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I am trying to make a shop for self learning but i have a small hiccup!

I have this list of items that has a USD value assigned to it. But I want to make it so if they buy a "100 GB/s" it should be cheaper than buying 100 "1 GB/s" or 10 "10 GB/s". I have been trying for ages but i can't seem to figure it out.

Currently if they buy 100 of "1 GB/s" it will cost $7200 which is the value of one "100 GB/s"

If I'm unclear please write down below so I can explain further.

How do I do it?

100 MB/s = $8
200 MB/s = $14
300 MB/s = $22
400 MB/s = $29
500 MB/s = $36
600 MB/s = $43
700 MB/s = $50
800 MB/s = $58
900 MB/s = $65
1 GB/s = $72
2 GB/s = $144 
3 GB/s = $216
4 GB/s = $288
5 GB/s = $360
6 GB/s = $432
7 GB/s = $504
8 GB/s = $576
9 GB/s = $648
10 GB/s = $720
11 GB/s = $792
12 GB/s = $864
13 GB/s = $936
14 GB/s = $1008
15 GB/s = $1080
20 GB/s = $1440
25 GB/s = $1800
30 GB/s = $2160
35 GB/s = $2520
40 GB/s = $2880
45 GB/s = $3240
50 GB/s = $3600
60 GB/s = $4320
70 GB/s = $5040
80 GB/s = $5760
90 GB/s = $6480
100 GB/s = $7200
Heinz
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  • What happens if you simply decrease the price for 100GB/s ??? – lmsteffan Sep 08 '14 at 17:32
  • You want the cost per GB to be a non-increasing function of the amount of GB per unit. – copper.hat Sep 08 '14 at 17:34
  • But if i decrease 100 GB/s with like 200 how much do I decrease all the other?, if i decrease all with 200 it's basically the same problem but lower price. – Heinz Sep 08 '14 at 17:35
  • First pick the cost per GB vs. amount of GB per unit curve. – copper.hat Sep 08 '14 at 17:35
  • copper.hat , could you explain that a little more, I don't understand. What is the "unit curve"? – Heinz Sep 08 '14 at 17:36
  • You have a collection of objects labelled $n$ B/s. For each object compute the cost per B/s, call it $c(n)$. You want to have $c(n)$ be a non-increasing function of $n$. Then price the smallest $n$ and work from there (or the other way around, if you prefer). – copper.hat Sep 08 '14 at 17:39
  • (Not 'unit curve', it was the curve generated by cost per B/s vs. the amount of B/s.) – copper.hat Sep 08 '14 at 17:44
  • I'm sorry, that is just too advanced for my math with c(n) and stuff. Is there any other way to explain it? – Heinz Sep 08 '14 at 17:44
  • Its just division. – copper.hat Sep 08 '14 at 17:47
  • Ok, I think it's just the fancy words that's messing with me – Heinz Sep 08 '14 at 17:48
  • @Heinz The way I'd set this up is to have price as a function of $x,GB$, such that $\mathrm{price}(n\cdot x)<n\cdot\mathrm{price}(x)$. What I mean by this is that the price of buying $n\cdot x$ GB is less than buying $x$ GB but $n$ times over. For example if $n=2$, you want $\mathrm{price}(2x)$ (buying double) to be cheaper than $2\mathrm{price}(x)$ (buying it twice). A function that comes to mind is $\mathrm{price}(x)=\log(x)$ but you'd probably need some tweaking to make sure you're not losing money. I'm not saying this is the best or easiest way, it's just how I'd do it. – Jam Sep 08 '14 at 17:48

1 Answers1

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One way to do this is to arrange that the unit price $p_N$ falls by a certain fraction $f$ when the quantity $N$ increases tenfold. This is a property of the function $$ p_N = p_1 N^{-k} $$ where $p_1$ is the price of one unit sold individually and $k$ is a constant.

$k$ is found by setting $$ 10^{-k} = 1 - f $$ which gives $$ k = -\log_{10}{(1-f)} $$

Example

$p_1 = \$72$ (the price of 1Gb/s)

$f=0.1$ (the unit price is reduced by 10% price for a tenfold increase in quantity)

With this data we find $$ k = -\log_{10}{(1-f)} = -\log_{10}{0.9} = 0.045757 $$

The unit price of a sale of 3Gb/s (N=3) is then $$ p_3 = p_1 3^{-k} = 72 \times 3^{-0.045757} = \$68.47005 $$ and the cost to the customer is $$ 3 p_3 = 3\times 68.47005 = 205.41 $$

As expected, the cost of 10Gb/s $(N=10)$ works out at \$648.00 (90% of \$720).

MartinG
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