Math Problem - Car park revenue

[psionl0]

The procedure is "follow the equation, ignore the text". But I concede I could perfectly ask BadBoy for additional information.

Give it up. Your attempt to appear to be a legend is failing miserably.

The OP erroneously said, "So lets assume it is nx2". You failed to notice that it didn't match the data provided (or you would have hauled him over the coals over it) so now you are attempting to sink the boots into him over this mistake to cover your deficiency.

 

[Ozzie]

I'd peg the 'sweet spot' at $10.00 an hour. That'll leave the lot about ¾ full, with enough spaces left over to cover the occasional user who needs a parking spot RIGHT GODDAMN NOW (as opposed to your regular customers).

I'd sweeten the pot by saying that anyone who arrives after a certain time (let's say 10AM) pays $15.00 an hour. I'd also throw in a monthly fee of $150.00 a month, as well as allowing reserved spots for another $50.00 a month (so a monthy parker with a reserved spot pays $200.00 a month).

 

[Brainster]

I put it into a spreadsheet. First line has 200 customers, at $7/day for a total of $1400, second line has 199 customers at 7.50 per day for a total of $1492.50. Continuing on like that (subtracting one customer and adding $.50 per customer per day, the max revenue is achieved at 107 customers at $53.50 per day, for a total of $5724.5.

Of course, as others have noted this assumes a straight-line demand curve. Given that the car park was initially unable to generate business at $18 per day it is very unlikely that there would be anybody parking there at $53.50/day.

ETA: Okay, I assume your formula for the decline in demand is n^2? Then your max revenue is at 175 customers at 9.50 per day for 1662.5, assuming we are sticking with 50 cent increments.

 

[WhatRoughBeast]

Of course, as others have noted this assumes a straight-line demand curve. Given that the car park was initially unable to generate business at $18 per day it is very unlikely that there would be anybody parking there at $53.50/day.

Yeah, but you made the same mistake AleCowan made. You overlooked the part that says

So if the price goes up my 50c then there is 1 new vacant space every day, increase of $1 is 2 vacant spaces, $1.50 is 4 spaces etc.

And to be fair, I overlooked it as well, the first time I read the post. It was only while I was composing retort to psion10, criticising him for getting the demand wrong, that I realised what was going on. BadBoy clearly meant the 2x part to refer to each step of .50.

 

[psionl0]

ETA: Okay, I assume your formula for the decline in demand is n^2?

4 data points were provided: (0,0) (1,1) (2,2) and (3,4). You can't fit all 4 points onto an n^2 function.

 

[psionl0]

4 data points were provided: (0,0) (1,1) (2,2) and (3,4). You can't fit all 4 points onto an n^2 function.

This is not exactly true. I originally excluded the (0,0) data point so that the remaining 3 points would fit an exponential equation (2^x-1). I did this because the OP had suggested that the number of empty spaces would double for each 50c price increase.

However, if I use only the last 3 data points, I can also get an n^2 equation. It turns out that the number of empty spaces for x >=1 is also given by 0.5(x² - x + 2).

By using that equation in the data sheet, I find that the "sweet" spot is $11 where the number of empty spaces would be 29 and the peak revenue would be $1,881.

 

[BadBoy]

But you had no good cause to do ignore the contradictory text. The question asked



BadBoy supplied enough information to make it clear that the proposed demand function is not linear, and psion10 quite accurately described the math.



But your own assumption, while slightly less unrealistic than BadBoy's, is even less justifiable given your background. You have assumed a linear demand function over the price range of 7 to 18 dollars, based on the observation of zero demand at 18 dollars. Really? On the one hand, assumptions of perfectly linear demand are commonly used in economics due to the mathematical simplicity which results, but real-world businesses know better. At the very least, very high asking prices can produce non-zero demands due to perceived status effects. Worse, even assuming a linear demand model, there is simply no evidence that, for instance, a cost of 17 dollars would not have likewise produced zero demand. In other words, your assumption that 18 dollars/zero demand establishes a real end point to the linear demand function is one you pulled out of a hat, and you should know better.



Neither do you.



Exactly. While I applaud the larger meaning, it also applies to your own assumptions.

With that said, BadBoy should be aware that his exponential demand function rises so sharply that for any price greater than 11 dollars, demand becomes negative, so the model function he proposed doesn't seem terribly likely.

But psion10's analysis is correct for the demand function given.

this.

I was going to reply to aleCcowaN myself then seen this post which basically states my case.

My question was more about how to formulate a solution given the simple parameters I provided. It was intended to be a vehicle to examine an applied math problem - that's all. I had a vague question in my head about how much the price would go up to, which led on to how I could work it out. But it was a question about math, not about marketing and economics, but that's fine - any response is good


The function for the vacancy increase against price was really arbitrary in some sense, though I made a mistake there (I meant a function like 2^n), but that was really for the sake of argument and of course in the real world other factors would contribute. I was imagining a curve (some kind of parabola) where at some point the price would start having a negative effect on turnover, and given some function (like 2^n) how would you work it out. And actually it tuns out that plotting it manually was the easiest solution to work out where the peak was. I though that perhaps there was a formula to figure out the max value in these cases, and how I would apply it in this simple case.

I guess if it is a quadratic then "a parabola that maximizes income/profit for X= -b/2a with a and b being the coefficients of the quadratic equation of income" would be the way to go though wouldn't it?

Cheers

 

[psionl0]

I guess if it is a quadratic then "a parabola that maximizes income/profit for X= -b/2a with a and b being the coefficients of the quadratic equation of income" would be the way to go though wouldn't it?

Cubic actually.

The only way that you would get a parabola for the revenue function is if the demand was a linear function of price (since revenue = price times demand).

 

[Monza]

The opening post reminded me of this...


A blonde walks into a bank in New York City and asks for the loan officer. She says she's going to Europe on business for two weeks and needs to borrow $5,000. The bank officer says the bank will need some kind of security for the loan, so the blonde hands over the keys to a new Rolls Royce.

The car is parked on the street in front of the bank; she has the title, and everything checks out. The bank agrees to accept the car as collateral for the loan. The bank's president and its officers all enjoy a good laugh at the blonde for using a $250,000 Rolls as collateral against a $5,000 loan. An employee of the bank then drives the Rolls into the bank's underground garage and parks it there.

Two weeks later, the blonde returns and repays the $5,000 and the interest, which comes to $15.41. The loan officer says, "Miss, we are very happy to have had your business, and this transaction has worked out very nicely; but we are a little puzzled. We checked you out and found that you are a multimillionaire. What puzzles us is - why would you bother to borrow $5,000?"

The blond replies....."Where else in New York City can I park my car for two weeks for only $15.41 and expect it to be there when I return?"

 

[WhatRoughBeast]

I guess if it is a quadratic then "a parabola that maximizes income/profit for X= -b/2a with a and b being the coefficients of the quadratic equation of income" would be the way to go though wouldn't it?

Cheers

In general, you do the following:

1) Determine the income function - demand times price.

2) Take the derivative of the income function.

3) Set the derivative to zero (which occurs at a peak) and evaluate.

4) The resulting value (a single value if you're lucky) must now be checked against your constraints. If it falls within the working interval, you're all set. If not, you have to check the limits and determine which limit is the value you want.

So, yeah, your equation is correct - maybe. In general, it's entirely possible that a demand function will produce a parabola of the wrong curvature; that is, one in which the zero derivative produces a minimum rather than a maximum. In which case step 4 applies.