I don't think the math question ever got answered. We all know that, when you want to reduce the concentration of some solute in the pool water, it is better to do larger exchanges of water (eg, 50% drain and refill) then to do smaller successive exchanges (eg., five 10% drains). But why?
Ci is the initial concentration
Ci+1 is the concentration after the next incremental dilution
Vr is the volume of water removed
Vtot is the total pool volume
Then, for the first dilution, you'd get -
Ci+1 = (1 - (Vr/Vtot)) x Ci
The second dilution would be -
Ci+2 = (1 - (Vr/Vtot)) x Ci+1
Ci+2 = (1 - (Vr/Vtot)) x (1 - (Vr/Vtot)) x Ci
Ci+2 = (1 - (Vr/Vtot))2 x Ci
For the third dilution you'd get -
Ci+3 = (1 - (Vr/Vtot)) x Ci+2 = (1 - (Vr/Vtot))3 x Ci
So on and so forth, such that you can write the compact form as -
Cfinal = (1 - (Vr/Vtot))n x Cinitial, where n is the number of successive dilutions.
As an example, say you started out with 100ppm CYA and you did five (5) 5% water removals. At the end of the five removals the CYA would be -
[CYA]final = (1 - (0.05))5 x 100ppm = (0.7738) x 100ppm = 77ppm
If you had removed 25% of the water initially as opposed to five 5% removals, then your new [CYA] would be 75ppm, not 77ppm. For small numbers of dilutions, the reduction in concentration is approximately close enough to the total dilution but when you try to do a lot of little dilutions, the differences are more significant. The amount you remove also affects the magnitude of the reduction -
One 30% removal = 30% reduction
Three 10% removals = 27.1% reduction
Six 5% removals = 25.6% reduction
In all cases, you removed 30% of your water but the concentration change is larger for largest amount of water removed in the fewest cycles. So, in the end, the more water you can remove and replace in the fewer number of cycles, the better off you will be in lowering concentrations.
Hope that helps.