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Old 10-21-2007, 02:22 AM
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Looks like a Simple Math Problem..

I am wondering if anyone might help me with this small math model I am trying to crack. It would help my FX money management program if I can solve this. It deals with an inversion of percents.

The typical percent problem..

Three kids do $1000 damage to a vehicle. Each says they acted independently and damaged in varying amounts. The court decides to charge them proportional to the damage.

Kid One 10% damage = $100 fine
Kid Two 30% damage = $300 fine
Kid Three 60% damage = $600 fine

ok.

BUT

Lets assume we now have an inverted situation. Lets assume that there is a new Youth Incentive Program which will offer a full $1000 stipend to the three kids related to the vehicle incident. The rules state that the child who did the least damage will receive the largest portion of the grant money and the child who was most destructive will receive the least , according to the above percentages.

How much does each child get? Formula required.

Thanks a million...
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Old 10-21-2007, 08:02 AM
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One approach would be to allot a proportion P[k] of grant money to each kid
such that the relative allotments are proportional to the relative damages.
That is, with D[k] being the damage by kid k, you'd get an system of
constraints for kids k and j by equations D[k]*P[k] = D[j]*P[j].

Plus that sum(P[i])=1.

I.e., in your example, kid 1 should receive 0.6/0.1=6 times more than kid
3, etc.

The general expression would be: P[k] = 1 / sum( i=1..N : D[k] /D[i] )

It results in the following allotments for your example:
P[1] = 1 / ( 0.1/0.1 + 0.1/0.3 + 0.1/0.6 ) = 67%
P[2] = 1 / ( 0.3/0.1 + 0.3/0.3 + 0.3/0.6 ) = 22%
P[3] = 1 / ( 0.6/0.1 + 0.6/0.3 + 0.6/0.6 ) = 11%

Unfortunately the approach falls to pieces if ever any one kid fails to
do some damage. (As if that could happen )

Another approach is to bring in a concept of "non-damage", such that when
100% damage is done, the N kids did (N-1) non-damage, in the proportions
(1-D[k]). By that approach, the allotment expression would be
P[k] = (1-D[k]))/(N-1)

For your example, the allotments would then be 45%, 35% and 20%.

The second approach deals well with kids not doing damage, but might
be seen as unfair in not preserving the relative damage proportion between
the kids.
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