Analytical solution to the radiotherapy fractionation problem including dose bound constraints
AANALYTICAL SOLUTION TO THE RADIOTHERAPYFRACTIONATION PROBLEM INCLUDINGDOSE BOUND CONSTRAINTS
LUIS A. FERN ´ANDEZ AND LUC´IA FERN ´ANDEZ
Dedicated to the memory of Juan Antonio Fern´andez (1957-2018)
Abstract.
This paper deals with the classic radiotherapy dose frac-tionation problem for cancer tumors concerning the following goals:a) To maximize the effect of radiation on the tumor, restricting theeffect produced to the organs at risk (healing approach).b) To minimize the effect of radiation on the organs at risk, whilemaintaining enough effect of radiation on the tumor (palliativeapproach).We will assume the linear-quadratic model to characterize the radia-tion effect and consider the stationary case (that is, without taking intoaccount the timing of doses and the tumor growth between them).The main novelty with respect to previous works concerns the pres-ence of minimum and maximum dose fractions, to achieve the minimumeffect and to avoid undesirable side effects, respectively.We have characterized in which situations is more convenient thehypofractionated protocol (deliver few fractions with high dose per frac-tion) and in which ones the hyperfractionated regimen (deliver a largenumber of lower doses of radiation) is the optimal strategy.In all cases, analytical solutions to the problem are obtained in termsof the data. In addition, the calculations to implement these solutionsare elementary and can be carried out using a pocket calculator. Introduction
According to the World Health Organization [13], “radiotherapy is oneof the major treatment options in cancer management. (...) Together withother modalities such as surgery and chemotherapy it plays an importantrole in the treatment of 40% of those patients who are cured of their can-cer. Radiotherapy is also a highly effective treatment option for palliationand symptom control in cases of advanced or recurrent cancer. The pro-cess of radiotherapy is complex and involves understanding of the principlesof medical physics, radiobiology, radiation safety, dosimetry, radiotherapy
Mathematics Subject Classification.
Key words and phrases.
Radiotherapy, fractionation, mixed and continuous optimiza-tion, linear quadratic model.The work of the first author was supported by the Spanish “Ministerio de Econom´ıa,Industria y Competitividad” under project MTM2017-83185-P. a r X i v : . [ m a t h . O C ] F e b LUIS A. FERN ´ANDEZ AND LUC´IA FERN ´ANDEZ planning, simulation and interaction of radiation therapy with other treat-ment modalities”.Mathematical modelling has played an important role in understandingand optimizing radiation delivery for cancer treatment. Since its formulationmore than 50 years ago, the linear-quadratic (LQ) model has become thepreferred method for characterizing radiation effects. Usually, it is stated asfollows: the survival probability S of a tumor cell after exposure to a singledose of radiation of d Gy is expressed as S = exp (cid:0) − α T d − β T d (cid:1) , where α T and β T are two positive parameters describing the radiosensitivityof the cell, [9]. It is well known that these parameters depend on the type ofradiation therapy chosen and also on the organ where the tumor is located[15]. More precisely, LQ model implies that if the initial size of the tumor is U , then it will be U · S after applying a d Gy dose. Let us recall that “Gray”( Gy ) is the unit of ionizing radiation dose in the International System ofUnits.LQ model has well documented predictive properties for fractionation/doserate effects in the laboratory and “it is reasonably well validated, experimen-tally and theoretically, up to about 10 Gy per fraction and would be reason-able for use up to about 18 Gy per fraction”, see [4]. Precisely, its range ofvalidity is a key point of controversy; although there is a general consensuson the existence of this range, significant disagreements remain on the exactvalues of its limits. Let us illustrate this fact with other recent quotes: from[9], “in vitro (...) some authors suggesting significant discrepancies at dosesof 5 Gy or above, while others report good agreement up to tens of Gy” andaccording to the French Society of Young Radiation Oncologists, “the dose/ fraction must be between 1 and 6 Gy ”, see [11].Given N doses, d , ..., d N , eventually different, if we consider the station-ary case (which means that neither the times of application of the doses northe growth of the tumor produced between them are taken into account),the probability of accumulated survival is given by(1) S N = exp (cid:32) − N (cid:88) i =1 (cid:0) α T d i + β T d i (cid:1)(cid:33) . From here it is clear that the effect of radiation on the tumor is determinedby the quadratic function(2) E T ( N, d ) = α T N (cid:88) i =1 d i + β T N (cid:88) i =1 d i . On the other hand, radiation also affects healthy organs and tissues nearthe tumor (which we will denote by OAR, organs at risk, hereafter). Ingeneral, healthy organs and tissues receive less radiation than the tumor,which we will denote by δd , with δ ∈ (0 ,
1] being the so-called “sparing
ADIOTHERAPY FRACTIONATION PROBLEM 3 factor”. The value of δ depends on factors such as the location and geometryof the tumor and also on the technology used to deliver the radiation, see [3].It can be seen as a measure of the accuracy of the radiotherapy: if clinicianscan keep the OAR almost unaffected by the radiation, δ will be about 0; ifnot, it will be larger, until reaching the value δ ≈ E OAR ( N, d ) = α δ N (cid:88) i =1 d i + β δ N (cid:88) i =1 d i , where α and β are the parameters associated to the healthy organs thatwe are trying to protect.Typical values for α , β , α T and β T can be found in the specialized lit-erature such as [15]. These data come from conducting experiments andthe corresponding adjustments (least squares regression) to achieve approx-imated values that best fit experimental data.Let us now introduce the most common strategies for fractionating radio-therapy treatments: • Hypofractionation : Deliver higher doses of radiation on few occa-sions. This strategy results in a significant reduction of its duration. • Hyperfractionation : Deliver a large number of lower doses of ra-diation that are given more than once a day.In this paper we study the classic radiotherapy dose fractionation problemrelated to the following goals:a) To maximize the effect of radiation on the tumor, restricting theeffect produced on the OAR (healing approach) in Section 2 andb) To minimize the effect of radiation on the OAR, maintaining enougheffect of radiation on the tumor (palliative approach) in Section 3.The first novelty with respect to previous works in this framework con-cerns the presence of dose fraction bounds of the type 0 < d min ≤ d ≤ d max .On one hand, these restrictions are connected to the range of validity of theaforementioned LQ model and can be estimated for each particular tumor;on the other hand, they also take into account the minimum and maximumdose fraction that can be applied in practical situations in order to achievea minimum effect and avoid undesirable side effects, respectively. It is wellknown that the dose per fraction value in most conventional treatments isaround 2 Gy, see for instance [12]. Depending on the tumor type, the valuesof d min and d max can be tunned, but the reference values d min = 1 Gy and d max = 6 Gy could be a valid generic choice. In this sense one can not findin [12] a single treatment recomendation with a dose fraction less than 1 Gy and very few larger than 6 Gy.
The counterpart for imposing a positive minimum dose fraction is thatthe total number of radiations N should not be fixed a priori and this is thesecond important novelty of this work: N will also be considered another LUIS A. FERN ´ANDEZ AND LUC´IA FERN ´ANDEZ unknown of the problem and we will study the dependence of the solutionwith respect to N . Among others, this approach was followed by [8], butonly for uniform dose treatments. Our approach here includes also the studyfor nonuniform protocols. A preliminary version of our results was presentedby the second author as part of the academic project [7], except for the studyof the dependence of the solution with respect to N which is new.Summarizing, new analytical solutions in terms of the data are obtainedfor both problems, improving known results in the literature to the best ofour knowledge, see for instance [10] and [14]. Moreover, the final calculationsto implement these solutions are elementary and can be carried out using apocket calculator.2. Maximizing the effect of radiation on the tumor
The aim of this first problem is to determine the best strategy to maximizethe effect of radiation on the tumor, while restricting the effect on the OAR(healing approach):( P ) Maximize E T ( N, d ) , subject to N ∈ N , d ∈ R N such that E OAR ( N, d ) ≤ γ OAR ,d min ≤ d i ≤ d max , i = 1 , ..., N, where E T ( N, d ) is given by (2), E OAR ( N, d ) by (3) and d min , d max and γ OAR are a priori known positive parameters, that should be provided by thespecialists. Roughly speaking, the restriction E OAR ( N, d ) ≤ γ OAR can beinterpreted in the sense that the percentage of survival cells of the OARshould be greater than or equal to 100 exp ( − γ OAR ) . This is the classic fractionation problem that has been studied (with somevariations) in several works, see for example the recent papers [2] and [14](where more than one OAR is considered) and the references therein. Thefirst novelty of our approach is that dose bound constraints are also included.Usually in the literature the lower bound 0 value is taken for d i and no upperbound is imposed; some exceptions are [5] and [6] where an upper bound isincluded, but not a positive lower bound. The danger of losing control ofthe tumor, due to the use of doses below a critical limit, has already beenpointed out by [8]. In addition, our approach to the problem is more usefulsince the number of doses N is not initially set as in [2] and [14]. The caseincluding repopulation was studied in [3], only assuming the non-negativityof d i .From a mathematical point of view, this is a mixed optimization probleminvolving a discrete variable, N ∈ N , which corresponds to the number ofradiation doses, and N continuous variables, d i ∈ R , ≤ i ≤ N , whichare the doses. In other words, this problem has the peculiarity of having avariable number of unknowns. ADIOTHERAPY FRACTIONATION PROBLEM 5
Along this paper, it will be denoted(4) ϕ ( r ) = α δr + β δ r , (5) λ = max (cid:26) , γ OAR ϕ ( d max ) (cid:27) , and(6) ρ = γ OAR ϕ ( d min ) . Also, we will denote by (cid:98) x (cid:99) the greatest integer less than or equal to x and by (cid:100) x (cid:101) the least integer greater than or equal to x . Finally, the notation d N = ( d , . . . , d ) means that d N ∈ R N having all the N components equalto d .2.1. Existence of solution for ( P ) .Theorem 1. Let us assume d min > and ρ ≥ . Then, the problem ( P ) has (at least) one solution.Proof. Taking into account the restrictions for ( P ) and that d min >
0, wehave N ≤ ρ . Hence, the set of feasible values for N is finite.If ρ = 1, the solution is ( N, d ) = (1 , d min ), because this is the onlyadmissible pair for ( P ).When ρ ∈ (1 , , the value N = 1 is still the only possible option.Consequently, we are faced with a maximizing problem of an increasing1 D function. Then, the solution will be given by the largest feasible value.In this case, it is quite easy to verify that the unique solution of ( P ) is thepair (1 , min { d max , d } ) , where d = − α + (cid:112) α + 4 β γ OAR β δ . Let us stressthat ϕ ( d ) = γ OAR .If ρ ≥ P ) to a finite collection of contin-uous optimization problems ( P N ) with fixed N given by:( P N ) Maximize ˜ E NT ( d ) = α T N (cid:88) i =1 d i + β T N (cid:88) i =1 d i , subject to d ∈ R N such that E OAR ( N, d ) ≤ γ OAR ,d min ≤ d i ≤ d max , i = 1 , ..., N. Firstly we will prove the existence of a solution for each problem ( P N ) (seeTheorem 2), for N running [1 , ρ ] ∩ N and denote it by d N . Then, it isenough to take the pair (cid:18) N , d N (cid:19) from the finite set (cid:110)(cid:16) N, d N (cid:17) : N ∈ [1 , ρ ] ∩ N (cid:111) , that maximizes the value of E T ( N, d ) as a solution to the problem ( P ). (cid:3) LUIS A. FERN ´ANDEZ AND LUC´IA FERN ´ANDEZ
The existence of a solution for each problem ( P N ) is proved below: Theorem 2.
Let us assume d min > , ρ ≥ and N ∈ [1 , ρ ] ∩ N . Thenthe problem ( P N ) has (at least) one solution.Proof. For small values of N , more precisely N ∈ [1 , λ ] ∩ N , it is easy toverify that the solution for ( P N ) is the trivial one with maximum doses, thatis, d N = ( d max , ..., d max ) . For other values, N ∈ ( λ , ρ ] ∩ N , the existenceof solution for ( P N ) follows from the classic Weierstrass Theorem, becausewe are maximizing a continuous objective function over a compact set. (cid:3) Remark 1. a) Let us point out that ( P ) is a nonconvex quadraticallyconstrained quadratic optimization problem (even ( P N ) with fixed N ), because the objective is to maximize a convex function. Typi-cally, this type of problems is computationally difficult to solve (see [14] ), but here we will see that it can be done analytically. b) Unless all the components of the solution are the same, the unique-ness of solution fails: it is enough to take two indices i, j ∈ { , ..., N } such that d i (cid:54) = d j and interchange these coordinates to generate a newsolution. c) Under the condition ρ < , it is apparent that the set of feasiblepoints is empty and hence, the existence of solution for ( P ) fails. d) The hypothesis d min > is also needed for proving the existence ofsolution for ( P ) , as it can be shown through the following example: Example 1. ( P ) Maximize E T ( N, d ) = N (cid:88) i =1 d i + N (cid:88) i =1 d i , subject to N ∈ N , d i ∈ R ,E OAR ( N, d ) = N (cid:88) i =1 d i + 2 N (cid:88) i =1 d i ≤ , ≤ d i ≤ , i = 1 , ..., N. It is clear that for all feasible points we have E T ( N, d ) ≤ E OAR ( N, d ) ≤ . Let us stress that here N can take any natural value, without re-strictions. Inspired by Theorem 4 below, let us consider the sequencegiven by d N = ( d N , ..., d N ) , with d N = −
14 + (cid:114)
116 + 5
N .
It is easy to check that it is feasible for N ≥ , E OAR ( N, d N ) = 10 , E T ( N, d N ) = 10 − N d N −→ , as N → + ∞ . ADIOTHERAPY FRACTIONATION PROBLEM 7
Hence, the problem ( P ) can not have solution ( N , d ) : the supre-mum value can not be attained since it should happen that
10 + N (cid:88) i =1 d i = E T ( N , d ) + N (cid:88) i =1 d i = E OAR ( N , d ) ≤ , which is clearly impossible. e) The deficiency of this type of models to produce solutions (when d min = 0 ) that prescribe infinite doses with fractions tending to zerowas pointed out by [8] . The following result provides a simpler version of the optimization prob-lem for the bigger values of N : Theorem 3.
Let us assume d min > , ρ ≥ and N ∈ ( λ , ρ ] ∩ N . Then,the inequality constraint of the problem ( P N ) has to be active at d N , with d N being a solution for ( P N ) .Proof. Arguing by contradiction, let us assume that the constraint is notactive, i.e.,(7) α δ N (cid:88) i =1 d i + β δ N (cid:88) i =1 d i < γ OAR . Since λ < N , we know that there exists some index j ∈ { , ..., N } such that d j < d max . Then, for sufficiently small (cid:15) >
0, the point ( d , ..., d j − , d j + (cid:15), d j +1 , ..., d N ) is feasible and satisfies˜ E NT ( d N ) < ˜ E NT (( d , ..., d j − , d j + (cid:15), d j +1 , ..., d N )) , but this contradicts the fact that d N is a solution for ( P N ). (cid:3) Hence, from now on, in this case we will consider the equality restriction α δ N (cid:88) i =1 d i + β δ N (cid:88) i =1 d i = γ OAR . Therefore, we deduce that N (cid:88) i =1 d i = 1 β δ (cid:34) γ OAR − α δ N (cid:88) i =1 d i (cid:35) , and the objective function can be written as(8) ˜ E NT ( d ) = (cid:20) α T − β T α β δ (cid:21) N (cid:88) i =1 d i + β T γ OAR β δ . Based on this identity, we can directly simplify the formulation of theproblem ( P N ) as follows: LUIS A. FERN ´ANDEZ AND LUC´IA FERN ´ANDEZ
Proposition 1.
Let us assume d min > , ρ ≥ , N ∈ ( λ , ρ ] ∩ N anddenote (9) ω δ = α T β T − α β δ . i) If ω δ > , then ( P N ) is equivalent to (10) ( P N, +1 ) Maximize N (cid:88) i =1 d i , subject to d ∈ K N , where (11) K N = { d ∈ R N : E OAR ( N, d ) = γ OAR , d min ≤ d i ≤ d max , ≤ i ≤ N } . ii) If ω δ < , then ( P N ) is equivalent to (12) ( P N, − ) Minimize N (cid:88) i =1 d i , subject to d ∈ K N . iii) If ω δ = 0 , then every feasible point for ( P N ) is a solution. Remark 2. a) The idea of this transformation can be found in [10] inthe context of the problem ( P ) that we will study in the next section. b) Let us note that for the majority of tumors α T /β T > α /β andtherefore, the case ω δ > is more frequent in clinical practice. c) As a consequence of Proposition 1, we can appreciate the great dif-ference between the cases ω δ > and ω δ < : in the first one, tomaximize the effect of radiation on the tumor we have to increasethe total dose, while in the second the total dose remains minimum(see also the Subsection 3.3 below). Solving ( P N ) . Let us begin by showing a 2 D -example of previousproblems that will inspire the general results of this section. Example 2.
Let us consider the following optimization problems: ( P , +1 ) Maximize d + d , subject to ( d , d ) ∈ R , d + d ) + d + d = 12 , ≤ d , d ≤ . ( P , − ) Minimize d + d , subject to ( d , d ) ∈ R , d + d ) + d + d = 12 , ≤ d , d ≤ . In the Figure 1 the points on the blue surface are those that satisfy theequality constraint and the intersection of blue and orange surfaces gives thecurve on which to maximize or minimize.
ADIOTHERAPY FRACTIONATION PROBLEM 9
Figure 1.
Visually one can guess that the unique solution to ( P , +1 ) is located onthe diagonal (more precisely, it is given by ( d , d ) with d = √ −
1) andthere are two solutions of ( P , − ) lying on the boundary (specifically, ( d , d )with d = 1, d = √ − d = √ − d = 1).2.2.1. Solving ( P N, +1 ) . In fact, what happens in previous example can beextended to the N − dimensional case. More precisely, we will prove that thesolution for ( P N, +1 ) is a vector with equal coordinates: Theorem 4.
Let us assume d min > , ρ ≥ and N ∈ ( λ , ρ ] ∩ N . Then,the unique solution to ( P N, +1 ) has the form d N = ( d , ..., d ) with (13) d = − α N + (cid:112) ( α N ) + 4 β N γ
OAR β δN . Proof.
By using the Cauchy-Schwarz inequality, we have (cid:32) N (cid:88) i =1 d i (cid:33) ≤ N (cid:32) N (cid:88) i =1 d i (cid:33) . Therefore, for each feasible point it follows that(14) γ OAR = α δ N (cid:88) i =1 d i + β δ N (cid:88) i =1 d i ≥ α δ N (cid:88) i =1 d i + β δ N (cid:32) N (cid:88) i =1 d i (cid:33) . Defining q ( z ) = β δ N z + α δz − γ OAR , previous inequality can be rewrittenas(15) q (cid:32) N (cid:88) i =1 d i (cid:33) ≤ . Taking into account that the polynomial q can be factorized in the form q ( z ) = β δ N ( z − z )( z − z ) with z < < z , we know that the relation (15)holds if and only if (cid:80) Ni =1 d i ∈ [0 , z ], because all the components d i have tobe positive.Now it is clear that the maximum value is achieved when (cid:80) Ni =1 d i = z .Combining this fact with (14), we deduce that(16) γ OAR = α δ N (cid:88) i =1 d i + β δ N (cid:88) i =1 d i ≥ α δ N (cid:88) i =1 d i + β δ N (cid:32) N (cid:88) i =1 d i (cid:33) = γ OAR . Hence(17) (cid:32) N (cid:88) i =1 d i (cid:33) = N N (cid:88) i =1 d i . In this case, Cauchy-Schwarz inequality becomes (in fact) an equality andthis is true if and only if all the components are equal, i. e., d = ... = d N . Therefore, d N = ( d , ..., d ) with d = z N and (13) holds.Let us emphasize that d satisfies d min ≤ d ≤ d max , thanks to thehypothesis N ∈ ( λ , ρ ]. (cid:3) Solving ( P N, − ) . Given d a solution of ( P N, − ), since the objectivefunction and the functions defining the restrictions are C , we can applythe Lagrange Multipliers Rule [1] to deduce the existence of real numbers α ∈ [0 , + ∞ ) , λ ∈ R and { µ i } Ni =1 ⊂ [0 , + ∞ ) verifying(18) α + | λ | + N (cid:88) i =1 µ i > , (19) α + λ α δ + 2 β δ d ... α δ + 2 β δ d N + µ N +1 − µ ... µ N − µ N = , (20) µ i ( d min − d i ) = 0 , µ i + N ( d i − d max ) = 0 , ≤ i ≤ N. Inspired by the 2D example, we will prove that d lies on the bound-ary of [ d min , d max ] N . Let us argue by contradiction assuming that d i ∈ ( d min , d max ), for all i ∈ { , ..., N } . Then, thanks to (20) we deduce that µ i = 0 , ∀ i ∈ { , ..., N } . In this case, (19) reads:(21) α + λ α δ + 2 β δ d ... α δ + 2 β δ d N = . ADIOTHERAPY FRACTIONATION PROBLEM 11 If λ = 0, the identity (21) implies that α = 0 , but this is not possible by(18). Therefore λ (cid:54) = 0 and from (21) we get d = ... = d N = − ( α + λα δ )2 λβ δ . In other words, we arrive to the solution of problem ( P N, +1 ), contradictingour initial hypothesis about d .Consequently, there exists (at least) one index j ∈ { , ..., N } such that d j ∈ { d min , d max } . Without loss of generality we can suppose that j = N .Let us see that in this case we can reduce the dimension of the optimizationproblem ( P N, − ) by means for the following auxiliary problem:( P N − , − ) Minimize N − (cid:88) i =1 d i + d N , subject to d ∈ R N − such that E OAR ( N − , d ) = γ OAR − α δd N − β δ d N ,d min ≤ d i ≤ d max , i = 1 , ..., N − . Proposition 2.
Assume that d = ( d , ..., d N ) is a solution of ( P N, − ) . Then ( d , ..., d N − ) is a solution of ( P N − , − ) .Proof. Every feasible point ( d , ..., d N − ) for the problem ( P N − , − ) satisfies E OAR ( N − , d ) = γ OAR − α δd N − β δ d N . This implies that ( d , ..., d N − , d N ) is a feasible point for ( P N, − ) . Hence,using that d is a solution of ( P N, − ), we get N − (cid:88) i =1 d i ≤ N − (cid:88) i =1 d i , which implies that ( d , ..., d N − ) is a solution of ( P N − , − ). (cid:3) Arguing exactly in the same form as before with the problem ( P N − , − ),we deduce that there must be an index j ∈ { , ..., N − } such that d j ∈{ d min , d max } and we can reduce again the dimension of the problem, obtain-ing a new problem with N − D problem:( P , − ) Minimize d + N (cid:88) i =2 d i , subject to d ∈ R such that α δd + β δ d = γ OAR − α δ N (cid:88) i =2 d i − β δ N (cid:88) i =2 d i ,d min ≤ d ≤ d max . Clearly, it is enough to solve the quadratic equation to get the solution.Summarizing previous results, given N ∈ ( λ , ρ ] ∩ N , the solution of( P N, − ) has one of the following structures:(22) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ) , or(23) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , with d ∗ ∈ ( d min , d max ) being the unique positive root of the quadratic equa-tion(24) ϕ ( d ∗ ) = γ OAR − Kϕ ( d min ) − ( N − K − ϕ ( d max ) , with ϕ defined in (4).We can characterize the unknown value K as follows:a) In the case (22), by using the equality restriction we derive that(25) K = N ϕ ( d max ) − γ OAR ϕ ( d max ) − ϕ ( d min ) . Of course, this holds if and only if the right hand side is a naturalnumber or zero.b) In the case (23), since ϕ is an strictly increasing function in [0 , + ∞ ),we know that ϕ ( d min ) < ϕ ( d ∗ ) < ϕ ( d max ) , and using (24) we get that K ∈ (cid:18) N ϕ ( d max ) − γ OAR ϕ ( d max ) − ϕ ( d min ) − , N ϕ ( d max ) − γ OAR ϕ ( d max ) − ϕ ( d min ) (cid:19) ∩ N , which means that(26) K = (cid:98) N ϕ ( d max ) − γ OAR ϕ ( d max ) − ϕ ( d min ) (cid:99) . Taking into account the conditions (25) and (26), it is easy to concludethat the latter structure (23) is more frequently found in practice than (22).Previous argumentations lead us to the following result:
Theorem 5.
Let us assume d min > , ρ ≥ and N ∈ ( λ , ρ ] ∩ N . Then,a solution to problem ( P N, − ) is given by a) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ) , when K defined by (25) belongsto N ∪ { } ; otherwise, b) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , with K defined by (26) and d (cid:63) satisfying (24). ADIOTHERAPY FRACTIONATION PROBLEM 13
Remark 3.
It is not difficult to show that a solution for ( P N, − ) is also asolution for the problem (27) Minimize N (cid:88) i =1 d i, subject to d ∈ R N such that E OAR ( N, d ) ≥ γ OAR ,d min ≤ d i ≤ d max , ≤ i ≤ N. We will use this property in the proof of Theorem 8 (see Appendix ). Analytical solution for ( P ) . As we pointed out, a solution of ( P )will be the pair (cid:18) N , d N (cid:19) , where d N denotes a solution of ( P N ) from thefinite set (cid:110)(cid:16) N, d N (cid:17) : N ∈ [1 , ρ ] ∩ N (cid:111) , maximizing the value of E T ( N, d ). In fact, combining previous results, wecan avoid the calculation of most solutions for ( P N ) by studying its depen-dence with respect to N . This is the goal of the next results. Let us startby studying the less frequent case: when (cid:98) λ (cid:99) = (cid:98) ρ (cid:99) . Theorem 6.
Let us assume d min > , ρ ≥ and (cid:98) λ (cid:99) = (cid:98) ρ (cid:99) . Then, theunique solution to problem ( P ) is given by the pair ( N , d N ) with N = (cid:98) λ (cid:99) and d N = ( d max , ..., d max ) . Proof.
In this case the set of feasible values for N is { , . . . , N } ⊂ N with N = (cid:98) ρ (cid:99) = (cid:98) λ (cid:99) . For those values of N , the solution for ( P N ) has the form d N = ( d max , ..., d max ). Among them, it is clear that in order to solve ( P )only the one with the largest number of components is of interest; this isattained at N . (cid:3) We will continue to analyze the most common case: when (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) .In the the trivial case ω δ = 0, the function to be minimized and the onedefining the restriction are proportional. Therefore, we can be derive thefollowing result: Proposition 3.
Let us assume d min > , ρ ≥ , (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) and ω δ = 0 . Then any feasible pair ( N , d ) with E OAR ( N , d ) = γ OAR is a solution toproblem ( P ) . In particular, the pairs ( N , d N ) with N ∈ {(cid:100) λ (cid:101) , . . . , (cid:98) ρ (cid:99)} and d N = ( d , ..., d ) , where (28) d = − α N + (cid:113) ( α N ) + 4 β N γ
OAR β δN , with N in the above set. Proof.
Due to the hypothesis ω δ = 0, we deduce straightforwardly thatproblem ( P ) is equivalent to( ˜ P ) Maximize E OAR ( N, d ) , subject to N ∈ N , d ∈ R N such that E OAR ( N, d ) ≤ γ OAR ,d min ≤ d i ≤ d max , i = 1 , ..., N, Obviously, the maximum value is reached when the restriction becomes anequality. This can be achieved in several ways, such as the treatments withequal doses described in the proposition statement. Let us emphasize that d ∈ [ d min , d max ] if and only if N ∈ [ λ , ρ ] . (cid:3) Theorem 7.
Let us assume d min > , ρ ≥ , (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) and ω δ > . Then, the unique solution to problem ( P ) is given by the pair (cid:18) N , d N (cid:19) with N = (cid:98) ρ (cid:99) and d N = ( d , ..., d ) , with d given by (28).Proof. Here, the set of feasible values for N is { , . . . , N } ⊂ N with N = (cid:98) ρ (cid:99) . Arguing as in previous theorem, among the small values (i. e. N ∈{ , . . . , N } , with N = (cid:98) λ (cid:99) , for solving ( P ) we only retain N = N and d N = ( d max , ..., d max ). For the other values, i. e. N ∈ { N +1 , . . . , N } , since ω δ >
0, the corresponding solution for ( P N ) is given by d N = ( d , ..., d ) with d defined in (13). In order to study the dependencewith respect to N for these values, thanks to Proposition 1, it is enough toconsider the auxiliary function ψ ( N ) = N d = − α N + (cid:112) ( α N ) + 4 β N γ
OAR β δ . Here, it follows easily that ψ ( N ) is an strictly increasing function and then,it will take its maximum value in { N + 1 , . . . , N } at N . Finally, we will derive that ( N , d N ) is the unique solution to problem( P ) by showing that(29) E T ( N , d N ) < E T ( N , d N ) . To that end, let us consider the linear function H ( x ) = N ( xd + d ) − N ( xd max + d max ) , x ∈ [ α β δ , + ∞ ) , with(30) d = − α N + (cid:112) ( α N ) + 4 β N γ OAR β δN . Using that N ϕ ( d max ) ≤ γ OAR = N ϕ ( d ) by the admissibility, weget that H (cid:18) α β δ (cid:19) ≥
0. Also, taking into account (30), it can also be
ADIOTHERAPY FRACTIONATION PROBLEM 15 checked that H (cid:48) ( x ) = N d − N d max > , because N < N . Then, fromthe assumption ω δ > H (cid:18) α T β T (cid:19) >
0, which isequivalent to (29). (cid:3)
In the case ω δ < , the situation is more complicated and it is detailed inthe next result. The proof is a little bit technical and is postponed to theAppendix 1: Theorem 8.
Let us assume d min > , ρ ≥ , (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) and ω δ < .Then, a solution to problem ( P ) is given by one of the following pairs: i) (cid:18) N , d N (cid:19) with N = (cid:98) λ (cid:99) and d N = ( d max , ..., d max ) , ii) (cid:18) N , d N (cid:19) with N = (cid:100) λ (cid:101) and d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ) , when K defined by (25) with N = N belongs to N ∪ { } , or iii) (cid:18) N , d N (cid:19) with N = (cid:100) λ (cid:101) and d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , where K is defined in (26) and d (cid:63) satisfies (24) with N = N .
Remark 4. a) Let us strongly highlight that all expressions included inTheorems 6-8 and Proposition 3 can be explicitly calculated from theinitial data of the problem. Moreover, the calculations to implementthese solutions are elementary and can be carried out using a pocketcalculator. b) On the other hand, when ω δ > , the optimal value of N is thelargest one within its range of possibilities (i.e. it is a hyperfraction-ated type treatment) with equal doses, while in the case ω δ < theoptimal value is the smallest one (i.e. it is a hypofractionated typetreatment). In this last case, let us stress that not all doses haveto be equal or large; in fact, some of them may be minimum. Asfar as we know, this structure is not usually cited in the specializedliterature. c) One interesting case appears when α T β T < α β , because then ω δ < forall δ ∈ (0 , and the optimal regimen is always of hypofractionatedtype, independent of the technology used and the geometry of thetumor. In practice this condition holds in some special cases, suchas the prostate tumor, where α T β T ≈ . Gy, while α β = 2 Gy , see [10] and [15] . d) After Remark 2-b) (see also the Subsection 3.3 below), it is clear thatthe hypofractionated case (associated with ω δ < ) is very convenientin the practice. Assuming that the other parameters are set, the con-dition ω δ < can always be achieved by taking δ close enough to . This last fact is related to increasing the precision of the radiotherapyprocess (for instance, by using cutting-edge technology). e) A related problem to ( P ) is studied in [3] and [6] , where the num-ber of dose fractions N is also an unknown, jointly with d . Theframework for that problems is more general, because a repopulationterm is included in the objective function, but only the lower bound d min = 0 is assumed. Furthermore, the determination of the optimalvalue for N is carried out in [6] by means of numerical simulations,while in [3, Theorem 2] it is done explicitly and the value N = 1 isobtained when ω δ < . In this last case, the single dose could be toolarge in practice (remember that no upper bound is imposed in [3] )and then more fractions would have to be tried until an acceptableone is found. Next, we illustrate the general process with a particular example:
Example 3.
Let us consider the following parameters taken from a typicalclinical situation: α T = 0 . Gy − , β T = 0 . Gy − , α = 0 . Gy − , β =0 . Gy − , see [10] , together with d min = 1 Gy, and d max = 6 Gy.
Then,the problem reads ( P ) Maximize E T ( N, d ) = 0 . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i , subject to N ∈ N , d i ∈ R ,E OAR ( N, d ) = 0 . δ N (cid:88) i =1 d i + 0 . δ N (cid:88) i =1 d i ≤ γ OAR , ≤ d i ≤ , i = 1 , ..., N. i) For δ = 0 . and γ OAR = 0 . , we have λ ≈ . , ρ ≈ . and ω δ ≈ . > . Among the values, N ∈ { , , , . . . , , , } ,we have proved (see Theorem 7) that the biggest one, N = 56 ,and the solution of ( P ) (that here is the hyperfractionated d =(1 . , ..., . ) provides the solution of ( P ) ; in fact, E T (56 , d ) ≈ . . We can easily check that with the standard protocol ˜ d S =(2 , ..., we get E T (25 , ˜ d S ) = 3 , and therefore there is about . gain in terms of effect on the tumor, while the efficiency regardingOAR is the same ( E OAR (56 , d ) = E OAR (25 , ˜ d S ) = 0 . ).On the other hand, the hypofractionated radiotherapy given by ˜ d = (2 . , ..., . produces E T (15 , ˜ d ) ≈ . , although the dam-age on OAR is also lower: E OAR (15 , ˜ d ) ≈ . . These last treat-ments are mentioned in [12, pg. 16] in connection with breast cancer.Of course, here we are only taking into account the mathematicalpoint of view. In clinical practice, other factors such as patient in-convenience and additional cost may advise the use of fewer doses,if the difference in terms of efficiency is considered small. ADIOTHERAPY FRACTIONATION PROBLEM 17 ii)
For δ = 0 . and γ OAR = 0 . , we calculate λ ≈ . , ρ ≈ . and ω δ = − < . In this case, the solution for ( P ) is given by ( N , d ) with N = 8 , d = (1 , d ∗ , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ) and d ∗ ≈ . Gy (see Theorem8-iii)), having E T (8 , d ) ≈ . . Recall that this is a hypofractionatedtype treatment. Just for comparison reasons, let us mention thatthe solution of ( P ) is ˜ d = (6 , ..., and the solution of ( P ) is ˜ d = (1 , , d ∗ , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ) with d ∗ ≈ . Gy producing E T (7 , ˜ d ) ≈ . and E T (9 , ˜ d ) ≈ . , that are smaller than E T (8 , d ) as expected. iii) Let us emphasize that the difference between “few” and “many”doses is relative to each particular problem and not an absolute clas-sification. For instance, in the problem ( P ) with δ = 0 . and γ OAR = 0 . , the solution is given by ( N , d ) with N = 7 , d =(1 . , . . . , . which corresponds to the hyperfractionated case(because N ∈ { , . . . , } ), although the number of delivered dosesis lower than in the previous hypofractionated treatment, see ii ) . For ω δ <
0, in most practical situations the solution is the one presentedin Theorem 8- iii ), but the alternatives i ) and ii ) can also appear as we showin the following example: Example 4.
Let us take the following parameters: α T = 0 . Gy − , β T =0 . Gy − , α = 0 . Gy − , β = 0 . Gy − , δ = 1 , d min = 1 Gy, and d max = 6 Gy.
Here, ω δ = − < and the problem under consideration is ( P ) Maximize E T ( N, d ) = 0 . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i , subject to N ∈ N , d i ∈ R , . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i ≤ γ OAR , ≤ d i ≤ , i = 1 , ..., N. For γ OAR = 0 . , we get that λ ≈ .
01 and ρ ≈ .
36. For N = (cid:98) λ (cid:99) = 10, the solution of ( P ) is given by d = (6 , ..., N = (cid:100) λ (cid:101) = 11, the solution of ( P ) is given by d = (1 , d ∗ , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ), with d ∗ ≈ . Gy . Since E T (10 , d ) = 12 > . ≈ E T (11 , d ), thesolution of ( P ) is given by N = 10 and d . This shows that sometimesthe option i ) is the valid one.Finally, taking γ OAR = 0 . , we have λ ≈ . , N = (cid:100) λ (cid:101) = 11 and K = N ϕ ( d max ) − γ OAR ϕ ( d max ) − ϕ ( d min ) = 1 . Then, the solution of ( P ) is given by d = (1 , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ) . Now, the solution of ( P ) is also given by N = 11 and d , because E T (11 , d ) = 12 .
1. So, in this case Theorem 8- ii ) holds.Once we have obtained the analytical expressions for the solution of prob-lem ( P ), we can deduce very easily its dependence with respect to the pa-rameters defining the problem. This will help us to know how to adjustthese parameters in order to achieve a desired solution. Let us show a resultin the direction: Corollary 1.
Let us assume d min > , ρ ≥ and ( N , d N ) is a solution of ( P ) . Then, N is an increasing function of γ OAR , decreasing with respect to α , β and δ and independent of α T and β T . Moreover, a) When ω δ > , N is also decreasing with respect to d min and inde-pendent of d max . b) When ω δ < , N is also decreasing with respect to d max and inde-pendent of d min .Proof. It is just a consequence of the expressions N = (cid:98) ρ (cid:99) with ρ givenby (6), when ω δ > , and N = (cid:98) λ (cid:99) or (cid:100) λ (cid:101) with λ given by (5), when ω δ < . (cid:3) In Table 1, we summarize the resolution of problem ( P ) in algorithmform, for the reader’s convenience.3. Minimizing the effect of radiation on the organs at risk
In this section, we will consider a problem closely related to that of theprevious section: the goal of this second issue will be to determine the beststrategy to minimize the effect of radiation on the organs at risk, whilemaintaining a minimum effect of radiation on the tumor. It is clear thatthis approach can be interesting (at least) for palliative therapies. Mathe-matically we formulate it in the following way:( P ) Minimize E OAR ( N, d ) , subject to N ∈ N ,d ∈ R N such that E T ( N, d ) ≥ γ T ,d min ≤ d i ≤ d max , i = 1 , ..., N, where E OAR ( N, d ) is given by (3), E T ( N, d ) is defined in (2) and γ T isa given positive parameter. Of course, this is also a mixed optimizationproblem with N + 1 unknowns: the number of radiation doses, N ∈ N , andthe value of the N doses, d i ∈ R , ≤ i ≤ N .This problem has recently been studied in the outstanding work [10],but with fixed N and only imposing the non-negativity constraint for thedoses. Moreover, in [10] it is also remarked that “The real interest of thepresent approach would be the determination of the optimum solution for ADIOTHERAPY FRACTIONATION PROBLEM 19
ALGORITHM FOR SOLVING ( P )DATA: α T , β T , α , β , d min , d max , δ and γ OAR (all positive, d min < d max and δ ≤ ω δ = α T β T − α β δ , λ = max (cid:26) , γ OAR ϕ ( d max ) (cid:27) and ρ = γ OAR ϕ ( d min ) , with ϕ ( r ) = α δr + β δ r . IF ρ < , ( P ) has NO SOLUTION.IF ρ = 1, the pair ( N , d N ) = (1 , d min ) is the UNIQUE SOLUTION of ( P ).IF ρ ∈ (1 , P ) is the pair( N , d N ) = (1 , min { d max , d } ) , with d = − α + (cid:112) α + 4 β γ OAR β δ .IF ρ ≥ (cid:98) λ (cid:99) = (cid:98) ρ (cid:99) , the UNIQUE SOLUTION of ( P )is the pair ( N , d N ) with N = (cid:98) λ (cid:99) and d N = ( d max , ..., d max ) . IF ρ ≥ , (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) and ω δ > , the UNIQUE SOLUTION of ( P )is the pair ( N , d N ) with N = (cid:98) ρ (cid:99) and d N = ( d , ..., d ) , where d = − α N + (cid:113) ( α N ) + 4 β N γ
OAR β δN .IF ρ ≥ , (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) and ω δ < , take N = (cid:100) λ (cid:101) andCALCULATE: M = N ϕ ( d max ) − γ OAR ϕ ( d max ) − ϕ ( d min ) .IF M ∈ N ∪ { } , take K = M and d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ).IF M (cid:54)∈ N ∪ { } , take K = (cid:98) M (cid:99) and d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , with d ∗ > ϕ ( d ∗ ) = γ OAR − Kϕ ( d min ) − ( N − K − ϕ ( d max ).Also take N = (cid:98) λ (cid:99) and d N = ( d max , ..., d max ) . CALCULATE: E T ( N , d N ) and E T ( N , d N ) . A SOLUTION of ( P ) is the pair ( N , d N ) that maximizes E T between them.IF ρ ≥ , (cid:98) λ (cid:99) < (cid:98) ρ (cid:99) and ω δ = 0 , ANY FEASIBLE PAIR (
N , d ) such that E OAR ( N , d ) = γ OAR is a SOLUTION for ( P ).In particular, the pairs ( N , d N ) with N ∈ {(cid:100) λ (cid:101) , . . . , (cid:98) ρ (cid:99)} and d N = ( d , ..., d ) , where d = − α N + (cid:113) ( α N ) + 4 β N γ
OAR β δN . Table 1.
Complete solution for problem ( P ) in algorithmic form. N in clinical practice”. As an intermediate step, we have achieved here theexpression of the optimal value for N in terms of the parameters of theproblem in this particular setting, see Table 2 for a detailed description,depending on the case.As we will see, the study for problem ( P ) can be carried out followingthe same argumentation to that of ( P ) with minor differences.3.1. Existence of solution for ( P ) . In the sequel we will denote(31) ϕ T ( r ) = α T r + β T r , (32) λ T = max (cid:26) , γ T ϕ T ( d max ) (cid:27) , and(33) ρ T = max (cid:26) , γ T ϕ T ( d min ) (cid:27) . Our first observation concerns the existence of solution for ( P ): Theorem 9.
Let us assume d min > . Then, the problem ( P ) has (at least)one solution.Proof. It is analogous to that of Theorem 1, although here there are in-finite feasible values for N : combining the restrictions, those such that N ∈ [ λ T , + ∞ ) ∩ N . We will begin by showing that for each fixed feasi-ble value N , the associated problem ( P N ) has a solution, where( P N ) Minimize ˜ E NOAR ( d ) = α δ N (cid:88) i =1 d i + β δ N (cid:88) i =1 d i , subject to d ∈ R N such that E T ( N, d ) ≥ γ T ,d min ≤ d i ≤ d max , i = 1 , ..., N. For large values of N , specifically for N ≥ ρ T , the solution of ( P N ) is thetrivial one with minimum values d min . Among them only the smallest valueof N could have practical interest, i.e., (cid:100) ρ T (cid:101) . For the other values, whenthey exist, that is for N ∈ [ λ T , ρ T ) ∩ N , the existence of solution for ( P N )is a consequence of Weierstrass Theorem, once more because the objectivefunction is continuous and the feasible set is compact. Therefore, for eachvalue of N in that interval, let us consider a global solution for the problem( P N ) that we will denote d N . Again, it is enough to take the pair (cid:18) N , d N (cid:19) from the finite set (cid:110)(cid:16) N, d N (cid:17) : N ∈ [ λ T , (cid:100) ρ T (cid:101) ] ∩ N (cid:111) , that minimizes the value of E OAR ( N, d ) as a solution to the problem ( P ). (cid:3) ADIOTHERAPY FRACTIONATION PROBLEM 21
Remark 5. a) Once more, except if all the coordinates of d N are equal,the solution will not be unique, because two different coordinates canbe permuted to generate a different solution. b) When ω δ > (see (9)), the hypothesis d min > is necessary forproving the existence of solution for ( P ) , as we can see through thefollowing example: Example 5. ( P ) Minimize E OAR ( N, d ) = N (cid:88) i =1 d i + N (cid:88) i =1 d i , subject to N ∈ N , d i ∈ R ,E T ( N, d ) = 2 N (cid:88) i =1 d i + N (cid:88) i =1 d i ≥ , ≤ d i ≤ , i = 1 , ..., N. Let us consider the sequence given by d N = ( d N , ..., d N ) , with d N = − (cid:114) N .
It is easy to check that it is feasible for N ≥ , E T ( N, d N ) = 10 , E OAR ( N, d N ) = 10 − N d N −→ , as N → + ∞ . Now, we can deduce that ( P ) can not have solution ( N , d ) becauseif E OAR ( N , d ) = 5 together with the restriction E T ( N , d ) ≥ weget (cid:80) Ni =1 d i ≥ , but this is incompatible with E OAR ( N , d ) = 5 . c) In contrast, when ω δ < , it is easy to show that problem ( P ) , withonly the lower bound constraints d i ≥ , has as solution ( N , d ) with N = 1 and d = − α T + (cid:112) ( α T ) + 4 β T γ T β T , see [10] . d) There are some particular cases in which the solution of ( P ) can bedetermined from previous argumentations very easily. For instance,when ρ T = 1 , because then (cid:18) N , d N (cid:19) = (1 , d min ) is the only admis-sible pair. Also when ρ T > and (cid:100) λ T (cid:101) = (cid:98) ρ T (cid:99) , because only thelarge values for N are feasible (i.e., those verifying N ≥ ρ T ) andconsequently ( N , d N ) with N = (cid:100) ρ T (cid:101) and d N = ( d min , ..., d min ) isthe solution of ( P ) . When N ∈ [ λ T , ρ T ) ∩ N , we know that d N = ( d min , ..., d min ) is not asolution of ( P N ), because it is not even feasible. Hence, we can simplify theproblem ( P N ) arguing in a similar way as in the proof of Theorem 3. Theorem 10.
Let us assume d min > and N ∈ [ λ T , ρ T ) ∩ N . Then, theinequality constraint of the problem ( P N ) has to be active at d N , being d N a solution of ( P N ) . From now on, the restriction will be taken as one of equality, that is, α T N (cid:88) i =1 d i + β T N (cid:88) i =1 d i = γ T . Again, applying the same procedure as for ( P ) in Section 2 we have N (cid:88) i =1 d i = 1 β T (cid:34) γ T − α T N (cid:88) i =1 d i (cid:35) , and the objective function will read(34) ˜ E NOAR ( d ) = (cid:20) α − β α T δβ T (cid:21) δ N (cid:88) i =1 d i + β δ γ T β T . Now, it is clear that we can simplify the formulation of the problem ( P N ),as follows: Proposition 4.
Let us assume d min > and N ∈ [ λ T , ρ T ) ∩ N . i) If ω δ > , then ( P N ) is equivalent to ( P N, +2 ) Maximize N (cid:88) i =1 d i , subject to d ∈ K N , where (35) K N = { d ∈ R N : E T ( N, d ) = γ T , d min ≤ d i ≤ d max , ≤ i ≤ N } . ii) If ω δ < , then ( P N ) is equivalent to ( P N, − ) Minimize N (cid:88) i =1 d i , subject to d ∈ K N . iii) If ω δ = 0 , then every feasible point for ( P N ) is a solution.Proof. It is enough to take into account that α − β α T δβ T = − β δω δ , where ω δ is defined in (9). (cid:3) Once we have seen that ( P N, +1 ) and ( P N, +2 ) are essentially the same prob-lem (resp. ( P N, − ) and ( P N, − )), we can “translate” the results obtained insection 2.2 to the current context as follows: Theorem 11.
Let us assume d min > and N ∈ [ λ T , ρ T ) ∩ N . Then, i) the unique solution to ( P N, +2 ) is given by d N = ( d , ..., d ) with (36) d = − α T N + (cid:112) ( α T N ) + 4 β T N γ T β T N .
ADIOTHERAPY FRACTIONATION PROBLEM 23 ii) a solution for ( P N, − ) has one of the following forms: (37) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ) , with (38) K = N ϕ T ( d max ) − γ T ϕ T ( d max ) − ϕ T ( d min ) ∈ N ∪ { } , or (39) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , with (40) K = (cid:98) N ϕ T ( d max ) − γ T ϕ T ( d max ) − ϕ T ( d min ) (cid:99) , and d ∗ ∈ ( d min , d max ) satisfying (41) ϕ T ( d ∗ ) = γ T − Kϕ T ( d min ) − ( N − K − ϕ T ( d max ) . Analytical solution for ( P ) . As a consequence of previous resultswe arrive to the main theorems of this section that completely clarifies thesituation concerning the problem ( P ). Recalling that ω δ = α T β T − α β δ (see(9)), we will see that ρ T and the sign of ω δ are the determinant factors inthis analysis.Again, the case ω δ = 0 is easily solved, because the function to be min-imized and the one defining the restriction are proportional. Hence, thefollowing result can be derived as Proposition 3: Proposition 5.
Let us assume d min > , ρ T > , (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ = 0 . Then any feasible pair ( N , d ) with E T ( N , d ) = γ T is a solution to problem ( P ) . In particular, the pairs ( N , d N ) with N ∈ {(cid:100) λ T (cid:101) , . . . , (cid:98) ρ T (cid:99)} and d N =( d , ..., d ) , where (42) d = − α T N + (cid:113) ( α T N ) + 4 β T N γ T β T N .
Let us now continue by studying the more frequent case ω δ >
0. Here, wehave to distinguish two different situations, depending on ρ T ∈ N or not: Theorem 12.
Let us assume that d min > , (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ > . a) If ρ T ∈ N , ρ T ≥ , then the unique solution to problem ( P ) is givenby N = ρ T and d N = ( d min , ..., d min ) , b) If ρ T (cid:54)∈ N , then the unique solution to problem ( P ) is given by (cid:18) N , d N (cid:19) , where: i) N = (cid:100) ρ T (cid:101) and d N = ( d min , ..., d min ) , or ii) N = (cid:98) ρ T (cid:99) and d N = ( d , ..., d ) , with d given by (42). The proof is similar to that of Theorem 7 and postponed to Appendix 2.Finally, the case ω δ < Theorem 13.
Let us assume d min > , ρ T > , (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ < .Then, a solution to problem ( P ) is given by (cid:18) N , d N (cid:19) , with N = (cid:100) λ T (cid:101) and a) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ) , when K defined by (38), with N = N , belongs to N ∪ { } ; otherwise, b) d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , with K defined by (40) and d (cid:63) satisfies (41), both with N = N .
Remark 6. a) Once more, let us emphasize that when ω δ > the opti-mal value of N is the largest one within its range of possibilities (i.e.it is a hyperfractionated type treatment), while in the case ω δ < the optimal value is the smallest one (i.e. it is a hypofractionatedtype treatment). This classification in terms of ω δ was described in [10] , considering nonnegative doses. b) For the hypofractionated case, the single exposure is chosen in [10] as the preferred one. But this dose could be too large in practiceand then two, three or more fractions would have to be tried untilan acceptable one is found. This fact is remarked in [3] by sayingthat the case ω δ ≤ “needs careful consideration since the validityof the model may be limited if N is small and the dose per fractionis large”. Under our approach, this difficulty is overcome and we getthe optimal number of dose fractions directly and its value, as in theother case. c) The uniqueness of solution fails when ω δ < because (as it was said)any permutation of the coordinates of the indicated solution providesa new one. In the following examples we will show that all the above possibilitiesmentioned in Theorems 12 and 13 can appear in practice:
Example 6.
Let us continue with the same parameters than in Example3, which are: α = 0 . Gy − , β = 0 . Gy − , α T = 0 . Gy − , β T =0 . Gy − , d min = 1 Gy and d max = 6 Gy.
ADIOTHERAPY FRACTIONATION PROBLEM 25 i) If δ = 1 , then ω δ = 8 > . Hence, we are considering the problem ( P ) Minimize E OAR ( N, d ) = 0 . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i , subject to N ∈ N , d i ∈ R such that E T ( N, d ) = 0 . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i ≥ γ T , ≤ d i ≤ , i = 1 , ..., N. When γ T = 4 , we can easily calculate that λ T ≈ . and ρ T ≈ . .Applying Theorem 12-b) the unique solution for ( P ) is one of thesetwo pairs: N = (cid:100) ρ T (cid:101) = 73 , d N = (1 , ..., with E OAR ( N , d N ) =4 . or N = (cid:98) ρ T (cid:99) = 72 , d N = ( d , ..., d ) with d ≈ . Gy and the objective function value E OAR ( N , d N ) ≈ . . Clearly, wechoose the second pair. ii)
Previous case is the most frequent in practice when ω δ > , but forsome specific values of γ T the alternative indicated in Theorem 12-b)occurs. For instance, taking γ T = 4 . in ( P ) , the values become λ T ≈ . and ρ T ≈ . . Now, the two candidates for being theunique solution for ( P ) are: N = (cid:100) ρ T (cid:101) = 73 , d N = (1 , ..., with E OAR ( N , d N ) = 4 . , as before, and N = (cid:98) ρ T (cid:99) = 72 , d N =( d , ..., d ) with d ≈ . Gy and E OAR ( N , d N ) ≈ . . It isapparent that here the solution is the first one. iii) If δ = 0 . , then ω δ = − < and the problem under study is ( P ) Minimize E OAR ( N, d ) = 0 . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i , subject to N ∈ N , d i ∈ R such that E T ( N, d ) = 0 . N (cid:88) i =1 d i + 0 . N (cid:88) i =1 d i ≥ γ T , ≤ d i ≤ , i = 1 , ..., N. For γ T = 4 . , we have that λ T ≈ . and ρ T ≈ . . Therefore,applying Theorem 13-b), we know that a solution for ( P ) is givenby ( N , d ) with N = (cid:100) λ T (cid:101) = 10 and d = (1 , d ∗ , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ) with d ∗ ≈ . Gy . In this case, E OAR (10 , d ) ≈ . . As in Example 3, forcomparison reasons, we can calculate that the solution of solutionof ( P ) is ˜ d = (1 , , d ∗ , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ) with d ∗ ≈ . Gy producing E OAR (11 , ˜ d ) ≈ . , that is slightly bigger than E OAR (10 , d ) , asexpected. iv) Again, it can be shown that, for some specific values of the parameter γ T , the alternative exposed in Theorem 13-a) is true. In particular,for γ T = 4 . , we have λ T ≈ . and ρ T ≈ . and a solutionfor ( P ) is given by ( N , d ) with N = (cid:100) λ T (cid:101) = 10 and d = (1 , , ..., (cid:124) (cid:123)(cid:122) (cid:125) ) . As we did for problem ( P ), here we can deduce quite easily how thesolution depends with respect to the parameters defining the problem ( P ). Corollary 2.
Let us assume d min > , ρ ≥ and ( N , d N ) is a solution of ( P ) . Then, N is an increasing function of γ T , decreasing with respect to α T and β T and independent of α , β and δ . Moreover, a) When ω δ > , N is also decreasing with respect to d min and inde-pendent of d max . b) When ω δ < , N is also decreasing with respect to d max and inde-pendent of d min .Proof. It follows from the expressions N = (cid:98) ρ T (cid:99) or (cid:100) ρ T (cid:101) with ρ T given by(33), when ω δ > , and N = (cid:100) λ T (cid:101) with λ T given by (32), when ω δ < . (cid:3) For the reader’s convenience, we have summarized the complete algorithmfor the resolution of the problem ( P ) in Table 2.3.3. Equivalent treatments.
We will finish this section by mentioninganother application of previous results. Let us begin by introducing bio-logically equivalent treatments. Two treatments of radiotherapy with doses d , ..., d N and ˜ d , ..., ˜ d ˜ N are said to be biologically equivalent for a certaintumor with characteristic parameters α T and β T when they have the sameeffect, that is, α T N (cid:88) i =1 d i + β T N (cid:88) i =1 d i = α T ˜ N (cid:88) i =1 ˜ d i + β T ˜ N (cid:88) i =1 ˜ d i . When all the doses are equal for both treatments, previous concept leadsto biologically equivalent doses (BED) (see [3]), that can be calculated veryeasily (see [11] for instance) from the equality N ( α T d + β T d ) = ˜ N ( α T ˜ d + β T ˜ d ) . A clearly interesting question is to determine (among all the equivalenttreatments) which one uses the lowest total dose. As in this case we arenot paying attention to the effect of radiation on the OAR, this can be
ADIOTHERAPY FRACTIONATION PROBLEM 27
ALGORITHM FOR SOLVING ( P )DATA: α T , β T , α , β , d min , d max , δ and γ T (all positive, d min < d max and δ ≤ ω δ = α T β T − α β δ , λ T = max (cid:26) , γ T ϕ T ( d max ) (cid:27) and ρ T = max (cid:26) , γ T ϕ T ( d min ) (cid:27) , with ϕ T ( r ) = α T r + β T r . IF ρ T = 1, the pair ( N , d N ) = (1 , d min ) is the UNIQUE SOLUTION of ( P ).IF ρ T > (cid:100) λ T (cid:101) = (cid:98) ρ T (cid:99) , the UNIQUE SOLUTION of ( P )is the pair ( N , d N ) with N = (cid:100) ρ T (cid:101) and d N = ( d min , ..., d min ) . IF ρ T ∈ N , ρ T ≥ (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ > , the pair ( N , d N )with N = ρ T and d N = ( d min , ..., d min ) is the UNIQUE SOLUTION of ( P ).IF ρ T (cid:54)∈ N , (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ > , take ( N , d N ) , with N = (cid:98) ρ T (cid:99) and d N = ( d , ..., d )where d = − α T N + (cid:113) ( α T N ) + 4 β T N γ T β T N .Also take N = (cid:100) ρ T (cid:101) and d N = ( d min , ..., d min ).CALCULATE: E OAR ( N , d N ) and E OAR ( N , d N ) . A SOLUTION of ( P ) is the pair ( N , d N ) that minimizes E OAR between them.IF ρ T > , (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ < , take N = (cid:100) λ T (cid:101) and CALCULATE M = N ϕ T ( d max ) − γ T ϕ T ( d max ) − ϕ T ( d min ) .IF M ∈ N ∪ { } , take K = M and d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K ).IF M (cid:54)∈ N ∪ { } , take K = (cid:98) M (cid:99) and d N = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) , with d ∗ > ϕ T ( d ∗ ) = γ T − Kϕ T ( d min ) − ( N − K − ϕ T ( d max ).A SOLUTION of ( P ) is the pair ( N , d N ) . IF ρ T > (cid:100) λ T (cid:101) < (cid:98) ρ T (cid:99) and ω δ = 0 , ANY FEASIBLE PAIR (
N , d ) such that E T ( N , d ) = γ T is a SOLUTION for ( P ).In particular, the pairs ( N , d N ) with N ∈ {(cid:100) λ T (cid:101) , . . . , (cid:98) ρ T (cid:99)} and d N = ( d , ..., d ) , where d = − α T N + (cid:113) ( α T N ) + 4 β T N γ T β T N .
Table 2.
Complete solution for problem ( P ) in algorithmic form. formulated in mathematical terms as the following optimization problem:( P ) Minimize N (cid:88) i =1 d i , subject to N ∈ N , d ∈ R N such that α T N (cid:88) i =1 d i + β T N (cid:88) i =1 d i = γ T ,d min ≤ d i ≤ d max , i = 1 , ..., N. After our study, we deduce that the solutions to this problem are treat-ments of type (37) or (39), see Theorem 11-ii). Let us mention that theseare the same than in the case ω δ <
0, see Theorem 13, and consequently thehypofractionated protocols are also optimal in this sense.4.
Conclusions
In this work, we have derived the analytical expressions for the optimaltotal number of radiations N and their specific doses d for problems ( P )and ( P ). They are presented in Tables 1 and 2 in algorithmic form (see alsoTheorems 6, 7, 8, 12 and 13) and there exists a clear parallelism betweenthe structure of the solutions for both problems. We have proved that theyessentially depend on the sign of the quantity ω δ = α T β T − α β δ . For fixed N , this fact is well known in the literature and it has been reportedseveral times in different frameworks (see for instance [10], [2] and [6]).Moreover, this is consistent with some clinical findings as noted in [10].When ω δ >
0, we have shown that the optimal number of doses N are (cid:98) ρ (cid:99) for ( P ) and (cid:98) ρ T (cid:99) or (cid:100) ρ T (cid:101) for ( P ), the upper values of their rangesof interest (i.e. hyperfractionated type treatments) with equal doses; whilein the case ω δ < , the optimal values of N are (cid:98) λ (cid:99) or (cid:100) λ (cid:101) for ( P ) and (cid:98) λ T (cid:99) for ( P ), the lower values of those ranges (i.e. hypofractionated typetreatments). In this last case, let us stress that not all doses have to bemaximum; in fact, some of them may be minimum and at most one ofthem can take an intermediate value. For non-uniform protocols, the lackof uniqueness for the solution can be used to our benefit, because the dosescan be administered in any order depending on various external factors suchas the condition of the patient. The study concerning the derivation of theoptimal number of doses N had already been performed for example in [8]in the hyperfractionated case, but (as far as we know) it is completely newfor the hypofractionated case.Let us emphasize again that the calculations to apply all these results areelementary and can be carried out using a pocket calculator from the initial ADIOTHERAPY FRACTIONATION PROBLEM 29 data. Of course, the algorithms described in Tables 1 and 2 can be im-plemented quite straightforwardly in any platform using any programminglanguage to make them more accessible.We hope that these theoretical results may provide useful insights toaddress more complete models (including repopulation terms and multipleOAR) and that, ultimately, will lead to some improvement (however small)in clinical practice, due to the impact it would have on the large number ofpatients who could benefit.
Acknowledgement
The authors would like to express their gratitude to Prof. Cecilia Pola(University of Cantabria) for fruitful discussions and helpful comments.
References [1] Bertsekas DP (2003) Nonlinear Programming. Athena Scientific, Belmont, Mas-sachusetts.[2] Bertuzzi A, Bruni C, Papa F, Sinisgalli C (2013) Optimal solution for a cancer radio-therapy problem, J Math Biol 66, 311-349.[3] Bortfeld T, Ramakrishnan J, Tsitsiklis JN, Unkelbach J (2015) Optimization of radia-tion therapy fractionation schedules in the presence of tumor repopulation, INFORMSJ Comput 27 (4), 788-803.[4] Brenner DJ (2008) The linear-quadratic model is an appropriate methodology fordetermining iso-effective doses at large doses per fraction. Semin Radiat Oncol 18 (4),234-239.[5] Bruni C, Conte F, Papa F, Sinisgalli C (2015) Optimal weekly scheduling in fraction-ated radiotherapy: effect of an upper bound on the dose fraction size, J Math Biol 71,361-398.[6] Bruni C, Conte F, Papa F, Sinisgalli C (2019) Optimal number and sizes of the dosesin fractionated radiotherapy according to the LQ model, Math Med Biol 36, 1-53.[7] Fern´andez L (2020) Problemas de optimizaci´on asociados a los tratamientos de ra-dioterapia. Final Undergraduate Project, University of Cantabria.[8] Jones B, Tan LT, Dale RG (1995) Derivation of the optimum dose per fraction fromthe linear quadratic model, Br J Radiol 68, 894-902.[9] McMahon SJ (2019) The linear quadratic model: usage, interpretation and challenges,Phys Med Biol 64, 1-24.[10] Mizuta M, Takao S, Date H, Kishimoto N, Sutherland KL, Rikiya O, Shirato H(2012) A mathematical study to select fractionation regimen based on physical dosedistribution and the linear-quadratic model. Int J Radiat Oncol Biol Phys 84, 829-833.[11] Radiation therapy dose calculator of the French Society of Young Radiation Oncolo-gists (2021) [12] Radiotherapy dose fractionation (2019) The Royal College of Radiologist, London.[13] Radiotherapy risk profile (2008). Technical Manual. World Health Organization,Geneva.[14] Saberian F, Ghate A, Kim M (2015) A two-variable linear program solves the standardlinear-quadratic formulation of the fractionation problem in cancer radiotherapy. OperRes Lett 43, 254-258.[15] Van Leeuwen CM, Oei AL, Crezee J, Bel A, Franken NAP, Stalpers LJA, Kok HP(2018) The alfa and beta of tumours: a review of parameters of the linear-quadraticmodel, derived from clinical radiotherapy studies. Radiat Oncol 13, 1-11.
Appendix : Proof of Theorem 8 The expression given in i ) is derived exactly as in Theorem 6 for thevalues N ≤ λ . Taking into account that d ∗ can be very close to d max or d min , item ii ) can be seen as a kind of special case of iii ). So, we willfocus on proving iii ) that it is the most complicated case. To that end, itis enough to show that if ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) is a solution for( P N ) and ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) ˜ K , ˜ d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − ˜ K ) is a solution for ( P N +11 ), with N > λ , then the following relation holds(43) Kd min + d ∗ + ( N − K − d max ≤ ˜ Kd min + ˜ d ∗ + ( N − ˜ K ) d max . Together with (8) and the assumption ω δ <
0, this implies iii ), because(43) means that the values of the objective function E T at the solutions aredecreasing with N and therefore, the maximum value will be attained at N = (cid:100) λ (cid:101) , the lowest value of N in the set ( λ , ρ ] ∩ N .Comparing their expressions in the form (26) with N + 1 and N , resp.,we conclude that ˜ K ≥ K + 1. Hence, if we denote K = ˜ K − K ∈ N , theinequality (43) can be written as(44) d ∗ − ˜ d ∗ ≤ K d min + (1 − K ) d max . Let us recall that d (cid:63) satisfies (24) and ˜ d ∗ verifies(45) ϕ ( ˜ d ∗ ) = γ OAR − ˜ Kϕ ( d min ) − ( N − ˜ K ) ϕ ( d max ) . We will show that (44) holds dividing the argumentation in three cases:Case 1.- Suppose that K d min + (1 − K ) d max ≤
0. We choose the point( d , ..., d N ) = ( K K − d min , ..., K K − d min (cid:124) (cid:123)(cid:122) (cid:125) K − , d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) ˜ K − K , ˜ d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − ˜ K ) , that under the assumption satisfies the bounds restrictions and E OAR ( N, d ) = α δ (cid:16) ˜ Kd min + ˜ d ∗ + ( N − ˜ K ) d max (cid:17) ++ β δ (cid:18) ( K ) K − d min + ( ˜ K − K ) d min + ( ˜ d ∗ ) + ( N − ˜ K ) d max (cid:19) ≥≥ ˜ Kϕ ( d min ) + ϕ ( ˜ d ∗ ) + ( N − ˜ K ) ϕ ( d max ) = γ OAR . This means that it is feasible for the problem (27). Taking into accountRemark 3, we get (44).Case 2.- Suppose now that K d min + (1 − K ) d max > K ϕ ( d min ) + (1 − K ) ϕ ( d max ) ≤
0. We can argue similarly choosing(46) ( d , ..., d N ) = ( d min , ..., d min (cid:124) (cid:123)(cid:122) (cid:125) K , ˜ d ∗ , d max , ..., d max (cid:124) (cid:123)(cid:122) (cid:125) N − K − ) . ADIOTHERAPY FRACTIONATION PROBLEM 31
Due to (45) and the hypothesis we have E OAR ( N, d ) = Kϕ ( d min ) + ϕ ( ˜ d ∗ ) + ( N − K − ϕ ( d max ) == γ OAR − K ϕ ( d min ) + ( K − ϕ ( d max ) ≥ γ OAR . Again, we have a feasible point for the problem (27) and therefore we deduce d ∗ ≤ ˜ d ∗ and hence (44), because d ∗ − ˜ d ∗ ≤ < K d min + (1 − K ) d max . Case 3.- Finally, suppose that K d min + (1 − K ) d max > K ϕ ( d min ) + (1 − K ) ϕ ( d max ) >
0. Here, we introduce the auxiliaryfunction defined for s ∈ [0 ,
1] by G ( s ) = (cid:114) α + 4 β (cid:16) γ OAR − ( ˜ K − sK ) ϕ ( d min ) − ( N − ˜ K + s ( K − ϕ ( d max ) (cid:17) . Solving the quadratic equations (24) and (45), it is easy to derive that d ∗ = − α + G (1)2 β δ , ˜ d ∗ = − α + G (0)2 β δ . Using the Mean Value Theorem, we deduce that there exists θ ∈ (0 ,
1) suchthat d ∗ − ˜ d ∗ = G (1) − G (0)2 β δ = G (cid:48) ( θ )2 β δ = K ϕ ( d min ) + (1 − K ) ϕ ( d max ) G ( θ ) δ . Therefore, the inequality (44) is equivalent to(47) K ϕ ( d min ) + (1 − K ) ϕ ( d max ) K d min + (1 − K ) d max ≤ G ( θ ) δ. Under the present hypotheses, the function G is strictly increasing and, since θ is an unknown value in (0 , θ = 0. On the other hand, the value K ∈ N is also unknown, but wecan verify that the function F ( m ) = mϕ ( d min ) + (1 − m ) ϕ ( d max ) md min + (1 − m ) d max == α δ + β δ (cid:18) md min + (1 − m ) d max md min + (1 − m ) d max (cid:19) , is strictly decreasing, because F (cid:48) ( m ) = β δ d min d max ( d min − d max )( md min + (1 − m ) d max ) < . Hence, the inequality (47) will be true if F (1) ≤ G (0) δ. We conclude bynoting that˜ d ∗ ∈ [ d min , d max ] ⇐⇒ ϕ ( ˜ d ∗ ) ∈ [ ϕ ( d min ) , ϕ ( d max )] ⇐⇒ (45) ⇐⇒ γ OAR − ˜ Kϕ ( d min ) − ( N − ˜ K ) ϕ ( d max ) ∈ [ ϕ ( d min ) , ϕ ( d max )] . Then, G (0) = (cid:114) α + 4 β (cid:16) γ OAR − ˜ Kϕ ( d min ) − ( N − ˜ K ) ϕ ( d max ) (cid:17) ≥≥ (cid:113) α + 4 β ϕ ( d min ) ≥ α + β δd min = F (1) δ , as asserted. Appendix : Proof of Theorem 12 It follows the same lines to that of Theorem 7.Case a).- Assume ρ T ∈ N , ρ T ≥ . As usual, we divide the interval for feasible values of N in two parts:[ λ T , ρ T ) ∩ N and [ ρ T , + ∞ ) ∩ N . In order to study the dependence with respect to N in the interval[ λ T , ρ T ), thanks to Proposition 4 (with ω δ >
0) and (36), it is enough toconsider the auxiliary function ψ ( N ) = N d = − α T N + (cid:112) ( α T N ) + 4 β T N γ T β T . Once more, it follows easily that ψ ( N ) is an strictly increasing function.Since we are assuming ρ T ∈ N and ρ T ≥
2, then ψ will take its maximumvalue in the set [ λ T , ρ T ) ∩ N at N = ρ T −
1. Therefore, the candidate forsolution to problem ( P ) is given by the pair ( N , d N ) with d N = ( d , ..., d ) , where d is given by(48) d = − α T N + (cid:112) ( α T N ) + 4 β T N γ T β T N . On the other hand, in the interval [ ρ T , + ∞ ), we know that the othercandidate for solution to problem ( P ) is given by the pair ( N , d N ) with N = ρ T and d N = ( d min , ..., d min ) . To derive that ( N , d N ) is the unique solution to problem ( P ), we willshow that(49) E OAR ( N , d N ) < E OAR ( N , d N ) . Following the same idea to that of the proof of Theorem 7, we introducethe auxiliary function H ( x ) = N ( xd + d ) − N ( xd min + d min ) , x ∈ [ α β δ , + ∞ ) . Taking into account (48) and that N ϕ T ( d min ) = γ T (by the definition of ρ T ), it can be checked that H (cid:48) ( x ) = N d − N d min < , since N < N . ADIOTHERAPY FRACTIONATION PROBLEM 33
Using that also γ T = N ϕ T ( d ), we get that H (cid:18) α T β T (cid:19) = 0 and fromthe assumption ω δ > H (cid:18) α β δ (cid:19) >
0, which isequivalent to (49).Case b).- Assume ρ T (cid:54)∈ N . Here, the optimal value of N in the inter-val [ λ T , ρ T ) is N = (cid:98) ρ T (cid:99) and d N = ( d , ..., d ) with d given by (48).In the interval [ ρ T , + ∞ ), the other candidate is N = (cid:100) ρ T (cid:101) with d N =( d min , ..., d min ) . When ρ T (cid:54)∈ N , any of them can provide the unique solution to problem( P ) (see for instance Example 6). Appendix : Proof of Theorem 13 When ω δ <
0, it is still true that N = (cid:100) ρ T (cid:101) and d N = ( d min , ..., d min ) . Arguing as in Appendix 1, the candidate when N runs [ λ T , ρ T ) ∩ N is N = (cid:100) λ T (cid:101) with d N given by Theorem 13- a ) or b ) and N = N , thanks to Theorem11 − ii ). We will conclude by showing that(50) E OAR ( N , d N ) ≤ E OAR ( N , d N ) . Let us argue with the expression b ) for d N , because (as we have pointed outbefore) the value d ∗ can be very close to d min or d max and hence item a ) canbe seen as a special case of b ). Therefore, the inequality (50) is equivalentto(51) Kϕ ( d min ) + ϕ ( d (cid:63) ) + ( N − K − ϕ ( d max ) ≤ N ϕ ( d min ) . For proving (51), we consider again a linear function such as H ( x ) = ( N − K )( xd min + d min ) − ( xd ∗ +( d ∗ ) ) − ( N − K − xd max + d max ) . By construction, we know that Kϕ T ( d min ) + ϕ T ( d (cid:63) ) + ( N − K − ϕ T ( d max ) = γ T ≤ N ϕ T ( d min ) . This is equivalent to say that H (cid:18) α T β T (cid:19) ≥ H is an increasing function, since ω δ <
0, we will have H (cid:18) α β δ (cid:19) ≥ H (cid:18) α T β T (cid:19) ≥ , which gives (51).So, taking into account that H (cid:48) ( x ) = ( N − K ) d min − d ∗ − ( N − K − d max , let us finish the proof by showing that H (cid:48) ( x ) ≥ N d min > N d max , this is true straightforwardly, because we know that N d max > Kd min + d ∗ + ( N − K − d max . When N d min ≤ N d max , we can argue as in the proof of Theorem 8,taking the point ˜ d = ( d , ..., d N ) = ( N N d min , ..., N N d min ) , that satisfies the bounds restrictions and E T ( N , ˜ d ) = α T N d min + β T ( N ) N d min ≥ N ϕ T ( d min ) ≥ γ T . This means that it is feasible for the problem ( P N , − ). Taking into accountthat ( N , d N ) is a solution for that problem, see Proposition 4- ii ) and Re-mark 3, we get H (cid:48) ( x ) ≥ Dep. Mathematics, Statistics and Computation, University of Cantabria(SPAIN)
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