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I'm trying to come up with a pricing algorithm for my product. I've already set some prices at low intervals, but I need the algorithm to calculate a reasonable for very large orders.

Here are the prices I've already set:

  • 20 units = $35 (1.75 per unit)

  • 40 units = $65 (1.625 per unit)

  • 60 units = $95 (1.583 per unit)

  • 80 units = $125 (1.562 per unit)

  • 100 units = $155 (1.55 per unit)

Is there a simple equation that can be used to carry on this type of trend? Ideally 100,000 units would work out to be about 1.00 per unit.

Leah Sapan
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3 Answers3

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Adding another answer to show a tool for a direct approach. Use Wolfram Alpha to get a fit suggestion. Logically the log form is better, it suggests $$ 2.10714-0.124745 \log(x) $$ which has a goodness of fit of $R^2 \approx 95\%$.

gt6989b
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So far, your prices fit perfectly cost=5+1.50*units. For a large number of units, you will average 1.50 with this. There are many ways to bend the curve to get to 1.00 at large quantities. Probably the simplest is to make another line, say cost=1005+1.00*units and charge the lower. The breakpoint is 2000 units here. The average cost for 100,000 will be 1.01 this way.

Ross Millikan
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  • I just want to say this is also a very valid answer. I just like the convenience of the using a curve. I was able to work my curve out to match my set prices within about a dollar's range, while still giving great discounts at high unit volumes. Thanks though! – Leah Sapan Feb 11 '14 at 21:30
  • Fair enough. You probably gave some reasonable thought to the prices you have. I would suggest they do not constrain what happens up high very well. There are many other curves that will do that. If you don't think about what you want, you may go broke. – Ross Millikan Feb 11 '14 at 21:46
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The easiest way to do this would be through a lookup table. If you want to offer a tiered discount, just define the quantity range for each discount level. Then, just look up the unit price for the given quantity, multiply by the quantity, and you've won.

John
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