Effect of Higher Saturation Flows and Wider Approaches: 16-Civ-B10 Traffic Engineering, Undated paper, Q3

Question 3: Effect of Higher Saturation Flows and Wider Approaches (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. Everything from Question 2, with two changes: every saturation flow is multiplied by $f = 1.10$ (left-through 2216.5, through-right 2475, left 2194.5, through 3382.5, left-through-right 2750 vphpl), and every approach width increases by 10 m — north and south to 30 m, east and west to 25 m. Volumes, pedestrian volumes, PHF, lost time and all-red are unchanged.

Find. The redesigned cycle length and phase lengths, and an explicit account of how each of the two changes affects the cycle — which is the part of the question that carries the marks.

Question 3 geometry: every approach 10 m wider (30 m north–south, 25 m east–west). The lane assignment is unchanged; the crosswalks each grow by 10 m, which is what drives the cycle length.

Approach. Show first that a uniform saturation-flow change leaves the lane balance and the green split algebraically untouched, so only $Y$ moves; then recompute the two controls separately — Webster's optimum, which responds to the saturation flows, and the pedestrian minimum green, which responds to the crosswalk lengths — and present a four-row decomposition (base, saturation only, width only, both) that names the governing control in every row. That decomposition is the answer to "how do these changes affect the cycle length".

Part (a of the analysis) — Show that the green split cannot change. Scaling every lane saturation flow by the same factor $f$ leaves the lane-balance equation $$\frac{v_L+x}{f\,s_{LT}}=\frac{v_T-x+v_R}{f\,s_{TR}}$$ unchanged, because $f$ cancels. The balancing split $x$ is therefore identical to Question 2, the same approaches (North and West) stay critical, and every flow ratio scales by exactly $1/f$. It follows that $y_i/Y$ is invariant: a changed green split in this question is an arithmetic error, not a result.
Recompute the sum of critical flow ratios and Webster's optimum. $$Y'=\frac{Y}{f}=\frac{0.73098}{1.10}=0.66453$$ $$C_o'=\frac{1.5L+5}{1-Y'}=\frac{20.0}{1-0.66453}=\frac{20.0}{0.33547}=\boxed{59.62\ \text{s}}$$ The 10% capacity gain drops the vehicle optimum by 14.7 s, from 74.35 s. The response is disproportionate because $C_o$ is hyperbolic in $Y$.
Recompute the pedestrian bound with the longer crosswalks. The approach widths are the crosswalk lengths, so the N–S phase must now clear a 25 m E–W carriageway and the E–W phase a 30 m N–S carriageway. With the split, lost time, pedestrian volumes, $S_p$ and $W_E$ all unchanged: $$0.41509\,(C-10)\;\ge\;3.2+\frac{25}{1.2}+0.15C\quad\Longrightarrow\quad C\ge 106.32\ \text{s}$$ $$0.58491\,(C-10)\;\ge\;3.2+\frac{30}{1.2}+0.15C\quad\Longrightarrow\quad C\ge 78.29\ \text{s}$$ The N–S phase still governs, and its bound has risen by 31.4 s.
Adopt the cycle and split the green. $$C'=\max(59.62,\ 106.32)=106.32\ \text{s}\;\longrightarrow\;\boxed{C'=110\ \text{s}}$$ $$g_{NS}'=0.41509(110-10)=41.51\ \text{s},\qquad g_{EW}'=0.58491(100)=58.49\ \text{s}$$ Pedestrian check at the adopted cycle: $G_{p,NS}' = 3.2+20.83+2.7(24.44)/4.0 = 40.53$ s ≤ 41.51 s and $G_{p,EW}' = 3.2+25.00+16.50 = 44.70$ s ≤ 58.49 s. Phase lengths are 46.51 s and 63.49 s, closing on 110.0 s exactly.

Answer the question asked — decompose the two effects. Because the adopted cycle is $\max(C_o, C_{\text{ped}})$, a change pays only when it relieves the bound that is currently binding:

Case	$C_o$ (s)	$C_{\text{ped}}$ (s)	Governing control	Adopted $C$
Base (Question 2)	74.35	74.88	pedestrian (marginally)	75 s
Saturation flows +10% only	59.62	74.88	pedestrian	75 s — no change
Widths +10 m only	74.35	106.32	pedestrian	110 s
Both changes (Question 3)	59.62	106.32	pedestrian	110 s

The conclusion is a complementarity argument and it is the answer worth marks: the extra saturation capacity buys nothing at all, because pedestrians were already the binding control and the capacity gain only lowers a bound that was not active. The entire increase from 75 s to 110 s is caused by the wider carriageways, which lengthen every crosswalk by 10 m and therefore add $10/1.2 = 8.3$ s to the pedestrian clearance in each phase — and that added clearance must be paid for out of a green share of only 0.415, so it costs $8.3/0.415 \approx 20$ s of cycle before the $N_{\text{ped}}$-versus-$C$ feedback amplifies it to 31 s.

Report what the longer cycle bought in capacity terms. $$x_{\text{crit}}'=\frac{Y'C'}{C'-L}=\frac{0.66453(110)}{100}=\boxed{0.731}$$ against 0.843 in Question 2. So the intersection is materially less saturated, but that improvement came from the saturation flows, not from the longer cycle. At 110 s the design is approaching the roughly 120 s practical maximum used in Canadian urban practice, and the honest engineering recommendation is that the widening should have been accompanied by a median pedestrian refuge: splitting a 25 m crossing into two 12.5 m stages would halve $L_c/S_p$, return control to the vehicles at $C_o' = 59.6$ s, and let the design run at roughly 65 s — a better outcome than either the base case or the widened one.

The two competing controls in each of the four cases. Blue is Webster's vehicle optimum, pink the pedestrian bound, and the red rule the adopted cycle. The saturation-flow gain shortens only the bar that is not binding, which is why row 2 changes nothing.

Two-phase plan at C = 110 s after the changes of Question 3. The green split (0.415 : 0.585) is provably identical to Question 2; only the absolute green times and the cycle grow.

Check — what "the approach width increases by 10 m" is taken to mean. The widening is taken to act through the crosswalk length and not to add running lanes: the lane assignment (inner left-through, kerb through-right) and the given per-lane saturation-flow table are held as the data of the problem. Two arguments support this reading. First, the question supplies the capacity change explicitly and separately as "+10% saturation flow rates" — that instruction would be redundant if the extra width were also meant to add lanes. Second, the saturation-flow table lists lane types, not an approach total, so it is silent about lane counts and cannot be re-derived from a width. For completeness, the alternative reading is quantified: a 30 m N–S carriageway takes eight 3.5 m lanes plus a 2 m median (four per approach) and a 25 m E–W carriageway takes six lanes plus a 4 m median (three per approach); redistributing the same demand over those lanes would drive every flow ratio down by roughly a further factor of two, take $Y'$ to about 0.33, and collapse $C_o'$ to about 30 s — which would leave the pedestrian bound of 106.3 s governing even more decisively, so the adopted cycle of 110 s is unchanged under either reading. Assumptions (2)–(4) from Question 2 ($W_E = 4.0$ m, $S_p = 1.2$ m/s, $A = 3.5$ s) carry over unchanged so that the comparison isolates the two stated changes.

Quantity	Symbol	Question 2	Question 3
Sum of critical flow ratios	$Y$	0.7310	0.6645
Green split (N–S : E–W)	$y_i/Y$	0.4151 : 0.5849	0.4151 : 0.5849 (unchanged)
Webster optimum cycle	$C_o$	74.35 s	59.62 s
Pedestrian bound, N–S phase	$C_{\text{ped}}$	74.88 s	106.32 s
Pedestrian bound, E–W phase	$C_{\text{ped}}$	59.13 s	78.29 s
Adopted cycle length	$C$	75 s	110 s
Effective green, N–S phase	$g_{NS}$	26.98 s	41.51 s
Effective green, E–W phase	$g_{EW}$	38.02 s	58.49 s
Critical degree of saturation	$x_{\text{crit}}$	0.843	0.731
Governing control	—	pedestrian	pedestrian (unchanged)

Topic. Sensitivity of a two-control signal design — how a capacity change and a geometric change propagate to the adopted cycle length when the binding constraint is pedestrian rather than vehicular.

Key relations. Scaling invariance: $s \to fs \Rightarrow y \to y/f$ with $x$ and $y_i/Y$ unchanged; $C_o = (1.5L+5)/(1-Y)$, hyperbolic in $Y$; the pedestrian bound $C \ge [3.2 + L_c/S_p + (y_i/Y)L]\,/\,[(y_i/Y) - 2.7v_{\text{ped}}/(3600W_E)]$, linear in $L_c$ and in $v_{\text{ped}}$; adopted $C = \max(C_o, C_{\text{ped}})$.

Why this works. The two levers act on different constraints and the adopted cycle is a maximum, not a sum, so the design responds only to whichever bound is active. That is why a 10% capacity gain that lowers $C_o$ by 15 s changes nothing, while a widening that raises $C_{\text{ped}}$ by 31 s changes everything. The four-row decomposition is the only presentation that makes this visible; a single before-and-after pair of cycle lengths hides the mechanism and cannot answer "how do these changes affect the cycle length". The invariance of the split is worth proving explicitly because it converts what looks like a full redesign into a one-parameter update.

Common pitfall. Recomputing the lane balance after scaling the saturation flows and reporting a different green split — it cannot change, and a change signals an arithmetic slip. The second pitfall is concluding that more capacity must shorten the cycle; when pedestrians govern it demonstrably does not, and recommending "more green" instead of a refuge island or a wider crosswalk is the wrong engineering answer.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §8.5; TRB Highway Capacity Manual, pedestrian minimum-green and crosswalk methodology; TAC Geometric Design Guide for Canadian Roads, Ch. 5 (intersections and pedestrian refuges).

Quantity	Symbol	Question 2	Question 3
Sum of critical flow ratios	\(Y\)	0.7310	0.6645
Green split (N–S : E–W)	\(y_i/Y\)	0.4151 : 0.5849	0.4151 : 0.5849 (unchanged)
Webster optimum cycle	\(C_o\)	74.35 s	59.62 s
Pedestrian bound, N–S phase	\(C_{\text{ped}}\)	74.88 s	106.32 s
Pedestrian bound, E–W phase	\(C_{\text{ped}}\)	59.13 s	78.29 s
Adopted cycle length	\(C\)	75 s	110 s
Effective green, N–S phase	\(g_{NS}\)	26.98 s	41.51 s
Effective green, E–W phase	\(g_{EW}\)	38.02 s	58.49 s
Critical degree of saturation	\(x_{\text{crit}}\)	0.843	0.731
Governing control	—	pedestrian	pedestrian (unchanged)

Question 3 of 7: Effect of Higher Saturation Flows and Wider Approaches

Question 3: Effect of Higher Saturation Flows and Wider Approaches (20 marks)