2
$\begingroup$

I'm an artist.

I'm trying to find a way to calculate the price of paintings of varying sizes.

I have tried to come up with some kind of equation to vary the price based on square cm.

The thing is, you cannot have a fixed price per cm2 for paintings, as a very small one would be too cheap, or a large one be too expensive.

For example, a painting that is 20x20 cm (400cm2) costs $15000.

That's a price of $37,5 per cm2.

If I apply that to a painting that is 175x200 cm, it would cost $1312500. wow.

In reality, a painting that size has an approx price of $125000.

So, I need some kind of equation, based on data from real world examples. I have tried to figure out how to do it, but I'm getting nowhere. My best guess is to try curve fitting a quadratic equation, but I can't figure out how to derive it.

Can you help? A smooth fit to the data is sufficient, as all data points are choosen inexactly up to this point. (i.e my gallerist and I have only made up approximate prices based on what is "reasonable" for a specific size of work).

I would like an equation where I only have to enter the dimensions of the painting (cm2) and I get the appropriate price/cm2.

Here's a few data points:

$\begin{array}{rRr} \text{Area (cm}^2\text{)} & \text{Price per area} & \text{Price} \\ \hline 400 & 37.500 & 15000 \\ 1849 & 16.225 & 30000 \\ 2107 & 14.238 & 30000 \\ 2907 & 12.040 & 35000 \\ 5600 & 8.036 & 45000 \\ 9801 & 5.101 & 50000 \\ 12000 & 4.660 & 55920 \\ 24000 & 3.958 & 95000 \\ 35000 & 3.571 & 125000 \\ 50000 & 3.000 & 150000 \\ \end{array}$

$\endgroup$
  • $\begingroup$ "my gallerist and I have only made up approximate prices" - This line really caught my attention. Why do you think that your gut feel on pricing will result in a smooth equation? It can just as easily result in 2-4 lines that have different slopes for a given range of painting sizes. A range of smaller paintings proportional to size, but beyond that, a 50% increase in size is only a 25% increase in price. $\endgroup$ – JTP - Apologise to Monica Feb 1 '14 at 21:23
1
$\begingroup$

I reworked your data considering this time the relation between the total price and the area. As you probably know, the standard sum of squares tend to favour the largest values. So, what I did was try to minimize the sum of the squares of relative errors on total price.

Still looking for simple formulas I arrived to

$\text{Total Price} = a + b * \text{area}^c$

for which I got $a = 9424.480, b = 162.5735, c = 0.619217$

As results, I obtained

           Area             Price            Predicted price
         400.000           15000.000           16066.299
        1849.000           30000.000           26563.665
        2107.000           30000.000           28007.525
        2907.000           35000.000           32105.958
        5600.000           45000.000           43464.442
        9801.000           50000.000           57564.893
       12000.000           55920.000           63993.445
       24000.000           95000.000           93244.667
       35000.000          125000.000          115303.900
       50000.000          150000.000          141471.770

which do not look too bad for me.

$\endgroup$
  • $\begingroup$ Very Nice results, thank you! $\endgroup$ – Perceptor Jan 23 '14 at 7:30
  • $\begingroup$ So, how do I formulate this equation? Price=(9424.480+162.5735)*area^0.618217 ? That is, x=9587.0535*y^0.619217, where x=price and y=area? $\endgroup$ – Perceptor Jan 23 '14 at 7:32
  • $\begingroup$ No, the formula is Price = 9424.480 + 162.5735 * area^0.618217. By the way, if you like this answe, you can accept it in order other people is this forum give a look at it. $\endgroup$ – Claude Leibovici Jan 23 '14 at 7:45
  • $\begingroup$ How about this? :) y=(322.18473+711.24516*x^0.5+(-9.5313808)*x+0.035322791*x^1.5)/(1+(-0.011088569)*x^0.5+2.7185466e-05*x+5.6040572e-08x^1.5) $\endgroup$ – Perceptor Feb 1 '14 at 11:36
  • $\begingroup$ @Perceptor.For sure, we could do better adding more parameters. I shall look at the problem tomorrow and I shall post the results as a new answer. $\endgroup$ – Claude Leibovici Feb 1 '14 at 12:04
1
$\begingroup$

I took the liberty of calculating the total price for the works of art, based on your data. Please correct them if I made mistakes.

Plugging those into Wolfram yields a good fitting quadratic equation, as you can see.

When I accidentaly missed the last data point, I even got a good fitting lineair equation of about $3x + 22707$ ($x$ being the area), which does make sense, as you will have setup costs that are independent of the size of the painting, plus a cost per square cm.

$\endgroup$
  • $\begingroup$ Interesting. A graph showing the total price instead. It works very well, except for the smallest paintings, for example my 20x20cm works, which gets a too high price (23907 instead of around 15000) $\endgroup$ – Perceptor Jan 22 '14 at 12:53
0
$\begingroup$

I just played a little with your numbers. If you do not like the results, just tell me and I should rework that tomorrow.

What I found is that your data can be represented not too bad (at least in my opinion) by

Log[price] = 6.75 - 0.53 Log[area]

Another one easier to handle : price = 1174.72 / area^0.574671

$\endgroup$
  • $\begingroup$ This is great feedback, thanks. It works very well up to about 10000 cm2, when the price starts to drop too much. hmm... $\endgroup$ – Perceptor Jan 22 '14 at 12:49
  • $\begingroup$ imgur.com/c8qoD7r $\endgroup$ – Perceptor Jan 22 '14 at 12:51
  • $\begingroup$ @Perceptor.I shall work this tomorrow morning. $\endgroup$ – Claude Leibovici Jan 22 '14 at 17:39
0
$\begingroup$

In my opinion, we must be careful when adding more and more parameters into a regression model. Remember that you only have ten data points.

The last model I proposed earlier was basically showing that the data could be represented quite well using $$\text{Total Price} = a + b * \text{area}^c$$

with $a = 9424.480, b = 162.5735, c = 0.619217$ and a sum of squares (SSQ) of relative errors equal to $0.0839$.

Exponent $c$ being not very far from $0.5$, we could effectively rethink the model as a polynomial of the square root of $area$. This is what I did, exploring the impact of the number of terms :

degree 1 : SSQ = 0.1070
degree 2 : SSQ = 0.0714
degree 3 : SSQ = 0.0371

For sure, the fit is better but the significance of the parameters decreases dramatically when the number of parameters increases.

Anyway, I give you below the results for a cubic polynomial of $\sqrt{\text{area}}$

           Area             Price            Predicted price  
         400.000           15000.000           15292.509
        1849.000           30000.000           28514.288
        2107.000           30000.000           30009.128
        2907.000           35000.000           33986.751
        5600.000           45000.000           43658.631
        9801.000           50000.000           54556.146
       12000.000           55920.000           59569.485
       24000.000           95000.000           86444.807
       35000.000          125000.000          114500.792
       50000.000          150000.000          160362.153

the coefficients being : 1129.12, 785.028, -4.1884 and 0.0172728. You may prefer this; for me, it is not better than the previous model and I should use it with care.

Your idea of using a rational approximation is interesting but, again, I should not recommend to use so many parameters. I tried with low order polynomials; for the same number of parameters the quality of the fit is almost the same.

$\endgroup$
  • $\begingroup$ Thank you very much ! I really appreciate your help. Most interesting. $\endgroup$ – Perceptor Feb 5 '14 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.