Greenshields Speed–Density Model by Least-Squares Regression: 16-Civ-B10 Traffic Engineering, December 2014, Q2

Question 2: Greenshields Speed–Density Model by Least-Squares Regression (15 + 5 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. Four simultaneous field observations of density and space-mean speed on one facility. Greenshields' model is to be fitted, and its capacity and the density at a stated operating speed extracted.

Observation	Density k (veh/km)	Speed u (km/h)
1	171	5
2	129	15
3	20	40
4	70	25
Operating speed of interest	30 km/h

Find. (a) the least-squares straight line $u = u_f - (u_f/k_j)k$ together with the free-flow speed and jam density it implies, and (c) the maximum flow (capacity) of the facility and the density at which the stream travels at 30 km/h.

Part (a) — the four observations and the fitted Greenshields line. The intercepts of the line are the two model parameters: the free-flow speed on the speed axis and the jam density on the density axis.

Approach. Greenshields' model is linear in density, so an ordinary least-squares regression of speed on density returns the model parameters directly as the intercept and the slope; capacity and the queried operating point then follow from the parabolic flow–density curve implied by that line.

Part (a) — recognise that the model is already linear. Greenshields assumed a linear speed–density relation: $$u = u_f - \frac{u_f}{k_j}\,k$$ so fitting the model means fitting a straight line $u = a + bk$ in which the intercept is the free-flow speed, $a = u_f$, and the slope is $b = -u_f/k_j$. No transformation of the data is needed.
Form the regression sums. With $n = 4$ observations, $\sum k = 390$ and $\sum u = 85$, so the centroid of the data is $$\bar{k} = \frac{390}{4} = 97.5\ \text{veh/km},\qquad \bar{u} = \frac{85}{4} = 21.25\ \text{km/h}$$ Taking deviations about that centroid, the two sums of products required by least squares are $$S_{ku} = \sum (k_i - \bar{k})(u_i - \bar{u}) = -2947.5, \qquad S_{kk} = \sum (k_i - \bar{k})^2 = 13157.0$$ The individual deviation products are $(+73.5)(-16.25)$, $(+31.5)(-6.25)$, $(-77.5)(+18.75)$ and $(-27.5)(+3.75)$, that is $-1194.375$, $-196.875$, $-1453.125$ and $-103.125$.
Solve for the slope and intercept. The least-squares estimates are $$b = \frac{S_{ku}}{S_{kk}} = \frac{-2947.5}{13157.0} = -0.22403\ \text{km/h per veh/km}$$ $$a = \bar{u} - b\,\bar{k} = 21.25 + 0.22403(97.5) = 43.09\ \text{km/h}$$ so the best-fit Greenshields line is $$\boxed{u = 43.09 - 0.2240\,k \quad (\text{km/h, with } k \text{ in veh/km})}$$
Read the two model parameters off the fitted line. The intercept is the free-flow speed and the density intercept is the jam density: $$u_f = a = 43.09\ \text{km/h}, \qquad k_j = -\frac{a}{b} = \frac{43.09}{0.22403} = 192.4\ \text{veh/km}$$ The fit is close: the coefficient of determination is $R^2 = 0.987$, so 98.7 % of the variation in the observed speeds is explained by density alone. Both parameters are physically sensible for an urban facility — a free-flow speed just above 43 km/h and a jam density near 192 veh/km, which corresponds to a jam spacing of about 5.2 m per vehicle.
Part (c) — convert the speed–density line into the flow–density parabola. Substituting the fitted line into $q = ku$ gives $$q = k\,u = u_f k - \frac{u_f}{k_j}k^2$$ a parabola through the origin and through $k = k_j$. Differentiating and setting $dq/dk = 0$ places the maximum at half the jam density: $$k_{\text{cap}} = \frac{k_j}{2} = 96.18\ \text{veh/km}, \qquad u_{\text{cap}} = \frac{u_f}{2} = 21.55\ \text{km/h}$$
Evaluate the maximum flow. Capacity is the product of those two coordinates, which is the familiar $u_f k_j/4$: $$q_{\max} = \frac{u_f\,k_j}{4} = \frac{43.09 \times 192.4}{4} = \boxed{2072\ \text{veh/h}}$$
Find the density corresponding to a speed of 30 km/h. Inverting the fitted line at $u = 30$ km/h: $$k = \frac{u - a}{b} = \frac{30 - 43.09}{-0.22403} = \boxed{58.44\ \text{veh/km}}$$ That operating point carries $q = u\,k = 30 \times 58.44 = 1753\ \text{veh/h}$, about 85 % of capacity, and it lies on the uncongested branch of the parabola because $58.44 < k_{\text{cap}} = 96.18$ veh/km. The same flow of 1753 veh/h would also be served on the congested branch, at $k = 133.9$ veh/km and $u = 13.1$ km/h — the same volume at half the speed, which is why flow alone never identifies an operating condition.

Part (c) — the flow–density parabola implied by the fitted line, with the capacity point and the 30 km/h operating point marked.

Check: the paper labels the two parts of this question “(a)” and “(c)” although only two parts are set and the grading scheme allots (15 + 5) marks; the second part is therefore answered as printed, under its printed label. The regression has been run on all four observations with no point rejected, which is the correct default; with only four points, discarding any one of them would swing $u_f$ by several km/h.

Quantity	Symbol	Value
Best-fit Greenshields line	—	u = 43.09 − 0.2240 k
Free-flow speed	u_f	43.09 km/h
Jam density	k_j	192.4 veh/km
Coefficient of determination	R²	0.987
Density and speed at capacity	k_cap, u_cap	96.18 veh/km, 21.55 km/h
Maximum flow (capacity)	q_max	2072 veh/h
Density at u = 30 km/h	k	58.44 veh/km
Flow at u = 30 km/h	q	1753 veh/h

Topic: calibrating a macroscopic speed–density model from field data and extracting capacity from it.

Key relations: the Greenshields line $u = u_f(1 - k/k_j)$; least squares $b = S_{ku}/S_{kk}$ and $a = \bar{u} - b\bar{k}$; the derived parabola $q = u_f k - (u_f/k_j)k^2$; and the capacity triple $k_{\text{cap}} = k_j/2$, $u_{\text{cap}} = u_f/2$, $q_{\max} = u_f k_j/4$.

Why this works: Greenshields' contribution was to notice that the observed speed–density scatter is close enough to linear that a two-parameter model captures it, and that those two parameters are both physically observable end states — the speed at which a driver travels with the road to themselves, and the density at which the stream stops. Because the model is linear in the variable being regressed, calibration reduces to ordinary least squares with no transformation, and every derived quantity is an algebraic consequence of the two intercepts rather than a new measurement. The capacity relations are worth memorising precisely because they are model-specific: capacity sits at exactly half the jam density only because the assumed speed–density relation is linear.

Common pitfall: regressing density on speed instead of speed on density, which returns the reciprocal slope and a different (and wrong) $u_f$; and quoting $q_{\max}$ without checking whether the queried operating point lies on the uncongested or the congested branch, since every sub-capacity flow occurs twice.

Source: Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 6 §6.3 (macroscopic speed–density models; Greenshields; capacity and the fundamental diagram).

Question 2 of 6: Greenshields Speed–Density Model by Least-Squares Regression

Question 2: Greenshields Speed–Density Model by Least-Squares Regression (15 + 5 marks)