πŸŽ“ CEE 350 Β· Traffic Analysis & Design Β· Spring 2026

Traffic Engineering
Explained Simply.

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 β€” 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.

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.
πŸ›£οΈ

Mobility

Moving efficiently from A to B at speed. Freeways maximize this β€” no stops, no driveways, no signals.

High speed β†’ Low access
🏠

Accessibility

Being able to pull in and out of properties. Local streets maximize this β€” driveways everywhere.

Low speed β†’ High access
βš–οΈ

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 β€” minimizing crashes, delays, and fuel waste simultaneously.
02 β€” HIGHWAY FUNCTIONAL CLASSIFICATION

Road Hierarchy β€” Not All Roads Are Equal

Roads are classified by their function β€” 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 β€” LIVE INTERSECTION VIEW: 4-WAY SIGNALIZED INTERSECTION
Phase: EW GREEN
Timer: 0s
Road TypeSpeed (mph)AccessPrimary FunctionExample
Freeway60–75None (controlled)Long-distance mobilityI-95, I-90
Principal Arterial45–55Very limitedRegional movementUS-1, State highways
Minor Arterial35–45LimitedConnecting to arterialsMajor city roads
Collector25–35ModerateCollect neighborhood trafficShopping center roads
Local Street15–25HighDirect property accessResidential 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)
πŸ”΄

Interrupted Flow

External devices (signals, stop signs) periodically halt traffic. Flow is at the mercy of timing.

All signalized streets, STOP-controlled roads
03 β€” 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.

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).
πŸ‹οΈ

Capacity (c)

The maximum vehicles a section can handle under typical conditions. A ceiling β€” you can't exceed it physically.

veh/hr/lane β€” a fixed physical limit
πŸ™‹

Demand

How many vehicles want to use the road. Demand can exceed capacity β€” that's when queues form.

veh/hr β€” what drivers want to do
πŸ“Š

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 β€” 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 β€” the true peak rate

Peak Hour Factor (PHF) β€” 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
PHF = V / (4 Γ— V₁₅,max)
V = hourly volume  |  V₁₅,max = max 15-min count  |  Range: 0.25 to 1.00
Maximum Flow Rate
v = V / PHF
Convert hourly volume to equivalent flow rate. Use this for capacity analysis, not raw volume.
⚑ SIM β€” PHF CALCULATOR: ENTER YOUR 15-MIN COUNTS
Practice Problems β€” 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?
β–Ά
πŸ’‘ HINTPHF = V_hourly / (4 Γ— 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 β€” Total hourly volume: V = 410 + 480 + 530 + 420 = 1,840 veh/hr Step 2 β€” Find maximum 15-minute count: V₁₅,max = 530 veh (the 5:00–5:15 period) Step 3 β€” Calculate PHF: PHF = 1,840 / (4 Γ— 530) = 1,840 / 2,120 = 0.868 Step 4 β€” Maximum flow rate: v = V / PHF = 1,840 / 0.868 = 2,120 veh/hr Step 5 β€” 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.
βœ“ 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?
β–Ά
Given: V = 900 veh/hr, PHF = 0.92 Maximum flow rate: v = V / PHF = 900 / 0.92 = 978 veh/hr Why this matters: Using 900 underestimates peak demand by ~8%. Using v = 978 correctly reflects the 15-min peak intensity.
βœ“ v = 978 veh/hr
04 β€” 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 β€” someone must yield.

An intersection is like a four-way conversation. Everyone can't talk simultaneously β€” someone facilitates (the signal) or people take turns (yield/stop control). Without any control, it's chaos.

Levels of Intersection Control

πŸ†“

No Control

Drivers adjust speed on their own. Only suitable for very low volume, good sight distance intersections.

πŸ”Ί

Yield

Minor road yields to major road. Driver must slow but can proceed if clear. Shared responsibility.

πŸ›‘

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 β€” TRAFFIC SIGNALS

Teaching Drivers to Share Space

Traffic signals are the most powerful tool for managing intersection conflicts. But they must be justified β€” signals aren't always the solution, and a poorly-timed signal can make things worse.

Live Signal Demo β€” Watch the Cycle

🚦 ANIMATION β€” 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): 70s Green 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 TypeShapeColorExamples
RegulatoryVertical RectangleBlack/WhiteSTOP, YIELD, Speed Limit, NO PARKING
WarningDiamondBlack/YellowCurve Ahead, Railroad, Bridge
GuideHoriz. RectangleWhite/GreenDestinations, Distance
06 β€” 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 β€” that's optimal timing.

Key Timing Terms β€” Explained One By One

πŸ”„ 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 β†’ 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

πŸ’š 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

βœ… Effective Green (g)

Time vehicles can actually discharge. Slightly longer than G because yellow allows some movement. Key formula input.

g = G + y + ar βˆ’ tL

⏱️ Lost Time (tL)

Wasted time per phase: startup (vehicles accelerating from rest) + clearance (last yellow vehicle). Usually β‰ˆ4 sec.

tL = l₁ + lβ‚‚ β‰ˆ 4 sec/phase

πŸ”΄ All-Red (AR)

Every approach red simultaneously. Safety clearance β€” the intersection empties before next phase starts.

Typically 1–2 seconds
Effective Green Time
g = G + y + ar βˆ’ tL
G=display green   y=yellow   ar=all-red   tL=lost time per phase (~4s)
⚑ SIM β€” SIGNAL TIMING ANIMATOR: WATCH THE CYCLE LIVE PHASE 1
CYCLE (s): 70s G1 (s): 37s
Practice Problems β€” 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?
β–Ά
Phase 1 β€” Effective Green: g₁ = G₁ + y + ar βˆ’ tL = 42 + 4 + 2 βˆ’ 4 = 44s Phase 2 Display Green: Gβ‚‚ = C βˆ’ G₁ βˆ’ y βˆ’ ar βˆ’ y βˆ’ ar = 80 βˆ’ 42 βˆ’ 4 βˆ’ 2 βˆ’ 4 βˆ’ 2 = 26s Phase 2 β€” Effective Green: gβ‚‚ = Gβ‚‚ + y + ar βˆ’ tL = 26 + 4 + 2 βˆ’ 4 = 28s Total Lost Time per Cycle: L = 2 Γ— tL = 2 Γ— 4 = 8s per cycle Cycle Efficiency: Efficiency = (C βˆ’ L)/C = (80 βˆ’ 8)/80 = 72/80 = 90% (Only 10% of cycle time is "wasted" on lost time)
βœ“ g₁=44s | gβ‚‚=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?
β–Ά
Effective Green: g = G + y + ar βˆ’ tL = 35 + 3.5 + 1.5 βˆ’ 4 = 36s g/C Ratio: g/C = 36 / 90 = 0.40 Meaning: This phase receives 40% of the cycle for vehicle discharge. Only 40% of the saturation flow rate becomes available capacity.
βœ“ g = 36s | g/C = 0.40
07 β€” 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 β€” 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 β€” Red β†’ Green

🎬 ANIMATION β€” QUEUE DISCHARGE: WATCH VEHICLES CLEAR ON GREEN
Queue Size: 8 veh Headway h (s): 2.1s
Saturation Flow Rate
s = 3600 / h
s = saturation flow rate (veh/h/ln)  |  h = saturation headway (s/veh)  |  Typical h β‰ˆ 2.0–2.5 s β†’ s β‰ˆ 1440–1800 veh/h
Typical values: For a standard approach lane, s β‰ˆ 1,800 veh/h/ln is the HCM baseline. Adjustments apply for heavy vehicles, grades, turning movements, etc.
08 β€” 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 β€” it only gets a fraction (g/C). So its effective capacity is reduced proportionally.

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 β€” its effective capacity is 1,000 Γ— (37/70) = 529 widgets per period. Same logic for lanes: full saturation Γ— green ratio = actual capacity.
Lane Group Capacity
cα΅’ = sα΅’ Γ— (gα΅’ / C)
sα΅’=saturation flow   gα΅’=effective green   C=cycle length
v/c Ratio (X) β€” The Key Performance Indicator
X = v / c = vα΅’ / [sα΅’ Γ— (gα΅’/C)]
X < 1.0 β†’ OK (demand < capacity). X = 1.0 β†’ at capacity. X > 1.0 β†’ FAILURE β€” queue grows indefinitely.
βš™οΈ CALCULATOR β€” LANE GROUP CAPACITY & v/c RATIO
Practice Problems β€” 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.
β–Ά
(a) Lane Group Capacity: cα΅’ = sα΅’ Γ— (gα΅’/C) = 1,800 Γ— (28/90) = 1,800 Γ— 0.311 = 560 veh/h (b) v/c Ratio: X = v / c = 450 / 560 = 0.804 (c) Assessment: X = 0.804 < 1.0 β†’ Approach is operating within capacity βœ“ However, X > 0.80 signals the approach is moderately loaded. As X β†’ 1.0, delays increase rapidly.
βœ“ 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?
β–Ά
X = 1.15 means: Demand (1.15Γ—) exceeds capacity (1.0Γ—) by 15%. Queue grows every cycle and NEVER clears β†’ breakdown β†’ LOS F Solving for required g at X = 0.90: X = v / [s Γ— (g/C)] 0.90 = 800 / [1,800 Γ— (g/80)] 1,800 Γ— g/80 = 800 / 0.90 = 888.9 g/80 = 888.9 / 1,800 = 0.494 g = 0.494 Γ— 80 = 39.5s β‰ˆ 40s Interpretation: Increasing effective green from current value to 40s will bring the approach within acceptable capacity.
βœ“ X > 1.0 β†’ Failure. Required g β‰ˆ 40s for X = 0.90
09 β€” 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).

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.
A
≀10 s/veh
B
>10–20
C
>20–35
D
>35–55
E
>55–80
F
>80 s or X>1
Uniform Delay d₁ (Webster / HCM)
d₁ = 0.5C(1 βˆ’ g/C)Β² / [1 βˆ’ min(1,X)Β·(g/C)]
C=cycle length   g=effective green   X=v/c ratio   Result in seconds/vehicle
Total Control Delay
d = d₁ + dβ‚‚ + d₃
d₁=uniform delay   dβ‚‚=incremental (random arrival) delay   d₃=preexisting queue delay
βš™οΈ CALCULATOR β€” UNIFORM DELAY + LOS
⚑ SIM β€” DELAY vs v/c: SEE HOW DELAY EXPLODES AS X β†’ 1.0
CYCLE C (s): 70s g/C RATIO: 0.53
Practice Problems β€” Delay & LOS
PROBLEM 7 β˜… (Exam Favorite)
A lane group has: C = 90s, g = 36s, v = 600 veh/h, s = 1,800 veh/h. Calculate (a) v/c ratio, (b) uniform delay d₁, and (c) LOS. Show all steps.
β–Ά
πŸ’‘ FORMULAd₁ = 0.5C(1 βˆ’ g/C)Β² / [1 βˆ’ min(1,X)Β·(g/C)]
Step 1 β€” g/C ratio: g/C = 36/90 = 0.400 Step 2 β€” Lane group capacity: c = s Γ— (g/C) = 1,800 Γ— 0.400 = 720 veh/h Step 3 β€” v/c ratio X: X = v/c = 600/720 = 0.833 Step 4 β€” Uniform delay numerator: 0.5 Γ— C Γ— (1 βˆ’ g/C)Β² = 0.5 Γ— 90 Γ— (1 βˆ’ 0.400)Β² = 45 Γ— (0.600)Β² = 45 Γ— 0.360 = 16.20 Step 5 β€” Uniform delay denominator: 1 βˆ’ min(1, 0.833) Γ— (g/C) = 1 βˆ’ 0.833 Γ— 0.400 = 1 βˆ’ 0.333 = 0.667 Step 6 β€” d₁: d₁ = 16.20 / 0.667 = 24.3 s/veh Step 7 β€” LOS: d₁ = 24.3 s/veh β†’ 20 < 24.3 ≀ 35 β†’ LOS C
βœ“ X = 0.833 | d₁ = 24.3 s/veh | LOS C
PROBLEM 8
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?
β–Ά
Same c = 720, g/C = 0.400. New X: X = 700/720 = 0.972 (much closer to capacity!) d₁ numerator (same): 0.5 Γ— 90 Γ— (0.600)Β² = 16.20 d₁ denominator: 1 βˆ’ 0.972 Γ— 0.400 = 1 βˆ’ 0.389 = 0.611 d₁ = 16.20 / 0.611 = 26.5 s/veh β†’ LOS C Comparison: 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!
βœ“ d₁ = 26.5 s/veh | LOS C | +2.2s delay for +100 veh/h
10 β€” 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.
ModeRight-of-WayCapacityBest For
Taxi/Ride-hailMixed traffic1–4 paxDoor-to-door, on-demand
JitneysMixed traffic5–15 paxFixed route, flexible schedule
Regular BusMixed traffic40–60 paxUrban corridors, frequent stops
BRT (Semi-rapid)Dedicated lanes60–160 paxHigh-volume urban corridors
Light Rail (LRT)Reserved ROW100–200 paxRegional corridors
Rapid Transit (Metro)Fully separated1,000+ pax/trainDense urban networks
11 β€” TRANSPORTATION PLANNING

Planning Before Building

Transportation planning is the process of deciding what infrastructure to build, where, when, and for whom β€” 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.

β†’ Productions & Attractions
2

Trip Distribution

Where do those trips go? Uses the Gravity Model.

β†’ O-D Matrix
3

Mode Split

Which mode do people use? Car, bus, train, walk? Uses Logit model based on utility.

β†’ Modal shares (%)
4

Traffic Assignment

Which specific routes do they take?

β†’ Link volumes
12 β€” 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

πŸ β†’πŸ’ HBW

Home-Based Work. Most time-sensitive. Drives morning/evening peak.

πŸ β†’πŸ›’ HBO

Home-Based Other. Home to shopping, school, recreation. More flexible timing.

πŸ’β†’πŸ’ NHB

Non-Home Based. Lunch trips, deliveries, errands between non-home stops.

Production vs Attraction
Production: trip end at RESIDENTIAL land use Attraction: trip end at NON-RESIDENTIAL land use
Key constraint: Ξ£P = Ξ£A. If not balanced, scale: Aα΅’(adj) = Aα΅’ Γ— (Ξ£P / Ξ£A)
13 β€” STEP 2: TRIP DISTRIBUTION (GRAVITY MODEL)

Where Do Those Trips Actually Go?

Trip distribution connects origin zones to destination zones using the Gravity Model β€” more trips go to zones that are larger and closer.

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.
Gravity Model Formula
Tα΅’β±Ό = Pα΅’ Γ— [Aβ±Ό Γ— F(t)α΅’β±Ό Γ— Kα΅’β±Ό] / [Ξ£β±Ό Aβ±Ό Γ— F(t)α΅’β±Ό Γ— Kα΅’β±Ό]
Tα΅’β±Ό=trips from i to j   Pα΅’=productions at i   Aβ±Ό=attractions at j   Kα΅’β±Ό=adj. factor (usually 1.0)
F = 1/W^c where W=travel time, c=calibration parameter
βš™οΈ CALCULATOR β€” GRAVITY MODEL: 3-ZONE TRIP DISTRIBUTION

Friction F = 1/tΒ² assumed. Enter productions (P), attractions (A), travel times (min).

ZONE P & A
TRAVEL TIMES (min)
FRICTION FACTORS
Click Calculate
Practice Problems β€” Gravity Model
PROBLEM 9 β˜…
Zone 1 has P₁ = 500 trips. Zones 2 and 3 have Aβ‚‚ = 300 and A₃ = 700. Travel times: t₁₂ = 5 min, t₁₃ = 10 min. Using F = 1/tΒ², find T₁₂ and T₁₃.
β–Ά
Step 1 β€” Friction factors: F₁₂ = 1/5Β² = 1/25 = 0.0400 F₁₃ = 1/10Β² = 1/100 = 0.0100 Step 2 β€” Weighted attractions: Aβ‚‚Β·F₁₂ = 300 Γ— 0.0400 = 12.00 A₃·F₁₃ = 700 Γ— 0.0100 = 7.00 Step 3 β€” Denominator (sum): Ξ£ = 12.00 + 7.00 = 19.00 Step 4 β€” Trip distribution: T₁₂ = P₁ Γ— (Aβ‚‚Β·F₁₂ / Ξ£) = 500 Γ— (12.00/19.00) = 500 Γ— 0.632 = 316 trips T₁₃ = P₁ Γ— (A₃·F₁₃ / Ξ£) = 500 Γ— (7.00/19.00) = 500 Γ— 0.368 = 184 trips Check: 316 + 184 = 500 = P₁ βœ“
βœ“ T₁₂ = 316 trips | T₁₃ = 184 trips
14 β€” STEP 3: MODE SPLIT (LOGIT MODEL)

Car, Bus, or Train? The Logit Model

Mode split predicts what fraction of trips will use each available mode. Based on utility β€” each mode has a perceived benefit/cost, and travelers pick the highest utility option.

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.
Utility Function
U = aβ‚€ + a₁X₁ + aβ‚‚Xβ‚‚ + ... + aα΅£Xα΅£
aβ‚€=mode-specific constant   X₁=cost (cents)   Xβ‚‚=travel time (min)   Coefficients are negative (more cost/time = lower utility)
Multinomial Logit β€” Probability of Choosing Mode i
P(i) = e^V(i) / Ξ£ e^V(r)
V(i)=computed utility of mode i   Higher probability = more people choose this mode
βš™οΈ CALCULATOR β€” LOGIT MODE SPLIT

Utility: U = a βˆ’ 0.002Β·X₁ βˆ’ 0.05Β·Xβ‚‚ (X₁=cost in cents, Xβ‚‚=time in min)

πŸš— AUTOMOBILE
🚌 BUS
🚈 LIGHT RAIL
Practice Problems β€” Logit Mode Split
PROBLEM 10 β˜…
Two modes are available: Auto and Bus. Utility function: U = a βˆ’ 0.025Β·(cost $) βˆ’ 0.04Β·(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.
β–Ά
Step 1 β€” Auto utility: V_auto = 0 βˆ’ 0.025(2.50) βˆ’ 0.04(20) = 0 βˆ’ 0.0625 βˆ’ 0.800 = βˆ’0.8625 Step 2 β€” Bus utility: V_bus = 0.50 βˆ’ 0.025(1.00) βˆ’ 0.04(40) = 0.50 βˆ’ 0.025 βˆ’ 1.600 = βˆ’1.125 Step 3 β€” Exponentiate: e^V_auto = e^(βˆ’0.8625) = 0.4222 e^V_bus = e^(βˆ’1.125) = 0.3247 Step 4 β€” Sum of exponentials: Ξ£ = 0.4222 + 0.3247 = 0.7469 Step 5 β€” 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.
βœ“ P(Auto) = 56.5% | P(Bus) = 43.5%
πŸ“ PRACTICE HUB

More Practice Problems β€” Mixed Topics

These problems mix multiple topics and are representative of exam-style questions.

MIXED PROBLEM A β€” 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.
β–Ά
Part 1 β€” PHF: V = 210+260+280+210 = 960 veh/hr V₁₅,max = 280 PHF = 960/(4Γ—280) = 960/1120 = 0.857 Part 2 β€” Adjusted flow rate: v = V/PHF = 960/0.857 = 1,121 veh/hr Part 3 β€” g/C ratio: g/C = 40/100 = 0.400 Part 4 β€” Capacity: c = 1,800 Γ— 0.400 = 720 veh/hr Part 5 β€” v/c ratio: X = 1,121/720 = 1.557 β†’ OVER CAPACITY β†’ LOS F immediately Part 6 β€” LOS F (X > 1.0): When X > 1.0, queue grows every cycle. d₁ formula becomes undefined (denominator β†’ 0 or negative). Report LOS F. The intersection MUST be redesigned.
βœ“ PHF=0.857 | v=1,121 veh/h | X=1.56 | LOS F (Failure)
MIXED PROBLEM B β€” 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?
β–Ά
(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 Γ— (g/C) = 1,636 Γ— (45/90) = 1,636 Γ— 0.500 = 818 veh/hr Vehicles per cycle = c Γ— C/3600 = 818 Γ— 90/3600 = 20.5 veh/cycle (c) Sufficient capacity? X = v/c = 750/818 = 0.917 < 1.0 β†’ YES, green is sufficient βœ“ The approach can handle 750 veh/hr with some headroom.
βœ“ s=1636 veh/hr | 20.5 veh/cycle | X=0.917 β†’ Sufficient
MIXED PROBLEM C β€” Planning (Gravity)
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.
β–Ά
F values (F = 1/t, not 1/tΒ²): F₁₂ = 1/3 = 0.333 F₁₃ = 1/5 = 0.200 Weighted attractions: Aβ‚‚ Γ— F₁₂ = 600 Γ— 0.333 = 200.0 A₃ Γ— F₁₃ = 400 Γ— 0.200 = 80.0 Sum: Ξ£ = 200.0 + 80.0 = 280.0 Trip distribution: T₁₂ = 1000 Γ— (200/280) = 1000 Γ— 0.714 = 714 trips T₁₃ = 1000 Γ— (80/280) = 1000 Γ— 0.286 = 286 trips Check: 714 + 286 = 1,000 = P₁ βœ“ Note: Zone 2 attracts more despite fewer attractions because it is much closer (3 min vs. 5 min).
βœ“ T₁₂ = 714 trips | T₁₃ = 286 trips
πŸ’‘ 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 β‰  G (effective β‰  display)

Effective green g = G + y + ar βˆ’ 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 β†’ queue grows every cycle.

04

Gravity Model: Check Balance

After computing Tα΅’β±Ό, sum each column (Dβ±Ό). If Dβ±Ό β‰  Aβ±Ό, 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) / Ξ£e^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 βˆ’ min(1,X)Β·(g/C)], not [1 βˆ’ XΒ·(g/C)]. When X > 1, use min(1, X) = 1, not X. This prevents the formula from returning negative delay (which is physically impossible).
πŸ“‹ QUICK REFERENCE

Formula Sheet β€” All CEE 350 Formulas

Traffic Volume
Peak Hour Factor
PHF = V / (4 Γ— 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 βˆ’ tL
tL = l₁ + lβ‚‚ β‰ˆ 4 s/phase typically
Saturation Flow
Sat. Flow Rate
s = 3600 / h
h = saturation headway (s/veh). Typical s β‰ˆ 1800 veh/h
β˜… Capacity
Lane Group Capacity
cα΅’ = sα΅’ Γ— (gα΅’/C)
sα΅’ = saturation flow, gα΅’ = eff. green, C = cycle length
β˜… Performance
v/c Ratio (X)
Xα΅’ = vα΅’ / cα΅’ = vα΅’ / [sα΅’(gα΅’/C)]
X < 1.0 = acceptable. X > 1.0 = failure (LOS F)
β˜… Delay
Uniform Delay (d₁)
d₁ = 0.5C(1βˆ’g/C)Β² / [1βˆ’min(1,X)Β·g/C]
Result in s/veh. Use for LOS determination
Delay
Total Control Delay
d = d₁ + dβ‚‚ + d₃
dβ‚‚=incremental, d₃=preexisting queue delay
β˜… Trip Distribution
Gravity Model
Tα΅’β±Ό = Pα΅’[Aβ±ΌFα΅’β±ΌKα΅’β±Ό] / [Ξ£Aβ±ΌFα΅’β±ΌKα΅’β±Ό]
Fᡒⱼ = 1/Wᢜ (friction factor). Kᡒⱼ = socioeconomic adj.
β˜… Mode Split
Logit Model
P(i) = e^V(i) / Ξ£e^V(r)
V = aβ‚€ + Ξ£aα΅£Β·Xα΅£ (utility function)
LOS Thresholds
Control Delay β†’ LOS
A≀10, B>10-20, C>20-35 D>35-55, E>55-80, F>80
Or X > 1.0 automatically = LOS F
Pedestrian
Min Ped Green
Gp = 3.2 + (2.7Β·Nped/WE) + (L/Sp)
For WE≀10ft: Gp = 3.2 + 0.27Β·Nped + L/Sp