Optimal weekly scheduling in fractionated radiotherapy: effect of an upper bound on the dose fraction size

C. Bruni; F. Conte; F. Papa; C. Sinisgalli

doi:10.1007/s00285-014-0805-8

Journal of Mathematical Biology

August 2015, Volume 71, Issue 2, pp 361–398 | Cite as

Optimal weekly scheduling in fractionated radiotherapy: effect of an upper bound on the dose fraction size

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Authors and affiliations

C. Bruni
F. Conte
F. Papa
C. SinisgalliEmail author

Article

First Online: 29 August 2014

Received: 23 December 2013

Revised: 09 June 2014

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1 Citations

Abstract

This work concerns the optimization of the dose fractionation for cancer radiotherapy schedules of the kind one fraction/day, five fractions/week, assuming a fixed overall treatment time. Constraints are set to limit the radiation damages to surrounding normal tissues, as well as the daily fraction size. The response to radiation of tumour and normal tissues is represented by the classical LQ model, including the exponential repopulation term. We provide a framework to analytically determine the optimal weekly scheme of radiation doses as a function of the tumour type, the fraction upper bound and the normal tissue parameters. For a comparison with the literature, we present some numerical examples of optimal treatment schedules for specific tumour types.

Keywords

Nonlinear programming Cancer radiotherapy Linear-quadratic model

Mathematics Subject Classification

90C30 90C90 92B05

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Notes

Acknowledgments

We wish to thank two anonymous referees for their constructive comments and stimulating suggestions.

Appendix

This appendix concerns the study of the intersection region of the constraints $g_{e}(d) \le 0,\,g_{l}(d) \le 0$ in the subset $d \ge 0$ of $R^{5}$. More precisely, as seen in Sect. 4, any possible extremal of Problem 1 must satisfy $g_{e}(d) = 0$ or $g_{l}(d) = 0$ or both, so we are actually interested in studying the systems (4.6) in $d \ge 0$. The first goal is to show that the quantities $k_{e} - k_{l},\,\rho _e k_{l} - \rho _{l} k_{e}$ and $v$ (functions of the normal tissue parameters) determine whether the boundaries $g_{e}(d) = 0$ and $g_{l}(d) = 0$ intersect each other in $d \ge 0$ or not. Furthermore, when there is not such an intersection, we prove that all the points lying on one of the two boundaries make the other constraint strictly satisfied (acceptable points), and that all the points of this second boundary violate the first constraint (not acceptable points). Then, all the extremals belong to the boundary of the same normal tissue constraint, so that Problem 1 has one prevalent constraint to be established again on the basis of the quantities $k_{e} - k_{l},\,\rho _e k_{l} - \rho _{l} k_{e}$ and $v$.

We start looking for the solutions of the system

$$\begin{aligned} \begin{array}{l} g_{e}(d) = 0, \\ g_{l}(d) = 0, \\ d_{k} \ge 0,\quad k=1,\ldots ,5. \end{array} \end{aligned}$$

(8.1)

Rather than the first two equations in (8.1), we can consider the differences $g_{e}(d) - g_{l}(d) = 0$ and $\rho _{e} g_{l}(d) - \rho _{l} g_{e}(d) = 0$, and we rewrite (8.1) as

$$\begin{aligned} \begin{aligned}&\sum _{k = 1}^{5} d_{k} = \frac{k_{e} - k_{l}}{\rho _{e} - \rho _{l}} = S, \\&\sum _{k = 1}^{5} d_{k}^{2} = \frac{\rho _e k_{l} - \rho _{l} k_{e}}{\rho _{e} - \rho _{l}} = Q, \\&d_{k} \ge 0,\quad k = 1, \ldots , 5. \end{aligned} \end{aligned}$$

(8.2)

From (8.2) it immediately appears that the original system (8.1) has no solution if $k_{e} - k_{l} \le 0$ or if $\rho _e k_{l} - \rho _{l} k_{e} \le 0$. Furthermore, we can prove that the solutions of (8.1) exist only if

$$\begin{aligned} k_{e} - k_{l} > 0,\quad \rho _e k_{l} - \rho _{l} k_{e} > 0, \quad 1 \le v \le 5. \end{aligned}$$

(8.3)

A way to prove the above statement is finding the minimal and maximal values of the total dose $\sum _{k = 1}^{5} d_{k}$ on the set

$$\begin{aligned} {\bar{D}}=\left\{ d\in R^{5} | \quad \sum _{k = 1}^{5} d_{k}^{2} = Q, \quad d_{k}\ge 0,\ k = 1, \ldots , 5\right\} . \end{aligned}$$

by means of a classical quadratic programming approach. Solving the optimality necessary conditions, we find the minimum point at $(\sqrt{Q}, 0, 0, 0, 0)$, with minimal total dose $\sum _{k = 1}^{5} d_{k} = \sqrt{Q}$, and the maximum point at $(\sqrt{Q/5}, \sqrt{Q/5}, \sqrt{Q/5}, \sqrt{Q/5}, \sqrt{Q/5})$, with maximal total dose $\sum _{k = 1}^{5} d_{k} = \sqrt{5 Q}$. This result implies that (8.2), and then (8.1), has solution if and only if

$$\begin{aligned} \sqrt{Q} \le S \le \sqrt{5 Q} \end{aligned}$$

(8.4)

which, recalling the definition (4.28) of $v$, amounts to say $1 \le v \le 5$.

When one of the conditions (8.3) is denied, no vector $d \ge 0$ exists such that $g_{e}(d) = g_{l}(d) = 0$. Then, we prove that one constraint is prevalent and, examining the values of the quantities $k_{e} - k_{l},\,\rho _e k_{l} - \rho _{l} k_{e}$ and $v$ in Table 2, we establish which is the prevalent constraint.

Let us start supposing $k_{e} - k_{l} \le 0$. For each $d \ge 0$, it follows $g_{l}(d) \le g_{e}(d)$, which implies that for each $d$ such that $g_{e}(d) \le 0$ it is $g_{l}(d) < 0$, whereas for each $d$ such that $g_{l}(d) = 0$ it is $g_{e}(d) > 0$. So, the early constraint is prevalent and no “late” vector is acceptable. Otherwise, if $\rho _e k_{l} - \rho _{l} k_{e} \le 0$, for each $d \ge 0$ it follows $\rho _{e} g_{l}(d) \ge \rho _{l} g_{e}(d)$, which implies that the late constraint is now prevalent and no “early” vector is acceptable.

Let us consider $k_{e} - k_{l} > 0,\,\rho _e k_{l} - \rho _{l} k_{e}>0$ and $v < 1$ proving that the early constraint is prevalent and that no “late” vector is acceptable. Let us proceed by contradiction, supposing that the two systems

$$\begin{aligned} \left\{ \begin{array}{l} g_{e}(d) \le 0, \\ g_{l}(d) > 0, \\ d_{k} \ge 0,\quad k=1,\ldots ,5, \end{array} \right. \quad \left\{ \begin{array}{l} g_{e}(d) < 0, \\ g_{l}(d) = 0, \\ d_{k} \ge 0,\quad k = 1, \ldots , 5. \end{array}\right. \end{aligned}$$

(8.5)

admit solutions so to get an absurd. Indeed, it is always true that $(\sum _{k=1}^{5} d_{k})^{2} \ge \sum _{k=1}^{5} d_{k}^{2}$ for $d \ge 0$, and since both the systems in (8.5) require $\sum _{k=1}^{5} d_{k} < S$ and $\sum _{k=1}^{5} d_{k}^{2} > Q$, it would follow $Q \le S^{2}$, which is in contrast with $v=S^{2}/Q < 1$. So, neither (8.1) nor (8.5) have solutions and it is easy to deduce that for each $d$ satisfying $g_{e}(d) \le 0$, it is $g_{l}(d) < 0$, and that for each $d$ satisfying $g_{l}(d) = 0$, it is $g_{e}(d) > 0$.

Finally, for $k_{e} - k_{l} > 0,\,\rho _e k_{l} - \rho _{l} k_{e} > 0$ and $v > 5$ we similarly prove that the late constraint prevails and no “early” vector is acceptable, by showing that the systems

$$\begin{aligned} \left\{ \begin{array}{l} g_{e}(d) > 0, \\ g_{l}(d) \le 0, \\ d_{k} \ge 0,\quad k = 1, \ldots , 5, \end{array} \right. \quad \left\{ \begin{array}{l} g_{e}(d) = 0, \\ g_{l}(d) < 0, \\ d_{k} \ge 0,\quad k = 1, \ldots , 5. \end{array} \right. \end{aligned}$$

(8.6)

cannot have solutions. Indeed, from the Cauchy–Schwartz inequality, we get $5 \sum _{k=1}^{5} d_{k}^{2} \ge (\sum _{k=1}^{5} d_{k})^{2}$ and since both the systems (8.6) require $\sum _{k=1}^{5} d_{k} > S$ and $\sum _{k=1}^{5} d_{k}^{2} < Q$, it would have to be $S^{2} \le 5 Q$, clearly in contrast with $v=S^{2}/Q > 5$. Thus, for each $d$ satisfying $g_{l}(d) \le 0$, it is $g_{e}(d) < 0$, and for each $d$ satisfying $g_{e}(d) = 0$, it is $g_{l}(d) > 0$.

The results proved in this Appendix are schematized in Table 2.

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Authors and Affiliations

C. Bruni
- 1
F. Conte
- 1
F. Papa
- 2
C. Sinisgalli
- 2
Email author

1.Dipartimento di Ingegneria Informatica, Automatica e Gestionale “A. Ruberti”Sapienza Università di RomaRomeItaly
2.Istituto di Analisi dei Sistemi ed Informatica “A. Ruberti” -CNRRomeItaly

Optimal weekly scheduling in fractionated radiotherapy: effect of an upper bound on the dose fraction size

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