Transcript Lecture 4 - Bandwidth Optimization

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
i
j
Time
Alternate
m
Simultaneous
Half-Integer Offset (Center Red)
Gm
Rm
Ri
Gj
Gi
Rj
Distance
CEE 764 – Fall 2010
Simultaneous Offset - Forward
(Lower Interference)
m
j
IL
Time
0.5 Rm  Tmj  I L  0.5 R j  NC   mj
NC   mj  0.5CK L
Gm
Gj
I L  0.5CK L  0.5( Rm  R j )  Tmj
0
Rm
Rj
Tmj  0.5(Tmj  T jm )
Distance
CEE 764 – Fall 2010
Simultaneous Offset - Forward
(Upper Interference)
m
j
 0.5Rm  Tmj  IU  0.5R j  NC   mj
Time
0
IU
NC   mj  0.5CKU
IU  0.5CKU  0.5( Rm  R j )  Tmj
Gm
Gj
Rm
Rj
Tmj  0.5(Tmj  T jm )
Distance
CEE 764 – Fall 2010
Simultaneous Offset - Forward
(No Interference – Slack Time)
m
j
Time
IS
Gm
Gj
0
Rm
Rj
Distance
CEE 764 – Fall 2010
Simultaneous Offset - Forward
(No Interference – Slack Time)
m
j
IS
Time
IS
Gm
Gj
Rm
Rj
Distance
CEE 764 – Fall 2010
Alternate Offset - Forward
(Lower Interference)
m
j
0.5 Rm  Tmj  I L  0.5 R j  NC   mj
NC   mj  0.5CK L
I L  0.5CK L  0.5( Rm  R j )  Tmj
Time
Tmj  0.5(Tmj  T jm )
IL
Gm
Rj
0
Rm
Gj
Distance
CEE 764 – Fall 2010
Alternate Offset - Forward
(Upper Interference)
m
j
IU
 0.5Rm  Tmj  IU  0.5R j  NC   mj
0
Time
NC   mj  0.5CKU
IU  0.5CKU  0.5( Rm  R j )  Tmj
Gm
Rj
Rm
Gj
Tmj  0.5(Tmj  T jm )
Distance
CEE 764 – Fall 2010
Alternate Offset - Forward
(No Interference)
m
j
Time
IS
Gm
Rj
Rm
Gj
Distance
CEE 764 – Fall 2010
Simultaneous Offset - Backward
(Lower Interference)
j
m
0.5 Rm  Tmj  I L  0.5 R j  NC   mj
NC   mj  0.5CK L
Time
I L  0.5CK L  0.5( Rm  R j )  (Tmj )
Tmj  0.5(Tmj  T jm )
Gj
Gm
IL
Rj
0
Rm
Distance
CEE 764 – Fall 2010
Summary of Equations
I L  0.5( Rm  R j )  0.5(Tmj  T jm )  0.5CK L
IU  0.5( Rm  R j )  0.5(Tmj  T jm )  0.5CKU
0   mj  0.5(Tmj  T jm )  0.5CK  C
 I S  Gm  G j  G j
 I S  IU  Gm  I max
 I S  I L  Gm  I max
* KL, KU, and K are integers
** K is even - simultaneous; K is odd - alternate
CEE 764 – Fall 2010
Summary of Equations
I L  0.5( Rm  R j )  0.5(Tmj  T jm )  0.5CK L
IU  0.5( Rm  R j )  0.5(Tmj  T jm )  0.5CKU
0   mj  0.5(Tmj  T jm )  0.5CK  C
 G j   I S  Gm  G j  0
 I S  IU  Gm  I max
 I S  I L  Gm  I max
* KL, KU, 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 IL and IU for each intersection
d. Perform total interference minimization
Min ITotal  Min[Max IU , j  Max I L, j ]
e. Identify the optimal progression band and offsets
Ba  Gm  I Total
f. Construct the time-space diagram
g. Adjust split of directional bandwidth if desired
CEE 764 – Fall 2010
Graphical Illustration
1
m
2
3
ITotal  Max ( IU ,1 , IU , 2 )  I L,3
IL,3
Time
IU,2
IU,1
Distance
CEE 764 – Fall 2010
Example
Speed is 40 mph
Int. #
Gj
Rj
Dmj
-0.5(Rm-Rj)
Tmj
0.5(Tmj+Tjm)
-IS
Imax
1
45
35
1500
-2.5
-25.56
-25.56
-5
40
2*
40
40
-
-
-
-
-
-
3
45
35
330
-2.5
5.62
5.62
-5
4
50
30
1830
-5.0
31.19
31.19
5
45
35
4830
-2.5
82.31
82.31
I L  0.5( Rm  R j )  0.5(Tmj  T jm )  0.5CK L
0   mj  0.5(Tmj  T jm )  0.5CK  C
CEE 764 – Fall 2010
KU
IU
KL
IL
θmj
-
-
-
-
-
0
1
-1
3.120
-36.88
43.12
0
1
-8.12
31.87
-5
IU  0.5( Rm  R j )  0.5(Tmj  T jm )  0.5CKU
 I S  Gm  G j  Gm
 I S  IU  Gm  I max
 I S  I L  Gm  I max
Illustration of Intersection #2 and #4
(Simultaneous Offset)
m=2
#4
Rj
Time
26.19
Gj
Gm
Rm
CEE 764 – Fall 2010
Distance
Illustration of Intersection #2 and #4
(Alternate Offset)
Time
m=2
#4
3.81
Gm
Rj
Gj
Rm
CEE 764 – Fall 2010
Distance
Illustration of Intersection #2 and #5
(Simultaneous – Slack Times)
m=2
#5
0.19
4.81
Time
Rj
Gj
Gm
Rm
CEE 764 – Fall 2010
Distance
Example (continued)
CEE 764 – Fall 2010
Int. #
Gj
IU
IL
1
45
11.94
23.06
2*
40
0
0
3
45
3.12
31.87
4
50
26.19
3.81
5
45
-0.19
-4.81
Rank #
Int. #
IU
IL
1
4
26.19
3.81
2
1
11.94
23.06
3
3
3.12
31.87
4
2=m
0
0
5
5
-0.19
-4.81
Example (continued)
Intersection by Rank
ITotal
1
2
3
4
5
U
U
U
U
U
26.19
L
U
U
U
U
11.94 + 3.81 = 15.75
L
L
U
U
U
L
L
L
U
U
L
L
L
L
U
L
L
L
L
L
Interference
Rank
Int. #
K-value
Offset
θmj
(Reference Start of Green)
L
1
4
1
Alternate
U
2
1
1
Alternate
U
3
3
0
Simultaneous
U
4
2=m
-
-
U
5
5
2
Simultaneous
CEE 764 – Fall 2010
Offset to Start of Green
(Alternate)
m
 G ,mj
Rj
Rm
  R ,mj 

2
2
Time
Rj
θG,mj
Gm
θR,mj
Rm
CEE 764 – Fall 2010
Distance
Offset to Start of Green
(Simultaneous)
m
 G ,mj
Rj
Rm
  R ,mj 
C 
2
2
Time
Rj
θG,mj
Gm
θR,mj = 0
Rm
CEE 764 – Fall 2010
Distance
Bandwidth with LT Phases
Time
Intersection 1
CEE 764 – Fall 2010
Intersection 2
Space
Major Terms
Phase B
OB left turn
Cycle
OB through
IB through
IB left turn
Link length in-between
CEE 764 – Fall 2010
Bandwidth Maximization
Outbound
Inbound Bandwidth
Outbound Bandwidth
CEE 764 – Fall 2010
Bandwidth Maximization
Inbound Bandwidth
Outbound Bandwidth
CEE 764 – Fall 2010
Phasing Sequence
Leading
Lagging
Lag-Lead
4 types of left turn sequence for each intersection
CEE 764 – Fall 2010
Lead-Lag
16 Combinations
Methodology
Maximum Bandwidth = Bo + Bi
Bo <= Gomin
Bi <= Gimin
Bmax = Go,min +Gi,min – Ii,min
j
X
CEE 764 – Fall 2010
Methodology
Bmax = Go,min +Gi,min – Ii,min
Interference
Bmax = Constant
Exception:
Bmax = Gi,min
CEE 764 – Fall 2010
Methodology
Bmax = Go,min +Gi,min – Ii,min
Interference
Bmax = Constant
Exception:
Bmax = Gi,min
CEE 764 – Fall 2010
Upper Interference
Iup
CEE 764 – Fall 2010
X – Intersection that has the smallest inbound green
Iujp = Gix - (-Rxn +Txj + Rjp + Gij + Tjx ) mod C
Gix+n*CL
(-Rxn +Txj + Rjp + Gij + Tjx )
Iujp
-Rxn +Txj + Rjp + Gij
Gix
-Rxn +Txj
-Rjp
-Rxn +Txj + Rjp
0
Rxn
CEE 764 – Fall 2010
-Rxn
Intersection X
Intersection j
Iujp = Gix - (-Rxn +Sox +Txj + Rjp + Gij + Tjx ) mod C
Gix+n*CL
(-Rxn +Sox+Txj + Rjp + Gij + Tjx )
Iujp
-Rxn +Sox+Txj + Rjp + Gij
Gix
-Rxn +Sox+Txj
-Rjp
-Rxn +Sox+Txj + Rjp
0
CEE 764 – Fall 2010
-Rxn+Sox
-Rxn
Rxn
Intersection X
Intersection j
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
CEE 764 – Fall 2010
Intersection X
Intersection j
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
41
j
30
55*
-30
CEE 764 – Fall 2010
0
x
30+34
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
Intersection j has the
smallest outbound green
41*
34-34+18+61-34=45
55*
34
CEE 764 – Fall 2010
x
Iu = 55-45=10
Iujp = Gix - (-Rxn + Sox –Txj + Rjp + Gij -Tjx ) mod C
30+45
T.T.=34 sec
20
45
55*
30+4-34+18+61
Sox
30+4
-(-30)=30
30
38
0
61
41*
30+4-34+18
18
30+4-34
20
j
Intersection j has the
smallest outbound green
45
30+4-34+18+61-34=45
Iu = 55-45=10
55*
30
CEE 764 – Fall 2010
x
ILjp = (-Rxn +Txj – Sj+ Rjp + Tjx) mod C
ILjp
Go,min
Slack Time, Sj
Sj = Goj – Go,min
0
Rxn
CEE 764 – Fall 2010
x
j
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
15
38+50-4
24
41
j
38
55*
-38
CEE 764 – Fall 2010
0
x
38+50
No valid lower interference
T.T.=34 sec
16
41
Iu = 24+55-61=18
Lower interference calculation causes upper
interference occurring. Intersection j must move
up by 4 sec to reduce upper interference.
Inbound band = 55-14=41
55*
30+34-4+22+61
30
38
30+34-4+22+34=116
61
24
140-116=24>>6
45
30+34-4+22
30+34-4
22
30+34
16
j
41
30
55*
-30
CEE 764 – Fall 2010
0
x
Lower Interference
Gij – Gix
Gij
C ILjp
C - ILjp <= (Gij – Gix)=Sj
CEE 764 – Fall 2010
ILjp >= C – Sj
ILjp = (-Rxn + Txj – Soj+ Rjp + Tjx) mod C
2-sec slack
T.T.=95 sec
30
55*
0
41*
-16+95-4-38+95=132
16
140-132=8<10
No lower interference
x
18
45
-16+95
65
30
38
-16+95-4-38
55*
j
41*
0
16
-16
CEE 764 – Fall 2010
-16+95-4
x
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
A
T.T.=56 sec
B
C
38
16
57
18
61*
50
41*
64
64
18
CEE 764 – Fall 2010
30
25
Example
T.T.=34 sec
A
T.T.=56 sec
B
C
38
16
57
18
61*
50
41*
64
64
18
30
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
CEE 764 – Fall 2010
CEE 764 – Fall 2010
Homework (C=60)
T.T.=36 sec
T.T.=51 sec
10
T.T.=27 sec
T.T.=41 sec
20
10
30
30
20
30
10
CEE 764 – Fall 2010
25*
15
25
25
26
15
12
10
30
25
13
15
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
20
25
61
64
45
57
41*
55*
18
CEE 764 – Fall 2010
30
18
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
Speed is 40 mph
Int. #
Gj
Rj
Dmj
-0.5(Rm-Rj)
Tmj
0.5(Tmj+Tjm)
-IS
Imax
1
45
35
1500
-2.5
-25.56
-25.56
-5
40
2*
40
40
-
-
-
-
-
-
3
45
35
330
-2.5
5.62
5.62
-5
4
50
30
1830
-5.0
31.19
31.19
5
45
35
4830
-2.5
82.31
82.31
IU  Gix  (Txj  G j  T jx ) mod C
 I S  IU  Gm  I max
CEE 764 – Fall 2010
-5
I L  (Txj  S j  T jx ) mod C
 I S  I L  Gm  I max
KU
IU
KL
IL
θmj
-
-
-
-
-
0
1
-1
3.120
-36.88
43.12
0
1
-8.12
31.87
0
-1
26.19
-13.81
0
1
-36.19
3.81
0
1
2
79.81
35.81
-0.19
1
2
-44.81
-4.81
CEE 764 – Fall 2010