Every concept in this course comes from real-world observation. This guide breaks down the math into plain English, with live simulations, animated traffic lights, practice problems with step-by-step solutions, and analogies you'll actually remember.
14CORE TOPICS
6CALCULATORS
5ANIMATIONS
20+PRACTICE PROBLEMS
01 — FOUNDATION
What is Traffic Engineering?
Traffic Engineering is the branch of transportation engineering that focuses on the safe, efficient movement of people and vehicles. It covers road design, signal timing, intersection control, and flow analysis.
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Every second of green you hand to one approach is a second you take from another. That is the sentence the whole course turns on. Traffic engineering almost never creates capacity — it allocates a fixed amount of space-time between competing users, and every formula ahead is a different way of deciding who gets it. Once you see that, the equations stop being arbitrary.
Think of the entire road network as a city's blood circulatory system. Roads are the arteries. Vehicles are blood cells. Traffic lights are valves. Too many cells (congestion) causes blockage. Traffic engineers are the cardiologists who keep the system healthy.
🛣️
Mobility
Moving efficiently from A to B at speed. Freeways maximize this — no stops, no driveways, no signals.
High speed → Low access
🏠
Accessibility
Being able to pull in and out of properties. Local streets maximize this — driveways everywhere.
Low speed → High access
⚖️
The Tradeoff
You can't have both at once. More access points = more friction = less mobility. This is the fundamental tension in road design.
The Principal Goal
Traffic Engineers design systems that provide a safe and efficient environment for roadway traffic — minimizing crashes, delays, and fuel waste simultaneously.
02 — HIGHWAY FUNCTIONAL CLASSIFICATION
Road Hierarchy — Not All Roads Are Equal
Roads are classified by their function — what job they do in the network. The classification determines design speed, access controls, and who uses them.
Think of the road system like an airport terminal. Freeways = runways (fast, no interruptions). Arterials = main terminal corridors (fast but with gates). Collectors = concourses (slower, connect to gates). Local streets = individual gates (slow, direct door access).
🎬 ANIMATION — LIVE INTERSECTION VIEW: 4-WAY SIGNALIZED INTERSECTION
Phase: EW GREEN
Timer: 0s
Road Type
Speed (mph)
Access
Primary Function
Example
Freeway
60–75
None (controlled)
Long-distance mobility
I-95, I-90
Principal Arterial
45–55
Very limited
Regional movement
US-1, State highways
Minor Arterial
35–45
Limited
Connecting to arterials
Major city roads
Collector
25–35
Moderate
Collect neighborhood traffic
Shopping center roads
Local Street
15–25
High
Direct property access
Residential streets
Interrupted vs Uninterrupted Flow
🟢
Uninterrupted Flow
No external controls stop traffic. Vehicles only interact with each other. Freeways are the ideal case.
Freeways, expressways, highways (>2mi between signals)
🔴
Interrupted Flow
External devices (signals, stop signs) periodically halt traffic. Flow is at the mercy of timing.
All signalized streets, STOP-controlled roads
03 — TRAFFIC OPERATIONS: IMPORTANT TERMS
The Language of Traffic Flow
Four critical terms that look similar but mean different things. Getting these right is the foundation of every HCM analysis.
The idea behind PHF
A road does not fail on average — it fails at its worst moment. If 1,600 vehicles arrive in an hour but 420 of them arrive in one 15-minute burst, the road briefly has to handle a rate of 1,680 veh/h, not 1,600. The peak hour factor is simply the ratio between the smooth average and that burst: PHF = V / (4 × V15,max). A PHF of 1.0 means traffic arrived perfectly evenly; 0.85 means it came in a lump. You design for the lump.
Imagine a coffee shop. Capacity = max customers per hour the baristas can serve. Demand = how many customers want to come. Volume = how many actually walked in this hour. Flow Rate = how fast they're arriving right now (the busy 15 minutes during morning rush).
🏋️
Capacity (c)
The maximum vehicles a section can handle under typical conditions. A ceiling — you can't exceed it physically.
veh/hr/lane — a fixed physical limit
🙋
Demand
How many vehicles want to use the road. Demand can exceed capacity — that's when queues form.
veh/hr — what drivers want to do
📊
Volume (V)
How many vehicles actually pass a point in a given time. Measured by counting. Equals demand if no congestion.
veh/hr or veh/day — measured count
⚡
Flow Rate (v)
Like volume but for a sub-hour period, expressed as if it lasted a full hour. Captures peak intensity.
v = V/PHF — the true peak rate
Peak Hour Factor (PHF) — Catching the Worst 15 Minutes
The full hour total might be 3,600 vehicles. But if 1,200 came in just one 15-minute window, you actually need to design for 4,800 veh/hr equivalent. PHF catches this spike. A PHF of 1.0 = perfectly even flow. PHF of 0.8 = significant peaking.
Peak Hour Factor
\[ \text{PHF} = \dfrac{V}{4 \cdot V_{15,\max}} \]
V = hourly volume | V₁₅,max = max 15-min count | Range: 0.25 to 1.00
Maximum Flow Rate
\[ v = \dfrac{V}{\text{PHF}} \]
Convert hourly volume to equivalent flow rate. Use this for capacity analysis, not raw volume.
⚡ SIM — PHF CALCULATOR: ENTER YOUR 15-MIN COUNTS
Practice Problems — Traffic Terms & PHF
PROBLEM 1
During the evening peak hour, vehicle counts for each 15-minute interval were: 410, 480, 530, and 420 vehicles. Calculate the PHF and the maximum flow rate. What does the PHF tell you about traffic peaking?
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💡 HINTPHF = V_hourly / (4 × V_max_15min). Maximum flow rate v = V / PHF. PHF close to 1.0 means even flow; close to 0.25 means extreme peaking.
Step 1 — Total hourly volume:
V = 410 + 480 + 530 + 420 = 1,840 veh/hrStep 2 — Find maximum 15-minute count:
V₁₅,max = 530 veh (the 5:00–5:15 period)
Step 3 — Calculate PHF:
PHF = 1,840 / (4 × 530) = 1,840 / 2,120 = 0.868Step 4 — Maximum flow rate:
v = V / PHF = 1,840 / 0.868 = 2,120 veh/hrStep 5 — Interpretation:
PHF = 0.868 means peaking is moderate. The worst 15 minutes
carries 15% more than the average 15-minute rate. Design
systems for 2,120 veh/hr, NOT 1,840 veh/hr.
✓ PHF = 0.868 | v_max = 2,120 veh/hr
PROBLEM 2
A signalized approach counts 900 vehicles in the peak hour. The PHF is known to be 0.92. What maximum flow rate should be used for HCM capacity analysis?
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Given: V = 900 veh/hr, PHF = 0.92
Maximum flow rate:
v = V / PHF = 900 / 0.92 = 978 veh/hrWhy this matters:
Using 900 underestimates peak demand by ~8%.
Using v = 978 correctly reflects the 15-min peak intensity.
✓ v = 978 veh/hr
04 — AT-GRADE INTERSECTIONS
Where Roads Meet
An at-grade intersection is where two or more roads cross at the same level. The key challenge: multiple streams of vehicles want to use the same space at the same time — someone must yield.
An intersection is like a four-way conversation. Everyone can't talk simultaneously — someone facilitates (the signal) or people take turns (yield/stop control). Without any control, it's chaos.
Levels of Intersection Control
🆓
No Control
Drivers adjust speed on their own. Only suitable for very low volume, good sight distance intersections.
🔺
Yield
Minor road yields to major road. Driver must slow but can proceed if clear. Shared responsibility.
🛑
STOP
Minor road must fully stop before entering. Must wait for a safe gap in major road traffic.
🚦
Signal Control
ALL approaches stop when red. Best for high-volume, multi-modal intersections.
05 — TRAFFIC SIGNALS
Teaching Drivers to Share Space
Traffic signals are the most powerful tool for managing intersection conflicts. But they must be justified — signals aren't always the solution, and a poorly-timed signal can make things worse.
Live Signal Demo — Watch the Cycle
🚦 ANIMATION — LIVE TRAFFIC LIGHT CYCLE WITH COUNTDOWN
EAST-WEST ↔
CURRENT PHASE
EW GREEN
TIME REMAINING
30
seconds
CYCLE PROGRESS
NORTH-SOUTH ↕
TIMING PLAN
Cycle C (s):70sGreen Split (%):53%
Why Install a Signal? The 9 MUTCD Warrants
A signal should only be installed if at least one warrant is met. Over-signalization increases delay and can increase rear-end crashes.
W1: 8-Hour Vehicular Volume
High consistent traffic volume throughout the day.
W2: 4-Hour Vehicular Volume
High volumes during multiple periods.
W3: Peak Hour
Excessive delay during peak hours only.
W4: Pedestrian Volume
High pedestrian demand crossing a busy road.
W5: School Crossing
Children crossing near a school.
W7: Crash Experience
Documented crash pattern that signals can reduce.
Traffic Sign Types
Sign Type
Shape
Color
Examples
Regulatory
Vertical Rectangle
Black/White
STOP, YIELD, Speed Limit, NO PARKING
Warning
Diamond
Black/Yellow
Curve Ahead, Railroad, Bridge
Guide
Horiz. Rectangle
White/Green
Destinations, Distance
06 — SIGNAL TIMING DESIGN
The Rhythm of an Intersection
Signal timing is about efficiently allocating the most limited resource at an intersection: time. Every second given to one phase is a second taken from another.
Signal timing is like baking multiple dishes using one oven. Each dish (phase) needs oven time. The full dinner cycle repeats. You lose time switching between dishes (lost time). You want everyone done as fast as possible — that's optimal timing.
Key Timing Terms — Explained One By One
🔄 Cycle Length (C)
Total time for ONE complete sequence of all phases. When the clock runs out, it starts over. Typically 60–120 seconds.
Longer C → more capacity, more max delay
🟢 Phase (Φ)
One direction's turn: Green + Yellow + All-Red. A 2-phase signal gives N-S their turn, then E-W.
Each approach gets at least 1 phase
💚 Green Interval (G)
The literal green light duration. This is what drivers see on the signal head. Design input.
G = what you set on the controller
✅ Effective Green (g)
Time vehicles can actually discharge. Slightly longer than G because yellow allows some movement. Key formula input.
g = G + y + ar − tL
⏱️ Lost Time (tL)
Wasted time per phase: startup (vehicles accelerating from rest) + clearance (last yellow vehicle). Usually ≈4 sec.
tL = l₁ + l₂ ≈ 4 sec/phase
🔴 All-Red (AR)
Every approach red simultaneously. Safety clearance — the intersection empties before next phase starts.
Typically 1–2 seconds
Effective Green Time
\[ g = G + y + a_r - t_L \]
G=display green y=yellow ar=all-red tL=lost time per phase (~4s)
y=flow ratio per phase ΣY=sum of critical flow ratios L=total lost time n=number of phases
Look at what the denominator is doing. ΣY is the fraction of the cycle the intersection is already committed to spending on traffic. If ΣY = 0.75, then only a quarter of every cycle is spare — so the lost time L has to be paid out of that quarter, and the cycle must stretch to afford it. Push ΣY toward 1.0 and the denominator collapses: the required cycle runs away to infinity. That is not a quirk of the algebra, it is the intersection telling you it is out of capacity, and no amount of signal timing will save it. You need another lane.
⚙️ CALCULATOR — WEBSTER CYCLE & GREEN SPLIT
The defaults are the signal-design problem from the practice exam at the bottom of this page — ΣY = 0.750, L = 8 s, C₀ = 68 s → use 70 s. Change one number at a time and watch which way the cycle moves.
⚡ SIM — SIGNAL TIMING ANIMATOR: WATCH THE CYCLE LIVEPHASE 1
CYCLE (s):70sG1 (s):37s
Practice Problems — Signal Timing
PROBLEM 3
A 2-phase signal has: C = 80s, G₁ = 42s, Y = 4s, AR = 2s, tL = 4s per phase. Calculate the effective green time for both phases and the total lost time per cycle. How efficient is this cycle?
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Phase 1 — Effective Green:
g₁ = G₁ + y + ar − tL = 42 + 4 + 2 − 4 = 44sPhase 2 Display Green:
G₂ = C − G₁ − y − ar − y − ar = 80 − 42 − 4 − 2 − 4 − 2 = 26s
Phase 2 — Effective Green:
g₂ = G₂ + y + ar − tL = 26 + 4 + 2 − 4 = 28sTotal Lost Time per Cycle:
L = 2 × tL = 2 × 4 = 8s per cycleCycle Efficiency:
Efficiency = (C − L)/C = (80 − 8)/80 = 72/80 = 90%
(Only 10% of cycle time is "wasted" on lost time)
✓ g₁=44s | g₂=28s | L=8s | Efficiency=90%
PROBLEM 4
What is the effective green time for a phase with G = 35s, Y = 3.5s, AR = 1.5s, and lost time tL = 4s? If the cycle is 90s, what fraction of the cycle does this phase effectively control?
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Effective Green:
g = G + y + ar − tL = 35 + 3.5 + 1.5 − 4 = 36sg/C Ratio:
g/C = 36 / 90 = 0.40Meaning:
This phase receives 40% of the cycle for vehicle discharge.
Only 40% of the saturation flow rate becomes available capacity.
✓ g = 36s | g/C = 0.40
07 — SATURATION FLOW RATE
How Fast Can Vehicles Actually Discharge?
When a queue of vehicles gets a green light, there's a maximum rate at which they can cross the stop bar. That's the saturation flow rate — the physical throughput limit of a lane.
Imagine a tube of toothpaste. The saturation flow rate is the maximum paste per second that can come out when you squeeze hard. Squeeze harder = same rate (it's saturated). The saturation headway is the time between each toothpaste "droplet" (vehicle) at that maximum rate.
Vehicle Queue Animation — Red → Green
🎬 ANIMATION — QUEUE DISCHARGE: WATCH VEHICLES CLEAR ON GREEN
Queue Size:8 vehHeadway h (s):2.1s
Saturation Flow Rate
\[ s = \dfrac{3600}{h} \]
s = saturation flow rate (veh/h/ln) | h = saturation headway (s/veh) | Typical h ≈ 2.0–2.5 s → s ≈ 1440–1800 veh/h
Typical values: For a standard approach lane, s ≈ 1,800 veh/h/ln is the HCM baseline. Adjustments apply for heavy vehicles, grades, turning movements, etc.
08 — LANE GROUP CAPACITY & v/c RATIO
Capacity, Demand, and the X Factor
At a signalized intersection, a lane group doesn't get the full hour of green — it only gets a fraction (g/C). So its effective capacity is reduced proportionally.
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X = 0.99 and X = 1.01 are not 2% apart in how they feel. Below 1.0 the queue empties every cycle, so delay is finite and the intersection recovers. Above 1.0 each cycle leaves a few cars behind, and those leftovers accumulate — delay is no longer a number, it is a slope. That is why v/c is a threshold and not a score, and why every capacity formula on this page exists to keep you on the right side of a single value.
If a factory can make 1,000 widgets/hour at full speed, but it's only allowed to run for 37 out of every 70 minutes — its effective capacity is 1,000 × (37/70) = 529 widgets per period. Same logic for lanes: full saturation × green ratio = actual capacity.
Lane Group Capacity
\[ c_i = s_i \cdot \dfrac{g_i}{C} \]
sᵢ=saturation flow gᵢ=effective green C=cycle length
v/c Ratio (X) — The Key Performance Indicator
\[ X = \dfrac{v}{c} = \dfrac{v_i}{s_i \cdot (g_i/C)} \]
X < 1.0 → OK (demand < capacity). X = 1.0 → at capacity. X > 1.0 → FAILURE — queue grows indefinitely.
⚙️ CALCULATOR — LANE GROUP CAPACITY & v/c RATIO
Practice Problems — Capacity & v/c
PROBLEM 5
A lane group has: s = 1,800 veh/h, g = 28s, C = 90s, v = 450 veh/h. Find (a) lane group capacity, (b) v/c ratio, and (c) whether the approach is operating acceptably.
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(a) Lane Group Capacity:
cᵢ = sᵢ × (gᵢ/C) = 1,800 × (28/90) = 1,800 × 0.311 = 560 veh/h(b) v/c Ratio:
X = v / c = 450 / 560 = 0.804(c) Assessment:
X = 0.804 < 1.0 → Approach is operating within capacity ✓
However, X > 0.80 signals the approach is moderately loaded.
As X → 1.0, delays increase rapidly.
✓ c = 560 veh/h | X = 0.804 | Acceptable (barely)
PROBLEM 6
An approach operates with X = 1.15. What does this mean physically? If you want to reduce X to 0.90, and current volume is 800 veh/h with s = 1,800 veh/h and C = 80s, what effective green time g is needed?
▶
X = 1.15 means:
Demand (1.15×) exceeds capacity (1.0×) by 15%.
Queue grows every cycle and NEVER clears → breakdown → LOS F
Solving for required g at X = 0.90:
X = v / [s × (g/C)]
0.90 = 800 / [1,800 × (g/80)]
1,800 × g/80 = 800 / 0.90 = 888.9
g/80 = 888.9 / 1,800 = 0.494
g = 0.494 × 80 = 39.5s ≈ 40sInterpretation:
Increasing effective green from current value to 40s will
bring the approach within acceptable capacity.
✓ X > 1.0 → Failure. Required g ≈ 40s for X = 0.90
09 — CONTROL DELAY & LEVEL OF SERVICE
How Bad is the Wait? The LOS System
Level of Service (LOS) is the report card of an intersection. It grades performance from A (smooth, fast) to F (failed, gridlock). The metric is average control delay per vehicle (seconds).
Why delay is the metric, not speed
Speed sounds like the natural way to grade an intersection, but a vehicle stopped at a red light has a speed of zero no matter how good the design is. What a driver actually experiences is how long they waited that they would not have waited on an empty road — control delay. It is measurable, it adds up along an approach, and it maps to something a non-engineer immediately understands: seconds of their life. LOS A–F is nothing more than that number cut into bands.
LOS is like restaurant wait times. LOS A = walk-in, seated immediately. LOS B = 5-min wait. LOS C = 20 min. LOS D = 45 min. LOS E = 1+ hour wait, people getting restless. LOS F = you're turned away.
Using the same approach as Problem 7 (g = 36s, s = 1,800 veh/h, C = 90s), if volume increases to v = 700 veh/h, recalculate d₁ and LOS. How much did delay increase with just a 100 veh/h increase?
▶
Same c = 720, g/C = 0.400. New X:
X = 700/720 = 0.972 (much closer to capacity!)
d₁ numerator (same):
0.5 × 90 × (0.600)² = 16.20
d₁ denominator:
1 − 0.972 × 0.400 = 1 − 0.389 = 0.611
d₁ = 16.20 / 0.611 = 26.5 s/veh → LOS CComparison:
At v=600: d₁ = 24.3 s/veh (LOS C)
At v=700: d₁ = 26.5 s/veh (LOS C, but getting worse)
Delta: +2.2 s/veh for +100 veh/h increase
This demonstrates that as X approaches 1.0,
small volume increases cause disproportionate delay increases!
✓ d₁ = 26.5 s/veh | LOS C | +2.2s delay for +100 veh/h
10 — PUBLIC TRANSPORTATION
Moving Many People Efficiently
Public transportation moves people collectively, reducing the number of vehicles on roads. It ranges from taxis all the way to fully grade-separated rapid rail systems.
Public transit is like a shared rideshare vs. individual cars. One bus at 40 people = 38 fewer cars on the road. The tradeoff: less flexibility but huge capacity gains.
Mode
Right-of-Way
Capacity
Best For
Taxi/Ride-hail
Mixed traffic
1–4 pax
Door-to-door, on-demand
Jitneys
Mixed traffic
5–15 pax
Fixed route, flexible schedule
Regular Bus
Mixed traffic
40–60 pax
Urban corridors, frequent stops
BRT (Semi-rapid)
Dedicated lanes
60–160 pax
High-volume urban corridors
Light Rail (LRT)
Reserved ROW
100–200 pax
Regional corridors
Rapid Transit (Metro)
Fully separated
1,000+ pax/train
Dense urban networks
11 — TRANSPORTATION PLANNING
Planning Before Building
Transportation planning is the process of deciding what infrastructure to build, where, when, and for whom — before spending billions of dollars. It requires predicting future travel demand.
Planning a transit system without demand forecasting is like opening a restaurant without checking if anyone wants to eat there.
The Four-Step Travel Demand Forecasting Model
TAZ (Traffic Analysis Zone) = The basic unit of analysis. Each zone has a centroid from which all trips originate/end. Zones are small (city blocks) in dense areas and larger in suburbs.
1
Trip Generation
How many trips start and end in each zone? Based on land use, income, household size.
→ Productions & Attractions
2
Trip Distribution
Where do those trips go? Uses the Gravity Model.
→ O-D Matrix
3
Mode Split
Which mode do people use? Car, bus, train, walk? Uses Logit model based on utility.
→ Modal shares (%)
4
Traffic Assignment
Which specific routes do they take?
→ Link volumes
12 — STEP 1: TRIP GENERATION
Who Makes How Many Trips?
Trip generation predicts how many trips start or end in each zone based on zone characteristics. It answers: "how much demand will this zone generate?"
Trip Purpose Categories
🏠→🏢 HBW
Home-Based Work. Most time-sensitive. Drives morning/evening peak.
🏠→🛒 HBO
Home-Based Other. Home to shopping, school, recreation. More flexible timing.
🏢→🏢 NHB
Non-Home Based. Lunch trips, deliveries, errands between non-home stops.
Production vs Attraction
\[ \text{Production: trip end at RESIDENTIAL land use} \]
\[ \text{Attraction: trip end at NON-RESIDENTIAL land use} \]
Key constraint: ΣP = ΣA. If not balanced, scale: Aᵢ(adj) = Aᵢ × (ΣP / ΣA)
13 — STEP 2: TRIP DISTRIBUTION (GRAVITY MODEL)
Where Do Those Trips Actually Go?
Trip distribution connects origin zones to destination zones using the Gravity Model — more trips go to zones that are larger and closer.
⚡
The name is the analogy, and it is a good one. Newton said attraction grows with mass and falls off with distance squared. Swap mass for how many jobs or shops a zone offers, and distance for travel time, and you have the gravity model unchanged: Tij ∝ Aj/tijc. The friction exponent c is the only thing you tune — it says how strongly this particular city dislikes travelling. Nothing about planets is being claimed; the shape of the relationship is just the same.
Gravity analogy: A huge mall far away competes with a small mall nearby. The gravity model formalizes this: More mass (attractions) + less distance = more trips attracted.
Mode split predicts what fraction of trips will use each available mode. Based on utility — each mode has a perceived benefit/cost, and travelers pick the highest utility option.
Why the exponential, and not just a ratio
You might expect mode share to be a simple ratio of utilities. The trouble is utilities can be negative (cost and time are penalties), and negative shares are meaningless. Taking eV fixes that: it is always positive, and it turns differences into ratios — only the gap between two modes' utilities matters, not where the zero sits. That is why P(i) = eV(i) / ΣeV(r) is the standard choice, and why adding a constant to every mode changes nothing.
Choosing between coffee shops: The closest is convenient (low travel time = high utility). The fancy one is expensive but better quality. You weigh cost vs. quality subconsciously. The logit model formalizes this weighting.
a₀=mode-specific constant X₁=cost (cents) X₂=travel time (min) Coefficients are negative (more cost/time = lower utility)
Multinomial Logit — Probability of Choosing Mode i
\[ P(i) = \dfrac{e^{V(i)}}{\sum_r e^{V(r)}} \]
V(i)=computed utility of mode i Higher probability = more people choose this mode
⚙️ CALCULATOR — LOGIT MODE SPLIT
Utility: U = a − 0.002·X₁ − 0.05·X₂ (X₁=cost in cents, X₂=time in min)
🚗 AUTOMOBILE
🚌 BUS
🚈 LIGHT RAIL
Practice Problems — Logit Mode Split
PROBLEM 10 ★
Two modes are available: Auto and Bus. Utility function: U = a − 0.025·(cost $) − 0.04·(time min). Auto: a=0, cost=$2.50, time=20min. Bus: a=0.50, cost=$1.00, time=40min. Calculate the probability of choosing each mode.
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Step 1 — Auto utility:
V_auto = 0 − 0.025(2.50) − 0.04(20)
= 0 − 0.0625 − 0.800 = −0.8625Step 2 — Bus utility:
V_bus = 0.50 − 0.025(1.00) − 0.04(40)
= 0.50 − 0.025 − 1.600 = −1.125Step 3 — Exponentiate:
e^V_auto = e^(−0.8625) = 0.4222
e^V_bus = e^(−1.125) = 0.3247
Step 4 — Sum of exponentials:
Σ = 0.4222 + 0.3247 = 0.7469
Step 5 — Probabilities:
P(Auto) = 0.4222 / 0.7469 = 0.565 = 56.5%
P(Bus) = 0.3247 / 0.7469 = 0.435 = 43.5%Note:
Even though auto costs more and bus has a mode constant bonus,
auto's shorter travel time (20 vs 40 min) makes it win.
✓ P(Auto) = 56.5% | P(Bus) = 43.5%
📝 PRACTICE HUB
More Practice Problems — Mixed Topics
These problems mix multiple topics and are representative of exam-style questions.
MIXED PROBLEM A — Multi-Step
An intersection approach has: V = 960 veh/hr (peak hour), 15-min counts are 210, 260, 280, 210. PHF? Use PHF-adjusted flow. Then with s=1800, g=40s, C=100s, find capacity, X, uniform delay d₁, and LOS.
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Part 1 — PHF:
V = 210+260+280+210 = 960 veh/hr
V₁₅,max = 280
PHF = 960/(4×280) = 960/1120 = 0.857Part 2 — Adjusted flow rate:
v = V/PHF = 960/0.857 = 1,121 veh/hrPart 3 — g/C ratio:
g/C = 40/100 = 0.400
Part 4 — Capacity:
c = 1,800 × 0.400 = 720 veh/hrPart 5 — v/c ratio:
X = 1,121/720 = 1.557 → OVER CAPACITY → LOS F immediately
Part 6 — LOS F (X > 1.0):
When X > 1.0, queue grows every cycle.
d₁ formula becomes undefined (denominator → 0 or negative).
Report LOS F. The intersection MUST be redesigned.
✓ PHF=0.857 | v=1,121 veh/h | X=1.56 | LOS F (Failure)
MIXED PROBLEM B — Saturation
Vehicles depart from a queue at saturation headway h = 2.2 sec/veh. (a) What is the saturation flow rate? (b) If the effective green is 45s, how many vehicles can depart in one green phase? (c) If demand is 750 veh/hr, is the green long enough?
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(a) Saturation flow rate:
s = 3600 / h = 3600 / 2.2 = 1,636 veh/hr(b) Vehicles per green phase:
(Need cycle length for capacity. Assume C = 90s)
c = s × (g/C) = 1,636 × (45/90) = 1,636 × 0.500 = 818 veh/hr
Vehicles per cycle = c × C/3600 = 818 × 90/3600 = 20.5 veh/cycle(c) Sufficient capacity?
X = v/c = 750/818 = 0.917 < 1.0 → YES, green is sufficient ✓
The approach can handle 750 veh/hr with some headroom.
Three zones: Zone 1 (P=1000, A=0), Zone 2 (P=0, A=600), Zone 3 (P=0, A=400). Travel times: t₁₂=3min, t₁₃=5min. Using F=1/t, distribute Zone 1's 1000 trips to Zones 2 and 3.
▶
F values (F = 1/t, not 1/t²):
F₁₂ = 1/3 = 0.333
F₁₃ = 1/5 = 0.200
Weighted attractions:
A₂ × F₁₂ = 600 × 0.333 = 200.0
A₃ × F₁₃ = 400 × 0.200 = 80.0
Sum:
Σ = 200.0 + 80.0 = 280.0
Trip distribution:
T₁₂ = 1000 × (200/280) = 1000 × 0.714 = 714 trips
T₁₃ = 1000 × (80/280) = 1000 × 0.286 = 286 tripsCheck: 714 + 286 = 1,000 = P₁ ✓
Note: Zone 2 attracts more despite fewer attractions because
it is much closer (3 min vs. 5 min).
✓ T₁₂ = 714 trips | T₁₃ = 286 trips
✓ QUICK CHECK
Do You Actually Believe These?
Six questions aimed at the misconceptions rather than at recall. Each one is a place students lose marks by reciting a formula they have not felt.
1. An approach carries 1,600 veh/h with PHF = 0.85. Which number goes into the capacity check?
2. Why does an optimum cycle length exist at all — why not just make the cycle very short?
3. Two lane groups have X = 0.98 and X = 1.02. How different are they in practice?
4. An intersection is graded on average control delay rather than average speed. Why?
5. In the gravity model, what does raising the friction exponent c actually say about a city?
6. You add 10 units of utility to every mode in a logit model. What happens to the mode shares?
💡 EXAM STRATEGIES
Tips to Ace the Exam
The most common mistakes and how to avoid them.
01
Always Adjust for PHF First
If given hourly volume V AND PHF, always compute v = V/PHF before any capacity analysis. Using raw V instead of v is the #1 mistake.
02
g ≠ G (effective ≠ display)
Effective green g = G + y + ar − tL. Never plug G directly into capacity formulas. Always convert first.
03
X > 1.0 = Automatic LOS F
If v/c > 1.0, you don't need to calculate d₁. The approach fails. Demand exceeds capacity → queue grows every cycle.
04
Gravity Model: Check Balance
After computing Tᵢⱼ, sum each column (Dⱼ). If Dⱼ ≠ Aⱼ, you need to iterate with corrected attraction values.
05
Logit: Show All 3 Steps
Always show: (1) Compute U for each mode, (2) Compute e^U, (3) P(i) = e^U(i) / Σe^U. Partial credit requires each step.
06
Units Are Everything
s is in veh/hr, g and C in seconds, v in veh/hr. If g/C is dimensionless, your units are right. If not, re-check.
⚠️ Common trap: The d₁ denominator is [1 − min(1,X)·(g/C)], not [1 − X·(g/C)]. When X > 1, use min(1, X) = 1, not X. This prevents the formula from returning negative delay (which is physically impossible).
📋 QUICK REFERENCE
Formula Sheet — All CEE 350 Formulas
Traffic Volume
Peak Hour Factor
PHF = V / (4 × V₁₅,max)
V = hourly volume, V₁₅,max = max 15-min count
Traffic Volume
Max Flow Rate
v = V / PHF
Convert hourly volume to peak flow rate for analysis
Signal Timing
Effective Green
g = G + y + ar − tL
tL = l₁ + l₂ ≈ 4 s/phase typically
Saturation Flow
Sat. Flow Rate
s = 3600 / h
h = saturation headway (s/veh). Typical s ≈ 1800 veh/h
5–6 total questions. 1 descriptive (up to intersection/signal design lecture) + 4–5 analytical problems across PHF, signal timing, and travel demand forecasting.
মোট ৫–৬টি প্রশ্ন। ১টি বর্ণনামূলক (ইন্টারসেকশন/সিগন্যাল ডিজাইন লেকচার পর্যন্ত) + ৪–৫টি বিশ্লেষণমূলক সমস্যা — PHF, সিগন্যাল টাইমিং এবং ট্রাভেল ডিমান্ড ফোরকাস্টিং।
Announced Breakdown (March 30, 2026):
Part A — Descriptive
Max 1 question · up to Intersection/Signal Design lecture
PHF & Related
Max 1 problem
Signal Timing Plan
Max 3 problems
Travel Demand
Step 2 (Gravity model) + Step 3 (Mode split) · Max 1 problem
Part A — Descriptive Question
Question 1 — Intersection & Signal Design
Q1
(a) Distinguish between at-grade and grade-separated intersections. Under what traffic volume conditions is each preferred?
(b) For a signalized intersection with heavy left-turn demand on one approach, explain the difference between a protected and a permitted left-turn phase. What is the implication for signal timing and pedestrian safety?
(c) Define Level of Service (LOS) A through F for a signalized intersection. What control delay threshold separates LOS C from LOS D?
(ক) অ্যাট-গ্রেড এবং গ্রেড-বিচ্ছিন্ন ইন্টারসেকশনের মধ্যে পার্থক্য করুন। কোন ট্রাফিক ভলিউম পরিস্থিতিতে প্রতিটি পছন্দনীয়?
(খ) সুরক্ষিত এবং অনুমতিপ্রাপ্ত বাম-মোড় ফেজের পার্থক্য ব্যাখ্যা করুন। সিগন্যাল টাইমিং এবং পথচারী নিরাপত্তায় এর প্রভাব কী?
(গ) সিগন্যালাইজড ইন্টারসেকশনের জন্য LOS A থেকে F সংজ্ঞায়িত করুন। LOS C ও D-এর মধ্যে নিয়ন্ত্রণ বিলম্বের সীমা কত?
KEY POINTS TO COVER — At-grade = all movements at same level, low-moderate volumes; grade-separated = flyover/underpass, high-speed/high-volume. Protected LTurn = dedicated phase, no conflict with oncoming; permitted = turns through gap in oncoming. LOS C/D boundary = 35 s/veh control delay.
মূল বিষয়: অ্যাট-গ্রেড = একই উচ্চতায় সব চলাচল, কম-মাঝারি ভলিউম; গ্রেড-বিচ্ছিন্ন = ফ্লাইওভার/আন্ডারপাস, উচ্চগতি/উচ্চ ভলিউম। সুরক্ষিত বাম-মোড় = নিবেদিত ফেজ, বিপরীত ট্রাফিকের সাথে কোনো দ্বন্দ্ব নেই। LOS C/D সীমা = ৩৫ s/veh।
Part (a): At-grade intersections have all movements at the same elevation — appropriate for urban/suburban roads with moderate volumes (AADT < 30,000 typically). Grade-separated interchanges physically separate crossing traffic — required for freeways, high-speed arterials, or where stopping traffic would cause severe delay/safety issues.
Part (b): Protected left-turn = dedicated signal phase where left-turners have a green arrow and no conflict with oncoming traffic. Permitted = left-turn allowed on circular green by finding gaps in oncoming flow. Protected is safer and needed when left-turn volume is high, but requires extra cycle time. Pedestrian safety: during protected phase, pedestrians crossing the conflicting street can also get a walk signal safely.
Part (c): LOS A ≤10 s/veh · LOS B 10–20 · LOS C 20–35 · LOS D 35–55 · LOS E 55–80 · LOS F >80 s/veh (or X > 1.0). LOS C/D boundary = 35 s/veh.
→ LOS C/D = 35 s/veh | Protected LTurn = no conflict with oncoming | Grade-separated for high speed/volume
Part B — PHF & Related (Max 1 Problem)
Problem 2 — Peak Hour Factor
P2
During the 4:00–5:00 PM peak hour, a traffic engineer recorded the following 15-minute counts at a signalized approach:
বিকাল ৪:০০–৫:০০ PM পিক আওয়ারে, একজন ট্রাফিক ইঞ্জিনিয়ার একটি সিগন্যালাইজড অ্যাপ্রোচে ১৫-মিনিটের গণনা রেকর্ড করেছেন:
Interval
Volume (veh)
4:00–4:15
310
4:15–4:30
385
4:30–4:45
420
4:45–5:00
345
(a) Calculate the Peak Hour Factor (PHF).
(b) Find the peak flow rate v for capacity analysis.
(c) The approach has a single lane with saturation flow rate s = 1,800 veh/h and cycle length C = 90 s. What minimum effective green time g is needed to maintain X ≤ 0.90?
(ক) পিক আওয়ার ফ্যাক্টর (PHF) হিসাব করুন।
(খ) ক্যাপাসিটি বিশ্লেষণের জন্য পিক ফ্লো রেট v নির্ণয় করুন।
(গ) অ্যাপ্রোচে একটি লেন আছে, s = ১,৮০০ veh/h এবং C = ৯০ s। X ≤ ০.৯০ বজায় রাখতে ন্যূনতম কার্যকর সবুজ সময় g কত হওয়া দরকার?
💡 HINT — V = sum of all 4 intervals. V₁₅,max = 420. PHF = V/(4×V₁₅,max). For part (c): X = v/c = v/[s×(g/C)] → solve for g.
💡 ইঙ্গিত — V = চারটি বিরতির যোগফল। V₁₅,max = ৪২০। PHF = V/(4×V₁₅,max)। (গ) অংশের জন্য: X = v/[s×(g/C)] → g-এর জন্য সমাধান করুন।
Step 1 — Total hourly volume: V = 310 + 385 + 420 + 345 = 1,460 veh/h
Round up to nearest 5 s → C₀ = 70 s. Note: ΣY = 0.750 is high — intersection is near capacity.
✓ y₁ = 0.300, y₂ = 0.450, ΣY = 0.750 → C₀ = 70 s
Problem 4 — Green Time Allocation
P4
Using the intersection data from Problem 3 (C = 70 s, L = 8 s, y₁ = 0.300, y₂ = 0.450), allocate the effective green times g₁ and g₂. Then find the actual displayed green G for each phase, given Y = 3 s (yellow) and AR = 1 s (all-red) per phase.
সমস্যা ৩-এর ডেটা ব্যবহার করে কার্যকর সবুজ সময় g₁ ও g₂ বরাদ্দ করুন। প্রতিটি ফেজের প্রকৃত প্রদর্শিত সবুজ G নির্ণয় করুন, Y = ৩ s (হলুদ) এবং AR = ১ s।
💡 HINT — Total effective green = C − L. Split proportionally to yᵢ/ΣY. Actual green: G = g − Y − AR + tL (since g = G + Y + AR − tL).
💡 ইঙ্গিত — মোট কার্যকর সবুজ = C − L। yᵢ/ΣY অনুপাতে ভাগ করুন। প্রকৃত সবুজ: G = g − Y − AR + tL।
Step 1 — Total effective green:
EFFECTIVE GREEN AVAILABLE
\[ g_\text{total} = C - L = 70 - 8 = 62 \text{ s} \]
Check: (G₁ + Y + AR) + (G₂ + Y + AR) = (25+3+1)+(37+3+1) = 29+41 = 70 s ✓
✓ g₁ = 25 s, g₂ = 37 s | G₁ = 25 s, G₂ = 37 s | Cycle check: 70 s ✓
Problem 5 — v/c Ratio, Delay & LOS
P5
Using C = 70 s, s = 1,800 veh/h, g₁ = 25 s, g₂ = 37 s, v₁ = 540 veh/h, v₂ = 810 veh/h:
(a) Calculate the v/c ratio (X) for each phase.
(b) Calculate the uniform delay d₁ for Phase 2 (the critical phase).
(c) Determine the LOS for Phase 2. (Assume d₂ = d₃ = 0.)
C = ৭০ s, s = ১,৮০০ veh/h, g₁ = ২৫ s, g₂ = ৩৭ s, v₁ = ৫৪০ veh/h, v₂ = ৮১০ veh/h ব্যবহার করে:
(ক) প্রতিটি ফেজের v/c অনুপাত (X) হিসাব করুন।
(খ) ফেজ ২ (ক্রিটিক্যাল ফেজ)-এর জন্য ইউনিফর্ম বিলম্ব d₁ হিসাব করুন।
(গ) ফেজ ২-এর LOS নির্ধারণ করুন। (ধরুন d₂ = d₃ = ০।)
💡 HINT — cᵢ = sᵢ × (gᵢ/C). Xᵢ = vᵢ/cᵢ. d₁ = 0.5C(1−g/C)² / [1−min(1,X)·(g/C)]. LOS from control delay table.
💡 ইঙ্গিত — cᵢ = sᵢ × (gᵢ/C)। Xᵢ = vᵢ/cᵢ। d₁ = 0.5C(1−g/C)² / [1−min(1,X)·(g/C)]। LOS নিয়ন্ত্রণ বিলম্ব সারণি থেকে।
Step 4 — LOS: d = 14.1 s/veh → LOS B (10–20 s/veh range)
✓ X₁ = 0.840, X₂ = 0.851 | d₁ = 14.1 s/veh | LOS B
Part D — Travel Demand Forecasting (Step 2 + Step 3)
Problem 6 — Gravity Model (Step 2) + Mode Split (Step 3)
P6
Step 2 — Trip Distribution (Gravity Model)
A study area has 3 zones. Productions and attractions have been balanced at a total of 800 trips. Friction factors Fᵢⱼ are given (K = 1.0 for all zone pairs):
Find the trip distribution matrix Tᵢⱼ for Zones 1 and 2.
Step 3 — Mode Split (Logit Model)
For the Zone 1→Zone 2 trips, a binary logit model gives utilities:
V(auto) = −0.40 − 0.025×(travel time, min) − 0.10×(cost, BDT)
V(transit) = 0 − 0.025×(travel time, min) − 0.10×(cost, BDT)
Auto: 22 min, cost = 30 BDT. Transit: 38 min, cost = 8 BDT.
Find the probability of choosing auto and transit. How many of the T₁₂ trips use each mode?
জোন ১ ও ২-এর জন্য ট্রিপ ডিস্ট্রিবিউশন ম্যাট্রিক্স Tᵢⱼ নির্ণয় করুন।
ধাপ ৩ — মোড স্প্লিট (লজিট মডেল)
জোন ১→জোন ২ ট্রিপের জন্য বাইনারি লজিট মডেল ইউটিলিটি:
V(অটো) = −০.৪০ − ০.০২৫×(ভ্রমণ সময়, মিনিট) − ০.১০×(খরচ, টাকা)
V(ট্রানজিট) = ০ − ০.০২৫×(ভ্রমণ সময়) − ০.১০×(খরচ)
অটো: ২২ মিনিট, খরচ = ৩০ টাকা। ট্রানজিট: ৩৮ মিনিট, খরচ = ৮ টাকা।
অটো ও ট্রানজিট বেছে নেওয়ার সম্ভাবনা নির্ণয় করুন। T₁₂ ট্রিপের মধ্যে কতটি প্রতিটি মোড ব্যবহার করবে?
💡 HINT — Tᵢⱼ = Pᵢ × (Aⱼ·Fᵢⱼ) / Σⱼ(Aⱼ·Fᵢⱼ). Then P(i) = eⱽ⁽ⁱ⁾ / Σeⱽ⁽ʳ⁾.