tag:blogger.com,1999:blog-6703779.post115827356097924615..comments2014-11-20T10:20:14.824+00:00Comments on scrofulous: no seriously, don't profilearam harrowhttp://www.blogger.com/profile/01272118188252697149noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6703779.post-1159473085093917672006-09-28T20:51:00.000+01:002006-09-28T20:51:00.000+01:00You're quite welcome.I'm certainly biased to drivi...You're quite welcome.<BR/><BR/>I'm certainly biased to driving being on highways, because I live close enough to work to walk, so driving is often to, say, Santa Fe, rather than a daily commute.<BR/><BR/>Yes, the DALYs do seem to be discounted -- but 300000 waking hours seems to be ~ 50 years average, which seems a bit high if it's discounted.<BR/><BR/>Yes, I had been assuming the v^4 was per accident, rather than per mile.<BR/><BR/>One sentence leapt out at me from http://www.dft.gov.uk/stellent/groups/dft_roads/documents/page/dft_roads_506880.pdf<BR/><BR/>> In the more generalised form (Baruya et al., 1999), this research identifies that the change in accident frequency for each 1km/h reduction in mean speed is inversely related to the current mean speed.<BR/><BR/>However this isn't clear to me whether this is per car-mile, per car-time, per road-mile, or what.<BR/>I think this can change the v^4 anywhere from v^4 to v^6, but I'm not quite sure.<BR/><BR/>Anyways, interesting problem.Aaron Denneyhttps://www.blogger.com/profile/06375678871775312955noreply@blogger.comtag:blogger.com,1999:blog-6703779.post-1159444762936886032006-09-28T12:59:00.000+01:002006-09-28T12:59:00.000+01:00Thanks Aaron! Doing those multiplications on a na...Thanks Aaron! Doing those multiplications on a napkin caused me to miss a factor of ten (now corrected). I've corrected the post accordingly to show your point that 65mph is close to optimal. Also I think DALY's use time-discounting, which is why the number of DALY's lost is only 33 times the number of deaths (from traffic accidents). On the other hand, if we think the future will be better than the present, then we'd like to spend more time there, and maybe should even have negative time-discounting.<BR/><BR/><BR/>I'm not sure I agree that A would be lower if we correctly modelled the actual distribution of speeds. (a) According to your formula, changing a delta-function on 65mph to a Gaussian with width of 10mph would only increase K by 14%, and a width of 15mph would increase K by 33%. (b) This assumes that everyone is always driving on the highway. <A HREF="http://www.eia.doe.gov/emeu/rtecs/appendb2.html" REL="nofollow">The EPA estimates</A> that Americans drive 45% of their miles on highways and 55% in cities, which would cut K by more than 50% if the average speed in cities is half the highway speed.<BR/><BR/>One other point I came across in the WHO report is that the rate of injury/accident scales as V^2, serious injury/accident as V^3 and death/accident as V^4. But we also need to know the rate of accidents/mile as a function of speed. Our calculations all assume this is constant; on the other, I think the results aren't too sensitive to this assumption.aram harrowhttps://www.blogger.com/profile/01272118188252697149noreply@blogger.comtag:blogger.com,1999:blog-6703779.post-1159401398548274582006-09-28T00:56:00.000+01:002006-09-28T00:56:00.000+01:00So we have: D(v) = # deaths/mile = A * v^4Let K = ...So we have: <BR/>D(v) = # deaths/mile = A * v^4<BR/>Let K = Integral(p(v) v^4 dv)<BR/>Current death rate R = Integral(p(v) D(v) dv) = Integral(p(v) A v^4 dv) = A K = 1.5 * 10^(-8) for current distribution over v.<BR/>T = Wasted hours per deaths.<BR/>Hours used per mile = 1/V<BR/>Total time per mile used is AT v^4 + 1/V. <BR/><BR/>Optimizing gives 4AT v^3 - 1/v^2 = 0, or v^5 = 1/(4 A T).<BR/><BR/>Using your velocity distribution of delta spike at 65, A = R / K = 8.4 * 10^(-16), and T = 300000.<BR/>This gives V = 63 MPH as least amount of time wasted. This isn't terribly far off from the 65 assumed.<BR/><BR/>Yes, this ignores accidents, speeding tickets, etc.<BR/><BR/>I would argue that A should be lower, as (a) It's not a spike. The v^4 weights higher speeds higher by quite a bit in the v^4 portion of the integral (replacing the delta by a gaussian of mean m and width w gives K = 3w^4 + 6w^2m^2 + m^4, rather than m^4), and (b) The p(v) is in terms of velocities weighted by miles travelled, not by cars on the road. Among other things, miles are traveled faster at higher speeds, so more miles are travelled by those going faster.<BR/><BR/>I'll also argue that we shouldn't optimize time, but rather current value of time. By the same argument that money in the future is not worth as much as money available now, I'll argue that time now is also worth more than time later. So we should replace T = 300000 with a discounted version. The easiest thing to do would be to assume a straight exponential decay of worth. Even a very small discount rate (a fraction of interest rates, say) can heavily discount future rates, <BR/><BR/>These both end up raising the optimal speed a bit. (Try 75 +- 5, which is what I see on my highways)Aaron Denneyhttps://www.blogger.com/profile/06375678871775312955noreply@blogger.comtag:blogger.com,1999:blog-6703779.post-1159311260972451372006-09-26T23:54:00.000+01:002006-09-26T23:54:00.000+01:00Interestingly, now when you google probability ter...Interestingly, now when you google probability terrorism dying your blog comes up #3. Now you know how to get to the first page on Google.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6703779.post-1159307922469752832006-09-26T22:58:00.000+01:002006-09-26T22:58:00.000+01:00One more about profiling from my new favorite blog...One more about profiling from <A HREF="http://www.overheardinnewyork.com/archives/006915.html" REL="nofollow">my new favorite blog</A>.aram harrowhttps://www.blogger.com/profile/01272118188252697149noreply@blogger.comtag:blogger.com,1999:blog-6703779.post-1159286589665623912006-09-26T17:03:00.000+01:002006-09-26T17:03:00.000+01:00Good point. The source I used is dated Jan 19, 200...Good point. The source I used is dated Jan 19, 2006. Since only 246 of the 9/11 fatalities were on the four planes, this means the 2000-6 risk is 20% higher than I had said. On the other hand, adding nine months to the denominator reduces the risk/flight. I get an adjusted result of 2 minutes and 8 seconds, unless I'm missing other crashes this year.<BR/><BR/>Another way to describe this risk, by the way, is that if each flight lasts three hours, then you'd have to fly 24 hours/day for 3000 years before you have a 50% chance of dying on a flight.aram harrowhttps://www.blogger.com/profile/01272118188252697149noreply@blogger.comtag:blogger.com,1999:blog-6703779.post-1159279984538819002006-09-26T15:13:00.000+01:002006-09-26T15:13:00.000+01:00"this was true in 1990-1999 entirely from accident...<EM><I>"this was true in 1990-1999 entirely from accidents, and in <B>2000-present</B>, entirely from 9/11 (!) as there have been <B>no fatal accidents</B> during that time ..."</I></EM><BR/><BR/><A HREF="http://www.washingtonpost.com/wp-dyn/content/article/2006/08/27/AR2006082700211.html" REL="nofollow">Commuter crash, August 27</A>correction?noreply@blogger.com