16-Civ-A6 Highway Design, Construction, and Maintenance · May 2014
Question 5 of 7: Singly-Constrained Gravity Distribution
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination, 98-Civ-A6 Transportation Planning & Engineering, May 2014 — 3 hours, closed book (one two-sided aid sheet). Seven questions; any five constitute a complete examination and all are of equal value (20 marks). All seven are solved below as a study resource.
Given. Productions $P_1,\dots,P_4 = 1200, 1500, 1000, 1700$ and distances to zone 5 $d_1,\dots,d_4 = 7.5, 5, 10, 12.5$ km; attraction to zone 5 is $A_5=500$ (part a). Friction is $F=1/d$ (attraction inversely proportional to distance).
Find. Trips $T_{i5}$ from each origin to zone 5, first with $A_5=500$, then with $A_5=1000$ and a new zone 6 ($P_6=700$, $d_6=2.5$ km); plus other distribution factors.
Attraction‑constrained star: all origin zones send trips to the single destination zone 5. Zone 6 (green) is added in part (b).
Approach. Use the singly (attraction)‑constrained gravity model: allocate the fixed attraction $A_5$ among origins in proportion to the weight $P_i/d_i$, so that the allocated trips sum exactly to $A_5$.
Model form. With friction $F_i=1/d_i$ and attraction constraint, $T_{i5}=A_5\,\dfrac{P_i/d_i}{\sum_j P_j/d_j}$. This normalisation guarantees $\sum_i T_{i5}=A_5$.
Weights (part a). $P_i/d_i$: zone 1 $=1200/7.5=160$; zone 2 $=1500/5=300$; zone 3 $=1000/10=100$; zone 4 $=1700/12.5=136$. Sum $=\sum P_j/d_j = 696$.
Allocate $A_5=500$ (part a). Multiply each weight by $500/696$: $T_{15}=500(160/696)=114.9$; $T_{25}=500(300/696)=215.5$; $T_{35}=500(100/696)=71.8$; $T_{45}=500(136/696)=97.7$. These sum to $\boxed{500}$.
Add zone 6 and raise $A_5$ (part b). New weight $P_6/d_6=700/2.5=280$, so the sum becomes $696+280=976$. With $A_5=1000$: $T_{15}=1000(160/976)=163.9$; $T_{25}=1000(300/976)=307.4$; $T_{35}=1000(100/976)=102.5$; $T_{45}=1000(136/976)=139.3$; $T_{65}=1000(280/976)=286.9$, summing to $\boxed{1000}$.
Interpretation. The nearby, high‑producing zones (2, and now 6) capture the largest shares, exactly as the $P/d$ weighting dictates; adding zone 6 redistributes the shares because the denominator grows.
Question 5 — trips to zone 5
Origin
(a) A₅=500
(b) A₅=1000, +zone 6
Zone 1
114.9
163.9
Zone 2
215.5
307.4
Zone 3
71.8
102.5
Zone 4
97.7
139.3
Zone 6
—
286.9
Total
500
1000
(c) Other factors affecting trip distribution. Besides travel distance, the impedance and interaction between zones depend on: travel time and generalised cost (fares, tolls, parking, fuel) rather than distance alone; the quality and mode of available service (congestion, transfers, comfort); the type and intensity of land use and the mix of activities (employment, retail, schools) that make a destination attractive; socio‑economic characteristics of the origin population (income, car ownership, household size); intervening opportunities and competition from other destinations offering the same activity; trip purpose and time of day; and physical or perceived barriers (rivers, safety, jurisdictional boundaries). Practical gravity models fold these into a calibrated deterrence function and zone‑pair K‑factors.
