Chapter 8, Problem 18P

To determine

Effect on cycle length due to $20\%$ higher pedestrian flow rates.

Answer to Problem 18P

Cycle length value increases with $4.76\%$, from $126$ to $132$.

Explanation of Solution

Given data:

Calculation:

Evaluating equivalent hourly flow −

  $\begin{array}{c}\text{Equivalent}\text{hourly}\text{flow}\text{=}\frac{\text{Traffic}\text{volume}}{\text{PHF}}\\ =\frac{133}{0.95}\\ \simeq 140\end{array}$

Similarly, evaluating equivalent hourly flow for all traffic movements −

Table 1

Approach (width)	N (56ft)	S (56ft)	E (68ft)	W (68ft)
Left turn	133/0.95 = 140	73/0.95 = 77	168/0.95 = 177	134/0.95 = 142
Through movement	443	393	593	544
Right turn	148	143	178	188
Conflicting pedestrian volume	948	1264	1264	948

Assuming lane configuration as one dedicated left turn and combined through and right lane −

Table 2

Approach	N	S	E	W
Left	140	77	177	142
Through + Right	589 (442+147)	535 (393+178)	771 (593+178)	730 (543+187)

Assume a phase scheme and find critical ratios( $Y_i$ ) and sum of critical ratios ( $\sum Y_i$ ). Assuming phase as follows −

Table 3

-	Phase lE-W (Left)	Phase llE-W (Through)	Phase lllN-S (Left)	Phase lVN-S (Through)
$q_{ij}$	177	771	140	589
$S_{ij}$	1000	3000 (1600+1400)	1000	3000 (1600+1400)
$Y_i=\frac{{\text{q}}_{ij}}{S_{ij}}$	0.177	0.257	0.140	0.196

Sum of critical ratios −

  $\begin{array}{c}\sum Y_i=Y_1+Y_2+Y_3+\mathrm{........}{Y}_n\\ =.177+0.257+0.140+0.196\\ =0.77\end{array}$

Assuming lost time per phase ( $l_i$ )= $3\mathrm{Sec}$

So, Total lost time −

  $\begin{array}{c}L=\sum l_i\\ =3+3+3+3\\ =12\mathrm{Sec}\end{array}$

Now, determining the optimum cycle length −

  $\begin{array}{c}{C}_o=\frac{1.5L+5}{1-\sum Y_i}\\ =\frac{1.5\times 12+5}{1-0.77}\\ =100\mathrm{Sec}\end{array}$

(Cycle lengths are generally multiple of $5\text{sec}\text{or}10\text{sec}$, Hence OK)

Finding Total effective green time −

  $\begin{array}{c}{G}_{ie}=C_o-L\\ =100-12\\ =88\text{Sec}\end{array}$

Effective time for phase $i$ can be calculated as −

  $G_{ei}=\frac{Y_i}{Y_1+Y_2+Y_3+\mathrm{........}{Y}_n}{G}_{ie}$

For Phase l

  $\begin{array}{c}{G}_{phasel}=\frac{0.177}{0.177+0.257+0.140+0.196}\times 88\\ =20.23\mathrm{Sec}\end{array}$

(Assuming yellow time as $4\text{sec}$ )

  $\begin{array}{c}{\left(G+Y\right)}_{phasel}=20.23+4\\ =24.23Sec\\ \simeq 25\mathrm{Sec}\end{array}$

For Phase ll

  $\begin{array}{c}{G}_{phasell}=\frac{0.257}{0.177+0.257+0.140+0.196}\times 88\\ =29.37\mathrm{Sec}\end{array}$

  $\begin{array}{c}{\left(G+Y\right)}_{phasell}=29.37+4\\ =33.37Sec\\ \simeq 34\mathrm{Sec}\end{array}$

For Phase lll

  $\begin{array}{c}{G}_{phaselll}=\frac{0.140}{0.177+0.257+0.140+0.196}\times 88\\ =16\mathrm{Sec}\end{array}$

  $\begin{array}{c}{\left(G+Y\right)}_{phaselll}=16+4\\ =20Sec\end{array}$

For Phase lV

  $\begin{array}{c}{G}_{phaselV}=\frac{0.196}{0.177+0.257+0.140+0.196}\times 88\\ =22.4\mathrm{Sec}\end{array}$

  $\begin{array}{c}{\left(G+Y\right)}_{phasel}=22.4+4\\ =26.4\text{Sec}\\ \simeq 27Sec\end{array}$

Table 4

Phase	Allocated green & yellow time (in sec)
${\left(G+Y\right)}_1$	$25$
${\left(G+Y\right)}_2$	$34$
${\left(G+Y\right)}_3$	$20$
${\left(G+Y\right)}_4$	$27$

Total cycle length

  $\begin{array}{c}C=\sum {\left(G+Y\right)}_{phasei}+\text{Total}\text{red}\text{time}\text{interval}\\ \text{=}\left(25+34+20+27\right)+4\left(1.5\right)\\ =112\mathrm{Sec}\end{array}$

Green time required for pedestrian crossing can be calculated as following formula:

(Assuming the crosswalk width as $9ft$< $10ft$ ).

  $G_p=3.2+\frac{L_{cc}}{S_p}+0.27N_{ped}$

Where,

  $L_{cc}$ = cross walk length

  $S_p$ = pedestrian speed (assuming as $4\frac{ft}{\sec }$ )

  $N_{ped}$ = number pedestrians crossing during an interval = $\frac{v_{pe{d}_i}}{3600}C$

  $v_{pe{d}_i}$ = pedestrian flow rate in the subject crossing for travel direction $i\left(p/h\right)$

  $C$ = total cycle length

Calculating $N_{ped}$ for each direction:

  $N_{ped}$ for N direction = $\frac{948}{3600}\times 112=29.49\simeq 30\sec$

  $N_{ped}$ for S direction = $\frac{1264}{3600}\times 112=39.32\simeq 40\sec$

  $N_{ped}$ for E direction = $\frac{1264}{3600}\times 112=39.32\simeq 40\sec$

  $N_{ped}$ for W direction = $\frac{948}{3600}\times 112=29.49\simeq 30\sec$

Calculating minimum time required ( $G_p$ ) for each approach:

Minimum time required for N approach ( $G_{P1}$ ): $3.2+\frac{56}{4}+0.27\times 30=25.3\simeq 26\sec$

Minimum time required for S approach ( $G_{P2}$ ): $3.2+\frac{56}{4}+0.27\times 40=28\sec$

Minimum time required for E approach ( $G_{P3}$ ): $3.2+\frac{68}{4}+0.27\times 40=31\sec$

Minimum time required for W approach ( $G_{P4}$ ): $3.2+\frac{68}{4}+0.27\times 30=28.3\simeq 29\sec$

Table 5

Phase	Minimum green time (in sec)
$G_{P1}$	$26$
$G_{P2}$	$28$
$G_{P3}$	$31$
$G_{P4}$	$29$

  $G_{P1}$ is greater than ${\left(G+Y\right)}_1$ so the allocated green and yellow time for phase $1$ is from the table 5 $26$ sec.

Sum of green and yellow time is given by,

  $\begin{array}{c}{G}_1+G_2+G_3+G_4=26\sec +28\sec +31\sec +29\sec \\ =114\sec \end{array}$

Total cycle length is given by,

  $\begin{array}{l}C=(\text{total}\text{green}\text{and}\text{yellow}\text{time})+(\text{total}\text{red}\text{time})\\ C=\left(26+34+31+29\right)+\left(4\times 1.5\right)\\ C=126\sec \end{array}$

Now increasing the pedestrian volume with $20\%$ and calculating the minimum time required by pedestrian for each approach:

  $\begin{array}{c}\text{New pedestrian volume = old pedestrian volume + old pedestrian volume}\\ =948+948\times \frac{20}{100}\\ =948+189.6\\ \simeq 1138\end{array}$

Table 6

New conflicting pedestrian volume

1138

1517

1517

1138

According to the new pedestrian volume calculating minimum time required by pedestrian for each approach:

Calculating $N_{ped}$ for each direction:

  $N_{ped}$ for N direction = $\frac{1138}{3600}\times 126=39.83\simeq 40\sec$

  $N_{ped}$ for S direction = $\frac{1517}{3600}\times 126=53.09\simeq 54\sec$

  $N_{ped}$ for E direction = $\frac{1517}{3600}\times 126=53.09\simeq 54\sec$

  $N_{ped}$ for W direction = $\frac{1138}{3600}\times 126=39.83\simeq 40\sec$

Calculating new minimum time required ( $G_p^{*}$ ) for each approach:

Minimum time required for N approach ( $G_{{}_{P1}}^{*}$ )= $3.2+\frac{56}{4}+0.27\times 40=28\sec$

Minimum time required for S approach ( $G_{{}_{P2}}^{*}$ ) = $3.2+\frac{56}{4}+0.27\times 54=31.78\simeq 32\sec$

Minimum time required for E approach ( $G_{P3}^{*}$ ) = $3.2+\frac{68}{4}+0.27\times 54=34.78=35\sec$

Minimum time required for W approach ( $G_{{}_{P4}}^{*}$ ) = $3.2+\frac{68}{4}+0.27\times 40=31\sec$

Comparing the $G_p^{*}$ values with table 4, and selecting new values as shown in table below:

Selecting greater values in between both $G_p^{*}$ and ${\left(G+Y\right)}_i$, which is taken as new minimum green time:

Table 7

Phase	New minimum green time (in sec)
$G_1^{*}$	$28$
$G_{{}_2}^{*}$	$32$
$G_3^{*}$	$35$
$G_{{}_4}^{*}$	$31$

Sum of green and yellow time is given by,

  $\begin{array}{c}{G}_1^{*}+G_2^{*}+G_3^{*}+G_4^{*}=28\sec +32\sec +35\sec +31\sec \\ =126\sec \end{array}$

Total new cycle length is given by,

  $\begin{array}{l}C=(\text{total}\text{green}\text{and}\text{yellow}\text{time})+(\text{total}\text{red}\text{time})\\ C=\left(28+32+35+31\right)+\left(4\times 1.5\right)\\ C=132\sec \end{array}$

Conclusion:

With using pedestrian volume flow rate $20\%$ higher than given, cycle length increases by $4.76\%$.