Signal Timing, Capacity and Delay for the Same Intersection: 16-Civ-B10 Traffic Engineering, December 2015, Q7

Question 7: Signal Timing, Capacity and Delay for the Same Intersection (20 marks — 4 each)

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. Everything carried forward from Question 6: two phases, intergreen \(I = 5.1\) s per phase, total lost time \(L = 10.2\) s, basic saturation flow \(s_0 = 2200\) pcu/h/lane, critical flow ratios \(y_{NS} = 0.470\) and \(y_{EW} = 0.364\) with \(Y = 0.834\), and approach demands of 1034, 620, 400 and 800 pcu/h for NB, SB, EB and WB. Each crosswalk is 3.0 m wide and, with no refuge, spans the full carriageway of \(2 \times 4.0 = 8.0\) m.

Find. (a) the pedestrian-governed minimum cycle; (b) the green split at \(C = 90\) s; (c) whether those greens satisfy pedestrians; (d) capacity and degree of saturation per approach; (e) the flow-weighted average delay.

Approach. Compute the pedestrian crossing requirement from the HCM crossing-time expression, then the vehicular split by Webster’s proportional rule \(g_i = (y_i/Y)(C-L)\), and finally capacity \(c = s_0 g/C\), degree of saturation \(x = q/c\), and Webster’s three-term delay for each approach.

Part (a) — pedestrian crossing time required. The HCM minimum pedestrian green for a crosswalk 3.0 m wide or narrower is \[ G_p = 3.2 + \frac{L_c}{S_p} + 0.27\,N_{\text{ped}} \] where 3.2 s is the start-up and clearance allowance, \(L_c\) is the kerb-to-kerb crossing length and \(S_p\) the walking speed. With no refuge the pedestrian must cross the full 8.0 m carriageway in one movement, and at \(S_p = 1.2\ \text{m/s}\), \[ G_p = 3.2 + \frac{8.0}{1.2} + 0 = 3.2 + 6.67 = 9.87\ \text{s} \] The paper gives no pedestrian volume, so the platoon term \(0.27 N_{\text{ped}}\) is taken as zero; its size is examined in the note below. Both crosswalk pairs are identical in length, so the same requirement applies in each phase.
Minimum cycle time for pedestrians. Each phase must provide its own pedestrian crossing time plus its intergreen, so the cycle cannot be shorter than the sum over the two phases: \[ C_{\min,\text{ped}} = \sum_{i=1}^{2}\left( G_{p,i} + I_i \right) = 2\,(9.87 + 5.1) = \boxed{29.9\ \text{s} \approx 30\ \text{s}} \] This is a low value, and it will not govern: the vehicular demand computed in Question 6 requires a far longer cycle. Webster’s optimum for this intersection is \(C_o = (1.5L+5)/(1-Y) = (15.3+5)/0.166 = 122\) s, so the 90 s cycle imposed in part (b) already sits well above the pedestrian floor.
Part (b) — effective green available at a 90 s cycle. Lost time is fixed at 10.2 s regardless of cycle length, so \[ \sum g = C - L = 90 - 10.2 = 79.8\ \text{s} \]
Allocate the green in proportion to the critical flow ratios. Webster’s rule gives each phase a share of the effective green equal to its share of \(Y\), which equalises the degree of saturation on the two critical movements: \[ g_i = \frac{y_i}{Y}\,(C - L) \] \[ g_{NS} = \frac{0.470}{0.834}(79.8) = \boxed{45.0\ \text{s}} \qquad g_{EW} = \frac{0.364}{0.834}(79.8) = \boxed{34.8\ \text{s}} \] The two greens plus the two intergreens close the cycle exactly: \(45.0 + 34.8 + 2(5.1) = 90.0\) s. Because no start-up lost time was assumed, the displayed green equals the effective green in each phase.
Part (c) — pedestrian check on the allocated greens. Pedestrians crossing a leg walk on the green of the conflicting phase: those crossing the north and south legs (the north–south carriageway) walk during the east–west green, and vice versa. The requirement in each case is \(G_p = 9.87\) s, and the available times are \[ g_{NS} = 45.0\ \text{s} \gg 9.87\ \text{s} \qquad g_{EW} = 34.8\ \text{s} \gg 9.87\ \text{s} \] Both phases exceed the pedestrian requirement by at least 24.9 s, and the 90 s cycle comfortably exceeds the 29.9 s pedestrian minimum, so the green intervals allocate ample time for pedestrians and no adjustment to the vehicular split is needed. Even a pedestrian walking at the slower 1.0 m/s value used for elderly and mobility-impaired users would need only 11.2 s.
Part (d) — lane capacity. Capacity is the saturation flow prorated by the green ratio \(\lambda = g/C\): \[ c = s_0\,\frac{g}{C} \] For the north–south approaches, \(\lambda_{NS} = 45.0/90 = 0.500\) and \(c = 2200(0.500) = 1100\ \text{pcu/h}\); for the east–west approaches, \(\lambda_{EW} = 34.8/90 = 0.387\) and \(c = 2200(0.387) = 851\ \text{pcu/h}\). Both NB and SB share the same capacity because they run on the same phase, and likewise EB and WB.
Degree of saturation for each approach. The degree of saturation is the demand-to-capacity ratio \(x = q/c\): \[ x_{NB} = \frac{1034}{1100} = 0.940 \qquad x_{SB} = \frac{620}{1100} = 0.564 \] \[ x_{EB} = \frac{400}{851} = 0.470 \qquad x_{WB} = \frac{800}{851} = \boxed{0.940} \] The two critical movements come out at exactly the same value, which is the whole point of the proportional split, and that common value is \(x_{\text{crit}} = YC/(C-L) = 0.834(90)/79.8 = 0.940\). All four approaches are below 1.0, so the timing is workable, but NB and WB are at 94 % of capacity and will overflow whenever demand fluctuates above its average.
Part (e) — Webster’s delay per approach. Webster’s formula combines a uniform-arrival term, a random-arrival (overflow) term and an empirical correction: \[ d = \frac{C(1-\lambda)^{2}}{2\,(1-\lambda x)} + \frac{x^{2}}{2q\,(1-x)} - 0.65\left(\frac{C}{q^{2}}\right)^{1/3} x^{\,(2+5\lambda)} \] with \(q\) in pcu/s. Taking the NB approach as the worked case, \(\lambda = 0.500\), \(x = 0.940\), \(q = 1034/3600 = 0.2872\) pcu/s: \[ d_{NB} = \frac{90(0.500)^{2}}{2(1-0.470)} + \frac{0.940^{2}}{2(0.2872)(0.060)} - 0.65\left(\frac{90}{0.0825}\right)^{1/3}(0.940)^{4.50} \] \[ d_{NB} = 21.24 + 25.73 - 5.07 = 41.9\ \text{s/pcu} \] The same substitution on the other three approaches gives \(d_{SB} = 17.1\), \(d_{EB} = 21.9\) and \(d_{WB} = 53.6\) s/pcu. The WB approach is worst because it combines the shorter green with a degree of saturation of 0.940, so its overflow term alone is 33.3 s.
Average overall delay for the intersection. The intersection average is the flow-weighted mean of the approach delays, \[ d_I = \frac{\sum q_i\,d_i}{\sum q_i} = \frac{1034(41.9) + 620(17.1) + 400(21.9) + 800(53.6)}{1034 + 620 + 400 + 800} \] \[ d_I = \frac{105\,580}{2854} = \boxed{37.0\ \text{s/pcu}} \] By the HCM signalised-intersection criteria (35 s < \(d\) ≤ 55 s) the intersection operates at level of service D — acceptable in an urban setting but with little margin, consistent with the critical degree of saturation of 0.94.

Check — the pedestrian platoon term. The paper gives no pedestrian volume, so \(N_{\text{ped}} = 0\) has been used in \(G_p\). The term is not negligible in principle: because \(N_{\text{ped}} = q_{\text{ped}}\,C/3600\) depends on the cycle, the pedestrian minimum cycle is a fixed point, \(C = (2a + L)/(1 - 2b)\) with \(a = 3.2 + L_c/S_p\) and \(b = 0.27\,q_{\text{ped}}/3600\). At an assumed 200 ped/h per crosswalk this yields \(C_{\min} = 30.9\) s instead of 29.9 s — still far below the vehicular requirement, so the conclusion in parts (a) and (c) is unaffected for any plausible pedestrian volume at this location.

Part	Quantity	Result
(a)	Pedestrian crossing time \(G_p\) (8.0 m, no refuge)	9.87 s per phase
(a)	Minimum cycle for pedestrians	29.9 s (say 30 s)
(b)	Effective green, N–S phase	45.0 s
(b)	Effective green, E–W phase	34.8 s
(c)	Pedestrian adequacy of those greens	Adequate — 45.0 s and 34.8 s against 9.87 s needed
(d)	Capacity, NB and SB	1100 pcu/h each
(d)	Capacity, EB and WB	851 pcu/h each
(d)	Degree of saturation NB / SB / EB / WB	0.940 / 0.564 / 0.470 / 0.940
(e)	Approach delay NB / SB / EB / WB	41.9 / 17.1 / 21.9 / 53.6 s/pcu
(e)	Average overall intersection delay	37.0 s/pcu — level of service D

Topic. Webster signal timing: pedestrian minimum cycle, green split, capacity, degree of saturation and delay.

Key relations. \(G_p = 3.2 + L_c/S_p + 0.27 N_{\text{ped}}\) (crosswalk ≤ 3.0 m); \(C_{\min,\text{ped}} = \sum (G_{p,i}+I_i)\); \(C_o = (1.5L+5)/(1-Y)\); \(g_i = (y_i/Y)(C-L)\); \(c = s\,g/C\); \(x = q/c = y C/g\); \(x_{\text{crit}} = YC/(C-L)\); Webster delay \(d = C(1-\lambda)^2/[2(1-\lambda x)] + x^2/[2q(1-x)] - 0.65 (C/q^2)^{1/3} x^{2+5\lambda}\).

Why this works. Splitting the available green in proportion to the critical flow ratios equalises \(x\) across the critical movements, and an equal-\(x\) split is the one that minimises total intersection delay — giving either phase more would push the other closer to saturation, where delay rises hyperbolically. That is visible in the numbers: the two critical approaches share \(x = 0.940\) exactly. Webster’s delay expression then separates the two physical sources of waiting. The first term is the deterministic delay of the queue accumulation polygon of Question 5, integrated over a cycle; the second is the extra delay caused by the randomness of arrivals, which is why it blows up as \(x \to 1\); and the third is an empirical correction of 5–15 % fitted to simulation, needed because the first two terms together overestimate delay. The pedestrian minimum cycle is a separate constraint that must be checked but rarely binds at a busy intersection.

Common pitfall. Dividing the cycle rather than the effective green, which double-counts the lost time and overstates both greens; and forgetting that pedestrians on a leg cross during the conflicting phase’s green, so the north-leg crosswalk must be checked against \(g_{EW}\), not \(g_{NS}\). In part (e) the frequent error is to average the four approach delays arithmetically instead of weighting them by flow, which here would give 33.6 s/pcu and the wrong level of service.

Source. Garber & Hoel, Traffic and Highway Engineering, Ch. 8, “Webster’s method for signal timing and delay”; HCM, signalised-intersection capacity, control delay and level-of-service thresholds; TAC, Manual of Uniform Traffic Control Devices for Canada, pedestrian clearance practice.

Question 7 of 7: Signal Timing, Capacity and Delay for the Same Intersection

Question 7: Signal Timing, Capacity and Delay for the Same Intersection (20 marks — 4 each)