Wednesday, February 01, 2006

Expected costs of a nuclear meltdown

Perhaps one of the topics to consider for our Energy/Environmental Policy project is whether or not there should be greater reliance on nuclear power plants. John Whitehead writes on an important issue with respect to the evaluation of nuclear power generation:
"Risk perception researchers have found that when dealing with low probability, high loss risks (like nuclear accidents), people don't handle expected values very well [E(loss) = p x Loss; where p is the probability of the event happening]. People tend to overestimate the probability of the bad event and focus on the high loss."

[ . . . . ]

A simple estimate of the annual risk of a nuclear meltdown (which is worse than a nuclear accident [think Homer Simpson]) is the ratio of the number of accidents divided by the product of the number of nuclear plants and the number of years of operation. There have been two nuclear accidents in the history of the world: Three Mile Island and Chernobyl (Wikipedia says that there have been five other potential meltdowns). The denominator of the ratio is about 12,000 (obtained from an "ignorance is bliss" naive Google search, better estimates are welcome). Therefore, one estimate of the probability of a nuclear accident is 0.000167 or a 1 in 6000 annual chance.

To put this in perspective, suppose there was a 1 in 6000 chance that the U.S. loses 1% of its annual GDP (1% of $12 trillion is $120 billion). The expected loss, given a 0.000167 probability is about $3.33 million (Dan Kolb's annual salary).

Let's hope that when we're considering the risk tradeoffs that we appropriately weight the losses by the probabilities. The expected loss of a nuclear plant near Raleigh is not a Chernobyl, as if it were a sure thing, but something much, much lower."

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