Modelling TCP & NTCP

Question 1

What does TCP stand for?

Answer 1

Tumour Control Probability

Question 2

What does NTCP stand for?

Answer 2

Normal Tissue Complication Probability

Question 3

What is the purpose of modelling TCP & NTCP?

Answer 3

1. Modern dose distributions are complex and we would like to assess their effects.
2. To understand and quantify the effects of uncertainties in dose or dose distribution. (what do DVHs really mean in terms of outcomes)
3. Ultimately to use TCP and NTCP calculations for treatment planning rather than merely physical dose (i.e. radiobiological planning).

Question 4

What is the general principle of modelling TCP & NTCP?

Answer 4

• Start with lots of data, e.g: a 3D dose distribution.
• Discard spatial, anatomical, and physiological data.
• Extract unambiguous data: either single point such as V20*, or global such as mean dose.
• Compute model based NTCP estimates.
• End with single numbers; chance of cure, and associated risks.

*V20 = % of organ getting more than 20 Gy

Question 5

What types of TCP & NTCP models are currently available, and what are the down sides of each?

Answer 5

Theoretical: this is easily influenced by the radiobiological parameters chosen to be input into the model, which have large uncertainties associated with them.
Empirical model (pick a function that looks like the data): these are only really applicable for large cohorts of patients and for the range of data selected.

Question 6

Why do TCP & NTCP models have to be used with caution?

Answer 6

• individual patients vary in their radiobiological responses to the same treatment regime
• some parameters have large uncertainties
• models are complex, so require expertise.

Question 7

Briefly describe how a TCP model works.

Answer 7

TCP models generally calculate the chance of no surviving clonogens in each DVH dose bin (Bin Control Probability) then multiply all the probabilities for each dose bin within the tumour to get an overall TCP.

Question 8

Briefly describe how an NTCP model works.

Answer 8

NTCP models generally transform the complex distribution through an Organ At Risk (OAR) into an Equivalent Uniform Dose (EUD) (a DVH reduction technique). Then calculate the NTCP based on clinical data for partial organ uniform dose irradiation (the volume effect).

Question 9

What is the linear quadratic rationale equation?

Answer 9

the mean number of lethal events per cell = αd + βd^2
Where α is the mean number of double strand breaks from a single hit, and β is the mean number of double strand breaks from more than one event

Question 10

What is the survival fraction?

Answer 10

The fraction of cells without lethal events:

exp -( αd + βd^2)

Question 11

What is the survival fraction for n number of fractions?

Answer 11

exp -n(αd + βd^2) = exp -D(α + βd)

Question 12

What is TCP?

Answer 12

The fraction of tumours with no active clonogenic cells after all treatment. (k = number of clonogenic cells).
i.e. the fraction with local control

Question 13

What is the TCP equation?

Answer 13

TCP = exp - ( k ) = exp - ( k0 * SF ) = exp - k0 * exp - ( α * BED )
where SF = exp -( αd + βd^2)

Question 14

What is the BED equation?

Answer 14

BED = - ln(SF) / α = D (1 + d / ( α / β ) )

Question 15

In the TCP vs Dose curve, what parameters affect the shape of the curve and how do they affect it?

Answer 15

k0 shifts the curve right as it increases

α increases the steepness and shifts the curve to the left

Question 16

Define each parameter in the equation to calculate TCP from a DVH, and state the method.

Answer 16

Principle: Calculate the product of the individual TCPs for each dose bin to get an overall TCP.

rho(c) is clonogenic cell density
D(i) is dose given to dose bin i
v(i) is volume receiving dose i
n is number of fractions.

Question 17

In terms of TCP, which is worse: hot or cold spots? Why?

Answer 17

Cold spots are worse than equivalent hot spots because they quickly drive TCP to zero. This equates to a geographical miss.

Question 18

To make a TCP calculation more sophisticated, which terms can be added in?

Answer 18

Variations in radiobiological parameters
usually a Gaussian spread in alpha caused by tumour cell heterogeneity. This has the effect of making TCP slopes shallower.

Variations in tumour cell density (e.g. decreasing radially from the centre of a tumour according to a plausible fall-off pattern).

Add the term for tumour proliferation .

Add term for spread of beta.

Question 19

What does a spread of alpha values do to the TCP curve?

Answer 19

A Gaussian spread of alpha causes the TCP curve across a cohort of patients to become shallower than a TCP curve with a unique alpha value. i.e. an averaging of unique α valued TCP curves.

Question 20

What difference does the simple and complex version of the TCP calculation make?

Answer 20

2-3%

Question 21

What sort of technique is used for NTCP models?

Answer 21

NTCP models usually employ a DVH reduction technique. This is where the true DVH shape is converted into a radiobiological equivalent DVH where only part of the volume is irradiated but at max dose of the real DVH.

Question 22

What are the main models used for NTCP modelling?

Answer 22

LKB (Lyman Kutcher Burman)

RS

Question 23

Describe the LKB NTCP model.

Answer 23

This is the most widely used model.
Step 1: DVH reduction
-Each volume element is considered to be subject to a power law dose volume relationship:
= delta V(i) ( D(i) / D(max) ] ^ 1 / n

-Calculate the sum of these elements to give a uniform partial organ irradiation at the max dose delivered that is radiobiologically equivalent to the true inhomogeneous DVH to the whole OAR. i.e. calculate an effective volume (Veff) of uniform irradiation.

Step 2: Calculate NTCP
-Based on an empirical function devised by Lyman to calculate NTCP tolerance dose as a function of % OAR volume uniformly irradiated.

Question 24

What is the equation for NTCP?

Answer 24

NTCP = 1 / SQRT ( 2 * pi ) * integral of exp ( - t ^2 / 2) dt
with the bounds of x and minus infinity.
OR
NTCP = 1 / 2 [ 1 + erf ( x / SQRT(2) ) ]
where erf is the error function used to create a sigmoidal shape

where:
x = D(max) - TD(50)(v) / m * TD(50)(v)
v = V(eff) / V(ref)
TD(50)(v) = TD(50)(1) * v ^ - n

Dmax is the max dose in the DVH
Veff is the effective volume
Vref is the total volume of the organ (where possible, otherwise an arbitrary reference volume)
TD50 is the tolerance Dose for 50% complications in the partial volume,
m is a variable that controls the slope of the NTCP vs dose curve.
n is a variable that controls the volume sensitivity of the NTCP.

Question 25

Define all the parameters in the NTCP equation.

Answer 25

Dmax is the max dose in the DVH
Veff is the effective volume
Vref is the total volume of the organ (where possible, otherwise an arbitrary reference volume)
TD50 is the tolerance Dose for 50% complications in the partial volume,
m is a variable that controls the slope of the NTCP vs dose curve.
n is a variable that controls the volume sensitivity of the NTCP.

Question 26

What is the m, n and TD(50) values for lung?

Answer 26

m = 0.18 
n = 0.87
TD(50) = 24.5

Question 27

What is the m, n and TD(50) values for the brain stem?

Answer 27

m = 0.14
n = 0.16
TD(50) = 65

Question 28

What is the m, n and TD(50) values for the esophagus?

Answer 28

m = 0.11
n = 0.06
TD(50) = 68

Question 29

What is the m, n and TD(50) values for liver?

Answer 29

m = 0.15
n = 0.32
TD(50) = 40

Question 30

What is the m, n and TD(50) values for rectum?

Answer 30

m = 0.15
n = 0.12
TD(50) = 80

Question 31

What is the m, n and TD(50) values for kidney?

Answer 31

m = 0.10
n = 0.70
TD(50) = 28

Question 32

What is the m, n and TD(50) values for bladder?

Answer 32

m = 0.11
n = 0.50
TD(50) = 80