CORRECTION FOR ERRORS

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CORRECTION FOR ERRORS

Dosimetric errors in delivering the prescribed dose per fraction made early in a treatment can be corrected by modifying the dose per fraction and total dose given subsequently to discovery of the error, using the LQ model to calculate the correcting doses which should be completed within the same overall time as originally prescribed.

The Mike Joiner showed how to calculate the dose per fraction used to bring the treatment back exactly to planned tolerance simultaneously for all tissues and tumour involved, following either hyperfractionated or hypofractionated errors made initially, without the need to know any α/β values.

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

Defining planned treatment as dp Gy per fraction to a total dose Dp Gy,suppose the initial error is de Gy per fraction given to a total of De Gy. Using the LQ model to describe all isoeffect relationships between total dose and dose per fraction for the tumour and the normal tissues, then the compensating dose per fraction of dc Gy to a total dose of Dc Gy are given by the simple formulae:

D c = D p - D e

\Rightarrow d c = \frac{D p \cdot d p - D e \cdot d e}{D p - D e}

Example:

According to the plan, a patient must have 30 fractions, each day with each dose being 1.8Gy. But due to an error, the patient received the first 5 fractions with each dose being 2Gy. How is the treatment regimen adjusted so that the patient still receives the same treatment effect on the tumor and protection on healthy organs as the initia treatment?

We have the formulae about correction for error in clinical practice:

d c = \frac{D p \cdot d p - D e \cdot d e}{D p - D e}

We also have:

D p = n \cdot d p = 30 \cdot 1.8 = 54 Gy

D e = n \cdot d e = 5 \cdot 2 = 10 Gy

\Rightarrow d c = \frac{D p \cdot d p - D e \cdot d e}{D p - D e} = \frac{54 \cdot 1.8 - 10 \cdot 2}{54 - 10} = 1.75 Gy

\Rightarrow D c = D p - D e = 54 - 10 = 44 Gy \Rightarrow n c = 25

Finally, we have the dose required to complete the course is dc=1,75 Gy ; nc= 25 fractions