Applications of Calculation in Real Estate Development

Calculus has many real-world uses and applications in the physical sciences, computer science, economics, business, and medicine. I will briefly touch on some of these uses and applications in the real estate industry.

Let’s start by using some calculation examples in speculative real estate development (ie the construction of new homes). Logically, a new home builder wants to make a profit after completing each home in a new home community. This builder will also need to be able to maintain (hopefully) positive cash flow during the construction process of each home, or each phase of the home’s development. There are many factors that go into calculating a profit. For example, we already know that the profit formula is: P = R – Cthat is, the gain (P) is equal to income (R) minus the cost (against). Although this main formula is very simple, there are many variables that can influence this formula. For example, low cost (against), there are many different cost variables, such as the cost of building materials, labor costs, real estate maintenance costs prior to purchase, utility costs, and insurance premium costs. insurance during the construction phase. These are some of the many costs to take into account in the formula mentioned above. Low income (R), variables such as the base sale price of the home, additional improvements or additions to the home (security system, surround sound system, granite countertops, etc.) could be included. Simply plugging in all these different variables in and of itself can be a daunting task. However, this is further complicated if the rate of change is not linear, which requires us to adjust our calculations because the rate of change of one or all of these variables is in the form of a curve (i.e. exponential rate of change) ? This is an area where calculation comes into play.

Let’s say last month we sold 50 houses with a median sales price of $500,000. Regardless of other factors, our income (R) is the price ($500,000) times x (50 houses sold) which equals $25,000,000. Let’s consider that the total cost to build the 50 houses was $23,500,000; therefore the gain (P) is 25,000,000 – $23,500,000 which equals $1,500,000. Now knowing these numbers, your boss has asked you to maximize profits for the next month. How do you do this? What price can you put?

As a simple example of this, let us first calculate the marginal profit in terms of X to build a house in a new residential community. We know that the income (R) is equal to the demand equation (p) times units sold (X). We write the equation as

R = pixels.

Suppose we have determined that the demand equation for selling a house in this community is

p = $1,000,000 – X/10.

With $1,000,000 you know you won’t sell any houses. Now, the cost equation (against) it is

$300,000 + $18,000X ($175,000 in fixed material costs and $10,000 per house sold + $125,000 in fixed labor costs and $8,000 per house).

From this we can calculate the marginal benefit in terms of X (units sold), then use the marginal profit to calculate the price we should charge to maximize profit. So the income is

R = pixels = ($1,000,000 – X/10) * (X) = $1,000,000X – x^2/10.

Therefore, the gain is

P = CR = ($1,000,000X – x^2/10) – ($300,000 + $18,000X) = 982,000x – (x^2/10) – $300,000.

From this we can calculate the marginal benefit by taking the derivative of the benefit

dp/dx = 982,000 – (X/5)

To calculate the maximum profit, we set the marginal profit equal to zero and solve

982,000 – (X/5) = 0

X = 4910000.

we plug X Go back to the demand function and get the following:

p = $1,000,000 – (4910000)/10 = $509,000.

So the price we need to set to make the maximum profit for each house we sell should be $509,000. The following month, you sell 50 more houses under the new price structure, earning a net profit increase of $450,000 from the previous month. Great job!

Now, for the next month, your boss asks you, the community developer, to find a way to reduce the costs of building homes. You already know that the cost equation (against) I was:
$300,000 + $18,000X ($175,000 in fixed material costs and $10,000 per house sold + $125,000 in fixed labor costs and $8,000 per house).

After astute negotiations with his building suppliers, he was able to reduce fixed material costs to $150,000 and $9,000 per house, and reduce labor costs to $110,000 and $7,000 per house. As a result, your cost equation (against) has changed to

against = $260,000 + $16,000X.

Because of these changes, you will need to recalculate the base benefit

P = CR = ($1,000,000X – x^2/10) – ($260,000 + $16,000X) = 984,000X – (x^2/10) – $260,000.

From this we can calculate the new marginal profit by taking the derivative of the new calculated profit

dp/dx = 984,000 – (X/5).

To calculate the maximum profit, we set the marginal profit equal to zero and solve

984,000 – (X/5) = 0

X = 4920000.

we plug X Go back to the demand function and get the following:

p = $1,000,000 – (4920000)/10 = $508,000.

So the price we need to set to get the maximum new profit for each house we sell should be $508,000. Now, even though we lowered the selling price from $509,000 to $508,000, and we still sell 50 units as in the previous two months, our profit has still increased because we reduced costs to the tune of $140,000. We can find out by calculating the difference between the first P = R – C and the second P = R – C containing the new cost equation.

1st P = CR = ($1,000,000X – x^2/10) – ($300,000 + $18,000X) = 982,000X – (x^2/10) – $300,000 = 48,799,750

2nd P = CR = ($1,000,000X – x^2/10) – ($260,000 + $16,000X) = 984,000X – (x^2/10) – $260,000 = 48,939,750

Taking the second win minus the first win, you can see a difference (increase) of $140,000 in earnings. Therefore, by reducing home construction costs, you can make the business even more profitable.

Let’s recap. By simply applying the demand function, marginal profit, and maximum profit calculation, and nothing else, he was able to help his company increase its monthly profit from the ABC Home Community project by hundreds of thousands of dollars. Through a little negotiation with his construction suppliers and labor leaders, he was able to lower his costs and through a simple readjustment of the cost equation (against), you could quickly see that by cutting costs, you increased profits once again, even after adjusting your maximum profit by reducing the selling price by $1,000 per unit. This is an example of the wonder of calculus when applied to real-world problems.