)_{+}^2\right] =\mu +\frac{\alpha }{\beta ^2}\mathbb {E}\left[ (t-s_m)_+^2\right] . \end{aligned}$$

(A4)

Substituting (A3) and (A4) into (A2), we obtain

$$\begin{aligned} Var[N_m(t)] =&\mu \left( 1-\mu \right) + \frac{\alpha (1+\alpha )}{\beta ^2}\int \limits _0^t (t-s)^2 f(s)ds. \end{aligned}$$

(A5)

The last integral can be written as $\int \limits _0^t (t-s)^2 f_s(s)ds=F_s(t)(t^2+\mathbb {E}[s_m^2|s_m\le t]-2t\mathbb {E}[s_m|s_m\le t]) =t\beta \mu /\alpha +F_s(t)(\mathbb {E}[s_m^2|s_m\le t]-t\mathbb {E}[s_m|s_m\le t])$, whence, once substituted in (A5) and after rearranging terms, we arrive at our result.

To show that the variance is greater than the mean, we consider the difference of (A5) and (A1):

$$\begin{aligned} Var[N_m(t)] - \mathbb {E}[N_m(t)]= & {} -\mu ^2 + \frac{\alpha (1+\alpha )}{\beta ^2} \int \limits _0^{t} (t-s)^2 f(s) ds > -\mu ^2 \\{} & {} + \frac{\alpha ^2}{\beta ^2} \int \limits _0^{t} (t-s)^2 f(s) ds. \end{aligned}$$

It suffices to show that the right-hand side of the above inequality is positive. Using (A1) to rewrite $\mu ^2$, further simplification yields

$$\begin{aligned} \int \limits _0^{t} (t-s)^2 f(s) ds > \left[ \int _0^{t} (t-s) f(s) ds\right] ^2, \end{aligned}$$

which is equivalent to showing $\mathbb {E}[(t-s_m)_+^2] > \mathbb {E}^2[(t-s_m)_+]$. This inequality is implied from Jensen’s inequality. $\square$

Proof of Lemma 5.1

By conditioning on the event that recruitment terminates before the time that the last site is activated, we can write $E[T_S(M)] = E[T_S(M)|N(s_{(M)}) \ge S] P\{N(s_{(M)}) \ge S\} + E[T_S(M)|N(s_{(M)}) \le S] P\{N(s_{(M)}) \le S\}.$ It is clear that $P\{N(s_{(M)}) \ge S\}$ is bounded above by ${\mathbb {E}[N(s_{(M)})]}/{S}$ by virtue of Markov’s inequality. As $S \rightarrow \infty$, $P\{N(s_{(M)}) \ge S\}$ approaches zero and we obtain that

$$\begin{aligned} \mathbb {E}\left[ T_S(M)\right] =\mathbb {E}\left[ s_{(M)}\right] +\frac{\beta (S-\mathbb {E}\left[ N\left( s_{(M)}\right) \right] }{\alpha M -1}. \end{aligned}$$

In addition, we have that $\mathbb {E}\left[ N\left( s_{(M)}\right) \right] =M\mathbb {E}\left[ N_m\left( s_{(M)}\right) \right] =M\alpha /\beta \left( \mathbb {E}\left[ s_{(M)}\right] -\mathbb {E}\left[ s_m\right] \right)$. Therefore the expected recruitment time is

$$\begin{aligned} \mathbb {E}\left[ T_S(M)\right] =\frac{\beta S}{\alpha M -1}+\mathbb {E}\left[ s_{(M)}\right] -\frac{\alpha M}{\alpha M -1}\left( \mathbb {E}\left[ s_{(M)}\right] -\mathbb {E}\left[ s_m\right] \right) . \end{aligned}$$

For relatively high cohort sizes, typical in Phase III, $\alpha M/(\alpha M -1)\approx 1$, and the expression above can be simplified as

$$\begin{aligned} \mathbb {E}\left[ T_S(M)\right] =\frac{\beta S}{\alpha M -1}+\mathbb {E}\left[ s_m\right] . \end{aligned}$$

In conclusion, if the target recruitment S and the number of sites M are large enough, we can conclude that the expected completion time of a clinical trial is approximately the expected time of completion when there are not activation delays plus the expected activation delay per site $\mathbb {E}\left[ s_m\right]$. $\square$

Proof of Theorem 5.2

Let us consider the continuous relaxation of Problem (4):

$$\begin{aligned} \min _{M\in (\frac{1}{\alpha },\infty )} \mathbb {E}\left[ \mathcal {C}(M)\right] , \end{aligned}$$

(A6)

If $\mathbb {E}\left[ s_m\right]$ is constant, $\mathbb {E}[\mathcal {C}(M)]$ is convex in the interval $(1/\alpha ,\infty )$, since $\mathbb {E}[\mathcal {C}(M)]''=2v\alpha ^2\beta S/(\alpha M-1)^3$. Consequently, this function attains its minimum at $M_R=1/\alpha + \sqrt{{v \beta S}/{(\alpha c)}}$. Since M is integer and $\mathbb {E}[\mathcal {C}(M)]$ is convex, it turns out that

$$\begin{aligned} M^*=\arg \min \left\{ \mathbb {E}\left[ \mathcal {C}\left( \lfloor M_R\rfloor \right) \right] ,\mathbb {E}\left[ \mathcal {C}\left( \lceil M_R\rceil \right) \right] \right\} . \end{aligned}$$

$\square$

Proof of Lemma 5.4

We will analyze the behavior of $\mathbb {E}[\tilde{\mathcal {C}}(M)]$ in the limits of the interval $(1/\alpha ,\infty )$. On the one hand, we have that $\lim _{M\rightarrow 1/\alpha }\mathbb {E}[\tilde{\mathcal {C}}(M)]'=-\infty$. On the other hand, $\lim _{M\rightarrow \infty }\mathbb {E}[\tilde{\mathcal {C}}(M)]'=c+v\cdot \lim _{M\rightarrow \infty }g'(M)>0$. Since $\mathbb {E}[\tilde{\mathcal {C}}(M)]$ is a continuous function, it follows that there is at least one critical point in $(1/\alpha ,\infty )$, and that it is a minimizer. This point, as well as any other critical point in the function, satisfies the first-order condition $g'(M)=\alpha \beta S/(\alpha M -1)^2 -c/v$. $\square$

Proof of Theorem 5.5

If the fist-order condition (9) has a unique solution, then Lemma 5.4 implies that the expected cost $\mathbb {E}[\tilde{C}(M)]$ is quasi-convex. A unique solution exists if and only if the function $g'(M)-\alpha \beta S/(\alpha M-1)^2+c/v$ has the single crossing ($\mathcal{S}\mathcal{C}$) property (Quah & Strulovici, 2012). This occurs if and only if the functions $g'(M)$ and $-\alpha \beta S/(\alpha M-1)+c/v$ are $\mathcal{S}\mathcal{C}$ and they obey signed ratio monotonicity. Indeed, these two functions are trivially $\mathcal{S}\mathcal{C}$, but the signed ratio monotonicity implies that in the interval $(1/\alpha , M_R]$

$$\begin{aligned} \frac{\frac{\alpha \beta S}{(\alpha M_1 -1)^2}-\frac{c}{v}}{g'(M_1)}\ge \frac{\frac{\alpha \beta S}{(\alpha M_2 -1)^2}-\frac{c}{v}}{g'(M_2)},\quad \text {for any }M_2>M_1, \end{aligned}$$

which occurs if the function

$$\begin{aligned} \Lambda (M) = \frac{\frac{\alpha \beta S}{(\alpha M -1)^2}-\frac{c}{v}}{g'(M)} \end{aligned}$$

is decreasing in this interval. Taking its first derivative yields $\Lambda '(M) = (-2\alpha ^2\beta S g'(M)/(\alpha M -1)^3 - g''(M)(\alpha \beta S/(\alpha M-1)^2 - c/v))/g'(M)^2$. Therefore $\Lambda '(M)\le 0$ if $-g''(M)/g'(M)\le h(M)$, where $h(M)=2\alpha ^2\beta S/((\alpha \beta S v - c(\alpha M -1)^2)(\alpha M -1))$.

For a bound that is independent of M, we analyze the function h(M) in the interval $(1/\alpha , M_R]$. We have that $\lim _{M\rightarrow 1/\alpha } h(M) = \infty$ and $\lim _{M\rightarrow 1/\alpha } h'(M) < 0$. Moreover, the equation $h'(M)=0$ yields $3c(\alpha M -1)^2-\alpha \beta S V = 0$ and has only one solution, $\hat{M}^*$, in the interval $(1/\alpha , M_R]$. Thus, this solution, $\hat{M}^*=1/\alpha + \sqrt{\beta S v/(3c\alpha )}$ is a minimum of h(M), and we can impose the bound

$$\begin{aligned} -\frac{g''(M)}{g'(M)}\le \frac{1}{\alpha } + \sqrt{\frac{\beta S v}{3c\alpha }},\quad \forall M\in \bigg (\frac{1}{\alpha },M_R\bigg ]. \end{aligned}$$

If this condition holds, then the function $g'(M)-\alpha \beta S/(\alpha M -1)^2 +c/v$ has the $\mathcal{S}\mathcal{C}$ property and the expected cost function is quasi-convex. $\square$

Proof of Proposition 5.7

From the proof of Proposition 5.2, we have that $M_R=1/\alpha +\sqrt{(v\beta S/(\alpha c))}$. In the case with site cohort size-dependent expected activation delays, any critical point of $\mathbb {E}[\tilde{\mathcal {C}}(M)]$ satisfies Equation (9). Solving for M as a function of $g'(M)$ yields

$$\begin{aligned} \tilde{M}_{R}=\frac{1}{\alpha }+\sqrt{\frac{\beta S}{\alpha (g'(\tilde{M}_R)+\frac{c}{v})}}. \end{aligned}$$

Since g(M) is an increasing function of the cohort size, it follows that ${M}_{R}\ge \tilde{M}_{R}$. $\square$

Proof of Proposition 6.1

We can use the Implicit Function Theorem to determine the direction of the change in $\tilde{M}_R$. In particular, from the first-order condition (9) we can define the functions

$$\begin{aligned} H_1(M,S)= & {} g'(M)-\frac{\alpha \beta S}{(\alpha M -1)^2}+\frac{c}{v},\\ H_2(M,c/v)= & {} g'(M)-\frac{\alpha \beta S}{(\alpha M -1)^2}+\frac{c}{v}, \end{aligned}$$

and calculate

$$\begin{aligned} \frac{d\tilde{M}_R}{dS}= & {} -\frac{\partial {H_1}/\partial {S}}{\partial {H_1}/\partial {M}}=\frac{\alpha \beta /(\alpha \tilde{M}_R-1)^2}{g''(\tilde{M}_R)+2\alpha ^2\beta S/(\alpha \tilde{M}_R-1)^3},\\ \frac{d\tilde{M}_R}{d\left( c/v\right) }= & {} -\frac{\partial {H_2}/\partial {(c/v)}}{\partial {H_2}/\partial {M}}=-\frac{1}{g''(\tilde{M}_R)+2\alpha ^2\beta S/(\alpha \tilde{M}_R-1)^3}. \end{aligned}$$

Since $\tilde{M}_R$ is a minimum in the range $(1/\alpha , \infty )$, it follows that $g''(\tilde{M}_R)\ge 0$ and consequently, $d\tilde{M}_R/dS>0$ and $d\tilde{M}_R/d(c/v)<0$. A similar derivation can be carried with $M_R$ by setting $g'(M)=0$ (no delays or constant expected delays), with equivalent results. $\square$

Proof of Proposition 6.2

Since both clinical trials only have one optimal solution, per Equation (9), the decreasing function $\alpha \beta S/(\alpha M-1)^2-c/v$ will cross both $g'_1(M)$ and $g'_2(M)$ exactly once. Since $CT_1$ is more efficient than $CT_2$, the function $g'_2(M)$ lies on top of $g'_1(M)$ for all M and, given that the right-hand side of (9) comes from $+\infty$ at $M=1/\alpha$, it will cross $g'_2$ before it crosses $g'_1$.

Appendix B: Recruitment process with exponentially distributed site activation delays

Proposition B.1

Let a site $m\in \mathcal {M}$ be initialized at time 0 and let $N_m(t)$ be a random variable denoting the number of recruits in this site at time t. Moreover, let $s_m\sim Exp(\lambda _s)$ be the random activation time of this site. Then, $N_m(t)$ has the following probability mass function:

$$\begin{aligned} P\{N_m(t) = j\} = {\left\{ \begin{array}{ll} e^{-t\lambda _s}+C_0\beta ^{\alpha }\lambda _s e^{-(t+\beta )\lambda _s},&{} j=0, \\ \lambda _s e^{-(t+\beta )\lambda _s} \displaystyle \left( {\begin{array}{c}j+\alpha - 1\\ \alpha -1\end{array}}\right) \displaystyle \sum _{k=0}^j \displaystyle \left( {\begin{array}{c}j\\ k\end{array}}\right) (-1)^{k} \beta ^{k+\alpha }C_k,&j=1,2,\dots , \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} C_k =\lambda _s^{k+\alpha -1}\int \limits _{\beta \lambda _s}^{(t+\beta )\lambda _s} e^{v} /v^{k+\alpha } dv,\quad k=0,1,\dots . \end{aligned}$$

Proof

Using (1) as a starting point, let us derive the probability mass function of $N_m(t)$ for the case where the activation delays are exponentially distributed. Let $p= \beta /(t-s + \beta )$. We will work first on the case for $j=1,2,\dots$. By virtue of the binomial formula, we can write

$$\begin{aligned} P\{N_m(t) = j\}&=\lambda _s {j+\alpha - 1\atopwithdelims ()\alpha -1} \displaystyle \int \limits _{0}^{t} p^{\alpha } (1-p)^{j} e^{-s\lambda _s} ds \nonumber \\&=\lambda _s {j+\alpha - 1\atopwithdelims ()\alpha -1} \displaystyle \int \limits _{0}^{t} \sum _{k=0}^{j} {j \atopwithdelims ()k} p^{k+\alpha } (-1)^k e^{-s\lambda _s} ds\nonumber \\&=\lambda _s {j+\alpha - 1\atopwithdelims ()\alpha -1}\displaystyle \sum _{k=0}^{j} {j \atopwithdelims ()k}(-1)^k\int \limits _{0}^{t} p^{k+\alpha } e^{-s\lambda _s} ds . \end{aligned}$$

(B7)

We can work on the last integral and write everything again as a function of s:

$$\begin{aligned} \int \limits _{0}^{t} p^{k+\alpha } e^{-s\lambda _s} ds&= \beta ^{k + \alpha } \int \limits _{0}^{t} \frac{e^{-s\lambda _s} }{(t-s + \beta )^{k+\alpha } } ds=\beta ^{k + \alpha } e^{-(t+\beta )\lambda _s} \int \limits _{0}^{t} \frac{e^{(t-s+\beta )\lambda _s}}{(t-s + \beta )^{k+\alpha } } ds\nonumber \\&=\beta ^{k + \alpha } e^{-(t+\beta )\lambda _s} C_k, \end{aligned}$$

(B8)

with $C_k=\int \limits _{0}^{t} e^{(t-s+\beta )\lambda _s}/(t-s + \beta )^{k+\alpha } ds$. Substituting (B8) in (B7) we obtain $P\{N_m(t) = j\}= \lambda _s{j+\alpha - 1\atopwithdelims ()\alpha -1}\sum _{k=0}^{j} {j \atopwithdelims ()k}(-1)^k \beta ^{k + \alpha } e^{-(t + \beta )\lambda _s} C_k$.

Finally, the expression for $j=0$ can be derived from (1) in a similar manner noting that, in this case, $j=0$ also implies that $k=0$. $\square$

Proposition B.2

Under the assumption of exponentially distributed site activation delays, the expected value of $N_m(t)$ is given by

$$\begin{aligned} \mathbb {E}[N_m(t)] = \frac{\alpha }{\beta } \left( t - \frac{1}{\lambda _s} + \frac{1}{\lambda _s} e^{-t\lambda _s}\right) , \end{aligned}$$

and the variance of $N_m(t)$ is given by

$$\begin{aligned} Var[N_m(t)]=\mu (1-\mu ) + \frac{\alpha (1+\alpha )}{\beta ^2} \left( t^2 - \frac{2t}{\lambda _s} - \frac{2}{\lambda _s^2} e^{-t\lambda _s} + \frac{2}{\lambda _s^2}\right) , \end{aligned}$$

with $\mu =\mathbb {E}[N_m(t)]$.

Proof

Using (2) as a starting point, the conditional expectation for the specific case of an exponential random variable is $\mathbb {E}[s_m|s_m\le t] = \int _0^{t} sf_s(s|s\le t)ds=\int _0^{t}sf_s(s)ds/F_s(t)=1/\lambda _s-e^{-t/\lambda _s}t/(1-e^{-t\lambda _s})$. Substituting in this same equation yields $\mathbb {E}[N_m(t)]=\alpha /\beta (t- 1/\lambda _s + e^{-t\lambda _s}/\lambda _s)$. With a similar approach, and starting from (3), we can prove that $Var [N_m(t)]=\mu (1-\mu ) + \alpha (1+\alpha )/\beta ^2 (t^2 - 2t/\lambda _s - 2e^{-t\lambda _s}/\lambda _s^2 + 2/\lambda _s^2)$.

Finally, it can be seen that $Var [N_m(t)] > \mu (1-\mu ) + \alpha ^2/\beta ^2 (t^2 - 2t/\lambda _s - 2e^{-t\lambda _s}/\lambda _s^2 + 2/\lambda _s^2)$. Thus, to prove that $Var [N_m(t)] > \mathbb {E}[N_m(t)],$ it suffices to show that $\mu (1-\mu ) + \alpha ^2\beta ^2 (t^2 - 2t/\lambda _s - 2e^{-t\lambda _s}/\lambda _s^2 + 2/\lambda _s^2) > \mu$, which can be further simplified to $1 \ge e^{-2t\lambda _s} + 2t\lambda _s e^{-t\lambda _s}.$ Let $x = t\lambda _s$. Note that $x\ge 0$, since $\lambda _s\ge 0$. Next, we can easily show that $e^{-2x} + 2x e^{-x}$ attains its maximum at 1 when $x=0$. The assertion follows. $\square$