16-Civ-A6 Highway Design, Construction, and Maintenance · May 2015
Question 5 of 7: Gravity Model Trip Distribution
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination — 98-Civ-A6 Transportation Planning & Engineering, May 2015. Closed book (one two-sided aid sheet), 3 hours. Seven questions of equal value (20 marks); any five constitute a complete paper. All seven are solved here as a study resource.
Given. A two-zone, production-constrained (singly constrained) gravity model with an inverse-square friction factor $F=1/d^2$.
Check (source reading): The printed model on page 5 divides by $d_{ij}^{2}$ in both numerator and denominator. The squared form is used here, as printed. (For reference, the $1/d$ reading would have given 208.3 intra / 91.7 inter in (a), so the exponent changes the answer materially.)
Given data
Scenario
$O_1$
$O_2$
$D_1$
$D_2$
$d_{ii}$ (intra)
$d_{12}$ (inter)
(a) base
150
150
100
200
2 km
5 km
(b) target
200
200
150
250
2 km
5 km
Find. The trip matrix $T_{ij}$, and the intra-zonal ($T_{11}+T_{22}$) and inter-zonal ($T_{12}+T_{21}$) totals, for both scenarios.
Approach. For each origin, weight every destination by $D_j/d_{ij}^2$, normalise so the row sums to $O_i$, and distribute the origin's productions accordingly.
(a) Origin 1. With $d_{11}^2=2^2=4$ and $d_{12}^2=5^2=25$, the attractiveness terms are $D_1/d_{11}^2=100/4=25$ and $D_2/d_{12}^2=200/25=8$, summing to 33. Thus
$$T_{11}=150\cdot\frac{25}{33}=113.6,\qquad T_{12}=150\cdot\frac{8}{33}=36.4.$$
(a) Origin 2. Terms $D_1/d_{21}^2=100/25=4$ and $D_2/d_{22}^2=200/4=50$, sum 54:
$$T_{21}=150\cdot\frac{4}{54}=11.1,\qquad T_{22}=150\cdot\frac{50}{54}=138.9.$$
Each row sums to its production (150). Grouping:
$$\boxed{\text{intra}=T_{11}+T_{22}=252.5,\quad \text{inter}=T_{12}+T_{21}=47.5.}$$
(b) Origin 1 (target). Terms $150/4=37.5$ and $250/25=10$, sum 47.5:
$$T_{11}=200\cdot\frac{37.5}{47.5}=157.9,\qquad T_{12}=200\cdot\frac{10}{47.5}=42.1.$$
(b) Origin 2 (target). Terms $150/25=6$ and $250/4=62.5$, sum 68.5:
$$T_{21}=200\cdot\frac{6}{68.5}=17.5,\qquad T_{22}=200\cdot\frac{62.5}{68.5}=182.5.$$
$$\boxed{\text{intra}=157.9+182.5=340.4,\quad \text{inter}=42.1+17.5=59.6.}$$
Because the inverse-square friction penalises the 5 km inter-zonal trip 6.25 times as heavily as the 2 km intra-zonal one, about 84–85 % of all trips stay within their own zone in both years. Of the 100 extra productions, 87.9 become intra-zonal and 12.1 inter-zonal trips.
Question 5 — Trip matrices and totals
Flow
(a) base
(b) target
$T_{11}$
113.6
157.9
$T_{12}$
36.4
42.1
$T_{21}$
11.1
17.5
$T_{22}$
138.9
182.5
Intra-zonal total
252.5
340.4
Inter-zonal total
47.5
59.6
(c) Other factors affecting trip distribution
Besides travel distance, trip distribution responds to: travel time and travel cost (the true impedance, of which distance is only a proxy), mode availability and level of service, zone attractiveness beyond raw attractions (retail floor space, employment mix, land-use type), socioeconomic characteristics of travellers (income, car ownership, household structure), intervening opportunities and competition among destinations, network connectivity and topology, and trip purpose (work versus shopping distributions differ). A more general model replaces $1/d^2$ with a calibrated friction function $F(c_{ij})$ of generalised cost, plus socioeconomic K-factors.
