CEE 764 – Fall 2010CEE 764 – Fall 2010

Topic 4Topic 4

Bandwidth OptimizationBandwidth Optimization

CEE 764 – Fall 2010CEE 764 – Fall 2010

Bandwidth Optimization MethodologyBandwidth Optimization Methodology

Half-IntegerHalf-Integer Algorithms by Algorithms by Brooks Brooks and and LittleLittle Based on minimizing Based on minimizing interferencesinterferences Providing equal bandwidth solution when Providing equal bandwidth solution when

speeds are the same for both directionsspeeds are the same for both directions Calculating interference for each individual Calculating interference for each individual

intersection based on a referencing intersection based on a referencing intersection, intersection, mm

The referencing intersection The referencing intersection mm is the one is the one that has the minimum greenthat has the minimum green

CEE 764 – Fall 2010CEE 764 – Fall 2010

Half-Integer Offset (Center Red)Half-Integer Offset (Center Red)

Rm

Distance

m j

Ri

Gm

Gi

Tim

e

Rj

Gj

i

Sim

ult

aneo

us

Alt

ern

ate

CEE 764 – Fall 2010CEE 764 – Fall 2010

Simultaneous Offset - Forward Simultaneous Offset - Forward (Lower Interference)(Lower Interference)

Rm

Distance

m j

Rj

Gm Gj

Tim

e

IL

mjjLmjm NCRITR 5.05.0

Lmj CKNC 5.0

mjjmLL TRRCKI )(5.05.0

)(5.0 jmmjmj TTT 0

CEE 764 – Fall 2010CEE 764 – Fall 2010

Simultaneous Offset - Forward Simultaneous Offset - Forward (Upper Interference)(Upper Interference)

Rm

Distance

m j

Rj

Gm Gj

Tim

e

IU

mjjUmjm NCRITR 5.05.0

mjjmUU TRRCKI )(5.05.0

)(5.0 jmmjmj TTT

Umj CKNC 5.00

CEE 764 – Fall 2010CEE 764 – Fall 2010

Simultaneous Offset - Forward Simultaneous Offset - Forward (No Interference – Slack Time)(No Interference – Slack Time)

Rm

Distance

m j

Rj

Gm Gj

Tim

e

IS

0

CEE 764 – Fall 2010CEE 764 – Fall 2010

Simultaneous Offset - Forward Simultaneous Offset - Forward (No Interference – Slack Time)(No Interference – Slack Time)

Rm

Distance

m j

Rj

Gm Gj

Tim

e

IS

IS

CEE 764 – Fall 2010CEE 764 – Fall 2010

Alternate Offset - Forward Alternate Offset - Forward (Lower Interference)(Lower Interference)

Rm

Distance

m j

RjGm

Gj

Tim

e IL

mjjLmjm NCRITR 5.05.0

Lmj CKNC 5.0

mjjmLL TRRCKI )(5.05.0

)(5.0 jmmjmj TTT

0

CEE 764 – Fall 2010CEE 764 – Fall 2010

Alternate Offset - Forward Alternate Offset - Forward (Upper Interference)(Upper Interference)

Rm

Distance

m j

RjGm

Gj

Tim

e

IU

mjjUmjm NCRITR 5.05.0

mjjmUU TRRCKI )(5.05.0

)(5.0 jmmjmj TTT

0

Umj CKNC 5.0

CEE 764 – Fall 2010CEE 764 – Fall 2010

Alternate Offset - Forward Alternate Offset - Forward (No Interference)(No Interference)

Rm

Distance

m j

RjGm

Gj

Tim

e

IS

CEE 764 – Fall 2010CEE 764 – Fall 2010

Simultaneous Offset - Backward Simultaneous Offset - Backward (Lower Interference)(Lower Interference)

Rm

Distance

j m

Rj

GmGjTim

e

IL

mjjLmjm NCRITR 5.05.0

Lmj CKNC 5.0

)()(5.05.0 mjjmLL TRRCKI

)(5.0 jmmjmj TTT

0

CEE 764 – Fall 2010CEE 764 – Fall 2010

Summary of EquationsSummary of Equations

LjmmjjmL CKTTRRI 5.0)(5.0)(5.0

UjmmjjmU CKTTRRI 5.0)(5.0)(5.0

CCKTT jmmjmj 5.0)(5.00

jjmS GGGI

* KL, KU, and K are integers** K is even - simultaneous; K is odd - alternate

maxIGII mUS

maxIGII mLS

CEE 764 – Fall 2010CEE 764 – Fall 2010

Summary of EquationsSummary of Equations

LjmmjjmL CKTTRRI 5.0)(5.0)(5.0

UjmmjjmU CKTTRRI 5.0)(5.0)(5.0

CCKTT jmmjmj 5.0)(5.00

0 jmSj GGIG

* KL, KU, and K are integers** K is even - simultaneous); K is odd - alternate

maxIGII mUS

maxIGII mLS

CEE 764 – Fall 2010CEE 764 – Fall 2010

Summary of Brooks’ AlgorithmSummary of Brooks’ Algorithm

a. Find intersection “a. Find intersection “mm” with smallest green” with smallest greenb. Travel times to right of “b. Travel times to right of “mm” (forward) are positive and to ” (forward) are positive and to the left (backward) are negativethe left (backward) are negativec. Calculate least allowable Ic. Calculate least allowable ILL and I and IUU for each intersection for each intersectiond. Perform total interference minimizationd. Perform total interference minimization

e. Identify the optimal progression band and offsetse. Identify the optimal progression band and offsets

f. Construct the time-space diagramf. Construct the time-space diagramg. Adjust split of directional bandwidth if desiredg. Adjust split of directional bandwidth if desired

][ ,, jLjUTotal IMaxIMaxMinIMin

Totalma IGB

CEE 764 – Fall 2010CEE 764 – Fall 2010

Graphical IllustrationGraphical Illustration

Distance

1 2

IU,1

Tim

e

m 3

IU,2IL,3

3,2,1, ),( LUUTotal IIIMaxI

CEE 764 – Fall 2010CEE 764 – Fall 2010

ExampleExample

LjmmjjmL CKTTRRI 5.0)(5.0)(5.0 UjmmjjmU CKTTRRI 5.0)(5.0)(5.0

CCKTT jmmjmj 5.0)(5.00 mjmS GGGI

Int. #Int. # GGjj RRjj DDmjmj -0.5(R-0.5(Rmm-R-Rjj)) TTmjmj 0.5(T0.5(Tmjmj+T+Tjmjm)) -I-ISS IImaxmax KKUU IIUU KKLL IILL θθmjmj

11 4545 3535 15001500 -2.5-2.5 -25.56-25.56 -25.56-25.56 -5-5 4040

2*2* 4040 4040 -- -- -- -- -- -- -- -- -- -- --

33 4545 3535 330330 -2.5-2.5 5.625.62 5.625.62 -5-5 00

11

-1-1

3.1203.120

-36.88-36.88

43.1243.12

00

11

-8.12-8.12

31.8731.87

44 5050 3030 18301830 -5.0-5.0 31.1931.19 31.1931.19

55 4545 3535 48304830 -2.5-2.5 82.3182.31 82.3182.31 -5-5

Speed is 40 mph

maxIGII mUS maxIGII mLS

CEE 764 – Fall 2010CEE 764 – Fall 2010

Illustration of Intersection #2 and #4 Illustration of Intersection #2 and #4 (Simultaneous Offset)(Simultaneous Offset)

Rm

Distance

m=2 #4

Rj

Gm

Gj

Tim

e

26.19

CEE 764 – Fall 2010CEE 764 – Fall 2010

Illustration of Intersection #2 and #4 Illustration of Intersection #2 and #4 (Alternate Offset)(Alternate Offset)

Rm

Distance

m=2 #4

RjGm

Gj

Tim

e 3.81

CEE 764 – Fall 2010CEE 764 – Fall 2010

Illustration of Intersection #2 and #5 Illustration of Intersection #2 and #5 (Simultaneous – Slack Times)(Simultaneous – Slack Times)

Rm

Distance

m=2 #5

Rj

Gm

Gj

Tim

e

4.81

0.19

CEE 764 – Fall 2010CEE 764 – Fall 2010

Example (continued)Example (continued)

Int. #Int. # GGjj IIUU IILL

11 4545 11.9411.94 23.0623.06

2*2* 4040 00 00

33 4545 3.123.12 31.8731.87

44 5050 26.1926.19 3.813.81

55 4545 -0.19-0.19 -4.81-4.81

Rank #Rank # Int. #Int. # IIUU IILL

11 44 26.1926.19 3.813.81

22 11 11.9411.94 23.0623.06

33 33 3.123.12 31.8731.87

44 2=m2=m 00 00

55 55 -0.19-0.19 -4.81-4.81

CEE 764 – Fall 2010CEE 764 – Fall 2010

Example (continued)Example (continued)

Intersection by RankIntersection by Rank IITotalTotal

11 22 33 44 55

UU UU UU UU UU 26.1926.19

LL UU UU UU UU 11.94 + 3.81 = 15.7511.94 + 3.81 = 15.75

LL LL UU UU UU

LL LL LL UU UU

LL LL LL LL UU

LL LL LL LL LL

InterferenceInterference RankRank Int. #Int. # K-valueK-value OffsetOffset θθmjmj

(Reference Start of Green)(Reference Start of Green)

LL 11 44 11 AlternateAlternate

UU 22 11 11 AlternateAlternate

UU 33 33 00 SimultaneousSimultaneous

UU 44 2=m2=m -- --

UU 55 55 22 SimultaneousSimultaneous

CEE 764 – Fall 2010CEE 764 – Fall 2010

Offset to Start of GreenOffset to Start of Green((AlternateAlternate))

Rm

Distance

m

Rj

Gm

θR,mj

Tim

e θG,mj

22,,mj

mjRmjG

RR

CEE 764 – Fall 2010CEE 764 – Fall 2010

Offset to Start of GreenOffset to Start of Green((SimultaneousSimultaneous))

Rm

Distance

m

Rj

Gm

θR,mj = 0

Tim

e θG,mj

22,,mj

mjRmjG

RC

R

CEE 764 – Fall 2010CEE 764 – Fall 2010

Bandwidth with LT PhasesBandwidth with LT Phases

Intersection 1 Intersection 2

Time

Space

CEE 764 – Fall 2010CEE 764 – Fall 2010

Major TermsMajor Terms

Cycle

Link length in-between

Phase B

OB through

IB left turn

IB through

OB left turn

CEE 764 – Fall 2010CEE 764 – Fall 2010

Bandwidth MaximizationBandwidth Maximization

Inbound Bandwidth

Outbound Bandwidth

Outbound

CEE 764 – Fall 2010CEE 764 – Fall 2010

Bandwidth MaximizationBandwidth Maximization

Inbound Bandwidth

Outbound Bandwidth

CEE 764 – Fall 2010CEE 764 – Fall 2010

Phasing SequencePhasing Sequence

Leading Lead-LagLag-LeadLagging

16 Combinations4 types of left turn sequence for each intersection

CEE 764 – Fall 2010CEE 764 – Fall 2010

MethodologyMethodology

Maximum Bandwidth = Bo + BiBo <= Gomin

Bi <= Gimin

Bmax = Go,min +Gi,min – Ii,min

X

j

CEE 764 – Fall 2010CEE 764 – Fall 2010

MethodologyMethodology

Bmax = Go,min +Gi,min – Ii,min

Exception:Bmax = Gi,min

Bmax = Constant

Interference

CEE 764 – Fall 2010CEE 764 – Fall 2010

MethodologyMethodology

Bmax = Go,min +Gi,min – Ii,min

Exception:Bmax = Gi,min

Bmax = Constant

Interference

CEE 764 – Fall 2010CEE 764 – Fall 2010

Upper InterferenceUpper Interference

Iup

X – Intersection that has the smallest inbound green

CEE 764 – Fall 2010CEE 764 – Fall 2010

Iujp = Gix - (-Rxn +Txj + Rjp + Gij + Tjx ) mod C

Iujp

0

-Rxn +Txj + Rjp + Gij

(-Rxn +Txj + Rjp + Gij + Tjx )

-Rxn +Txj

-Rxn +Txj + Rjp

-Rxn

Gix

Gix+n*CL

Intersection X Intersection j

-Rjp

Rxn

CEE 764 – Fall 2010CEE 764 – Fall 2010

Iujp = Gix - (-Rxn +Sox +Txj + Rjp + Gij + Tjx ) mod C

Iujp

0

-Rxn +Sox+Txj + Rjp + Gij

(-Rxn +Sox+Txj + Rjp + Gij + Tjx )

-Rxn +Sox+Txj

-Rxn +Sox+Txj + Rjp

-Rxn

Gix

Gix+n*CL

Intersection X Intersection j

-Rjp

Rxn-Rxn+Sox

CEE 764 – Fall 2010CEE 764 – Fall 2010

No valid upper interference

Iujp

(-Rxn +Sox+Txj + Rjp + Gij + Tjx )

Intersection X Intersection j

-Rjp

If Gix - (-Rxn +Sox+Txj + Rjp + Gij + Tjx )<-Sij, no valid upper interference.

CEE 764 – Fall 2010CEE 764 – Fall 2010

T.T.=34 sec

61

38

45

16

55*

41

30

2216

55*

41

-30

0

30+34+22

30

30+34

30+45-34

Iu = 55-41=14

x

j

30+34+22+61

30+34+22+61+34=181=181-140=41

Iujp = Gix - (-Rxn + Txj + Rjp + Gij + Tjx ) mod C

CEE 764 – Fall 2010CEE 764 – Fall 2010

T.T.=34 sec

61

34

45

20

55*

41*

34

18

20

55*

41*

34

0

-(-34)=34

34-34

34-34+18

34-34+18+61-34=45

Iu = 55-45=10

x

j

34-34+18+61

Iujp = Gix - (-Rxn + Sox –Txj + Rjp + Gij -Tjx ) mod C

Intersection j has the smallest outbound green

Sox

CEE 764 – Fall 2010CEE 764 – Fall 2010

T.T.=34 sec

61

38

41*

20

55*

45

30

18

20

55*

45

30

0

-(-30)=30

30+4-34

30+4-34+18

30+4-34+18+61-34=45

Iu = 55-45=10

x

j

30+45

30+4-34+18+61

Iujp = Gix - (-Rxn + Sox –Txj + Rjp + Gij -Tjx ) mod C

Intersection j has the smallest outbound green

Sox30+4

CEE 764 – Fall 2010CEE 764 – Fall 2010

ILjp = (-Rxn +Txj – Sj+ Rjp + Tjx) mod C

ILjp

Slack Time, Sj

Sj = Goj – Go,minRxn

0

Go,min

x j

CEE 764 – Fall 2010CEE 764 – Fall 2010

T.T.=50 sec

60

30

45

24

55*

41

38

15

24

55*

41

-38

0

38+50-4+15

38

38+50

IL = 9

x

j

38+50-4+15+50=149

ILjp = (-Rxn +Txj – Sj+ Rjp + Tjx) mod C

38+50-4

CEE 764 – Fall 2010CEE 764 – Fall 2010

T.T.=34 sec

61

38

45

16

55*

41

30

22

16

55*

41

-30

0

30+34-4

30

Iu = 24+55-61=18

x

j

140-116=24>>6

No valid lower interference

30+34

30+34-4+22+61

30+34-4+22+34=116

30+34-4+22

Lower interference calculation causes upper interference occurring. Intersection j must move up by 4 sec to reduce upper interference. Inbound band = 55-14=41

24

CEE 764 – Fall 2010CEE 764 – Fall 2010

Lower InterferenceLower Interference

C - ILjp <= (Gij – Gix)=Sj

C ILjp

Gij

Gij – Gix

ILjp >= C – Sj

CEE 764 – Fall 2010CEE 764 – Fall 2010

T.T.=95 sec

30

41*

55*

16

18

45

38

0

2-sec slack

-16

-16+95-4-38+95=132

x

j

65

ILjp = (-Rxn + Txj – Soj+ Rjp + Tjx) mod C

-16+95

30

41*

55*

16

0

x

-16+95-4

-16+95-4-38

140-132=8<10No lower interference

CEE 764 – Fall 2010CEE 764 – Fall 2010

ExampleExample

There are three coordinated intersections A, B, and C with a cycle length of 140 sec. Travel times and phasing sequence and splits are shown below. Determine the maximum bandwidth and offsets.

T.T.=34 sec T.T.=56 sec

61*

25

38

41*

16

64

18

64

50

30

57

18

A B C

CEE 764 – Fall 2010CEE 764 – Fall 2010

ExampleExampleT.T.=34 sec T.T.=56 sec

61*

25

38

41*

16

64

18

64

50

30

57

18

A B C

B: Iu =61-(-18+34-30+64+34)=61-84=-23, not valid (23>3); IL = (-18+34-9-30+34)=11*

C: Iu =61-(-18+90-25+64+90)=61-61=0*, IL = (-18+90-16-25+90)=121, not valid (140-121=19>3)

Therefore, it has a 11 lower interference caused by B and zero interference by C. Bandwidth = 41+(61-11)=91

CEE 764 – Fall 2010CEE 764 – Fall 2010

CEE 764 – Fall 2010CEE 764 – Fall 2010

Homework (C=60)Homework (C=60)

T.T.=36 sec T.T.=51 sec

30

15

25

10

30

30

1010

10

25*

20

15

20

25

30

15

25

12

26

13

T.T.=27 sec T.T.=41 sec

CEE 764 – Fall 2010CEE 764 – Fall 2010

HomeworkHomework

There are three coordinated intersections A, B, and C with a cycle length of 140 sec. Travel times and phasing sequence and splits are shown below. Determine the maximum bandwidth and offsets.

T.T.=34 sec T.T.=56 sec

61

38

41*

20

55*

45

3018

25

18

64

57

CEE 764 – Fall 2010CEE 764 – Fall 2010

HomeworkHomework

There are two intersections A and B, running coordinated control for the major street, with the cycle length of 60 sec.

For intersection A, The outbound through time is 15 sec, and the inbound through time is 20 sec; the outbound left-turn interval is 10 sec, and the inbound left-turn interval is 15 sec.

For intersection B, The outbound through time is 20 sec, and the inbound through time is 25 sec; the outbound left-turn interval is 15 sec, and the inbound left-turn interval is 20 sec.

The distance between the two intersections is 1320 feet. Questions: Questions: Assuming intersection A is running leading left turns, and Intersection B is running leading left Assuming intersection A is running leading left turns, and Intersection B is running leading left

turn for outbound direction and lagging left turn for inbound direction. At the speed limit of 30 mph turn for outbound direction and lagging left turn for inbound direction. At the speed limit of 30 mph for both directions, what is the maximum total bandwidth (of both directions) between the two for both directions, what is the maximum total bandwidth (of both directions) between the two intersections?intersections?

If the speed limit is 20 mph, what is the maximum bandwidth when intersection A is running If the speed limit is 20 mph, what is the maximum bandwidth when intersection A is running leading left turn, and intersection B is running lagging left turn?leading left turn, and intersection B is running lagging left turn?

All red and yellow time can be ignored for the calculation. All red and yellow time can be ignored for the calculation. Please list all your calculations and adjustments if any (Tips: use Synchro to optimize bandwidth Please list all your calculations and adjustments if any (Tips: use Synchro to optimize bandwidth

manually to check your results)manually to check your results)

CEE 764 – Fall 2010CEE 764 – Fall 2010

Brooks’ Algorithm – A Special CaseBrooks’ Algorithm – A Special Case

CTGTGI jxjxjixU mod)(

Int. #Int. # GGjj RRjj DDmjmj -0.5(R-0.5(Rmm-R-Rjj)) TTmjmj 0.5(T0.5(Tmjmj+T+Tjmjm)) -I-ISS IImaxmax KKUU IIUU KKLL IILL θθmjmj

11 4545 3535 15001500 -2.5-2.5 -25.56-25.56 -25.56-25.56 -5-5 4040

2*2* 4040 4040 -- -- -- -- -- -- -- -- -- -- --

33 4545 3535 330330 -2.5-2.5 5.625.62 5.625.62 -5-5 00

11

-1-1

3.1203.120

-36.88-36.88

43.1243.12

00

11

-8.12-8.12

31.8731.87

44 5050 3030 18301830 -5.0-5.0 31.1931.19 31.1931.19 00

-1-1

26.1926.19

-13.81-13.81

00

11

-36.19-36.19

3.813.81

55 4545 3535 48304830 -2.5-2.5 82.3182.31 82.3182.31 -5-5 00

11

22

79.8179.81

35.8135.81

-0.19-0.19

11

22

-44.81-44.81

-4.81-4.81

Speed is 40 mph

maxIGII mUS maxIGII mLS

CTSTI jxjxjL mod)(

CEE 764 – Fall 2010CEE 764 – Fall 2010