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16-Civ-A6 Highway Design, Construction, and Maintenance · December 2013

Question 5 of 7: Gravity-model trip distribution from a single origin

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 (December 2013). Closed book, one two‑sided aid sheet, 3 hours. Seven questions; any five constitute a complete examination and each is of equal value (20 marks). All seven are solved below as a study resource.

Reference texts (subject).

Question 5: Gravity-model trip distribution from a single origin (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. All 1,000 trips produced by zone 1 are distributed to zones 2–5, with a friction factor inversely proportional to travel time, $F_{1j}=1/t_{1j}$.

Given data
Destination zone j2345
Travel time $t_{1j}$ (min)5102030
Attraction $A_j$ (present)5020075450
Attraction $A_j$ (future)100225100600

Find. The zone‑1 trip distribution now ($P_1=1000$) and in future ($P_1=1250$), and the zone with the largest increase.

Approach. Use the singly‑constrained gravity model $T_{1j}=P_1\dfrac{A_j/t_{1j}}{\sum_k A_k/t_{1k}}$ so that the distributed trips sum exactly to the production $P_1$.

  1. Present weights. Compute $w_j=A_j/t_{1j}$: $w_2=50/5=10$, $w_3=200/10=20$, $w_4=75/20=3.75$, $w_5=450/30=15$; sum $\sum w=48.75$.
  2. Present distribution (a). $T_{1j}=1000\,w_j/48.75$ gives $$\boxed{T_{12}=205,\ T_{13}=410,\ T_{14}=77,\ T_{15}=308}$$ (sum = 1000, as required).
  3. Future weights. With the new attractions, $w_2=100/5=20$, $w_3=225/10=22.5$, $w_4=100/20=5$, $w_5=600/30=20$; sum $\sum w=67.5$.
  4. Future distribution (b). $T_{1j}=1250\,w_j/67.5$ gives $$\boxed{T_{12}=370,\ T_{13}=417,\ T_{14}=93,\ T_{15}=370}$$ (sum = 1250).
  5. Increases (c). $\Delta T_{12}=+165$, $\Delta T_{13}=+6$, $\Delta T_{14}=+16$, $\Delta T_{15}=+63$ (changes taken from the unrounded trip numbers, so they can differ by 1 from the rounded columns). The largest increase is to zone 2.
Final results — Question 5 (trips from zone 1)
Destination(a) present(b) futureIncrease
Zone 2205370+165
Zone 3410417+6
Zone 47793+16
Zone 5308370+63
Total10001250+250

(c) Why zone 2 gains the most

Zone 2 has by far the largest increase (+165 trips, about +80%). Two effects reinforce each other. First, its attraction doubles (50 → 100), the biggest proportional attraction gain of any zone. Second, zone 2 is the closest destination (5 min), so its friction factor $1/t$ is the highest — every unit of added attraction there is weighted more heavily than the same unit added to a distant zone. The combination of a large attraction increase and the shortest travel time makes zone 2 capture a disproportionate share of both the redistributed trips and the extra 250 trips of production.