BED Applications in Practice

BED Applications in Practice. The main application of the BED model is to design and/or compare different fractionation or dose-rate schemes It can also be used for correction errors rest periods.

Examples of the use of the BED model. Simple fractionation changes Correction for errors Conversion to 2 Gy/fraction equivalent dose Effect of change in overall treatment time Correction for rest periods Change in dose rate.

Example 1: simple change in fractionation. Question: what dose/fraction delivered in 25 fractions will give the same probability of late normal tissue damage as 60 Gy delivered in 30 fractions at 2 Gy/fraction? The L-Q equation is:.

Solution (cont’d). Assuming α/β for late reacting normal tissues is 3 Gy, the BED for 60 Gy at 2 Gy/fraction is 60(1 + 2/3) = 100.

Solution (cont’d). Then the dose/fraction, d, is given by: 100 = 25d(1 + d/3) Solving this quadratic equation for d gives: d = 2.27 Gy/fraction.

sing the L-Q model to correct for errors. Int. J. Radiat. Oncol. Phys. Biol., Vol. 58, No.3, pp. 871-875, 2004.

The Mike Joiner method. Joiner found that if several fractions are delivered at the wrong dose/fraction, you can derive a dose/fraction to use for the remainder of the course that will result in the planned BEDs being delivered to all tissues it is independent of the α/β of the tissue.

The Mike Joiner method: definitions. The planned total dose is: Dp Gy at dp Gy/fraction The dose given erroneously is: De Gy at de Gy/fraction The dose required to complete the course is: Dc Gy at dc Gy/fraction in Nc fractions.

The Joiner equations.

Example 2: dose below prescribed for 1st two fractions.

Example 2 (cont’d.). The extra dose required is: Dc = 42 – 6 = 36 Gy Hence the number of fractions required is: Nc = 36/7.67 = 4.7 Since we cannot deliver 0.7 of a fraction, complete the treatment with 5 fractions of 36/5 = 7.2 Gy/ fraction • always round out the number of fractions up, since increased fractionation spares normal tissues.

Additional benefit of the Joiner model. The solution is not only independent of α/β but it is also independent of any geometrical sparing of normal tissues.

Conversion to 2 Gy/fraction equivalent dose. 2.

Example 3. What total dose given at 2 Gy/fraction is equivalent to 50 Gy delivered at 3 Gy/fraction for (a) cancers with α/β = 10 Gy? (b) normal tissues with α/β = 3 Gy? Answers (a) D2 = 50(1 + 3/10)/(1 + 2/10) = 54.2 Gy (b) D2 = 50(1 + 3/3)/(1 + 2/3) = 60.0 Gy.

Example 4: change in fractionation accounting for repopulation.

Solution I: assume no repopulation and no geometrical sparing.

Solution I (cont’d.): effect on late-reacting normal tissues.

Solution II: assume a geometrical sparing factor of 0.6.

Solution III: assume geometrical sparing and repopulation (at k = 0.3/day).

Solution III (cont’d.): effect on late reactions.

What does this mean?. Decreasing the number of fractions, i.e. hypo fractionation, does not necessarily mean increasing the risk of normal tissue damage when keeping the effect on tumor constant This is why we may be using far more hypofractionation in the future, especially since it will be more cost effective.

Example 5: Rest period during treatment. Problem: a patient planned to receive 60 Gy at 2 Gy/fraction over 6 weeks is rested for 2 weeks after the first 20 fractions How should the course be completed at 2 Gy/fraction if the biological effectiveness is to be as planned?.

Solution I: for late reacting normal tissues. Since late-reacting normal tissues probably do not repopulate during the break, they do not benefit from the rest period so the dose should not be increased Complete the course in 10 more fractions of 2 Gy.

Solution II: for cancer cells. Assume that the cancer is repopulating at an average rate, so k = 0.3 BED units/day and α/β = 10 Gy For a rest period of 14 days, the BED needs to be increased by 14 x 0.3 = 4.2 The BED for the additional N fractions of 2 Gy is then: 2N(1 + 2/10) – (7/5)N x (0.3) which must equal 4.2 Solution is N = 2.12 i.e. instead of 10 fractions you need about 12 fractions of 2 Gy But remember, the effect on normal tissues will increase.

Summary. The BED model is useful for the solution of radiotherapy problems with changes in fractionation and/or dose rate But remember, this equation must be just an approximation for the highly complex biological changes that occur during radiotherapy the model is approximate the parameters are approximate But the model is useful!.