Quote:

1. After subtracting the whole number spots, we're left with all the fractional spots for each age group -- I then allocated them, highest portion first. Meaning, M35-39's 0.60 beat out M60-64's 0.52.

I'm not sure what happened here but both you and transitionfour thought that 2012 IMCA allocated 6 Kona spots in M40-44. I am here to say emphatically that they only awarded 5. How did you both arrive at the same incorrect conclusion?

I'm not sure what happened here but both you and transitionfour thought that 2012 IMCA allocated 6 Kona spots in M40-44. I am here to say emphatically that they only awarded 5. How did you both arrive at the same incorrect conclusion?

My guess is what has happened is that their calculator was written back in the days when IMs had 100 slots to give out. And their algorithm, though flawed, was still pretty close. Because after giving out the 25ish freebies, there were still 75 left to prorate. But now with 40-50 slots to distribute, when they give out the 25 freebies there are only 20 or so to prorate, so the error in their model is exacerbated. As you see here. 5 vs 7 is big deal. 12 vs 14 less big deal.

Again, I could care less how they distribute. Give them all to 24 and under to encourage youth in the sport for all I care. But, show some transparency. This qualifying for Kona shit matters a lot to some people. There should be nothing to hide.

I thought you thought that 6 were awarded last year in M40-44 because you published my time. You should have published 5th place's time since only 5 were handed out.

Right, yes, I published your time as an estimate of what you'd have to run this year -- assuming there are 6 spots in 40-44 for the 2013 race. I was actually surprised when you said you just missed it -- I didn't know there was one fewer in the AG last year (that almost hurts more!)

Russ Brandt wrote:

see the picture I took here (posted by Bryan Dunn):
http://forum.slowtwitch.com/...ing=bryancd;#4014462

It looks like there are no longer directly proportional slots as it seems they must split some directly to the women then on to the AGs in a gender proportion.

Don't worry, that hurt is exactly was has motivated me tremendously for the last year. Digging all this up from last year has only made me triple down on my strategy to not let this happen again! Thanks!

You got 6th in M40-44 at 2012 IMCDA? You got screwed. M40-44 had 15.8% of the field, but only received 10% of the slots. WTC could have met it's criteria of all AGs with a participant receiving a slot and given M40-44 a more representative 14.0% of the slots. Too bad for you, huh?

I can't describe the highs and lows I felt over two days. I just barely eeked out 6th place after a final half mile footrace with 7th place Art Sosa Jr. I was feeling pretty good that 6 was good enough and I slept well that night. The next day, I saw the Kona allocation and about shit my stomach out of my pants. Now, I needed someone to not take their spot and I was hoping first place (none other than the great Tom Evans, previous overall winner of the race) would not be interested in going to Kona as an amateur. I was wrong. All spots taken. I was first loser.

And, while you didn't know it at the time, your AG was actually first in the queue to get a rollout slot, but there must have been none of those. Talk about rough luck. WTC's suboptimal slot allocation implementation robs you of an arguably deserved slot. Then all you need is a single rolldown and don't get one. Then all you need is a single roll-out slot because your AG will get it. And, no luck.

You deserve a break. You've got a free registration to SavageMan if you want it.

transitionfour wrote:

I'd like to figure out how exactly the Kona slot calculation works -- I've read what it says on the Kona FAQ. 1 slot per age group starting, then remaining slots allocated proportional to the number of racers. But, how exactly do they assign the slots proportionally.

Method #1
I estimated them this way for CDA first:

1. Allocate 1 slot for every age group where there is a starter.
2. Take the remaining 29 spots and figure what each age group's share of spots would be if you could sub-divide them exactly on proportion of total racers.
   
   That gives :
   M35-39 has 3.605 additional kona spots
   M60-64 has 0.522 spots
   
   
3. So, I allocate the whole number portion as additional spots -- so here 35-39 gets another three along with their original 1, so they are up to 4. In total that used up another 20 of the 29 remaining spots, so 9 left.
4. After subtracting the whole number spots, we're left with all the fractional spots for each age group -- I then allocated them, highest portion first. Meaning, M35-39's 0.60 beat out M60-64's 0.52.

Which gave me these results:






But then, Dave Latourette pointed out that he has a spreadsheet used to do the actual calculations for an awards ceremony, and :




So by his spreadsheet 30-34 loses one, and 60-64 gains one.

---

Ok, I have accepted the mission to replicate the WTC allocation spreadsheet precisely so that it is out there in the public domain with no mystery, intrique, or confusion. And I have to say, despite all my efforts, there is something strange and inexplicable (and definitely suboptimal) going on in that magical spreadsheet and all I can conclude is that it is doing something faulty as the result it produces cannot be replicated by any reasonable implementation of their stated process, and the result it produces is hard to justify when scrutinized.

The WTC allocation implementation is simple, at least as it is commonly understood and described on the Ironman.com FAQ:

1. Every AG with starters receives a single slot. For IMCDA this is 21 slots, leaving 29 slots unallocated.
2. Distribute the remaining slots (29) in a pro-rated distribution across all age groups.

The details of step 2 are unknown, however. While performing a prorated distribution for each AG is trivial: slots_for_AG = total_slots * AG_starters / total_starters, it gets more complex when you think about how to handle the fractional component of each distribution. Obviously you cannot split up a single Kona slot across AGs, so somehow the algorithm must result in whole numbers only. There are a few ways this can be dealt with, and it is not known how WTC does it. So, let's try them all.

1. Rounding: Simply round up or down. 2.4 slots = 2. 2.6 slots = 3. This is not a mathematically proper way to deal with the fractional component, but it is a way to deal with it. If a small AG deserves 1.5 slots this will round up to 2 slots, while another AG that deserves 15.49 will round down to 15 slots. Fair? No, it's not. If there were more slots to give away from the get-go, the larger AG would cross the 16 mark before the smaller AG crosses the 2, so it is wrong for the smaller AG to receive a 2nd before the larger AG receives a 16th via rounding. Thus, rounding creates a false distribution. Additionally, rounding can result in distributing fewer or more slots than are available. If more AGs round down than up, you will distribute fewer slots than available. If more AGs round up than down, you more slots will be distributed than are available to give away.
2. Floor and Rank: Always truncate the fractional component (floor). 2.01 slots = 2 slots. 2.99 slots = 2 slots. This will result in distributing fewer slots than are available because you are always rounding down. Hand out the remainder by ranking the decimal component that was truncated from each AG. So, an AG with 2.99 slots will get a 3rd slot before an AG with 5.67 slots gets a 6th, because 99 > 67. For the same reason as the 1.50 vs 15.49 example in the Rounding above, this is mathematically incorrect way of dealing with the fractional component. 50>49 means the 1.5 AG gets the next before the 15.49 AG, and that is wrong.
3. Floor and Repeat: As above, truncate the fractional component. Rather than ranking the decimal component to divvy out all the remaining slots that truncating leaves, simply run the whole algorithm again and again while increasing the total number of slots available until the total of the fractionally truncated allocated slots meets the number of slots actually available to distribute. So, if you have 50 slots to allocate, you may actually have to attempt to distribute 62 before the total of the truncated slots reaches 50. This is the proper way to do it, mathematically speaking.

So, I've implemented all 3 methods above plus an "optimal solution". The optimal solution is: rather than giving out a slot to all AGs with a starter (21) and prorating the rest (29), to give a slot to all AGs that earn less than one on their merits (8) and prorate the rest (42). This results in every AG with a starter receiving at least one slot and the optimally closest slot distribution to participant distribution, which it was is usually believed to be the intent of the WTC allocation process.

None of these 4 methods actually precisely replicate the actual WTC allocation, assuming Dave is correct in his description of what the WTC spreadsheet provides for the 2013 IMCDA numbers. The Rounding and Floor and Rank approaches are close, but not precise. The more mathematically proper, Floor and Repeat, is definitely not what WTC's implementation does as it produces a very different result, although it is closer to the optimal method than the actual WTC distribution is.


Lists below are 13 male AGs (a-m), then 13 female AGs (A-M) starting at 18-24 (a,A) going through 80+ (m,M) for each. Actual is as determined by WTC spreadsheet result produced by Dave Latourette.


So, what gives with M35-39 only receiving 4 slots in the WTC spreadsheet (column d) ? Using any of the other described approaches that AG receives either 5 or 6 slots. That AG has 12.1% participant contribution to the total field. 6 slots is 12% of the 50 slots available. 5 slots is 10%. 4 slots is 8%. 6 slots is best. 4 slots is way off.

Any way you slice it I cannot see how WTC arrives at just 4 slots for M35-39 in this case. There is some mystery going on in that spreadsheet and if I were in the M35-39 AG I would want someone to explain it to me. And, if I were in any AG and cared about Kona, I would want someone to explain it to me. Because 2013 IMCDA's M35-39 is going to be some other races some other AG.