2010-10-17, 12:14 PM
Truncating the percentage weakens it by a sometimes noticeable amount. It also complicates things, so you'll need an auxiliary value.
For a given value of d, let p = [√(d/4)] be the percentage. Then you have
percentage = flat
pa/100 = d/4
pa = 25d
a = 25d/p.
For instance:
d = 20000
p = [√(20000/4)] = 70
a = 25*20000/70 = 7142.85 <-- this is when flat = percentage.
7142 should favor flat by a marginal amount; 7143 should favor percentage. Note that to find the "breakaway point" as you defined it, you have to take the ceiling of 25d/p. (This also explains the rounding errors in your first post. You took the floor, not the ceiling; same thing applies.)
Verify:
flat: 20000/4 = 5000
percentage: 70*7142/100 = 4999.4 (= 4999)
percentage: 70*7143/100 = 5000.1 (= 5000)
But given that you have to calculate the percentage anyway, I don't see any advantage of this over just comparing the two reductions directly.
For a given value of d, let p = [√(d/4)] be the percentage. Then you have
percentage = flat
pa/100 = d/4
pa = 25d
a = 25d/p.
For instance:
d = 20000
p = [√(20000/4)] = 70
a = 25*20000/70 = 7142.85 <-- this is when flat = percentage.
7142 should favor flat by a marginal amount; 7143 should favor percentage. Note that to find the "breakaway point" as you defined it, you have to take the ceiling of 25d/p. (This also explains the rounding errors in your first post. You took the floor, not the ceiling; same thing applies.)
Verify:
flat: 20000/4 = 5000
percentage: 70*7142/100 = 4999.4 (= 4999)
percentage: 70*7143/100 = 5000.1 (= 5000)
But given that you have to calculate the percentage anyway, I don't see any advantage of this over just comparing the two reductions directly.