16-Civ-A6 Highway Design, Construction, and Maintenance · December 2015

Question 5 of 7: Gravity Model Trip Distribution

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: 98-Civ-A6 Transportation Planning & Engineering, National Examination, December 2015. Seven questions, each of equal value (20 marks); any five constitute a complete examination. Closed book, one two-sided aid sheet permitted. Three hours. All seven questions are solved below as a study resource.

Reference texts. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis (Wiley); Papacostas & Prevedouros, Transportation Engineering and Planning (Prentice Hall); Roess, Prassas & McShane, Traffic Engineering (Pearson); Ortúzar & Willumsen, Modelling Transport (Wiley) for the demand-model chapters (trip generation, distribution, mode choice, assignment).

Question 5: Gravity Model Trip Distribution (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. A two-zone system with a production-constrained gravity model and an inverse-square friction factor $F_{ij}=1/d_{ij}^{2}$ (as printed on page 4 of the paper).

Given data
Item	Base year	Target year
Productions $P_1,\,P_2$	450, 550	600, 800
Attractions $A_1,\,A_2$	700, 300	950, 450
Intra-zonal distance	5 km	5 km
Inter-zonal distance	10 km	10 km

Find. Intra-zonal and inter-zonal trips in the base year (a) and the target year (b); other distribution factors (c).

Approach. Compute the friction factors $F_{ij}=1/d_{ij}^{2}$, then apply the production-constrained gravity model $T_{ij}=P_i\,\dfrac{A_jF_{ij}}{\sum_k A_kF_{ik}}$ so that each origin's trips sum exactly to its production.

Friction factors. With $d_{\text{intra}}=5$ km and $d_{\text{inter}}=10$ km: $$F_{11}=F_{22}=\tfrac{1}{5^2}=0.04,\qquad F_{12}=F_{21}=\tfrac{1}{10^2}=0.01.$$ Halving the distance quadruples the friction factor, so an intra-zonal trip is four times as attractive as an inter-zonal one, all else equal.
(a) Zone 1 row ($P_1=450$). Denominator $\sum_k A_kF_{1k}=700(0.04)+300(0.01)=28+3=31$: $$T_{11}=450\cdot\frac{28}{31}=406.5,\qquad T_{12}=450\cdot\frac{3}{31}=43.5.$$
(a) Zone 2 row ($P_2=550$). Denominator $700(0.01)+300(0.04)=7+12=19$: $$T_{21}=550\cdot\frac{7}{19}=202.6,\qquad T_{22}=550\cdot\frac{12}{19}=347.4.$$ Each row sums to its production ($450$ and $550$), confirming the constraint.
(a) Intra- and inter-zonal totals. $$\boxed{\text{intra}=T_{11}+T_{22}=406.5+347.4=753.8,\qquad \text{inter}=T_{12}+T_{21}=43.5+202.6=246.2}$$ The column sums ($609.1$ and $390.9$) do not reproduce the attractions ($700$, $300$) — expected, because only the production constraint is enforced.
(b) Target year, Zone 1 ($P_1=600$). Denominator $950(0.04)+450(0.01)=38+4.5=42.5$: $$T_{11}=600\cdot\frac{38}{42.5}=536.5,\qquad T_{12}=600\cdot\frac{4.5}{42.5}=63.5.$$
(b) Target year, Zone 2 ($P_2=800$). Denominator $950(0.01)+450(0.04)=9.5+18=27.5$: $$T_{21}=800\cdot\frac{9.5}{27.5}=276.4,\qquad T_{22}=800\cdot\frac{18}{27.5}=523.6.$$ $$\boxed{\text{intra}=536.5+523.6=1060.1,\qquad \text{inter}=63.5+276.4=339.9}$$ Total travel grows from 1000 to 1400 trips; the inter-zonal share is essentially unchanged (24.6 % → 24.3 %), because the steep $1/d^2$ deterrence keeps roughly three trips in four inside their zone of origin in both years.

(c) Other factors affecting trip distribution. Distance (or, more generally, generalized travel time and cost, not merely airline distance) is only one determinant. Trip distribution also responds to: the socio-economic character and attractiveness of the destination zone (employment, retail floor area, school/service capacity); traveller income and auto ownership; trip purpose (work trips tolerate longer distances than shopping trips); the quality and cost of the available modes and the level of congestion; land-use mix and the presence of intervening opportunities between origin and destination; and physical or perceived barriers (rivers, jurisdictional boundaries, safety). These are what a calibrated friction (deterrence) function and $K$-factors are meant to absorb.

Question 5 — gravity-model trips
Trips	(a) Base year	(b) Target year
$T_{11}$ / $T_{12}$	406.5 / 43.5	536.5 / 63.5
$T_{21}$ / $T_{22}$	202.6 / 347.4	276.4 / 523.6
Intra-zonal ($T_{11}+T_{22}$)	753.8	1060.1
Inter-zonal ($T_{12}+T_{21}$)	246.2	339.9

Topic: Production-constrained (singly constrained) gravity model for trip distribution.

Key relations: $T_{ij}=P_i\,\dfrac{A_jF_{ij}}{\sum_k A_kF_{ik}}$ with deterrence $F_{ij}=f(d_{ij})$; here $F_{ij}=1/d_{ij}^{2}$ (the classical inverse-square gravity form).

Why this works: The gravity analogy makes trips between two zones grow with the "masses" (production and attraction) and fall with a friction of separation. Normalising each origin row by $\sum_k A_kF_{ik}$ forces the trips leaving zone $i$ to total exactly its production $P_i$, which is the singly-constrained balancing. Because $F$ here penalises distance steeply ($1/d^2$), the intra-zonal cells (half the distance, four times the friction) capture about three-quarters of all trips.

Common pitfall: Misreading the friction exponent — using $1/d$ where the paper prints $1/d^2$ changes the inter-zonal total from 246 to 376 trips. Also forgetting to balance — writing $T_{ij}\propto P_iA_jF_{ij}$ without the row denominator leaves each origin's trips not summing to $P_i$. Also using the same denominator for both rows: each origin has its own $\sum_k A_kF_{ik}$.

Source: Papacostas & Prevedouros, Transportation Engineering and Planning, Ch. 12 (trip distribution); Ortúzar & Willumsen, Modelling Transport, Ch. 5.

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