The Gold Standard of Urban Systems | Why an Added Link Can Raise Equilibrium Journey Time

Imagine a map with a start S, a destination T and two intermediate points A and B. Initially, travellers can follow S–A–T or S–B–T. A proposal adds a one-way connection from A to B. That creates a third route, S–A–B–T. Nothing in the example allows travel from B back to A.

The proposal appears generous. Travellers retain the original choices and gain another one. If a coordinator could preserve yesterday’s route allocation, yesterday’s journey times would remain feasible. Yet a traveller choosing their own route may find the newly available alternative attractive. Once many people make that choice, the link loads change.

There are therefore two questions. What allocations can the network accommodate? Which allocation is stable under the stipulated route-choice behaviour? Adding a link enlarges the first set without guaranteeing a better answer to the second question. This distinction is the centre of the paradox.

Draw the three routes on paper and name the links they share. The upper original route and the new route both use S–A. The lower original route and the new route both use B–T. That sharing is the mechanism we will calculate. A shortcut that no one shares with another route would describe a different problem.

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Let the total demand be four normalised flow units over a fixed modelled period. A flow unit is a chosen scale for the calculation, not one indivisible driver. The example treats travellers as a continuous population, so the effect of one traveller changing route is negligible relative to the total flow.

Two links have flow-dependent journey times. If S–A carries q units, its time is 6q minutes. If B–T carries q units, its time is also 6q minutes. The other two original links, A–T and S–B, each take 30 minutes regardless of their assigned flow in this model.

The added A–B link takes zero minutes. This is a simplifying assumption that isolates the route-choice effect; it is not a claim that a physical road takes no time. All link times are nonnegative on the allowed domain. Route time is the sum of its link times.

Keep the functions visible beside the drawing. Writing 6q is different from labelling a link “six minutes”. At q = 2, that link takes twelve minutes; at q = 4, it takes twenty-four. A common mistake is to carry yesterday’s time into tomorrow’s changed flow allocation. The function remains the same while its input changes.

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First leave A–B unavailable. Let a units take S–A–T and the remaining 4 − a take S–B–T. The upper route takes 6a + 30 minutes. The lower route takes 30 + 6(4 − a) minutes.

If both routes are used at an equilibrium, neither can be quicker than the other. Otherwise an infinitesimal traveller on the slower route could improve by switching. Equating the times gives 6a + 30 = 30 + 24 − 6a. Hence 12a = 24, so a = 2.

Each original route therefore carries two units and takes 42 minutes. The demand check is 2 + 2 = 4. The link-time check is 6 × 2 = 12 on each variable link. Each full route combines that twelve-minute component with a thirty-minute component.

An endpoint allocation is not stable here. If everyone used the upper route, its time would be 54 minutes, while the unused lower route would offer 30 minutes to an infinitesimal switcher. The equal split follows from the stated incentives and functions. We have not assumed that travellers generously divide themselves into two equal groups.

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Now add A–B but initially keep the two-and-two allocation. S–A and B–T still each carry two units, so the old routes still take 42 minutes. The new route would combine twelve minutes on S–A, zero on A–B and twelve on B–T, giving 24 minutes.

That unused alternative defeats the old allocation as an equilibrium. A traveller can improve by switching from a 42-minute route to a route that currently takes 24 minutes. The possibility matters even though the flow on A–B was initially zero.

Do not conclude that everyone will continue to enjoy 24 minutes. That number belongs to the old link loads. The people attracted to the shortcut also join the two flow-sensitive links. Their choices change the inputs to both journey-time functions.

This is a useful point to stop the calculation and explain it aloud. Which times were evaluated before anyone switched? Which links gain traffic when someone uses S–A–B–T? If those answers are unclear, another page of algebra will hide the confusion. Marking the shared links gives the next calculation a physical interpretation.

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Consider the assignment in which all four units take S–A–B–T. S–A now carries four units, as does B–T. Each variable link takes 6 × 4 = 24 minutes. The shortcut route therefore takes 24 + 0 + 24 = 48 minutes.

At these same link loads, the upper original route takes 24 + 30 = 54 minutes. The lower original route takes 30 + 24 = 54 minutes. An infinitesimal traveller on the used route cannot improve by switching to either original alternative.

The all-shortcut assignment is thus an equilibrium under the model. Its 48-minute journey is six minutes longer than the original equilibrium’s 42 minutes. The increase is 6/42, approximately 14.3%. All demand is still served; the comparison concerns journey time rather than an increase in the number of trips.

Notice the simultaneous comparison. We evaluate each possible route using the link loads of the candidate equilibrium. We do not compare a new congested route with an old route’s uncongested historical time. The consistent calculation explains how a new option can attract everyone even though the resulting equilibrium leaves everyone worse off than before.

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A compact ledger makes the shared loads easier to audit. Call the flow on the upper original route x, the lower original route y, and the shortcut route z. They satisfy x + y + z = 4, with all three nonnegative.

The flow on S–A is x + z. The flow on B–T is y + z. The fixed-time links carry x and y respectively, while A–B carries z. A route flow belongs to one complete path; a link flow can combine traffic from several paths.

The three route times are therefore U = 6(x + z) + 30, L = 30 + 6(y + z), and C = 6(x + z) + 6(y + z). These formulas use the same underlying link loads. They cannot be independently adjusted as if each route had its own private copy of the shared roads.

Read each row horizontally before calculating times. In the last row, the shortcut does not carry all the traffic on either variable link. Adding route flows and adding link flows are related operations with different meanings.

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For a candidate assignment, first check conservation and nonnegativity. Then calculate every available route’s time from the same link loads. Each used route must have the minimum available time. An unused route may be slower or tied, but it must not offer a strict improvement.

The all-shortcut row passes: its used route takes 48 minutes, while both unused originals take 54. The original equal split fails once the link is open: its used routes take 42 minutes, while the unused shortcut offers 24. That is enough to distinguish feasibility from equilibrium.

We can also establish why the all-shortcut assignment is the only equilibrium for this four-unit example. Across all feasible assignments, x + z ≤ 4 and y + z ≤ 4. Consequently, each variable-link time is at most 24 minutes, below the fixed alternative’s thirty minutes.

If any traveller uses the upper original route, replacing its A–T segment with A–B–T offers the currently shorter B–T segment instead. If any uses the lower original route, the alternative S–A–B replaces its thirty-minute S–B segment with the currently shorter S–A. Thus neither original route can carry positive flow at equilibrium. All four units must use the shortcut.

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The equilibrium rule is a mathematical description of incentives. It does not call a traveller inconsiderate or imply that a real person thinks through these equations. Each traveller is assumed to prefer a lower modelled journey time, with the same origin, destination and available routes.

A person considering a switch compares alternatives at the current aggregate flow. In the continuous-population model, their own contribution is negligible. This is why the comparison uses the current link times. A finite game with a handful of large vehicles could require a different switching calculation.

The model also assumes the relevant information and route access. Real travellers may care about fares, reliability, transfers, accessibility, familiarity, childcare or whether a route is actually permitted. A forty-minute route is not automatically preferable to everyone simply because another takes forty-two minutes.

Keep those limits attached to the interpretation. The example demonstrates that individually improving choices need not produce the smallest total journey time. It does not explain every observed congestion pattern or rank the motives of road users. The benefit of the model is a precise question about feedback, not a moral label for the people in the network.

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An average can conceal the accounting unless we first calculate the total. Before the new link, four units each experience 42 minutes. Total modelled journey time is 4 × 42 = 168 flow-unit minutes. After the new equilibrium forms, it is 4 × 48 = 192.

The increase is 24 flow-unit minutes. Dividing by the unchanged four units gives the same six-minute increase per traveller. These are scaled quantities because flow was normalised. Converting them into actual person-minutes would require the number of travellers represented by one unit.

For a separate scaling interpretation, suppose one flow unit represents one hundred travellers in the modelled period. The four units then represent four hundred travellers. The equilibrium increase of twenty-four flow-unit minutes corresponds to 2,400 person-minutes, while dividing by four hundred still gives six minutes per traveller. State the chosen scaling rather than presenting the normalised total as an observed population quantity.

Another check adds time over links. At the all-shortcut assignment, S–A contributes 4 × 24 = 96 flow-unit minutes and B–T contributes another 96. The zero-time connector contributes zero, giving 192. This agrees with the route-based account.

This double calculation catches a common error: adding three route times without weighting them by route use. Unused routes do not contribute experienced journey time merely because their times can be evaluated. The route ledger and the link ledger should agree when they describe the same assignment, demand and cost definition.

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Opening a link does not erase the original routes. With the link open, the old two-and-two allocation still produces a total of 168. It is feasible even though it is no longer stable under the stated individual route-choice rule.

We can do slightly better in this particular example by assigning x = 1.5, y = 1.5 and z = 1. Each variable link carries 2.5 units and takes fifteen minutes. Each original route takes 45 minutes, while the shortcut route takes thirty.

The experienced total is 1.5 × 45 + 1.5 × 45 + 1 × 30 = 165 flow-unit minutes. Dividing by four gives 41.25 minutes. The open network therefore admits an allocation better than both the original equilibrium and the new equilibrium.

That allocation is not a user equilibrium: someone on an original route would prefer the currently thirty-minute shortcut. Its purpose is to show the distinction cleanly. A better coordinated outcome exists, but the assumed individual incentives do not hold it in place. The paradox concerns the equilibrium selected by behaviour, not a physical impossibility of using the old routes.

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For readers ready for a deeper check, total time at any feasible assignment is G = 6(x + z)² + 6(y + z)² + 30(x + y). The square appears because flow q experiences a time 6q, contributing 6q² to the total.

For a fixed shortcut flow z, we have x + y = 4 − z. The two squared terms are smallest when x = y. One way to see this is that, for a fixed sum, the sum of two squares increases with the square of their difference. Unequal sharing loads one variable link more heavily.

Substitute x = y = (4 − z)/2. Each variable link then carries (4 + z)/2. The total becomes G = 3(4 + z)² + 30(4 − z), or 168 − 6z + 3z². Completing the square gives G = 165 + 3(z − 1)².

The square cannot be negative. Its minimum is zero at z = 1, which lies inside the allowed interval 0 ≤ z ≤ 4. Thus the coordinated minimum is exactly 165, achieved at x = y = 1.5 and z = 1. We have supplied both an impossibility argument for anything lower and an assignment attaining the bound.

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The same map does not imply the same result at every demand. Replace four by a positive demand D while keeping the functions 6q and thirty minutes. Without the shortcut, the equal split gives each route a time of 30 + 3D.

When D < 5, each variable-link time is at most 6D < 30. The same route comparison as before makes the all-shortcut assignment the equilibrium. Its time is 12D. At D = 5, the all-shortcut assignment still satisfies the equilibrium inequalities, with ties in the alternative comparisons.

Within this range, the shortcut increases equilibrium time when 12D > 30 + 3D. Rearranging gives 9D > 30, or D > 10/3. Below that threshold it improves time; at the threshold the two times agree.

Do not extend the all-shortcut formula to every larger demand. Beyond five units, a variable link can become slower than the thirty-minute alternative, and the equilibrium conditions must be reconsidered. A correct formula has a domain, not just a memorable shape.

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In the four-unit comparison, demand is held fixed. The example does not require more trips to appear after the road opens. The increase follows from changed route allocation among the same stipulated flow units.

Induced demand concerns changes in travel demand in response to changed conditions. Departure-time changes, mode changes and destination choices can also matter in real networks. Those processes may coexist with route-choice feedback, but they are not needed for this demonstration.

A useful explanation names which quantities the model holds fixed. Here they include total demand, origins and destinations, link functions, route availability apart from the new connector, and the journey-time objective. Behaviour changes through route choice inside that boundary.

If a real study finds more traffic and longer journeys after an opening, the timing alone does not identify this particular mechanism. Demand growth, construction, incidents and other changes may offer competing explanations. The paper network gives a testable conceptual possibility. Establishing what happened at an actual site requires appropriate data and a model suited to that site.

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Suppose an exercise adds a fifteen-minute-equivalent perceived charge to the shortcut route. This is a defined cost used for route choice; it is not fifteen minutes of extra physical travel. At the coordinated assignment, original routes take 45 minutes, while the shortcut takes thirty plus fifteen in perceived cost.

All three perceived costs are then 45. Under this modified choice rule, the coordinated assignment satisfies the equilibrium condition. The physical journey-time total remains 165 because the exercise did not change the travel-time functions.

The example shows how changing incentives can affect which feasible allocation is stable. It does not establish a suitable real toll, permission system or transport policy. A monetary payment would need a stated conversion to perceived time, and travellers need not share one valuation.

Also keep objectives separate. A payment is not automatically a physical delay or a destroyed social resource. If the exercise asks for total travel time, adding payments to that total changes the quantity being measured. A real appraisal must state how costs, revenue, distribution and access enter the decision. The algebra can expose the question while leaving the actual policy assessment to the relevant evidence.

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“More choices cannot hurt because everyone can keep their old route” proves that an old assignment remains feasible. It does not prove that travellers will continue to choose it. Repair the sentence by naming the behavioural rule and recalculating the candidate equilibrium.

“The shortcut is zero minutes, so the new journey must be quicker” looks at one component while ignoring its shared approaches. Repair it by writing the full path and the new flows on S–A and B–T. Zero connector time does not imply zero cost of reaching and leaving it.

“The best possible total got worse” confuses equilibrium with optimisation. The open network permits the old assignment and also the 165-unit coordinated minimum. Repair the claim to say that the modelled user-equilibrium total increased from 168 to 192.

“Therefore removing roads always improves traffic” converts one conditional counterexample into a universal rule. Repair it with the demand comparison: at D = 2, this very connector improves equilibrium time. The value of the example is that a local intuition is insufficient. The next useful action is a whole-network analysis under stated conditions, not automatic enthusiasm for either adding or removing a link.

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Begin with the proposed outcome: which travellers, journeys and periods should improve? Distinguish journey time from vehicle throughput, person throughput, reliability, accessibility and environmental effects. A result about one measure should retain its name when reported.

Next identify the relevant route alternatives and shared links. The exercise’s four nodes were deliberately simple. A real network may involve junction capacity, queues that block other movements, signal timing, public transport and route restrictions. Its time functions require evidence rather than convenient coefficients copied from a teaching example.

Ask how route choice and demand are represented. What observations support those assumptions? Has the model been checked against conditions separate from those used to set its parameters? Which uncertainties could change the comparison? An impressive-looking map does not answer these questions by itself.

Finally, preserve the before-and-after boundary. Compare matching periods and state other changes affecting the network. School learners can analyse supplied datasets and published diagrams. The useful educational task is to interrogate assumptions and reproduce a calculation; operating traffic controls or collecting observations in hazardous road positions is unnecessary for this lesson.

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Start by reading link functions at a fixed flow. Ask a learner what 6q means at q = 0, 2 and 4. If they reply six minutes in every case, repair function reading before introducing equilibrium.

Then give the route ledger with x, y and z already supplied. Ask for the two shared-link loads and the three route times. Keep conservation visible. This separates arithmetic practice from the later decision about which flows are stable.

Next remove the equilibrium answer and ask the learner to test two candidate allocations. They should identify a profitable alternative at the original split after the connector opens, and show why none exists at the all-shortcut assignment. Reduce prompts once they explain the comparison reliably.

For extension, ask for a coordinated assignment and an argument that it is optimal. The completed-square calculation provides one route. A learner who is not ready for that algebra can still understand the feasible 165-unit comparison and accurately refrain from claiming a proved minimum. Progress means matching the conclusion to the proof actually supplied, then transferring the distinction to another demand.

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Use a new network with the same directed shape but different coefficients. The two variable links now take 4q minutes, the fixed links take twenty-four minutes each, the connector takes zero, and demand is five normalised units.

First find the equilibrium without the connector. Then test the all-shortcut assignment after it opens. Calculate the experienced total in both cases. Finally determine whether the open network admits a coordinated assignment with a lower total than the original equilibrium.

Work through the route and link ledgers before reading the answer. State why a candidate is feasible and why it is or is not an equilibrium. If you calculate an average, write the total that it comes from. Those steps make the reasoning visible enough to correct.

A useful second variation lowers demand to two units while keeping the new coefficients. Decide whether the connector still harms equilibrium time. This extra change prevents the exercise from becoming a rule that every connector must be harmful. The mechanism depends on both the network and the demand acting through it.

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Without the connector, the five units split equally, producing 24 + 4 × 2.5 = 34 minutes per route and a total of 170. With all five on the connector route, each variable link takes twenty minutes. The used route takes forty, while each original alternative takes forty-four. This is an equilibrium, with total 200.

For a coordinated assignment, symmetry gives x = y = (5 − z)/2. Total journey time is 2(5 + z)² + 24(5 − z), which simplifies to 170 − 4z + 2z² = 168 + 2(z − 1)². Its minimum is 168 at z = 1 and x = y = 2. Each original route then takes 36 minutes and the shortcut takes 24, giving 2 × 36 + 2 × 36 + 24 = 168.

At demand two, the original equilibrium takes 28 minutes and the all-shortcut route takes sixteen. The connector improves the result. Its original alternatives take thirty-two at that assignment, so the equilibrium check passes.

If your five-unit shortcut time was twenty, you counted only one shared variable link. If your optimum total was 144, you omitted the shortcut users or misweighted the original routes. Repair the ledger at that specific point, then use a different demand to check independent understanding.

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The zero-minute connector is useful for exposing the mechanism, but the effect does not depend on that exact value. Keep demand at four and give A–B a fixed journey time c. At an all-shortcut assignment, the used route takes 48 + c minutes, while each original alternative takes 54. Thus this assignment remains an equilibrium when 0 ≤ c ≤ 6.

For example, c = 4 gives an equilibrium journey of 52 minutes, still above the original forty-two. The connector now consumes real modelled travel time, yet the shared-link feedback remains. This variation answers the reasonable objection that a physically impossible zero-minute link might be doing all the explanatory work.

The domain matters again. At c = 9, putting everyone on the shortcut would produce 57 minutes against an original alternative of 54, so that candidate fails. A symmetric mixed assignment instead has z = 3 and x = y = 0.5. Each variable link carries 3.5 units and takes twenty-one minutes. Original routes take 51; the shortcut takes 21 + 9 + 21 = 51. All used routes tie, so the equilibrium check passes.

Within the mixed range, the equality can be derived rather than guessed. With x = y = (4 − z)/2, original-route time is 42 + 3z. Shortcut time is 24 + 6z + c. Equating them gives z = (18 − c)/3. This lies between zero and four when 6 ≤ c ≤ 18. At c = 18, the connector can be unused and the original split again satisfies the condition.

These are model comparisons across different networks, not instructions to slow a real connection. They demonstrate why the effect of a link cannot be read from its travel time alone. Changing that time changes route choices, which then change shared congestion. Even a directionally sensible statement such as “make this segment quicker” needs the rest of the model when the intended claim concerns everyone’s final equilibrium journey.

A practical learner check is to distinguish three quantities: the connector’s own time, a complete route’s time at a specified assignment, and the equilibrium time after choices adjust. Writing all three on separate lines prevents a true statement about one segment from becoming an unsupported statement about the complete network.

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Does equilibrium mean the network is doing its best?

No. It means no infinitesimal traveller can improve their own stipulated cost by switching alone. The coordinated objective considers total cost across travellers. The 192-unit equilibrium and 165-unit minimum demonstrate the distinction within the same open network.

Does the paradox require demand growth?

No. The main comparison holds demand at four units. Changed demand is another possible feedback in real transport, but it is not needed for this mathematical example.

Must every unused route be slower at equilibrium?

It may be tied. What matters is that it cannot offer a strict improvement to a traveller on a used route. The demand-five boundary illustrates why equality should be kept separate from a strict inequality.

Can a navigation app guarantee the coordinated result?

Nothing in this model establishes such a guarantee. Recommendations, compliance, information and the objective being optimised would need their own assumptions and evidence. A proposed route is part of the system when many people respond to it.

What is the smallest reliable explanation?

Name the shared links, recalculate their flows, test all available routes at the same assignment, and distinguish a stable individual-choice outcome from the best feasible total. If one of those steps is missing, return to it before drawing a wider conclusion.

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A compact report can still retain the full logic. Write the demand and model first, then give the original equilibrium, the new equilibrium and the separate coordinated minimum. State that the comparison holds demand fixed. That order prevents the word improvement from referring to three different objectives within one paragraph.

For the main example, the report can say that the new connection preserves the old feasible allocation but changes individual incentives. Its equilibrium rises from forty-two to forty-eight minutes, while the open network’s minimum average is 41.25. These are compatible results because equilibrium and coordination impose different conditions. A reader should be able to find the route ledger supporting each number.

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For the wider city system, return to The Gold Standard of Urban Systems. For a different network question, Testing Neighbourhood Service Demands Against Shared Route Capacity tests whether destination demands can be jointly delivered; capacity feasibility is distinct from this route-choice equilibrium problem.

For traffic quantities and their interpretation, read Traffic Flow, Density and Moving Jams. For the interaction between traffic control and behaviour, continue with Making Traffic Lights Cooperate Instead of Compete. These existing explanations provide the broader setting for the worked equilibrium ledger developed here.

The primary historical source is Braess, Nagurney and Wakolbinger’s English translation of On a Paradox of Traffic Planning, published in Transportation Science in 2005 from Braess’s original work. The network arithmetic, demand comparisons and transfer exercise above are independently developed teaching cases. The source was reviewed on 22 October 2026.

Continue with Learning That Transfers into Life for the educational purpose, or How X Works for another system. Carry forward one practical habit: when an intervention changes people’s choices, recompute the relationships those choices depend on.

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