Traffic Flow Analysis: (traffic stream and shockwave)

Dr. Gang-Len Chang Professor and Director of Traffic Safety and

Operations Lab. University of Maryland, College Park, MD 20742

1

Q, K, V, Q = K. V (space mean speed)

K and V relation? Q and V relation? Kmax?

Qmax

Vmax

2

k

q

3

4

5

6

7

8

9

10

11

12

13

14

15

16

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Shockwave

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Shock wave - a rapid change in traffic conditions (speed, density, and flow)

A moving boundary between two different traffic states

Forward, backward, and stationary shock waves

Shockwave Analysis A illustrative video: shockwave at signalized

intersections

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u1, k1

u2, k2

u3, k3

20

u1, k1 u2, k2

Low density High density

Shock wave speed

w

a b

12

12 )()(kk

kqkqws −−

=

Shockwave Analysis – General Equation

21

q2

k1

w=(q2-q1)/(k2-k1) 1 Low density

k

q Space

Time

0

w

u1

u2

k2

q1

2 High density

k 1

k 2

u2=q2/k2

u1=q1/k1

22

Traffic Interruption

kA kC=kJ/2 kJ

qC qA

wAC

k

q

A

C

J

wAJ

wCJ

uC uA

uJ= 0

Space

Time

0

C J A A

J

C

uC

wAC

wAJ

wCJ

uA

uJ=0

Traffic interaction between two density levels

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u1, k1 u2, k2

Low density High density

Shock wave speed

w

a b

11 )( kwuqa −=

w u k u kk k

w q qk k

=−−

=−−

2 2 1 1

2 1

2 1

2 1

Since qa = qb, then (u1 - w)k1 = (u2 - w)k2

22 )( kwuqb −=

24

)1(

)1()1(

)1(and

and

21

12

11

22

12

1122

12

12

jf

jf

jf

jf

kkkuw

kk

kkkuk

kku

w

kkuu

kkkukuw

ukqkkqqw

+−=

−

−−−

=

−=−−

=

=−−

=

Using the flow-density diagram, determine the speeds of the following shock waves: (1) Jam conditions change to capacity conditions (2) Density 3kj/4 rapidly decreases to kj/4

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k1= kA k2= kJ

J

AfAJ

J

JAfAJ

Jf

kku

wk

kkuw

kkkuw

−=+

−=

+−=

)1(

)1( 2112

Arrival conditions Jam conditions

Waves of Stopping

Wave of Starting

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k1 = kj k2 = kC = kj /2

2)2/1(

)1( 2112

fCJ

J

JJfCJ

Jf

uw

kkkuw

kkkuw

−=+

−=

+−=

Shockwave Theory- Congestion formulation and Dissipation Definition:

if the traffic stream is stationary over time and distance, then: k and n (and therefore λ) are independent of x (distance) and t

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tttxM

ttt

q ixix ii

∆∆

=∆

Φ−Φ=

),,()()( 0

)(]1)([lim

0t

tMP

xt

λ=∆

≥⋅→∆

xxtxN

xtt

k ixx ii

∆∆

=∆Φ−Φ

=),,()()(

)()],,([lim

0xk

xxtxNP

tx

=∆

∆→∆

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Shockwave Theory- Congestion formulation and Dissipation

using a plane in n-x-t space, n(x,t) : number of cars at time t location x ------- assumed relation which satisfy:

akxt +−= λ

)tan(),(),( tconskx

txnconstt

txn−=

∂∂

===∂

∂ λ

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λ

v

∆λ∆x

u

Shockwave Theory- Congestion formulation and Dissipation

when vehicles or traffic flows change from (λ1, µ1) to (λ2, µ2) then, ⇒ along this intersection

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111111 axktn +−= λ

222222 axktn +−= λ

tn

tn

∂∂

=∂∂ 21

dtdxk

dtdxk 2211 −=− λλ⇒

kkkdtdx

∆∆

=−−

==λλλµ

21

21∴ (shockwave speed)

Shockwave Theory- Congestion formulation and Dissipation

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t t

x

u3

u2

u1

Shockwave Theory- Congestion formulation and Dissipation

An Example on Freeway Incident Impact Analysis

On a section of freeway, an accident occurs at 10:00am at point B. First, both lanes are blocked, then after 15 minutes, one lane is cleared.

Traffic Sensor Data: A: q = 2700 vph, s = 90km/h B: one-lane – q = 1500vph, s = 7.5km/h B: two-lane – q = 3600vph, s = 60km/h Kjam = 300veh/lane/km

Questions: 1. Where is the end of queue at 10:15am? 2. When will the queue stop to increase? 3. How long is the max queue length? 4. By what time must the second lane be cleared to avoid the impact from the accident extending to A? 5. What is the total incident impact period? 33

A B

Incident

14.17km

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Q1: Where is the end of the queue at 10:15 AM? Speed of the shockwave:

hkmkkqq /10

30090

270002700

21

211 −=

−

−=

−−

=µ

(queue formulation speed) k2 = kmax , q2 = 0 After 15 min, X1(t = 15 min) = µ1×t = -10×(15/60) = -2.5 km (negative toward upstream)

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Q2: When are vehicles last forced to stop by the queue?

After the partial clearning: Traffic flow q3 (one lane) = 1500 vph Traffic able changes again form a second shockwave k3 = 1500/7.5 = 200 , q2 = 0 µII = (queue dissipation speed) |µII| < |µ1| queue will be dissolved

The second wave meets the back of the first wave XI = X (t = 15) +µI×t = -2.5 - 10×t (Initial queue at 10 min) XII = µII×t = -15×t XI = XII -2.5-10×t = -15×t t* = 0.5 h = 30 min ⇒ 10:45 (30 min. after the start of the shockwave II)

hkmkkqq /15

20030015000

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32 −=−

−=

−−

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Q3: The maximum queue

by the time the congestion dissipation move commences at 10:15 AM, X1 = max. queue = 2.5 km

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Q4: What is the max. distance of the end of the queue

from the site of the accident? The max. distance at time 10:45 AM (30 min) ∴ XI (t = 30) = -2.5 - 10×(0.5 hr) = -7.5 km

At 10:45 AM, shock waves I and II meet and form a new shockwave: q1 = 2700 (coming flow), q2 = 1500 (outgoing) k1 = 30 k2 = 200 to reach X = 14.17 = µIII⋅t + X0 = -7.06t – 7.5 = 14.17 t = 57 min ⇒ the distribution should have reached A at 11:42 AM if the blockage is completely removed: ⇒ the 4th shockwave:

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Q5: By what time must the second lane be cleared if the disturbance to the traffic flow resulting from the accident is not to extend to entrance A?

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The 4th shockwave is a congestion dissipation wave: It takes 0.95h = 14.17/15 =57 min to reach A. Note: |µIV| > |µIII| ⇒ meet at A, not to disturb traffic at A This wave should begin 10:45 AM After wave IV meet wave III, a new shock wave will form: q1 = 2700 (coming flow), q5 = 3600 (outgoing) k1 = 30, k5 = 60 ∴

⇒ increasing speed

hkmkkqq

IV /1560/3600200

36001500

43

43 −=−−

=−−

=µ

hkmV /30603036002700

=−−

=µ

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H

h

t∆

u1

x

1 2 nN

u2

v1

v2

t∆ta ∆tb

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max1

11

0kk

q−−

=µmax2

2

2max

22

0kk

qkk

q−

=−

−=µ

1µ=∆th

2µ=−∆ Hth

)(21 Htt −∆=∆ µµ

21

2

µµµ−

−=∆

Ht ⇒

N = kmax ⋅ h = mile

nilevehicle

⋅ = vehicles Total vehicle

∴

Note: shockwave is usually negative

)()(

11

max21max12

max21

21

max

max

kkqkkqHkqq

kH

khN

−−−⋅⋅⋅

=

−

⋅=

⋅=

µµ

Shockwave use the negative sign to indicate the direction ⇒the actual queue is

)()( 2max11max2

max21

max

kkqkkqHkqq

khN

−−−⋅⋅⋅

=

⋅−=−

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Homework (see p. 163): Please prove: The total stopped time of all N vehicles

−−

−⋅−

−⋅=2

2max

1

1max

max2)1(

qkk

qkk

kNNHNTH