CEE 764 – Fall 2010 Topic 4 Bandwidth Optimization

CEE 764 – Fall 2010 Bandwidth Optimization Methodology  Half-Integer Algorithms by Brooks and Little  Based on minimizing interferences  Providing equal bandwidth solution when speeds are the same for both directions  Calculating interference for each individual intersection based on a referencing intersection, m  The referencing intersection m is the one that has the minimum green

CEE 764 – Fall 2010 Half-Integer Offset (Center Red) RmRm Distance mj RiRi GmGm GiGi Time RjRj GjGj i Simultaneous Alternate

CEE 764 – Fall 2010 Simultaneous Offset - Forward (Lower Interference) RmRm Distance mj RjRj GmGm GjGj Time ILIL 0

CEE 764 – Fall 2010 Simultaneous Offset - Forward (Upper Interference) RmRm Distance mj RjRj GmGm GjGj Time IUIU 0

CEE 764 – Fall 2010 Simultaneous Offset - Forward (No Interference – Slack Time) RmRm Distance mj RjRj GmGm GjGj Time ISIS 0

CEE 764 – Fall 2010 Simultaneous Offset - Forward (No Interference – Slack Time) RmRm Distance mj RjRj GmGm GjGj Time ISIS ISIS

CEE 764 – Fall 2010 Alternate Offset - Forward (Lower Interference) RmRm Distance mj RjRj GmGm GjGj Time ILIL 0

CEE 764 – Fall 2010 Alternate Offset - Forward (Upper Interference) RmRm Distance mj RjRj GmGm GjGj Time IUIU 0

CEE 764 – Fall 2010 Alternate Offset - Forward (No Interference) RmRm Distance mj RjRj GmGm GjGj Time ISIS

CEE 764 – Fall 2010 Simultaneous Offset - Backward (Lower Interference) RmRm Distance jm RjRj GmGm GjGj Time ILIL 0

CEE 764 – Fall 2010 Summary of Equations * K L, K U, and K are integers ** K is even - simultaneous; K is odd - alternate

CEE 764 – Fall 2010 Summary of Equations * K L, K U, and K are integers ** K is even - simultaneous); K is odd - alternate

CEE 764 – Fall 2010 Summary of Brooks’ Algorithm a. Find intersection “m” with smallest green b. Travel times to right of “m” (forward) are positive and to the left (backward) are negative c. Calculate least allowable I L and I U for each intersection d. Perform total interference minimization e. Identify the optimal progression band and offsets f. Construct the time-space diagram g. Adjust split of directional bandwidth if desired

CEE 764 – Fall 2010 Graphical Illustration Distance 12 I U,1 Time m 3 I U,2 I L,3

CEE 764 – Fall 2010 Example Int. # GjGjGjGj RjRjRjRj D mj -0.5(R m -R j ) T mj 0.5(T mj +T jm ) -I S I max KUKUKUKU IUIUIUIU KLKLKLKL ILILILIL θ mj * Speed is 40 mph

CEE 764 – Fall 2010 Illustration of Intersection #2 and #4 (Simultaneous Offset) RmRm Distance m=2#4 RjRj GmGm GjGj Time 26.19

CEE 764 – Fall 2010 Illustration of Intersection #2 and #4 (Alternate Offset) RmRm Distance m=2#4 RjRj GmGm GjGj Time 3.81

CEE 764 – Fall 2010 Illustration of Intersection #2 and #5 (Simultaneous – Slack Times) RmRm Distance m=2#5 RjRj GmGm GjGj Time

CEE 764 – Fall 2010 Example (continued) Int. # GjGjGjGj IUIUIUIU ILILILIL * Rank # Int. # IUIUIUIU ILILILIL =m

CEE 764 – Fall 2010 Example (continued) Intersection by Rank I Total UUUUU26.19 LUUUU = LLUUU LLLUU LLLLU LLLLL InterferenceRank Int. # K-valueOffset θ mj (Reference Start of Green) L141Alternate U211Alternate U330Simultaneous U42=m-- U552Simultaneous

CEE 764 – Fall 2010 Offset to Start of Green ( Alternate ) RmRm Distance m RjRj GmGm θ R,mj Time θ G,mj

CEE 764 – Fall 2010 Offset to Start of Green ( Simultaneous ) RmRm Distance m RjRj GmGm θ R,mj = 0 Time θ G,mj

CEE 764 – Fall 2010 Bandwidth with LT Phases Intersection 1Intersection 2 Time Space

CEE 764 – Fall 2010 Major Terms Cycle Link length in-between Phase B OB through IB left turn IB through OB left turn

CEE 764 – Fall 2010 Bandwidth Maximization Inbound Bandwidth Outbound Bandwidth Outbound

CEE 764 – Fall 2010 Bandwidth Maximization Inbound Bandwidth Outbound Bandwidth

CEE 764 – Fall 2010 Phasing Sequence LeadingLead-LagLag-LeadLagging 16 Combinations 4 types of left turn sequence for each intersection

CEE 764 – Fall 2010 Methodology Maximum Bandwidth = Bo + Bi Bo <= Go min Bi <= Gi min B max = G o,min +G i,min – I i,min X j

CEE 764 – Fall 2010 Methodology B max = G o,min +G i,min – I i,min Exception: B max = G i,min B max = Constant Interference

CEE 764 – Fall 2010 Methodology B max = G o,min +G i,min – I i,min Exception: B max = G i,min B max = Constant Interference

CEE 764 – Fall 2010 Upper Interference I up X – Intersection that has the smallest inbound green

CEE 764 – Fall 2010 I ujp = G ix - (-R xn + T xj + R jp + G ij + T jx ) mod C I ujp 0 -Rxn +Txj + Rjp + Gij (-Rxn +Txj + Rjp + Gij + Tjx ) -Rxn +Txj -Rxn +Txj + Rjp -Rxn Gix Gix+n*CL Intersection XIntersection j -R jp R xn

CEE 764 – Fall 2010 I ujp = G ix - (-R xn + S ox + T xj + R jp + G ij + T jx ) mod C I ujp 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 XIntersection j -R jp R xn -Rxn+Sox

CEE 764 – Fall 2010 No valid upper interference I ujp (-Rxn +Sox+Txj + Rjp + Gij + Tjx ) Intersection XIntersection j -R jp If Gix - (-Rxn +Sox+Txj + Rjp + Gij + Tjx ) <-Sij, no valid upper interference.

CEE 764 – Fall 2010 T.T.=34 sec * * I u = 55-41=14 x j =181 = =41 I ujp = G ix - (-R xn + T xj + R jp + G ij + T jx ) mod C

CEE 764 – Fall 2010 T.T.=34 sec * 41* * 41* (-34)= =45 I u = 55-45=10 x j I ujp = G ix - (-R xn + S ox – T xj + R jp + G ij - T jx ) mod C Intersection j has the smallest outbound green S ox

CEE 764 – Fall 2010 T.T.=34 sec * 20 55* * (-30)= =45 I u = 55-45=10 x j I ujp = G ix - (-R xn + S ox – T xj + R jp + G ij - T jx ) mod C Intersection j has the smallest outbound green S ox 30+4

CEE 764 – Fall 2010 I Ljp = (-R xn +T xj – S j + R jp + T jx ) mod C I Ljp Slack Time, S j S j = G oj – G o,min R xn 0 G o,min x j

CEE 764 – Fall 2010 T.T.=50 sec * * I L = 9 x j =149 I Ljp = (-R xn +T xj – S j + R jp + T jx ) mod C

CEE 764 – Fall 2010 T.T.=34 sec * * I u = =18 x j =24>>6 No valid lower interference = 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 2010 Lower Interference C - I Ljp <= (G ij – G ix )=S j C I Ljp G ij G ij – G ix I Ljp >= C – S j

CEE 764 – Fall 2010 T.T.=95 sec 30 41* 55* sec slack =132 x j 65 I Ljp = (-R xn + T xj – S oj + R jp + T jx ) mod C * 55* 16 0 x =8<10 No lower interference

CEE 764 – Fall 2010 Example 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 secT.T.=56 sec 61* * ABC

CEE 764 – Fall 2010 Example T.T.=34 secT.T.=56 sec 61* * ABC B: I u =61-( )=61-84=-23, not valid (23>3); I L = ( )=11* C: I u =61-( )=61-61=0*, I L = ( )=121, not valid ( =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 2010

Homework (C=60) T.T.=36 secT.T.=51 sec * T.T.=27 secT.T.=41 sec

CEE 764 – Fall 2010 Homework 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 secT.T.=56 sec * 20 55*

CEE 764 – Fall 2010 Homework 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:  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 for both directions, what is the maximum total bandwidth (of both directions) between the two intersections?  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?  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 manually to check your results)

CEE 764 – Fall 2010 Brooks’ Algorithm – A Special Case Int. # GjGjGjGj RjRjRjRj D mj -0.5(R m -R j ) T mj 0.5(T mj +T jm ) -I S I max KUKUKUKU IUIUIUIU KLKLKLKL ILILILIL θ mj * Speed is 40 mph

CEE 764 – Fall 2010