When traffic is bad, the obvious solution seems simple: build more roads.

More roads mean more choice. More choice should mean faster journeys.

But does it always?

Every morning, 4,000 drivers travel from Start (S) to Finish (T). They have two possible routes:

Each road has a travel time:

For example, if 2,000 drivers use S → A, that stretch takes 20 minutes. If 3,000 drivers use B → T, that stretch takes 30 minutes.

The diagram is showing a road network: S to A takes drivers ÷ 100 minutes, A to T 45 minutes, S to B 45 minutes, B to T drivers ÷ 100 minutes. A dashed new road from A to B takes 0 minutes.

Assume every driver wants to minimise their own journey time, knows how long each road takes, and can choose their route freely.

Q1: What happens before the new road is built?

Let x drivers take the upper route, and the rest take the lower route.

Write down the travel time on each route as a function of x. How will the 4,000 drivers divide themselves between the two routes? How long does each journey take?

Q2: What happens when the city builds a shortcut?

The city opens a new one-way road from A to B, which takes zero minutes to travel. Drivers now have three possible routes:

Everyone still chooses whichever route minimises their own journey time.

Where will the drivers end up, and how long will their journeys take?

Q3: What has gone wrong?

Compare the outcomes in Q1 and Q2. Why hasn’t the extra road made journeys faster?

What is the difference between what is best for an individual driver and what is best for all drivers together?

Q4: What would a planner do?

Suppose the shortcut stays open, but a traffic planner can tell each driver which route to take, to minimise the total time spent travelling by all 4,000 drivers.

Should the planner send anyone along the shortcut? If so, how many? How does the total compare with Q1 and Q2? Is the planner’s plan fair to every driver?

Q5: Can a toll fix it?

Instead of giving orders, the city charges a toll for using the shortcut. Suppose drivers treat the toll like extra minutes of travel time.

How large would the toll need to be for drivers’ own choices to match the planner’s plan in Q4? What other policies might help? Can individually rational decisions be relied on to produce the best outcome for society?

Bonus: Is this really about roads?

Think about internet connections, electricity grids, supply chains or organisations.

Can you imagine a situation where adding a new connection makes the whole network perform worse? What would have to be true about how people use the network for that to happen?

P.S. Roll up your sleeves and explore the questions. Once you’ve grappled with them for a while, if you’re still stuck, there are graduated hints below, but challenge yourself first! (The answer comes in the next issue. Subscribe to get it straight to your inbox.)

That's how many megacities — urban areas with at least 10 million inhabitants — existed worldwide in 2025, according to the United Nations.

In 1975, there were just eight. Today, 19 of the 33 are in Asia. The world's largest is Jakarta, with nearly 42 million people.

Why do cities keep getting bigger?

Economists call it agglomeration economies. When people and businesses cluster together, they benefit from larger labour markets, specialised suppliers, shared infrastructure and the exchange of ideas. Proximity makes people more productive. A large city is more than simply a collection of people; the connections between them create economic value.

But there's a catch.

As cities expand, housing becomes more expensive, roads more congested, pollution more severe and infrastructure increasingly stretched. These are diseconomies of agglomeration.

And there's an important distinction between what makes sense for an individual and what makes sense for society. One more person moving to a city may benefit from its opportunities, while imposing additional congestion and housing pressures on everyone already there.

That raises a deceptively simple question:

If bigger cities generate greater productivity, is there a point at which adding more people makes everyone worse off?

Last time, three students arriving at university could become friends, at a cost, and we asked: who pays to be the person in the middle?

Start by listing the friendships that could exist. With three students, there are only three possible pairs:

Think of each friendship as a light switch: it’s either on (they’re friends) or off (they’re not). A network is just a setting of the three switches.

Now count the settings:

So there are 8 possible networks. Here they all are (✓ = friends, ✗ = not friends):

These really are 8 different friendship groups. Network 5 is a different social world from network 7: in one, Asha and Cara don’t know each other; in the other, Asha and Ben don’t.

But some of them have the same shape. Networks 2, 3 and 4 each have one friendship, with one student left out. Networks 5, 6 and 7 each have one student who is friends with both others, while those two aren’t friends. We’ll call this shape a line, because you can draw it as one: for example, Asha – Ben – Cara, with Ben in the middle.

From now on, we’ll write a line as A–B–C, meaning A and B are friends, B and C are friends, but A and C are not. The student in the middle is the one with two friends.

So the 8 networks fall into 4 shapes:

(Check: 1 + 3 + 3 + 1 = 8.)

Start with the line A–B–C: Asha and Ben are friends, Ben and Cara are friends, but Asha and Cara are not. Ben is in the middle.

Now the other two lines:

These are three different networks: a different pair aren’t friends in each, and a different student is in the middle. But in our model they happen to give the same total, because every friendship is worth the same and costs the same, whoever it’s between.

The other shapes:

So no single network is best: the three lines tie, each with a total of 0.9. But notice who does worst in each: the student in the middle, holding the network together, earns less than either of the students on the edge. The students on the edge get their friend-of-a-friend benefit for free. The one in the middle pays for two friendships.

Take the line A–B–C, and check each possible move:

So nobody wants to change anything. Economists call a network like this pairwise stable: no one wants to end a friendship, and no two people both want to start one. By the same reasoning, the other two lines are stable too.

But could the students get stuck in one of the other shapes instead? Check each one:

Where do the students end up?

The triangle has one friendship too many; the other shapes have too few.

So at a cost of 0.4, a line is the only stable shape: wherever the students start, they end up in one. Individual choices lead the group to the best outcome. But the theory can’t tell us which line forms: in other words, who ends up as the hub. And since the middle is the worst-paid position, nobody is keen to volunteer.

With c = 0.55, the line’s total becomes 0.5 + 0.5 + 0.75 + 0.75 − 4 × 0.55 = 0.3. The triangle’s total is now −0.3 and a single friendship gives −0.1. So the lines are still best for the group.

But look at Ben, in the line A–B–C. His payoff is now 1 − 1.1 = −0.1:

The only stable network is the empty one. The students end up with no friendships at all, even though a line would give the group a total payoff of 0.3.

This is the central idea of the problem. What is best for the group is not what individuals choose, because the hub bears the costs while the students on the edge enjoy the benefits. Asha and Cara each gain an indirect connection through Ben, without compensating him for the cost of maintaining those links. This is a positive network externality.

Start small, with 4 students: one hub (H) and three on the edge (call them 1, 2 and 3).

In the star, the hub is friends with 1, 2 and 3, and there are no other friendships.

The complete network is the star plus three extra friendships: 1–2, 1–3 and 2–3.

So instead of working out every payoff again, ask: what does one extra friendship do?

Take the friendship between 1 and 2:

Alt text: Before: a star with hub H linked to students 1, 2 and 3; 1 and 2 are connected only through the hub. After: the same star plus a direct friendship between 1 and 2. Each of 1 and 2 gains δ − δ² and pays c; nobody else is affected.

So each extra friendship changes the group’s total by:

2(δ − δ² − c)

It’s worth adding if δ − δ² > c, and not worth adding if c > δ − δ².

As long as the hub’s friendships remain in place, every extra friendship between two students on the edge has exactly this effect, because every such pair starts out as friends-of-friends through the hub. So either all of them are worth adding, or none of them are.

That gives the answer: the star beats the complete network if and only if c > δ − δ².

Check it against Q2. With three students, the line A–B–C is a star (Ben is the hub), and the triangle is the complete network: the line plus one extra friendship (Asha–Cara). With δ = 0.5 and c = 0.4, that friendship changes the total by 2(0.5 − 0.25 − 0.4) = −0.3. And indeed the line gives 0.9 and the triangle 0.6. ✓

Why does n disappear? Because the size of the group only changes how many extra friendships there are, not what each one does. With n students, there are n − 1 on the edge, so there are (n − 1)(n − 2) ÷ 2 extra friendships. Each changes the total by 2(δ − δ² − c), so:

W_complete − W_star = (n − 1)(n − 2)(δ − δ² − c)

For n ≥ 3, the factor (n − 1)(n − 2) is always positive, so it never changes which network wins. Only the sign of δ − δ² − c matters.

(For the very keen: you can also get this by adding up every student’s payoff in each network. The hub in the star gets (n − 1)(δ − c); each of the n − 1 students on the edge gets δ − c + (n − 2)δ²; in the complete network, each of the n students gets (n − 1)(δ − c). Subtract, and the same formula drops out.)

Last part: what’s the highest cost at which the complete network could ever be best?

The complete network can only be best if c ≤ δ − δ². So we need the largest possible value of δ − δ². Call it f(δ) = δ − δ²:

If a friendship costs more than ¼, the complete network can never be best for the group, whatever the value of δ.

Why is the peak in the middle? Remember that δ − δ² is the gain from turning a friend-of-a-friend into a direct friend:

This is the connections model of the economists Matthew Jackson and Asher Wolinsky (1996), one of the founding models of network economics.

The hub loses c − δ on each of its n − 1 friendships. So each student on the edge must pay at least t = c − δ for maintaining their friendship.

After paying that fee, each student on the edge gets:

δ − c + (n − 2)δ² − (c − δ) = 2(δ − c) + (n − 2)δ²

This is non-negative exactly when W_star ≥ 0, which is when:

c ≤ δ + (n − 2)δ²/2

So whenever the star creates non-negative total value, transfers can, in principle, make it worthwhile for everyone to participate. But agreeing on those payments may itself be difficult.

There are plenty of real-world examples:

Would real students form the star? Laboratory experiments suggest it’s harder than the theory implies. Goeree, Riedl and Ule (2009) found that star networks were much harder to establish among identical participants, but formed frequently when participants differed in their costs or benefits.

In our model, the hub pays most and gains least: it bears the cost of every friendship, while the people on the edge enjoy connections they don’t pay for. When friendships are cheap, the network survives anyway. When they get expensive, the hub walks away, and a network that would have benefited everyone never forms.

Connections create value for people who don’t pay for them. That gap between private incentives and social value is why we see hubs charging fees, platforms taking commission and governments subsidising transport links.

The Human Network (Matthew O. Jackson, 2019): an accessible introduction to network economics by one of the field’s founders.

Marginal gains… Answer for today’s question is out in two days… Come back then or subscribe to get it direct to your inbox!

P.P.S.

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