Taking volume contraction into account when mixing water with ethanol
Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!))

First, calculate the total mass  in kg of the volume of solution needed; divide the volume by the density of the specific ethanol-water solution. You can look up this density in a handbook or Google it. One simple table is at https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html, although many tables are available, even to tenths of a percent.

Second, calculate the mass of 96% ethanol needed:  = (∗/100)×1/0.96. The ratio / will be higher than , the % desired, because the ethanol is only 96%. Convert this to volume in liters, if you desire, by dividing by 0.80138, the density of 96% ethanol.

Third, calculate the amount of pure water needed:  =  − .   / will be less than 100 -   because of the 4% water already in the ethanol. But the total mass will give you the volume required, and the % ethanol will be correct - BY MASS. 

Commercial isopropanol solutions are frequently sold as 70% by volume; this calculates out to 64.7% by weight. Your procedure, if you desire to make variable % solutions BY VOLUME, will need to do a calculation before you even do the first step: convert the % ethanol by volume to % ethanol by mass.