A volume count at a point on a road records 960 vehicles per hour. Which of the following can be concluded?

Nothing about how well the road is operating — the same flow occurs both in free flow and in congestion

The road is operating below capacity, since capacity for a typical lane exceeds 960 veh/h

The density can be found directly, since flow determines density uniquely

The road is congested, because 960 veh/h in a single lane implies queuing

Which of the following statements about traffic studies and signals are correct?

The 85th percentile spot speed is taken as the safe or design speed for a road

An origin–destination study can justify a bypass in a way that a volume count cannot

Webster’s optimum cycle length increases without limit as the sum of critical flow ratios approaches unity

The 85th percentile spot speed is taken as the safe or design speed for a road

An origin–destination study can justify a bypass in a way that a volume count cannot

Webster’s optimum cycle length increases without limit as the sum of critical flow ratios approaches unity

Regulatory traffic signs are triangular and warning signs are circular

A junction is signalled and Webster’s method returns an optimum cycle length of 230 s. The correct engineering response is to

treat it as a signal that the junction is near saturation and add capacity or reduce demand, rather than implement it

implement the 230 s cycle, since it is by definition the delay-minimising value

halve the lost time, since L is the only variable the designer controls

add a third phase, which will reduce Y and shorten the cycle

Traffic Engineering: Studies, Flow Theory, Capacity and Signal Design

Traffic engineering is the part of this paper where the same measured number can describe two opposite conditions, and getting used to that is most of the subject. A count of 960 vehicles an hour past a point is compatible with a road running freely at 48 km/h and with the same road crawling at 12 km/h — the worked numbers below show both, on one relationship — because flow is a product of density and speed, and a product does not tell you its factors. That single fact reorganises everything else. It is why a traffic study measures speed and density and not just volume; why capacity turns out to be a maximum rather than a limit, occurring at a middling density with everything above it worse in both directions; why congestion is not a bigger version of free flow but the other branch of the same curve; and why a junction that is already at capacity cannot be improved by any signal timing at all. The second thread is that traffic is stochastic — arrivals are random, drivers differ, and no design can be right for every minute — so the tools are statistical: a design hour rather than a peak minute, an 85th-percentile speed rather than a maximum, and a signal cycle that minimises average delay while accepting that some cycles will overflow. Reading the subject that way makes Webster’s formula intelligible rather than arbitrary: it is the cycle length that minimises expected delay, and its behaviour near saturation is the honest warning built into it.

Traffic studies, and the two speeds that are not the same

Volume study — vehicles past a point, classified by type and converted to PCU so that a bus and a scooter can be added. Its output is the design hour volume, conventionally the 30th highest hour of the year rather than the highest, because designing for the single worst hour of 8760 is not economic.
Speed study — spot speeds at a point, from which the 85th percentile is taken as the design or safe speed, and the 15th as the lower limit for a minimum-speed sign. The 85th percentile is used because most drivers are reasonable, so the speed that 85% do not exceed is a better statement of what the road actually invites than any legal number.
Origin–destination study — where trips begin and end, by roadside interview, licence-plate matching or the return-postcard method. It answers a question no volume count can: whether the traffic on a road wants to be there, which decides between widening the road and building a bypass.
Parking and accident studies — the first gives accumulation and turnover; the second gives rates, normalised per million vehicle-kilometres so that a busy road and a quiet one can be compared honestly.

⚠️ Time mean speed and space mean speed are different quantities

Measure the speeds of three vehicles as 30, 40 and 50 km/h. The time mean speed — the arithmetic average of speeds observed at a point — is (30 + 40 + 50)/3 = 40.00 km/h. The space mean speed — the average over a length of road, which is the harmonic mean — is 3/(1/30 + 1/40 + 1/50) = 38.30 km/h. They are not the same, the space mean is always the smaller, and the gap widens with the spread of speeds: here it is 4.44%. The reason is a sampling bias, and it is worth seeing clearly. A point observer counts every vehicle once, so fast vehicles are represented in proportion to how many there are. An observer looking at a stretch sees a snapshot in which slow vehicles occupy the road for longer and are therefore over-represented in the sample at any instant — so the average over space is pulled down. This matters because only the space mean speed belongs in q = kv. Substituting the time mean overstates the flow, and a question that gives you spot speeds and asks for flow is testing exactly that step. One more caution: the often-quoted relation v_t = v_s + σ²/v_s is an approximation, not an identity — for these three speeds it gives 40.11 rather than 40.00 — so use it to estimate, never to convert an answer you are expected to compute exactly.

Flow theory: one product, two branches

Traffic flow theory rests on three quantities that describe a stream and are tied by one identity: flow q in vehicles per hour, density k in vehicles per kilometre, and space mean speed v, with q = kv. Any model of traffic is a statement about how two of them are related, and the simplest is Greenshields: speed falls linearly with density, v = v_f(1 − k/k_j), from the free-flow speed v_f at an empty road to zero at the jam density k_j. Substituting into the identity gives q = v_f k(1 − k/k_j) — a parabola in density, and its shape is the whole subject.

Greenshields with v_f = 60 km/h and k_j = 100 veh/km
Density k	Speed v	Flow q = kv	State of the road
20 veh/km	48 km/h	960 veh/h	Free flow — comfortable, well below capacity
50 veh/km	30 km/h	1500 veh/h — capacity	The maximum the road can pass, at half the free speed
80 veh/km	12 km/h	960 veh/h	Congested — the same flow, a quarter of the speed

🎯 Capacity is a maximum, not a ceiling — and flow alone is ambiguous

Look at the first and third rows. The same flow of 960 veh/h describes two completely different roads — one running at 48 km/h with room to spare, one crawling at 12 km/h and about to break down. A volume count alone cannot distinguish them, and that is why a traffic study measures speed or density as well, and why "the road carries 960 vehicles an hour" is not by itself a statement about how well it is working. The peak of the parabola is the capacity, q_max = v_f k_j/4 = 1500 veh/h, and it occurs at k = k_j/2 and v = v_f/2 — capacity is reached at half the free-flow speed, not near it. Three consequences follow, and each is examined. First, a road at capacity is at the top of a hill: any further increase in density moves it onto the falling branch, where flow decreases as density rises. Second, that branch is unstable — this is why a jam, once formed, does not simply clear when the extra demand goes away, and why flow measured during recovery differs from flow at the same density during build-up (the hysteresis real data shows). Third, and most usefully, it explains ramp metering: deliberately holding vehicles back at an entry keeps the mainline density near k_j/2 instead of letting it slip past, so restricting inflow can increase throughput. Greenshields itself is a crude model — real speed does not fall linearly with density — but every one of these conclusions survives in the better models, because they follow from the parabola having a peak rather than from its exact shape.

Signal design by Webster’s method

A signalised junction has a fixed amount of time to give away each cycle, and some of it is unavoidably lost: the start-up lost time while a queue accelerates, and the all-red clearance between phases. Webster derived, from a delay model with random arrivals, the cycle length that minimises average delay per vehicle: C₀ = (1.5L + 5)/(1 − Y), where L is the total lost time per cycle and Y = Σy_i is the sum of the critical flow ratios — the ratio, for the worst approach in each phase, of the arriving flow to the saturation flow. The effective green is then shared out in proportion: g_i = (y_i/Y)(C₀ − L).

A worked two-phase design: L = 12 s, y₁ = 0.35, y₂ = 0.30
Step	Relation	Value
Sum of critical ratios	`Y = y₁ + y₂`	0.65
Optimum cycle	`C₀ = (1.5L + 5)/(1 − Y)`	65.71 s
Total effective green	`C₀ − L`	53.71 s
Green, phase 1	`(y₁/Y)(C₀ − L)`	28.92 s
Green, phase 2	`(y₂/Y)(C₀ − L)`	24.79 s

⚠️ The (1 − Y) denominator is a warning, not just algebra

Keep the lost time at 12 s and watch what the formula does as the junction fills: at Y = 0.65 it gives C₀ = 65.7 s; at 0.75, 92 s; at 0.85, 153 s; at 0.90, 230 s. The cycle length diverges as Y approaches 1, and that is the model telling the truth rather than breaking down. Y = 1 means the sum of the demands equals the sum of the capacities with no time left over for the lost time, so the junction is saturated and no timing plan can serve it — the queue grows without limit whatever you do. Three practical rules follow directly. First, an optimum cycle above about 120 seconds is not implemented; delay is flat near the optimum, so a shorter cycle costs very little while a long one is intolerable for pedestrians and for the side road, and drivers begin to disobey it. Second, when Y is too high the answer is not a longer cycle but less demand or more capacity: an extra lane, a turning restriction, a flyover, or diverting traffic — the geometry has to change, not the timing. Third, the standard sanity check is that the greens and the lost time must reconstruct the cycle: 28.92 + 24.79 + 12 = 65.71 = C₀ ✓. And note that delay near the optimum is genuinely flat, which is why Webster’s cycle is a starting point that is then rounded to a convenient 5-second multiple rather than a value to be held to two decimals.

Traffic signs complete the section and divide into three families whose shapes carry the meaning independently of the legend, which is deliberate: regulatory signs are circular and mandatory, and disobeying one is an offence; warning (cautionary) signs are triangular and tell the driver about a hazard ahead; informatory signs are rectangular and give direction, distance and facilities. The shape convention exists so that a sign is readable at a distance, at speed, in poor light and by a driver who cannot read the language — which is the same design logic as the 85th-percentile speed: assume the driver is reasonable and give them what they need in the form they can use.

Key takeaways

q = kv, so the same flow occurs twice — once free-flowing and once congested. Flow alone is ambiguous.
Greenshields capacity q_max = v_f k_j/4, at k = k_j/2 and v = v_f/2 — capacity comes at half the free-flow speed.
Space mean speed (harmonic) is always smaller than time mean speed, and only the space mean belongs in q = kv.
Webster: C₀ = (1.5L + 5)/(1 − Y), g_i = (y_i/Y)(C₀ − L). Check that the greens plus lost time rebuild C₀.
C₀ diverges as Y → 1: a saturated junction cannot be signalled out of trouble, it needs capacity or less demand.
Sign shapes carry meaning: circular regulatory, triangular warning, rectangular informatory.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A traffic stream follows the Greenshields model with a free-flow speed of 60 km/h and a jam density of 100 veh/km. The capacity of the road in veh/h is ______
Numerical answer — type the value.
Show answer
Answer: 1500
1500 veh/h. Greenshields sets v = v_f(1 − k/k_j), and q = kv = v_f k(1 − k/k_j) is a parabola in k with zeros at k = 0 (an empty road carries nothing) and k = k_j (a jammed road carries nothing). Its peak sits midway between them, at k = k_j/2 = 50 veh/km, where v = v_f/2 = 30 km/h, giving q_max = v_f k_j/4 = 60 × 100/4 = 1500 veh/h. Two features of that result are the reason the model is taught despite being crude. First, capacity occurs at half the free-flow speed, not near it — a road running at its maximum throughput is already travelling at half the speed of an empty one, which is why "the road is at capacity" and "the road is running well" are different statements. Second, and more usefully, the flow is ambiguous: q = 960 veh/h occurs at k = 20 (with v = 48 km/h, free-flowing) and at k = 80 (with v = 12 km/h, congested). A volume count of 960 tells you nothing about which of those two roads you are looking at, which is exactly why traffic studies measure speed or density as well. It also explains ramp metering: holding vehicles back at an entry keeps the mainline near k_j/2 instead of letting it slip onto the falling branch, so restricting inflow can genuinely increase throughput. Greenshields itself is too simple — real speed does not fall linearly with density — but these conclusions follow from the curve having a peak at all, so they survive in better models.
A two-phase signal has a total lost time of 12 s per cycle. The critical flow ratios are 0.35 and 0.30. By Webster’s method, the optimum cycle length in seconds is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 65.71
65.71 s. Y = 0.35 + 0.30 = 0.65, so C₀ = (1.5L + 5)/(1 − Y) = (1.5 × 12 + 5)/(1 − 0.65) = 23/0.35 = 65.71 s. The greens follow: total effective green is C₀ − L = 53.71 s, shared as g₁ = (0.35/0.65) × 53.71 = 28.92 s and g₂ = (0.30/0.65) × 53.71 = 24.79 s, and the check 28.92 + 24.79 + 12 = 65.71 = C₀ ✓ confirms the arithmetic. What the formula is saying is more important than the substitution. It minimises average delay per vehicle, balancing two opposing effects: a longer cycle wastes proportionally less time on the fixed lost time each cycle, but makes every stopped vehicle wait longer for its turn. The optimum is where those balance, and the delay curve near it is genuinely flat — which is why the answer is rounded to a convenient 5-second multiple in practice rather than held to two decimals. Now watch the denominator, which is the part that carries the engineering. Keeping L = 12 s: at Y = 0.75 the cycle is 92 s, at 0.85 it is 153 s, at 0.90 it is 230 s. C₀ diverges as Y → 1, and that is not a defect of the model but the model being honest — Y = 1 means the demands exactly consume the capacity with nothing left for the lost time, so the junction is saturated and no timing plan can serve it. In practice a cycle above about 120 s is not implemented, and when Y is that high the answer is more capacity or less demand — a lane, a turn ban, a flyover — not a longer cycle.
Three vehicles are observed with speeds 30, 40 and 50 km/h. The space mean speed in km/h is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 38.3
38.30 km/h. The space mean speed is the harmonic mean: v_s = n/Σ(1/vᵢ) = 3/(1/30 + 1/40 + 1/50) = 3/0.078333 = 38.30 km/h. The time mean speed is the arithmetic mean, (30 + 40 + 50)/3 = 40.00 km/h, so the two differ by 1.70 km/h — 4.44% — on the same three vehicles. The space mean is always the smaller of the two, and the gap grows with the spread of speeds: it vanishes only when every vehicle travels at the same speed. The reason is a sampling bias, and seeing it clearly is what stops the two from being confused. An observer at a point counts every vehicle once, so each contributes equally. An observer photographing a length of road catches whatever happens to be on it at that instant, and slow vehicles spend longer on the section and are therefore over-represented in the snapshot — pulling the average down. This distinction earns marks because only the space mean speed satisfies the fundamental relation q = kv. If a question gives spot speeds and asks for flow, the harmonic mean is the one to use, and substituting the arithmetic mean overstates the flow. Two cautions on the shortcut. The often-quoted v_t = v_s + σ²/v_s is an approximation: for these three speeds it returns 40.11 rather than 40.00, so it will not reproduce an exact answer. And the variance in it is the variance of the space distribution, not of the observed spot speeds — another reason to compute the harmonic mean directly rather than convert.
A volume count at a point on a road records 960 vehicles per hour. Which of the following can be concluded?
1. Nothing about how well the road is operating — the same flow occurs both in free flow and in congestion
2. The road is operating below capacity, since capacity for a typical lane exceeds 960 veh/h
3. The density can be found directly, since flow determines density uniquely
4. The road is congested, because 960 veh/h in a single lane implies queuing
Show answer
Answer: A — Nothing about how well the road is operating — the same flow occurs both in free flow and in congestion
Nothing about how well the road is operating. This is the central lesson of flow theory, and it is a lesson about a product. Flow is q = kv, and the same product arises from many different pairs of factors. On the Greenshields curve used in this chapter — v_f = 60 km/h, k_j = 100 veh/km — a flow of 960 veh/h occurs at k = 20 veh/km with v = 48 km/h, an entirely comfortable road running at four-fifths of the free-flow speed, and at k = 80 veh/km with v = 12 km/h, a road crawling and close to breakdown. The flow–density parabola is symmetric about its peak, so every flow below capacity has two density solutions, one on each branch. A count of 960 cannot tell you which branch you are on. That is why option (c) is wrong: the mapping from flow to density is two-to-one, not one-to-one, and inverting it requires a second measurement. It is also why every real traffic study collects speed or density alongside volume — an occupancy detector, a floating-car run, a travel-time measurement — and why level-of-service is defined on density, which is single-valued, rather than on flow, which is not. Options (b) and (d) both leap to a conclusion the data cannot support, in opposite directions, and they illustrate the same error: treating a flow figure as if it were a measure of how full the road is. The practical form of this lesson is one every engineer meets: when a jam clears, the measured flow rises as conditions improve, so a low flow is as consistent with severe congestion as with an empty road.
For the signal of question 2 (L = 12 s, y₁ = 0.35, y₂ = 0.30, C₀ = 65.71 s), the effective green time for phase 1 in seconds is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 28.92
28.92 s. The green available to be shared is not the whole cycle but what remains after the lost time: C₀ − L = 65.71 − 12 = 53.71 s. Webster then divides it in proportion to the critical flow ratios, because each phase should be given green in proportion to the demand it must clear: g₁ = (y₁/Y)(C₀ − L) = (0.35/0.65) × 53.71 = 0.53846 × 53.71 = 28.92 s, and by the same rule g₂ = 24.79 s. Always finish with the reconstruction check — 28.92 + 24.79 + 12 = 65.71 = C₀ ✓ — because it catches the two commonest errors at once: sharing the whole cycle instead of C₀ − L (which gives 35.38 s and a total that overruns the cycle by the lost time), and normalising by something other than Y. Two points of substance sit behind the arithmetic. First, y = q/s is the ratio of arriving flow to saturation flow for the worst approach in that phase, not the average of the approaches — a phase is only as good as its critical movement, and using an average silently under-greens the busiest lane. Second, the effective green is not the green shown on the lamp: it is the displayed green plus the amber that drivers actually use, minus the start-up lost time while a stopped queue accelerates. That bookkeeping is what L collects, and it is why lost time appears as a subtraction here and again inside C₀. In practice these values are rounded to convenient whole seconds, which costs almost nothing because delay near the Webster optimum is very flat.
Which of the following statements about traffic studies and signals are correct?
1. The 85th percentile spot speed is taken as the safe or design speed for a road
2. An origin–destination study can justify a bypass in a way that a volume count cannot
3. Webster’s optimum cycle length increases without limit as the sum of critical flow ratios approaches unity
4. Regulatory traffic signs are triangular and warning signs are circular
Show answer
Answer: A — The 85th percentile spot speed is taken as the safe or design speed for a road; B — An origin–destination study can justify a bypass in a way that a volume count cannot; C — Webster’s optimum cycle length increases without limit as the sum of critical flow ratios approaches unity
(a), (b) and (c). (a) is correct, and the reasoning behind it is worth stating because it is not a statistical convention but a design philosophy: most drivers choose a speed appropriate to the road, so the speed that 85% of them do not exceed is a better description of what the road actually invites than any figure chosen in advance. Setting a limit far below the 85th percentile produces widespread disobedience and a larger spread of speeds, which is itself a hazard — speed variance, not speed alone, correlates with crash risk. The 15th percentile serves the opposite purpose, as a floor for a minimum-speed sign. (b) is correct and is the sharpest illustration of why different studies answer different questions. A volume count tells you how much traffic is on a road; only an O–D study tells you how much of it wants to be there. If most trips merely pass through the town, a bypass removes them and widening the high street does not; if most trips start or end in the town, a bypass is money spent on the wrong road. No amount of counting distinguishes those two cases. (c) is correct: C₀ = (1.5L + 5)/(1 − Y) has (1 − Y) in the denominator, so with L = 12 s the cycle runs 65.7 s at Y = 0.65, 92 s at 0.75, 153 s at 0.85 and 230 s at 0.90, diverging as Y → 1. That is the formula reporting saturation honestly, not failing — and it is why cycles above about 120 s are not implemented and a high Y calls for capacity rather than timing. (d) is wrong, with the shapes exactly swapped: regulatory signs are circular (mandatory, an offence to disobey) and warning signs are triangular; informatory signs are rectangular.
An 8 km stretch of highway carrying an average daily traffic of 12,000 vehicles records 45 accidents in a year. The accident rate per million vehicle-kilometres is ______ (round to three decimal places, taking 365 days)
Numerical answer — type the value.
Show answer
Answer: 1.284
1.284 per million vehicle-km. Total exposure is ADT × length × days = 12,000 × 8 × 365 = 35,040,000 vehicle-km = 35.04 million, so the rate is 45/35.04 = 1.284 per million vehicle-km. The reason accident studies insist on this normalisation rather than a raw count is that a raw count measures traffic as much as danger. A busy urban arterial will always record more accidents than a quiet rural road, and comparing the two on absolute numbers would direct every rupee of safety spending to the busiest roads regardless of whether they are unusually dangerous. Dividing by exposure asks the right question: given how much travel happens here, how often does something go wrong? Note the two forms and keep them apart, because a question will use either. Rate per million vehicle-kilometres is for a length of road, and is what the arithmetic above computes. Rate per million entering vehicles is for a spot — a junction, a level crossing — where there is no length to travel over and exposure is counted as arrivals. Using the wrong one is dimensionally inconsistent and produces a number that cannot be compared with anything. One honest limitation is worth carrying: this rate treats all accidents alike, so a stretch with many minor collisions can outrank one with a few fatal ones. That is why severity-weighted indices exist alongside it, and why a bare rate is a screening tool for identifying black spots rather than the final basis for spending.
A junction is signalled and Webster’s method returns an optimum cycle length of 230 s. The correct engineering response is to
1. treat it as a signal that the junction is near saturation and add capacity or reduce demand, rather than implement it
2. implement the 230 s cycle, since it is by definition the delay-minimising value
3. halve the lost time, since L is the only variable the designer controls
4. add a third phase, which will reduce Y and shorten the cycle
Show answer
Answer: A — treat it as a signal that the junction is near saturation and add capacity or reduce demand, rather than implement it
Treat it as a saturation warning and change the capacity or the demand. A cycle of 230 s means Y is around 0.90 — the critical flows are consuming ninety per cent of the saturation flows, leaving almost nothing for the lost time, and C₀ = (1.5L + 5)/(1 − Y) is diverging. The formula has not failed; it is reporting that the junction is close to the point where no timing plan can serve the demand, and at Y = 1 the queue grows without bound whatever the signal does. Implementing 230 s, as option (b) suggests, fails for reasons the delay model does not even capture: pedestrians will not wait four minutes and will cross against the signal, side-road drivers will conclude the signal is broken and enter anyway, queues will grow long enough to block upstream junctions, and the calculation’s own assumption of orderly random arrivals stops holding. In practice cycles are capped near 120 s, and the cost of that cap is small because delay is flat near the Webster optimum. The genuine remedies all change Y rather than C₀: an extra lane or a widened approach raises the saturation flow, banning a turning movement removes a conflicting demand, a flyover or a bypass diverts traffic away, and coordinating with adjacent signals can improve arrival patterns. Option (c) misidentifies the lever — reducing lost time is worth doing but the numerator moves linearly while the denominator is collapsing, so trimming L from 12 s to 6 s at Y = 0.90 still leaves a 140 s cycle. Option (d) has it backwards: adding a phase increases total lost time and usually increases Y as well, lengthening the cycle rather than shortening it. Phases are added for safety — to separate conflicting movements — and are paid for in capacity.