Question 3 of 7: Webster signal design (20 marks)

Question 3: Webster signal design (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The peak-hour turning-movement table, the pedestrian volumes, the lane saturation flows, and a lost time of 3.5 s plus an all-red of 1.5 s at each phase change.

Find. An appropriate phasing system with its justification, the intersection geometry and lane assignment, the cycle length by the Webster method, and the phase lengths — together with the degree of saturation and average delay the design delivers.

Approach. Convert the peak-hour volumes to design flow rates, settle the geometry from the stated widths, price and reject the four-phase and single-lane alternatives, balance the two lanes on each approach to find the flow ratios, sum the critical ones, then take the larger of Webster’s vehicle optimum and the HCM pedestrian minimum as the cycle.

1. Convert peak-hour volumes to design flow rates. Webster works on the flow rate within the peak, so every movement is divided by the peak-hour factor: $q = V/\text{PHF} = V/0.9$. For the South approach, for example, $369/0.9 = 410.00$, $582/0.9 = 646.67$ and $360/0.9 = 400.00$ veh/h.

Design flow rates, volume divided by the 0.9 peak-hour factor

Approach	Left (veh/h)	Through (veh/h)	Right (veh/h)	Approach total
North	127.78	222.22	213.33	563.33
South	410.00	646.67	400.00	1456.67
East	237.78	705.56	444.44	1387.78
West	205.56	457.78	405.56	1068.89

2. Settle the geometry from the stated widths. The tabulated “Approach (Width)” is the full curb-to-curb width shared by both directions, not a one-direction width. At a 3.5 m urban lane, $15 = 4(3.5) + 1.0$ and $18 = 4(3.5) + 4.0$, so both streets are four-lane cross-sections — two lanes per approach — the N–S street with a 1.0 m median strip and the E–W street with a 4.0 m median. The reading is confirmed by the loadings it produces: the eight lanes carry between 272 and 752 veh/h, which is ordinary urban arterial duty. Reading 15 m as a one-direction width would give four lanes on the North approach carrying 141 veh/h each, which no engineer would build.
3. Price the four-phase alternative, and reject it. With only two lanes per approach, protecting the left turns means dedicating the inner lane to lefts ($s = 2002$) and putting the whole through-plus-right demand on the single outer lane ($s = 1895$). The critical flow ratio of each of the four phases is then $$Y_4 = \frac{410.00}{2002} + \frac{1046.67}{1895} + \frac{237.78}{2002} + \frac{1150.00}{1895} = 0.2048 + 0.5523 + 0.1188 + 0.6069 = \boxed{1.483}$$ Since $Y_4 > 1$ the four-phase plan is infeasible at any cycle length: there is not enough time in an hour, let alone enough after lost time. Protected left turns are therefore ruled out by the geometry.
4. Price the single-shared-lane alternative, and reject it too. The saturation-flow table offers a left-through-right lane at 2366 vphpl, which would apply if each approach had one lane. Then $$Y_1 = \frac{1456.67}{2366} + \frac{1387.78}{2366} = 0.6157 + 0.5866 = \boxed{1.202} > 1$$ also infeasible. Two lanes per approach with permitted lefts is the only workable plan, and that is the answer to “determine an appropriate phasing system”: two phases, North–South then East–West, with permitted left turns, justified both by the geometry and by the arithmetic above.
5. Balance the two lanes on each approach. Drivers distribute themselves so that both lanes reach the same degree of saturation, so the through demand splits with $x$ vehicles in the inner (left-through) lane and the rest in the outer (through-right) lane. Setting the two $v/s$ ratios equal, $$\frac{v_L + x}{s_{LT}} = \frac{v_T - x + v_R}{s_{TR}} \;\Longrightarrow\; x = \frac{s_{LT}(v_T + v_R) - s_{TR}\,v_L}{s_{LT} + s_{TR}}$$ For the South approach, $x = [1775(646.67 + 400.00) - 1895(410.00)]/(1775 + 1895) = 294.52$ veh/h, giving an inner-lane flow of $410.00 + 294.52 = 704.52$ and an outer-lane flow of $646.67 - 294.52 + 400.00 = 752.15$ veh/h.

Lane balance and flow ratios; the two critical approaches are shown in bold

Approach	Through split x (veh/h)	Inner lane (veh/h)	Outer lane (veh/h)	Flow ratio y
North	144.68	272.46	290.88	0.1535
South	294.52	704.52	752.15	0.3969
East	433.42	671.20	716.58	0.3781
West	311.41	516.97	551.92	0.2913

6. Identify the critical movements and sum them. Each phase is sized by its heaviest approach, and it is the flow ratio, not the volume, that decides: $y_S = 0.3969 > y_N = 0.1535$ and $y_E = 0.3781 > y_W = 0.2913$, so South governs the N–S phase and East governs the E–W phase. Hence $$Y = y_S + y_E = 0.3969 + 0.3781 = \boxed{0.7750}$$
7. Total lost time. With no start-up lost time quoted separately, the whole of each intergreen is taken as lost: $$L = n(\ell + R_{ar}) = 2(3.5 + 1.5) = \boxed{10.0\ \text{s}}$$
8. Webster optimum cycle. $$C_o = \frac{1.5L + 5}{1 - Y} = \frac{1.5(10.0) + 5}{1 - 0.7750} = \frac{20.0}{0.2250} = \boxed{88.91\ \text{s}}$$ Note how steeply this depends on $Y$: the relation is hyperbolic, so the last few per cent of capacity are very expensive in cycle length.
9. Pedestrian minimum cycle — the constraint that usually governs. The table gives 850 to 1,200 conflicting pedestrians per hour, which is heavy, so the cycle must also be long enough for each phase to clear its crosswalk. The HCM minimum pedestrian green for a crosswalk wider than 3.0 m is $$G_p = 3.2 + \frac{L_c}{S_p} + 2.7\,\frac{N_{ped}}{W_E},\qquad N_{ped} = \frac{v_{ped}\,C}{3600}$$ Two assumptions are needed and are declared under the paper’s Note 2: a walking speed $S_p = 1.2$ m/s and an effective crosswalk width $W_E = 4.0$ m, both standard for a signalised urban intersection carrying this pedestrian volume. The crosswalk a phase must clear is the one whose pedestrians walk parallel to that phase’s vehicles, so the N–S vehicle phase clears the east and west legs, which span the 18 m E–W carriageway at 1,200 ped/h, and the E–W vehicle phase clears the north and south legs, spanning the 15 m N–S carriageway at 905 ped/h.
10. Solve the pedestrian constraint as a fixed point in C. Because $N_{ped}$ itself grows with $C$, this is not a one-shot evaluation. The green a phase receives is $(y_i/Y)(C - L)$, so the requirement per phase is $$\frac{y_i}{Y}(C - L) \;\ge\; a + b\,C,\qquad a = 3.2 + \frac{L_c}{S_p},\qquad b = \frac{2.7\,v_{ped}}{3600\,W_E}$$ which is linear in $C$ and solves to $C \ge (a + kL)/(k - b)$ with $k = y_i/Y$. For the N–S phase $a = 3.2 + 18/1.2 = 18.20$, $b = 2.7(1200)/[3600(4.0)] = 0.2250$ and $k = 0.3969/0.7750 = 0.51211$, so $$C \ge \frac{18.20 + 0.51211(10.0)}{0.51211 - 0.2250} = \frac{23.32}{0.28711} = 81.23\ \text{s}$$ and for the E–W phase, with $a = 3.2 + 15/1.2 = 15.70$, $b = 0.16969$ and $k = 0.48789$, $C \ge 64.67$ s. The governing pedestrian bound is therefore $\boxed{81.23\ \text{s}}$, the larger of the two.
11. Adopt the cycle. Both bounds must be satisfied, so the design cycle is the larger, rounded up to a practical 5 s increment. Here the vehicle optimum governs: 88.91 s against 81.23 s, a margin of 7.68 s. Hence $$C = \boxed{90\ \text{s}}$$
12. Split the green in proportion to the flow ratios. The total effective green is $g_{tot} = C - L = 90 - 10 = 80.0$ s, apportioned as $y_i/Y$: $$g_{NS} = \frac{0.3969}{0.7750}(80.0) = \boxed{40.97\ \text{s}},\qquad g_{EW} = \frac{0.3781}{0.7750}(80.0) = \boxed{39.03\ \text{s}}$$ and these sum to 80.0 s exactly, as they must.
13. Convert to displayed greens and close the cycle. The displayed green returns the 3.5 s of acceleration/deceleration lost time to each phase, leaving the all-red as the only unassigned interval: $$G_{NS} = 40.97 + 3.5 = \boxed{44.47\ \text{s}},\qquad G_{EW} = 39.03 + 3.5 = \boxed{42.53\ \text{s}}$$ Check: $44.47 + 1.5 + 42.53 + 1.5 = 90.0$ s, so the cycle closes exactly.
14. Check that the pedestrians really are served at C = 90 s. With $N_{ped} = 1200(90)/3600 = 30.0$ pedestrians on the E–W walk, $$G_p = 3.2 + \frac{18}{1.2} + 2.7\frac{30.0}{4.0} = 3.2 + 15.0 + 20.25 = 38.45\ \text{s} \;\le\; g_{NS} = 40.97\ \text{s}$$ and with $N_{ped} = 905(90)/3600 = 22.63$ on the N–S walk, $G_p = 3.2 + 12.5 + 15.27 = 30.97$ s against $g_{EW} = 39.03$ s. Both pass, with 2.52 s and 8.06 s of slack respectively.
15. Capacity and degree of saturation. The critical lane on each phase is the shared left-through lane, so $$c_{NS} = s_{LT}\frac{g_{NS}}{C} = 1775\frac{40.97}{90} = 807.99\ \text{veh/h},\qquad c_{EW} = 1775\frac{39.03}{90} = 769.78\ \text{veh/h}$$ Because the green was split in proportion to the flow ratios, both critical lanes must land on the same degree of saturation, $x_{crit} = YC/(C - L) = 0.7750(90)/80 = 0.8719$. And they do: $704.52/807.99 = 0.8719$ on South and $671.20/769.78 = 0.8719$ on East. That one line validates the flow ratios, the lane assignment and the split together.
16. Average delay by Webster’s formula. With $\lambda = g/C$ and $x$ as above, $$d = \frac{C(1-\lambda)^{2}}{2(1 - \lambda x)} + \frac{x^{2}}{2q(1-x)} - 0.65\left(\frac{C}{q^{2}}\right)^{1/3} x^{\,2+5\lambda}$$ giving $\boxed{32.5\ \text{s}}$ per vehicle on the critical South lane and $\boxed{34.1\ \text{s}}$ on the critical East lane — HCM level of service C to D, which is what a degree of saturation of 0.87 should deliver.