Traffic Flow Theory — Greenshields Model & Level of Service
Speed-flow-density relationships, Greenshields linear model, maximum flow capacity, optimum density, macroscopic and microscopic flow models, and LOS — with GATE CE worked examples
Last Updated: April 2026
- Greenshields model assumes a linear speed-density relationship: u = uf(1 – k/kj), making it the simplest and most widely tested traffic flow model.
- Maximum flow (capacity): qmax = uf kj/4 — occurs at optimum density ko = kj/2 and optimum speed uo = uf/2.
- Flow-density parabola: q = uf k(1 – k/kj) — a downward-opening parabola with maximum at k = kj/2.
- Speed-flow relationship: for each flow q < qmax, there are two possible speeds — one subcritical (high speed, low density) and one overcritical (low speed, high density).
- Greenberg’s logarithmic model and Underwood’s exponential model are alternative flow models for better fitting at high densities.
- Level of Service (LOS A–F) classifies traffic quality from free-flow to breakdown — LOS C is the standard design target for most Indian highways.
- Traffic flow models are used for freeway/highway capacity analysis and traffic simulation.
1. Traffic Flow Variables — q, u, k Revisited
Traffic flow theory describes the collective behaviour of vehicles on a road using three macroscopic state variables:
| Variable | Symbol | Unit | Definition | Boundary Values |
|---|---|---|---|---|
| Flow (volume) | q | vehicles/hour | Number of vehicles passing a fixed point per unit time | 0 ≤ q ≤ qmax |
| Space mean speed | u (or us) | km/h | Average speed of vehicles over a road section at an instant | 0 ≤ u ≤ uf |
| Density (concentration) | k | vehicles/km | Number of vehicles per unit length of road at an instant | 0 ≤ k ≤ kj |
⭐ Fundamental flow equation: q = u × k
Boundary conditions:
When k = 0 (empty road): u = uf (free-flow speed); q = 0
When k = kj (jam density): u = 0 (all stopped); q = 0
At intermediate k: flow q > 0; speed decreases as density increases
A traffic flow model specifies the relationship between any two of these three variables — the third is determined from q = uk. The key model to specify is the speed-density relationship u(k), from which q(k) and q(u) are derived.
2. Greenshields Linear Model
Greenshields (1935) proposed the simplest possible speed-density relationship: a straight line decreasing from free-flow speed uf at zero density to zero speed at jam density kj.
⭐ Speed-density relationship (Greenshields):
u = uf (1 – k/kj)
or equivalently: u = uf – (uf/kj) × k
This is the equation of a straight line with:
- y-intercept: uf (free-flow speed when k = 0)
- x-intercept: kj (jam density when u = 0)
- Slope: –uf/kj (negative)
Parameters:
- uf = free-flow speed (km/h) — driver’s desired speed in empty road conditions
- kj = jam density (vehicles/km) — maximum packing density when all vehicles are stopped
2.1 Physical Justification
As density increases, drivers perceive shorter headways and reduce speed to maintain safety. The linear model assumes this reduction is proportional to density — each additional vehicle per km causes the same reduction in average speed. While this is a simplification (in reality, speed drops more sharply near jam conditions), the Greenshields model fits field data reasonably well at low to medium densities and has the important advantage of yielding simple closed-form expressions for capacity and optimal conditions.
3. Derived q–k and q–u Relationships
3.1 Flow-Density Relationship
Substituting u = uf(1 – k/kj) into q = uk:
⭐ q = uf k (1 – k/kj)
This is a downward-opening parabola in the q–k plane:
- q = 0 at k = 0 (empty road)
- q = 0 at k = kj (jam — all vehicles stopped)
- q is maximum at k = kj/2 (optimum density)
3.2 Speed-Flow Relationship
From u = uf(1 – k/kj): k = kj(1 – u/uf)
Substituting into q = uk:
q = u × kj(1 – u/uf) = kju – kju²/uf
⭐ q = kj u (1 – u/uf)
Also a parabola — q is maximum at u = uf/2
3.3 The q–u–k Relationship Diagram
The three pairwise relationships (u–k, q–k, and q–u) form a consistent system. Key insight: for any given flow q < qmax, there are two possible states:
- Uncongested (free-flow) branch: High speed, low density — stable operation
- Congested branch: Low speed, high density — unstable, prone to stop-and-go
Traffic can be on either branch for the same flow rate — a fundamental implication for traffic management and incident response.
4. Maximum Flow, Optimum Speed, Optimum Density
To find maximum flow from q = uf k(1 – k/kj):
dq/dk = uf(1 – 2k/kj) = 0
→ Optimum density: ko = kj/2
Optimum speed:
uo = uf(1 – ko/kj) = uf(1 – 1/2) = uf/2
⭐ Maximum flow (capacity):
qmax = uo × ko = (uf/2) × (kj/2)
qmax = uf kj / 4
Summary of critical (optimum) conditions:
| Parameter | Value | In Terms of Model Parameters |
|---|---|---|
| Optimum density ko | kj/2 | Half the jam density |
| Optimum speed uo | uf/2 | Half the free-flow speed |
| Maximum flow qmax | uf kj/4 | Product of uo and ko |
Physical meaning: Maximum capacity is achieved when vehicles are travelling at exactly half their free-flow speed and half the road is “used up” by vehicles. Driving faster (lower density) or slower (higher density) both reduce the throughput below the maximum.
5. Other Traffic Flow Models
5.1 Greenberg’s Logarithmic Model
u = uo × ln(kj/k)
where uo = optimum speed (= speed at maximum flow)
Maximum flow: qmax = uo × kj/e (e = 2.718)
Optimum density: ko = kj/e
Limitation: u → ∞ as k → 0 (unrealistic for low densities); better fit at high densities
5.2 Underwood’s Exponential Model
u = uf × e–k/ko
where ko = optimum density
Maximum flow: qmax = uf ko/e
Limitation: u → uf × e–kj/ko > 0 as k → kj (u never reaches zero — jam density is technically infinite)
Better fit at low densities (free-flow region) than Greenberg
5.3 Model Comparison
| Model | u–k Relationship | Optimum ko | Best Fit Region |
|---|---|---|---|
| Greenshields | Linear | kj/2 | Low–medium density |
| Greenberg | Logarithmic | kj/e | Medium–high density |
| Underwood | Exponential | ko (given) | Low density / free flow |
| Northwestern (2-regime) | Different for free/congested | At breakpoint | Matches real data best |
6. Microscopic Flow Models — Headway and Following Distance
6.1 Safe Following Distance
Microscopic models describe the behaviour of individual vehicles — particularly how a following driver maintains a safe gap from the lead vehicle.
Safe following distance (stopping sight distance based):
dsafe = u × tr + u²/(2a) + Lv
where tr = reaction time; a = deceleration; Lv = vehicle length
Headway: h = dsafe/u = tr + u/(2a) + Lv/u
Relationship between safe headway and density:
k = 1/dsafe = 1/(u × tr + u²/(2a) + Lv)
6.2 Jam Density from Vehicle Length
At jam conditions (u = 0): all vehicles stationary, bumper-to-bumper
kj = 1/Lv
For average vehicle length Lv = 5 m = 0.005 km: kj = 1/0.005 = 200 vehicles/km
For a mix of cars and trucks (average L ≈ 6 m): kj ≈ 167 vehicles/km
7. Traffic Shockwave
A traffic shockwave is the boundary between two different traffic states (different q, u, k values) that propagates along the road. It occurs when a sudden change in traffic conditions — an accident, a signal turning red, a sudden speed reduction — creates a boundary between a dense slow region and a less dense fast region.
Shockwave speed:
w = (q₂ – q₁) / (k₂ – k₁)
where states 1 and 2 are the conditions on either side of the shockwave
Sign convention:
- w > 0: shockwave moves in the direction of traffic (forward)
- w < 0: shockwave moves against traffic (backward — e.g., back of a queue propagating upstream)
- w = 0: stationary shockwave (fixed queue, e.g., at a red signal)
Example: Traffic suddenly stops at an incident (state 2: q₂ = 0, k₂ = kj = 200 veh/km) from a free-flowing state (state 1: q₁ = 2000 veh/hr, k₁ = 40 veh/km, u₁ = 50 km/h):
w = (0 – 2000)/(200 – 40) = –2000/160 = –12.5 km/h (queue propagates upstream at 12.5 km/h)
8. Level of Service (LOS)
Level of Service is a qualitative measure of traffic operating conditions — it grades the quality of traffic flow from A (best) to F (worst). LOS is used to evaluate existing road performance and set design targets for new roads.
8.1 LOS for Basic Freeway Segments (HCM Framework)
| LOS | Density (vehicles/km/lane) | Speed (km/h) | v/c | Description |
|---|---|---|---|---|
| A | ≤ 7 | ≥ 100 | ≤ 0.35 | Free flow; drivers select desired speed; complete mobility |
| B | 7–11 | 90–100 | 0.35–0.54 | Reasonably free flow; slightly restricted freedom |
| C | 11–16 | 80–90 | 0.55–0.77 | Stable flow; freedom to manoeuvre noticeably restricted |
| D | 16–22 | 70–80 | 0.78–0.90 | Approaching unstable; speed and freedom affected; tolerable for short periods |
| E | 22–28 | ≥ 60 | 0.91–1.00 | At or near capacity; low speed; no freedom to manoeuvre; any incident causes breakdown |
| F | > 28 | < 60 | > 1.00 | Forced flow; breakdown; stop-and-go conditions; demand exceeds supply |
8.2 LOS for Two-Lane Rural Highways (IRC context)
For Indian two-lane highways (IRC:64), LOS is determined by the volume-to-capacity (v/c) ratio using PCU flow:
| LOS | v/c Ratio | Indian Context |
|---|---|---|
| A | < 0.35 | Very low traffic; rural areas |
| B | 0.35–0.54 | Light traffic; comfortable driving |
| C | 0.55–0.77 | Design standard for most NH/SH |
| D | 0.78–0.90 | Acceptable for urban roads |
| E | 0.91–1.00 | Near capacity; unacceptable for design |
| F | > 1.00 | Over capacity; breakdown |
8.3 Design LOS Standards
- National and State Highways (rural): LOS C or better (v/c ≤ 0.77)
- Urban arterials and expressways: LOS C or D (v/c ≤ 0.90)
- Intersection approaches: LOS D acceptable under peak hour conditions
9. Worked Examples (GATE CE Level)
Example 1 — Maximum Flow from Greenshields Model (GATE CE 2022 type)
Problem: A road has a free-flow speed uf = 80 km/h and a jam density kj = 160 vehicles/km. Using the Greenshields model, find (a) the optimum speed, (b) the optimum density, and (c) the maximum flow (capacity).
Given: uf = 80 km/h; kj = 160 veh/km
(a) Optimum speed:
uo = uf/2 = 80/2 = 40 km/h
(b) Optimum density:
ko = kj/2 = 160/2 = 80 vehicles/km
(c) Maximum flow:
qmax = uf kj/4 = 80 × 160/4 = 12,800/4 = 3200 vehicles/hour
Alternatively: qmax = uo × ko = 40 × 80 = 3200 veh/hr ✓
Answer: uo = 40 km/h; ko = 80 veh/km; qmax = 3200 veh/hour
Example 2 — Speed at a Given Density (GATE CE 2021 type)
Problem: Using the Greenshields model with uf = 60 km/h and kj = 120 veh/km, find the space mean speed when the density is 30 vehicles/km, and compute the corresponding flow rate.
Given: uf = 60 km/h; kj = 120 veh/km; k = 30 veh/km
Space mean speed (Greenshields):
u = uf(1 – k/kj) = 60 × (1 – 30/120) = 60 × (1 – 0.25) = 60 × 0.75 = 45 km/h
Flow rate:
q = u × k = 45 × 30 = 1350 vehicles/hour
Check using q–k formula:
q = uf k(1 – k/kj) = 60 × 30 × (1 – 30/120) = 1800 × 0.75 = 1350 veh/hr ✓
Answer: u = 45 km/h; q = 1350 vehicles/hour
Example 3 — Finding Free-Flow Speed and Jam Density from Data (GATE CE type)
Problem: Traffic observations on a highway show that at a density of 20 veh/km the speed is 70 km/h, and at a density of 60 veh/km the speed is 30 km/h. Using the Greenshields model, find uf and kj.
Greenshields model: u = uf – (uf/kj) × k
This is a straight line: u = uf – m × k, where m = uf/kj
Two equations from two data points:
Point 1: 70 = uf – 20m …(1)
Point 2: 30 = uf – 60m …(2)
Subtract (2) from (1):
40 = 40m → m = 1.0 km/h per veh/km = 1.0 km²/(h·veh)
Substitute m = 1.0 into (1):
70 = uf – 20 × 1.0 → uf = 90 km/h
Jam density:
kj = uf/m = 90/1.0 = 90 vehicles/km
Maximum flow:
qmax = uf kj/4 = 90 × 90/4 = 8100/4 = 2025 vehicles/hour
Answer: uf = 90 km/h; kj = 90 veh/km; qmax = 2025 veh/hour
Example 4 — Greenberg Model Application (GATE CE type)
Problem: For a road following Greenberg’s model with uo = 35 km/h and kj = 140 veh/km, find (a) the speed at k = 40 veh/km, (b) the maximum flow, and (c) the optimum density.
Greenberg model: u = uo × ln(kj/k)
Given: uo = 35 km/h; kj = 140 veh/km
(a) Speed at k = 40 veh/km:
u = 35 × ln(140/40) = 35 × ln(3.5) = 35 × 1.2528 = 43.8 km/h
(b) Maximum flow:
qmax = uo × kj/e = 35 × 140/2.718 = 4900/2.718 = 1803 vehicles/hour
(c) Optimum density:
ko = kj/e = 140/2.718 = 51.5 vehicles/km
Optimum speed: uo = 35 km/h (given directly — it’s the parameter in Greenberg’s model)
Answer: u(k=40) = 43.8 km/h; qmax = 1803 veh/hr; ko = 51.5 veh/km
Example 5 — Two Possible States for a Given Flow (GATE MCQ type)
Problem: For a Greenshields model with uf = 100 km/h and kj = 200 veh/km, find the two possible densities at which q = 4000 vehicles/hour, and determine which state is the stable (uncongested) operating condition.
Given: uf = 100 km/h; kj = 200 veh/km; q = 4000 veh/hr
qmax = uf kj/4 = 100 × 200/4 = 5000 veh/hr
Since q = 4000 < qmax = 5000 ✓ — two solutions exist.
q = uf k(1 – k/kj)
4000 = 100k(1 – k/200) = 100k – 100k²/200 = 100k – 0.5k²
0.5k² – 100k + 4000 = 0
k² – 200k + 8000 = 0
k = [200 ± √(200² – 4 × 8000)] / 2 = [200 ± √(40000 – 32000)] / 2 = [200 ± √8000] / 2
√8000 = 89.44
k₁ = (200 – 89.44)/2 = 110.56/2 = 55.3 veh/km (uncongested, k < ko = 100)
k₂ = (200 + 89.44)/2 = 289.44/2 = 144.7 veh/km (congested, k > ko = 100)
Corresponding speeds:
u₁ = q/k₁ = 4000/55.3 = 72.3 km/h (fast, uncongested — stable)
u₂ = q/k₂ = 4000/144.7 = 27.6 km/h (slow, congested — unstable)
Answer: k₁ = 55.3 veh/km (u = 72.3 km/h, uncongested, stable); k₂ = 144.7 veh/km (u = 27.6 km/h, congested, unstable)
Example 6 — Shockwave Speed (GATE CE type)
Problem: Free-flowing traffic (State 1) has q₁ = 1800 veh/hr and k₁ = 30 veh/km. An accident reduces traffic to State 2 with q₂ = 600 veh/hr and k₂ = 120 veh/km. Find the speed and direction of the resulting shockwave.
Given: q₁ = 1800 veh/hr; k₁ = 30 veh/km; q₂ = 600 veh/hr; k₂ = 120 veh/km
Shockwave speed:
w = (q₂ – q₁)/(k₂ – k₁) = (600 – 1800)/(120 – 30) = –1200/90 = –13.33 km/h
Interpretation:
w = –13.33 km/h → the shockwave propagates upstream (against traffic direction) at 13.33 km/h.
This is the rate at which the back of the queue grows upstream from the accident site.
After 1 hour, the back of the queue will be 13.33 km upstream of the incident.
Answer: w = –13.33 km/h — queue propagates upstream at 13.33 km/h
10. Common Mistakes
Mistake 1 — Using qmax = uf kj/2 Instead of uf kj/4
Error: Writing qmax = uf kj/2 (forgetting that both u and k are at half their maximum values).
Root Cause: qmax = uo × ko = (uf/2) × (kj/2) = ufkj/4. Multiplying two halves gives one-quarter — not one-half. A common error is taking only one factor of 1/2.
Fix: Derive it explicitly: qmax = uo × ko = (uf/2) × (kj/2) = ufkj/4. The factor is 1/4, not 1/2. Alternatively, substitute k = kj/2 into q = ufk(1–k/kj) and confirm q = uf(kj/2)(1–1/2) = ufkj/4.
Mistake 2 — Thinking There is Only One Density for a Given Flow Rate
Error: Given a flow q < qmax, computing only one density using q = ufk(1–k/kj) and missing the second solution.
Root Cause: The flow-density relationship q = ufk(1–k/kj) is a quadratic in k — for q < qmax, it has two real roots (one < ko, one > ko). Students often find only the smaller root (uncongested state) without noticing the quadratic gives two answers.
Fix: Set up q = ufk(1–k/kj) as a quadratic in k, solve using the quadratic formula, and always interpret both solutions: smaller k = uncongested (stable, high speed); larger k = congested (unstable, low speed).
Mistake 3 — Using Time Mean Speed Instead of Space Mean Speed in the Greenshields Model
Error: Treating the “speed” u in the Greenshields model as time mean speed when comparing model predictions with field data measured as average of spot speeds (TMS).
Root Cause: The Greenshields model uses space mean speed us (correct for q = u × k). Field spot speed studies measure instantaneous speeds, giving TMS. TMS ≥ SMS — using TMS in the model overestimates the flow.
Fix: For traffic flow theory calculations, always use SMS in the fundamental equation and the Greenshields model. If TMS is given, convert: us = ut² – σ²/ut (approximate) or use the harmonic mean directly from individual speed data.
Mistake 4 — Wrong Sign for Shockwave Speed Direction
Error: Computing |w| = |Δq/Δk| and not keeping track of the sign — concluding the queue moves forward when it actually moves backward.
Root Cause: When a congestion forms behind an incident, the change in flow Δq = qcongested – qfree is negative (flow drops), and Δk = kcongested – kfree is positive (density increases). w = Δq/Δk < 0 → the shockwave moves upstream (against traffic). Taking the absolute value loses this crucial directional information.
Fix: Always compute w = (q₂–q₁)/(k₂–k₁) with sign. Positive w = shockwave moves with traffic; negative w = moves against traffic (upstream). The sign determines whether a queue grows upstream (backward shockwave, dangerous) or dissipates (forward shockwave).
Mistake 5 — Confusing Optimum Density ko in Greenshields vs Greenberg Models
Error: Using ko = kj/2 (Greenshields result) when solving a Greenberg model problem.
Root Cause: Each model has different optimum conditions. Greenshields: ko = kj/2. Greenberg: ko = kj/e. Underwood: ko is a model parameter. Mixing models and results gives wrong answers.
Fix: Identify the model being used first. Greenshields: all “halves” (ko = kj/2, uo = uf/2, qmax = ufkj/4). Greenberg: ko = kj/e; qmax = uokj/e. Derive the optimum conditions afresh for each model from dq/dk = 0.
11. Frequently Asked Questions
Q1. Why does the Greenshields model predict two possible traffic states for the same flow rate?
This dual-state phenomenon is a fundamental feature of traffic flow, not just a mathematical artefact of the parabolic q–k curve. For any flow rate below capacity (q < qmax), traffic can operate in either a high-speed/low-density state (free-flow or uncongested) or a low-speed/high-density state (congested). The two states exist on either side of the maximum of the q–k parabola. In practice, the uncongested state is stable: if a small perturbation increases density slightly, the resulting decrease in speed sends a corrective signal upstream, and traffic returns to free-flow. The congested state is unstable: small perturbations grow into stop-and-go waves that propagate upstream. This instability explains why urban motorways can suddenly transition from steady flow to stop-and-go congestion — a single braking event on the congested side of capacity can trigger an upstream-propagating shockwave. Traffic management strategies (variable speed limits, ramp metering, incident detection) try to keep traffic on the stable uncongested branch and prevent the transition to the congested regime.
Q2. What are the limitations of the Greenshields model in describing real traffic flow?
Greenshields’ linear u–k relationship has several well-known deficiencies when compared with field data. First, it overestimates speed at high densities — real traffic tends to reduce speed more sharply than linearly as density approaches the jam value, forming a more concave u–k curve. Second, it predicts that at very low densities (k → 0) speed approaches uf gradually, whereas real drivers quickly reach their desired speed and maintain it over a wide low-density range (a flat plateau in the u–k curve). Third, the model gives a symmetric q–k parabola with maximum at exactly kj/2, whereas observed data typically show the maximum at lower densities (ko ≈ 0.2–0.35 kj). Despite these limitations, the Greenshields model remains widely used for educational purposes and analytical work because it yields simple closed-form results for capacity, optimum conditions, and shockwave speeds. More accurate but mathematically complex alternatives include the Northwestern two-regime model, the Van Aerde model, and the LWR continuum theory with more realistic fundamental diagrams.
Q3. How is traffic flow theory used in practice for Indian highway planning?
In Indian practice, traffic flow theory is applied through the Level of Service framework (IRC:64) and traffic simulation for corridor analysis. The fundamental equation q = uk and the q–k parabola inform the design of capacity improvements: if current flow is approaching qmax (LOS E), widening from two to four lanes approximately doubles capacity. Traffic flow theory also underpins the design of toll plazas (how many toll lanes are needed to prevent queue buildup), weaving sections on expressways (where merging and diverging flows interact), and ramp designs on urban flyovers. For the Bharatmala Pariyojana expressway corridor projects, NHAI requires traffic demand forecasts using four-step travel demand models, with the fundamental flow equation used to assign traffic to individual highway links. Microsimulation (VISSIM) is used for complex interchange design on NHs with heavy mixed traffic. The LOS concept is being integrated into India’s Highway Capacity Manual (under development by CSIR-CRRI), which will replace the current IRC:64 tables with more comprehensive Indian-specific capacity models.
Q4. How does traffic flow theory explain why adding capacity sometimes makes congestion worse (Braess’s paradox)?
Braess’s paradox (1968) is a counterintuitive result in network traffic theory: adding a new road (or increasing capacity on one link) to a network can sometimes increase average travel times for all users. It occurs when individually rational route-choice decisions collectively produce a worse outcome. Each driver chooses the route that minimises their own travel time — but if many drivers switch to the new road, the resulting congestion on the new link may exceed the benefit it offered. This is not captured by simple link-capacity analysis (which assumes traffic distributes evenly) but emerges from network equilibrium models (Wardrop’s user equilibrium principle). In the q–u–k framework, the instability of the congested branch plays a role: a new road may attract enough traffic to push the network onto the congested branch of multiple links simultaneously, increasing overall delay. Real-world examples include cases in Seoul, Stuttgart, and New York where removing or closing roads temporarily improved traffic flow. For Indian planning, this means that simply widening an inner-city corridor can attract induced demand that fills the additional capacity and may worsen network-level conditions unless demand management measures (parking pricing, public transport investment, staggered work hours) are implemented simultaneously.