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Topic 4 Bandwidth Optimization

Topic 4 Bandwidth Optimization. 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

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Topic 4 Bandwidth Optimization

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  1. Topic 4 Bandwidth Optimization

  2. 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

  3. Half-Integer Offset (Center Red) m i j Simultaneous Alternate Time Gm Gj Ri Rm Rj Gi Distance

  4. Simultaneous Offset - Forward (Lower Interference) m j IL Time Gm Gj 0 Rm Rj Distance

  5. Simultaneous Offset - Forward (Upper Interference) m j 0 IU Time Gm Gj Rm Rj Distance

  6. Simultaneous Offset - Forward (No Interference – Slack Time) m j IS Time Gm Gj 0 Rm Rj Distance

  7. Simultaneous Offset - Forward (No Interference – Slack Time) m j IS IS Time Gm Gj Rm Rj Distance

  8. Alternate Offset - Forward (Lower Interference) m j IL Time Gm Rj 0 Rm Gj Distance

  9. Alternate Offset - Forward (Upper Interference) m j IU 0 Time Rj Gm Rm Gj Distance

  10. Alternate Offset - Forward (No Interference) m j IS Time Rj Gm Rm Gj Distance

  11. Simultaneous Offset - Backward (Lower Interference) j m Time Gm Gj IL 0 Rj Rm Distance

  12. Summary of Equations * KL, KU, and K are integers ** K is even - simultaneous; K is odd - alternate

  13. Summary of Equations * KL, KU, and K are integers ** K is even - simultaneous); K is odd - alternate

  14. 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 IL and IU 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

  15. Graphical Illustration 1 m 2 3 IL,3 IU,2 Time IU,1 Distance

  16. Example Speed is 40 mph

  17. Illustration of Intersection #2 and #4 (Simultaneous Offset) m=2 #4 Rj 26.19 Gj Time Gm Rm Distance

  18. Illustration of Intersection #2 and #4 (Alternate Offset) m=2 #4 3.81 Time Rj Gm Gj Rm Distance

  19. Illustration of Intersection #2 and #5 (Simultaneous – Slack Times) m=2 #5 0.19 4.81 Rj Gj Time Gm Rm Distance

  20. Example (continued)

  21. Example (continued)

  22. Offset to Start of Green (Alternate) m Rj θG,mj Time Gm θR,mj Rm Distance

  23. Offset to Start of Green (Simultaneous) m Rj θG,mj Time Gm θR,mj = 0 Rm Distance

  24. Bandwidth with LT Phases Time Intersection 1 Intersection 2 Space

  25. Major Terms Phase B OB left turn Cycle OB through IB through IB left turn Link length in-between

  26. Bandwidth Maximization Outbound Inbound Bandwidth Outbound Bandwidth

  27. Bandwidth Maximization Inbound Bandwidth Outbound Bandwidth

  28. Phasing Sequence Leading Lagging Lag-Lead Lead-Lag 4 types of left turn sequence for each intersection 16 Combinations

  29. Methodology Maximum Bandwidth = Bo + Bi Bo <= Gomin Bi <= Gimin Bmax = Go,min +Gi,min – Ii,min j X

  30. Methodology Bmax = Go,min +Gi,min – Ii,min Interference Bmax = Constant Exception: Bmax = Gi,min

  31. Methodology Bmax = Go,min +Gi,min – Ii,min Interference Bmax = Constant Exception: Bmax = Gi,min

  32. Upper Interference Iup X – Intersection that has the smallest inbound green

  33. Iujp = Gix - (-Rxn +Txj + Rjp + Gij+ Tjx) mod C Gix+n*CL Iujp (-Rxn +Txj + Rjp + Gij + Tjx ) -Rxn +Txj + Rjp + Gij Gix -Rxn +Txj -Rjp -Rxn +Txj + Rjp 0 Rxn -Rxn Intersection X Intersection j

  34. Iujp = Gix - (-Rxn +Sox +Txj + Rjp + Gij+ Tjx) mod C Gix+n*CL Iujp (-Rxn +Sox+Txj + Rjp + Gij + Tjx ) -Rxn +Sox+Txj + Rjp + Gij Gix -Rxn +Sox+Txj -Rjp -Rxn +Sox+Txj + Rjp 0 Rxn -Rxn+Sox -Rxn Intersection X Intersection j

  35. No valid upper interference (-Rxn +Sox+Txj + Rjp + Gij + Tjx ) Iujp If Gix - (-Rxn +Sox+Txj + Rjp + Gij + Tjx ) <-Sij, no valid upper interference. -Rjp Intersection X Intersection j

  36. Iujp = Gix - (-Rxn + Txj + Rjp + Gij+ Tjx) mod C T.T.=34 sec 16 Iu = 55-41=14 41 30+34+22+61+34=181 =181-140=41 55* 30+34+22+61 30 38 61 30+45-34 45 30+34+22 22 16 30+34 41 j 30 55* -30 0 x

  37. Iujp = Gix - (-Rxn + Sox –Txj + Rjp + Gij -Tjx) mod C T.T.=34 sec 20 41* -(-34)=34 55* 34-34+18+61 34 34 0 61 Sox 45 34-34+18 18 34-34 20 j 41* Intersection j has the smallest outbound green 34-34+18+61-34=45 Iu = 55-45=10 55* 34 x

  38. Iujp = Gix - (-Rxn + Sox –Txj + Rjp + Gij -Tjx) mod C 30+45 T.T.=34 sec 20 45 30+4 55* Sox 30+4-34+18+61 -(-30)=30 30 38 0 61 41* 30+4-34+18 18 30+4-34 20 j 45 Intersection j has the smallest outbound green 30+4-34+18+61-34=45 Iu = 55-45=10 55* 30 x

  39. ILjp = (-Rxn+Txj – Sj+ Rjp + Tjx) mod C ILjp Go,min Slack Time, Sj Sj = Goj – Go,min 0 Rxn j x

  40. ILjp = (-Rxn+Txj – Sj+ Rjp + Tjx) mod C T.T.=50 sec 24 41 55* 38+50-4+15+50=149 38 30 IL = 9 60 45 38+50-4+15 38+50 15 38+50-4 24 41 j 38 55* -38 0 x

  41. No valid lower interference T.T.=34 sec 16 Lower interference calculation causes upper interference occurring. Intersection j must move up by 4 sec to reduce upper interference. Inbound band = 55-14=41 41 Iu = 24+55-61=18 55* 30+34-4+22+61 30 38 61 30+34-4+22+34=116 24 140-116=24>>6 45 30+34-4+22 30+34-4 22 30+34 16 j 41 30 55* -30 0 x

  42. Lower Interference Gij – Gix Gij C ILjp C - ILjp <= (Gij – Gix)=Sj ILjp >= C – Sj

  43. ILjp = (-Rxn+ Txj – Soj+ Rjp + Tjx) mod C 2-sec slack T.T.=95 sec 30 55* 0 41* 140-132=8<10 No lower interference -16+95-4-38+95=132 16 x 18 45 -16+95 65 -16+95-4 30 38 -16+95-4-38 55* j 41* 0 16 -16 x

  44. 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 sec T.T.=56 sec A B C 38 18 16 57 61* 50 41* 64 64 30 18 25

  45. Example T.T.=34 sec T.T.=56 sec A B C 38 18 16 57 61* 50 41* 64 64 30 18 25 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

  46. Homework (C=60) T.T.=36 sec T.T.=51 sec T.T.=27 sec T.T.=41 sec 20 10 10 30 25 25 26 30 30 20 25 30 25* 15 12 10 13 15 10 15

  47. 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 sec T.T.=56 sec 38 25 20 61 64 45 57 41* 55* 18 30 18

  48. 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)

  49. Brooks’ Algorithm – A Special Case Speed is 40 mph

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