The Urban Housing Market: Structures, Attributes, and Density

A developer owns one piece of land. Should it hold a few spacious homes or a taller building with many smaller apartments? Buyers may pay more for space and privacy, but land is scarce and tall buildings cost more to construct. Urban housing economics explains how these competing forces determine what gets built, how densely it gets built, and when an old building is worth replacing.

Learning goals and study method

Explain a home as a bundle of attributes, interpret hedonic regression coefficients, calculate optimal Floor Area Ratio, evaluate a redevelopment proposal, and compare competing land bids. Learn the economic reasoning before memorising a formula.

1. A home is a bundle, not a standard unit

Two homes with the same floor area can sell for different prices because one has better transport access, more bathrooms, a better layout, or a quieter street. Buyers purchase the bundle of these features together.

Diminishing marginal utility means an additional unit of an attribute can add less satisfaction when you already have more of it. A first bathroom may be much more useful than a fourth. This is a useful modelling idea, not a claim that every feature has diminishing value in every setting: attributes can also complement one another.

A hedonic model estimates how market rent or price varies with observed attributes. Its implicit or shadow price is the change in the predicted outcome associated with a small attribute change, holding the model’s other variables fixed. It is not a separately invoiced price for the bathroom.

For example, suppose an illustrative fitted model for annual rent is:

R = 6,000 + 100A + 1,500B - 200d,

where A is floor area in square metres, B is the number of bathrooms, and d is distance in kilometres. At A = 60, B = 1, and d = 5, predicted annual rent is 12,500. One more bathroom adds 1,500 in this linear model, other included attributes unchanged.

That coefficient is an association conditional on the model. It does not establish that adding a bathroom causes rent to increase by exactly 1,500. Omitted renovation quality, school access, measurement errors, and selection can affect estimates. Bedrooms and floor area are also correlated; a bedroom coefficient at fixed total area is not the value of adding both a room and extra area.

2. Linear, log-linear, and log-log models

Choose a specification that matches the economic question and inspect whether it describes the data well.

Specification	Example	Interpretation, holding other variables fixed
Linear	R = a + bX	One more X changes predicted R by b currency units
Log-linear	ln(R) = a + bX	One more X changes the fitted rent ratio by exp(b); approximately 100b percent for small b
Log-log	ln(R) = a + b ln(X)	b is the elasticity: a 1% increase in X implies approximately b% higher R

In a log-linear model, b = 0.08 means an exact fitted percentage change of 100[exp(0.08)-1] = 8.33%, rather than exactly 8%. In a log-log model with b = 0.6, a 10% area increase changes fitted rent by 100[(1.10)^0.6-1], approximately 5.89%.

Logs require positive variables. A log-log model cannot directly take the log of zero bathrooms or zero distance. Also, exponentiating a fitted log outcome does not automatically recover the conditional mean of rent in levels; the error distribution and retransformation adjustment matter.

Within-city models compare properties using structure and neighbourhood attributes. Cross-city models can relate metropolitan median prices to population, growth, and construction costs. A relationship between city medians is not automatically an individual property’s attribute price. Different housing mixes and local institutions can influence those medians.

Teach it back: Why is a bathroom coefficient not automatically the profit from installing a bathroom?

Self-check: value, cost, and causality

It estimates a conditional market association. The actual price effect can differ, and construction has a cost. If the outcome is annual rent, compare the value of the net rental stream with the upfront construction cost rather than comparing one year’s rent directly with that cost.

3. Developers compare marginal value with marginal cost

Imagine adding a feature costs 8,000 and increases sale value by 12,000. The change adds 4,000 of project surplus before other costs. If the next enhancement costs another 8,000 but adds only 5,000 of value, it reduces surplus.

At an unconstrained interior optimum, the marginal increase in value equals marginal cost. Discrete features, regulations, financing, and interactions between attributes can prevent exact equality. A developer is choosing an entire configuration, not independently maximising every regression coefficient.

Now apply the same reasoning to the amount of floor area placed on one site.

4. FAR measures intensity, not simply height

Floor Area Ratio (FAR), F, is total built floor area divided by land area. A 1,000-square-metre site supporting 3,000 square metres of floor area has F = 3.

FAR does not uniquely specify the number of storeys. A three-storey building covering the whole lot and a six-storey building covering half the lot can both have FAR 3, ignoring local counting rules and design constraints.

In this stylised model, sale price per unit of floor area falls with development density:

P(F) = alpha - beta F.

Alpha reflects location and structural appeal before the modelled density discount. Positive beta represents buyers’ discount for additional density, which might capture less privacy or crowding. This is an assumed relationship, not a universal empirical law.

Average construction cost per unit of floor area rises with density:

C(F) = mu + tau F.

Mu is baseline construction cost. Positive tau captures more expensive structural systems, lifts, foundations, and related costs. Here C is cost per floor area, not the annual construction flow C used in the first lecture. Symbols acquire meaning from their units and context.

5. A one-site development calculation

Normalise land area to one unit. F then gives the floor area per unit of land. Revenue per land unit is F times P, and construction cost per land unit is F times C. The amount left to pay for land is:

p(F) = F[P(F)-C(F)] = F(alpha-mu) - F^2(beta+tau).

This is a residual land bid, before paying for the land. It is not the developer’s final profit after acquiring the site. It equals a supportable land value only when all relevant non-land costs and the required development return have been accounted for consistently. The simple formula abstracts from costs not explicitly included.

Choose F to maximise that residual. The marginal change is:

dp/dF = (alpha-mu) - 2F(beta+tau).

Setting it to zero gives:

*F = (alpha-mu)/[2(beta+tau)]**,

p = (alpha-mu)^2/[4(beta+tau)].*

These are interior solutions requiring alpha > mu and beta+tau > 0. The second derivative is -2(beta+tau), so the stationary point is a maximum. If alpha is no greater than mu, the nonnegative-F optimum is F = 0 in this simplified development option, and the squared expression for interior p* must not be used. Agricultural use or another outside option can also dominate development.

Worked example

Use generic currency units per square metre of floor area: alpha = 600, mu = 200, beta = 40, and tau = 60. These are invented teaching inputs.

Then F = 400/(2 x 100) = 2*. Price is 520, average construction cost is 320, and residual value per square metre of land is 2 x (520-320) = 400.

FAR	Price per floor area	Cost per floor area	Residual per land area
1	560	260	300
2	520	320	400
3	480	380	300
4	440	440	0

Building more initially spreads the opportunity over more floor area. Beyond FAR 2, the extra area no longer compensates for the density discount and rising costs. Adding floor area also changes the value and cost of the rest of the project in this assumed schedule, which explains the factor of two in the derivative.

Teach it back: Why does the developer stop before the point where P equals C?

Self-check: average spread versus marginal surplus

P-C is the average spread per floor unit. Maximising total residual F(P-C) requires accounting for how higher F changes that spread across the project. At FAR 4 the spread is zero; building that much destroys the positive residual available at FAR 2.

6. Better locations and binding density limits

Increase alpha from 600 to 800 while holding the other worked-example parameters fixed. Optimal FAR rises from 2 to 3, and the land residual rises from 400 to 900. Better locations support both more intensive building and higher land bids.

This is substitution of capital for land: when access is valuable, using more construction per land unit can be worthwhile. It does not require every expensive location to have tall buildings; regulation, heritage, design, and construction conditions can restrict the response.

If a binding rule imposes F no greater than 1.5 in the original example, the constrained choice is F = 1.5, producing residual 400(1.5)-100(1.5)^2 = 375. Under these concave schedules, a simple upper limit gives F chosen = min(F, F maximum)* when the interior conditions hold. A nonbinding limit has no effect in this calculation.

7. Why redevelopment happens in jumps

An existing building is durable. It cannot generally be converted into a new optimal building one infinitesimal unit at a time. Replacement sacrifices an income-producing asset and incurs construction and demolition costs. Existing construction expenditure is sunk, but the building’s current income value is an opportunity cost.

Let F0 be the existing FAR, p0 the acquisition or opportunity value of the existing land-and-building asset per land unit, and delta the demolition cost per unit of existing floor area. Here delta is a demolition-cost parameter, not the annual stock-loss rate in the first lecture.

Redevelopment is worthwhile in the simplified comparison if:

p - p0 > delta F0.*

New-building construction costs are already deducted in p*. Do not subtract them twice. Do not also add p0 as a second land cost if it already includes both the existing land and structure.

Under the further approximation p0 = F0(alpha0-beta F0), the comparison becomes:

F(alpha-beta F) - F0(alpha0-beta F0) > delta F0 + F*(mu+tau F*).**

That expanded expression assumes the old property’s value can be represented by its sale value per floor unit times existing floor area. In practice, p0 should reflect the actual asset value, including relevant income and option value. Alpha0 is the old structure’s demand intercept, which may be lower because of physical or functional obsolescence.

Example: New development supports residual p* = 400 per land unit. The existing property costs p0 = 350 to acquire, F0 = 1, and demolition costs delta = 30 per existing floor unit. The redevelopment surplus is 400-350-30 = 20 per land unit. At p0 = 380, it becomes -10, so replacement is unattractive under these inputs.

Obsolete structures and a large feasible increase in density can make replacement more attractive, but do not guarantee it. Financing, disruption, approvals, taxes, and the option to wait can alter the decision. Lumpy replacement helps explain redevelopment waves; this static inequality does not by itself determine their timing.

8. Highest house price does not mean highest land bid

Suppose two buyer groups compete for the same land. Group 1 will pay more for low-density housing but dislikes crowding strongly. Group 2 accepts greater density. Use the same cost parameters mu = 200 and tau = 50 for both.

Input or outcome	Group 1: strong density aversion	Group 2: more density tolerant
Alpha	800	700
Beta	150	25
Optimal FAR	1.5	3.3333
Sale price per floor unit at chosen FAR	575	616.67
Maximum residual per land unit	450	833.33

At a common low FAR of 0.5, Group 1 offers 725 per floor unit, above Group 2’s 687.50. Yet after each group’s density is optimised, Group 2 generates the higher land residual. The group willing to pay more at one common housing configuration need not be the group that supports the highest land bid.

Notice that the ranking of housing prices can itself reverse when FAR changes. The key comparison is each feasible project’s total revenue less non-land costs per land unit, rather than a single observed price per square metre. Different schedules can also produce a lower optimal housing price alongside a higher land bid; that outcome is possible, not required.

To see that latter case directly, retain Group 1 above and give Group 2 alpha = 600 and beta = 25. Group 2 then chooses FAR approximately 2.6667, sells at approximately 533.33 per floor unit, and supports land residual approximately 533.33 per land unit. Its housing price is below Group 1’s 575, while its land bid exceeds Group 1’s 450. More floor area makes the difference.

This extends the previous lecture’s sorting argument. Commuting preferences matter, but density preferences, lot consumption, and construction economics also determine who can outbid whom for a location. Competitive land value measures ability to pay under the model; it is not a complete measure of social welfare.

Feynman study lab

Explain it simply

A home is a package of features. Buyers value those features, while developers pay to supply them. More floor area creates more units to sell, but excessive density can lower the price buyers pay and raise construction costs. The best project leaves the most value for the land. An old building is replaced only if the new project’s value compensates for what must be bought or sacrificed and demolished.

Teach it back without looking

Explain the FAR 2 example to a beginner in two minutes. Define floor area, land area, average construction cost, and residual land value. Then answer:

Why can a project selling cheaper floor space afford to pay more for the same land?

Check your explanation after you try

It may create enough additional floor area, at acceptable construction cost, to produce a larger total residual per land unit. The final Group 2 example sells at about 533.33 per floor unit but supports land value of about 533.33 per land unit, exceeding Group 1’s 450 despite Group 1’s higher housing price of 575.

Transfer questions

1. A 2,000-square-metre site supports 5,000 square metres of floor area. What is FAR?
2. Let alpha = 700, mu = 300, beta = 25, and tau = 75. Find F*, P, C, and p*.
3. Under those inputs, what happens if a binding FAR limit is 1?
4. Does the interior squared-value formula apply if alpha = 250 and mu = 300?
5. If p* = 500, p0 = 440, F0 = 2, and demolition costs 35 per existing floor unit, should redevelopment proceed under the simple rule?
6. A log-linear bathroom coefficient is 0.1. Is the exact fitted increase 10%?

Check the transfer questions

1. FAR is 2.5.
2. F* = 2; P = 650; C = 450; p* = 400 per land unit.
3. F = 1 and residual value is 400-100 = 300 per land unit.
4. No. With nonnegative F and positive beta+tau, the model chooses F = 0 for this development option; compare with other uses separately.
5. No. Demolition costs 70, so the surplus is 500-440-70 = -10 per land unit.
6. No. It is 100[exp(0.1)-1], approximately 10.52%, for the fitted ratio with other variables fixed.

Find the gap and try again

If your explanation became vague, locate the gap: regression association versus causation, floor area versus land area, average spread versus marginal surplus, or redevelopment value versus sunk cost. Redo that example. Close the notes, explain it again, and change one input before predicting the result. Tomorrow, derive F* from the residual-value expression without looking and state when the interior formula fails.