2
$\begingroup$

I'm trying to work out a basic formula for calculating a volume discount for volume product purchasing (i.e: the per-unit price decreases more and more as the number of units purchased increases). The simple way to go is to set pricing tiers (i.e: <100 = $40.00 per unit, 100-199 = $30.00 per unit, 200-299 = $25.00 per unit etc.), but if this were to be plotted on a graph displaying per-unit prices as the number of units purchased increased, it would show a stepped trend, i.e:

enter image description here

Therefore, a similar graph showing total cost as the number of units purchased would also show a stepped trend.

I was hoping there is a formula I can use that dispenses with the lazy tiering method, and can provide a smooth trend at a per-unit resolution, suiting the below example:

enter image description here

What would be a good way to achieve this?

Any help would be much appreciated.

$\endgroup$
3
  • $\begingroup$ What is the "lazy tiering method?" $\endgroup$
    – saulspatz
    Commented Sep 12, 2018 at 3:46
  • $\begingroup$ Try $C/n$ for some $C$ you choose. $\endgroup$ Commented Sep 12, 2018 at 3:46
  • $\begingroup$ @saulspatz "The simple way to go is to set pricing tiers (i.e: <100 = $40.00 per unit, 100-199 = $30.00 per unit, 200-299 = $25.00 per unit etc.)" $\endgroup$ Commented Sep 12, 2018 at 3:49

1 Answer 1

-1
$\begingroup$

You would need to use the X to the Y power function on the calculator and use a decimal for the power, .95 or something like that to represent the discount for volume discounts. For example $40 to the .95 power. X would be the number of units they are purchasing and the $40 is your starting price and the .95 power would do the discounting (cheaper per unit the more you buy). Apologize for the autoformatting I can't seem to fix it so I included a written equation in the link.

Click on the link to view the equation written out

x($40) to the .95power

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .