Quantified Tree Risk Assessment (QTRA) Version 5 - Questions & Answers

[Acer ventura]

Hi

 

Awhile back I had a few off forum emails about whether I would discuss the evolvement of QTRA version 5 after it was released last year, in a similar manner to the QTRA Q&A thread I started here;

 

http://arbtalk.co.uk/forum/general-chat/53668-quantified-tree-risk-assessment-qtra-questions-answers.html

 

Fresh back from a QTRA & VTA tour downunder, and now mentally and physically back in the right time zone, I’m going to start one here. Version 5 of QTRA is it's most significant evolution since it was put it together in the 1990s, and the main improvements are;

 

Monte Carlo Simulations

The Risk of Harm is calculated using Monte Carlos simulations, which are built into the engine of the system. In short, Monte Carlo simulations mean QTRA calculators work out the ‘most likely’ Risk of Harm from 10 000 possible outcomes for each combination of Target, Size, and Probability of Failure range. The benefits of calculating the most likely outcome are threefold. Risk aversion (worst case scenario) is avoided and valuable resources are therefore not wasted on unnecessary tree work. Being risk prone (best case scenario) is avoided, and its associated vulnerability to claims of a breach in the duty of care. Perhaps most importantly, the inherent uncertainty that is ever present when assessing the risk from tree failure is embraced and accounted for.

 

Risk of Harm

Risk of Harm outputs are colour-coded around an easy to understand traffic light theme, red, amber, yellow, and green. The colours relate to the long-standing and internationally recognised tolerable and acceptable thresholds for risk in the Tolerability of Risk (ToR) Framework. With version 5, a ‘yellow risk’ can be considered tolerable and imposed on the public if it is As Low As Reasonably Practicable (ALARP). A ‘green risk’ is acceptable and it is not necessary to demonstrate any benefits to the risk.

 

Risk Manager Guidance

This has been beefed up in the Practice Note to enable risk managers to consider not only the level of risk from their trees, but the costs of reducing the risk and the benefits lost from felling trees. By adopting advisory risk thresholds within the Practice Note, risk managers are able to pursue simple and robust tree risk management principles.

 

Probability of Failure Calibration Exercises

Selecting a Probability of Failure range is the most uncertain and assessor dependent part of a tree risk assessment. For a number of years now, during the field exercises of both the QTRA and VTA days, we have spent time calibrating Probability of Failure ranges with the delegates. Along with improved ‘benchmark’ guidance in the user manual, more time is now spent in the field with these exercises and greater consistency is being achieved. The VTA day, in particular, focuses on the underpinning principles and practical application of selecting a Probability of Failure range for a tree or branch with confidence.

 

New Calculators

There are new manual and software (windows and android) calculators. Both have been re-designed to make them easier to use.

 

Easier to Use

With Monte Carlo simulations driving the Risk of Harm outputs there’s no scope to mess about trying to refine the inputs and potentially ending up in a maths cul-de-sac. The assessor merely needs to choose the correct Target, Size, and Probability of Failure range on the new improved calculator, and the colour-coded Risk of Harm is generated for them. The colour of risk then informs the guidance to the risk manager.

 

The Practice Note is available here.

 

Quantified Tree Risk Assessment

 

If anyone wants to discuss any the above, other elements of QTRA v5, or indeed any other tree risk issues, then please feel free to fire away with any questions you may have, and I’ll do my best to help out.

 

Cheers

 

Acer ventura

Edited April 15, 2014 by Acer ventura
Missed out android software

[daltontrees]

Some definite improvements. Will look it over later.

[Acer ventura]

I've moved this probability debate over because it's getting way off topic, and I was going to drag one more subject in that thread because it relates to Monte Carlo simulations in QTRA v5.

 

It starts here.

 

http://arbtalk.co.uk/forum/general-chat/73829-isa-best-management-practices-tree-risk-assessment-traq.html

 

 

To put what the Risk BMP risk rankings are saying in terms of tangible numbers in relation to say schools, which I opened the thread with.

 

Risk BMP Low Risk <1/2

 

If you have a class of 30 children and a year later 14 of them are dead, this would be ranked as a Low risk.

 

Sorry, I don't follow this. Have 14 children been killed by one tree? Doesn't the risk arise from a hazard and the harm is done when a tree fails and then once failed that hazard cannot fail again?

 

I was looking to express what the upper value of a risk ranking of less than <1/2 per annum for a fatality would mean in language outside of probabilities. Another way. Consider there are 30 trees in the school with a Risk BMP ranking of Low risk <1/2 for a fatality per annum. What’s the worst that could happen and the risk ranking still be correct?

 

Yes that clarifies that there is more than one tree. However, isn't the risk of 14 of the trees failing and killing a child each is 1/2 to the power of 14 or 0.000061, following the multiplication theorem of probability?

 

Pedant’s hat on - it would be less than <0.000061 for a Low risk rating of less than <1/2 for a fatality. But that would only be the case if you pick the 14 out of the 30 trees that all have an equal Low risk rating of less than <1/2 of failing and causing a fatality, AND then identify the correct sequence in which each one of them fail and cause a fatality over the year. What you’ve done is calculate the odds of picking the correct sequence of heads or tails with 14 consecutive tosses from an unbiased coin.

 

Nope.Each tree either fails or ti doesn't. The sequence is irrelevant. If tree 1 fails and tree 2 fails and tree 4 fails and tree 7 fails etc. te probability of each is 1/2 and the probability of any 14 failures is 1/2 x 1/2 x 1/2 etc. 14 times, which is 0.000061. But since you were setting out initially to express the risk of this in 'language outside of probabilities', it would suffice to express it as freakishly bad luck. Like a coin toss coming up tails evey time 14 times in a row, it's unlikely but in practice and in theory it's possible.

 

Jules...In the meantime, see whether you can get your head around the difference between ‘independent’ and ‘dependent’ probabilities and why you don’t cumulatively multiply the former.

 

Is that meant to be humour? The independent failure at 50:50 of a number of trees is independent probability. The probability of all of them failing can only be obtained by multiplying the individual probabilities. That is damn near the definition of the multiplication theorem. Please don't unnecessarily cast a shadow of discreditability over the othehrwise interesting and thorough thread by suggesting that one of the very few commentators (me) is wrong on something like this when I am patently right. Not because I have any particular insight, it's just because the scenario is no more complex than repeated coin-tossing.

 

Hi Jules

 

You almost answered your own question with your contradiction in your penultimate post.

 

Nope.Each tree either fails or ti doesn't. The sequence is irrelevant.

 

Like a coin toss coming up tails evey time 14 times in a row.

 

Arsing about with probabilities can all too easily mess with people’s heads, and I include myself here.

 

Try this;

 

Take 30 unbiased coins.

 

Toss each one in turn.

 

On the table you now have 30 coins with 15 coins tails up and 15 coins heads up.

 

Is the probability of 15 of the 30 coins being tails up?

 

15/30 = 1/2?

 

Or is it

 

1/2 x 1/2 etc 15 times = 1/33 000?

 

Cheers

 

Acer ventura

[Gary Prentice]

Take 30 unbiased coins.

 

Toss each one in turn.

 

On the table you now have 30 coins with 15 coins tails up and 15 coins heads up.

 

Is the probability of 15 of the 30 coins being tails up?

 

15/30 = 1/2?

 

Or is it

 

1/2 x 1/2 etc 15 times = 1/33 000?

 

Cheers

 

Acer ventura

 

Whats the prize for the correct answer?

[daltontrees]

Yeah, if there's a prize I'll spend time on it. It's certainly an easier one than the Trees of Damocles scenario.

[Amelanchier]

Is the probability of 15 of the 30 coins being tails up?

 

I would imagine that as the event has already happened the probability is 1.

 

As an aside (like you guys need an aside right?), I can show you someone who will correctly guess the outcome of ten consecutive coin tosses in the order that they happen.

 

All I need are 1024 (2^10) volunteers and a ten round tournament. Winner correctly guesses the toss and moves on - loser weeps into his beer over what could have been. The probability of any one sequence of tosses is 0.0009765625 but the probability that the winner (one person in just over a thousand) correctly guesses that sequence is 1...

 

Thirty rounds is possible but a little impractical (2^30) - you'd need just over a billion people.

[arb culture]

I'm with Tony - it's already happened so the answer is P=1.

 

I've just had a read through v5 practice note. Thanks for bringing it to my attention, it looks much better than previous versions.

 

After reading practice note 5 I am very tempted to do a U-turn and sign up to QTRA (if you'll still have me), but I'd want a bit more info on how the calculations work first. Any chance of some info on the monte carlos method used?

 

As for Tony's aside - are you sure you know someone who can accurately predict coin tosses as they happen? Or do you mean you know someone who can predict the probabilities of the results of coin tosses as they happen. I'm intrigued either way :-)

[arb culture]

Oh, and Acer is right about the dependent and independent thing. If you have a population of 30 trees and one falls over, then the next probability scenario will only have a population of 29 trees, and the next will have 28 etc...

 

This is also different to calculating the probability of all the trees falling over in a single event, or calculating the probability of all (or several) of the trees falling over in one year (or any other period of time), or there being several casualties from a single tree failure event.

 

This is obviously of less concern when dealing with extremely large populations and rare events such as national tree populations and serious tree failure casualties, or where total population numbers are large but not accurately known (eg LA or estate tree populations).

 

So tree failures in a limited population are different to coin tosses. :-)

[arb culture]

 

As an aside (like you guys need an aside right?), I can show you someone who will correctly guess the outcome of ten consecutive coin tosses in the order that they happen.

 

All I need are 1024 (2^10) volunteers and a ten round tournament. Winner correctly guesses the toss and moves on - loser weeps into his beer over what could have been. The probability of any one sequence of tosses is 0.0009765625 but the probability that the winner (one person in just over a thousand) correctly guesses that sequence is 1...

 

Thirty rounds is possible but a little impractical (2^30) - you'd need just over a billion people.

 

Oops, misread that the first time - sorry.

 

I think the probability of 1 person out of 1024 correctly predicting a sequence of ten coin tosses would be closer to P=0.63 :-)

 

However, post event, I would agree that the probability of the winner correctly predicting the sequence of ten coin tosses would be 1. That would be because; in order to win they would have correctly predicted the sequence, therefore it would have already happened, therefore the P=1 :-)

[Amelanchier]

Oops, misread that the first time - sorry.

 

I think the probability of 1 person out of 1024 correctly predicting a sequence of ten coin tosses would be closer to P=0.63 :-)

 

P=0.63 would mean that your guy would be more likely to guess ten tosses than one toss?

 

However, post event, I would agree that the probability of the winner correctly predicting the sequence of ten coin tosses would be 1. That would be because; in order to win they would have correctly predicted the sequence, therefore it would have already happened, therefore the P=1 :-)

 

But there's the rub. The chances that one person in 1024 (ten rounds of one-on-one) can correctly predict the outcome of ten consecutive coin tosses is pretty far out there P=0.00000095367431640625 ((2^10*(1/1024) I think?) but structuring the event in a tournament means that the probability of one person out of the competing 1024 doing so is P=1...

 

The artifice re-frames the probability - someone has to win (all other things being equal) so the otherwise unlikely outcome is certain.

 

It makes me wonder if we don't make that subtle shift sometimes with trees.