Question 5 of 7: Two-Zone Trip Distribution by the Gravity Model

Question 5: Two-Zone Trip Distribution by the Gravity Model (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.

Find. The four cells of the trip matrix in each year, aggregated into intra-zonal trips ($T_{11}+T_{22}$) and inter-zonal trips ($T_{12}+T_{21}$), and an explanation of how the friction factor shapes that split.

Approach. Apply the singly constrained (production-constrained) gravity model, in which each zone's productions are distributed among the destination zones in proportion to the product of attraction and friction factor. Compute the two friction factors once, form the denominator for each production zone, and allocate.

1. Evaluate the friction factors. There are only two distinct distances, so only two factors are needed: $$F_{\text{intra}} = \exp\left(-0.1 \times 10\right) = e^{-1} = 0.36788,\qquad F_{\text{inter}} = \exp\left(-0.1 \times 20\right) = e^{-2} = 0.13534$$ Their ratio, $F_{\text{intra}}/F_{\text{inter}} = e = 2.718$, is the quantity that drives the whole answer: per unit of attraction, a destination 10 km away is 2.72 times as likely to be chosen as one 20 km away.
2. Write the gravity model. In production-constrained form, $$T_{ij} = P_i\,\frac{A_j F_{ij}}{\displaystyle\sum_{k}A_k F_{ik}}$$ which guarantees by construction that each row of the trip matrix sums to its production, $\sum_j T_{ij}=P_i$.
3. (a) Distribute the trips produced by zone 1. The denominator is $$\sum_k A_k F_{1k} = A_1F_{11} + A_2F_{12} = 100(0.36788) + 200(0.13534) = 36.788 + 27.067 = 63.855$$ so $$T_{11} = 150\left(\frac{36.788}{63.855}\right) = 150(0.57606) = 86.42\ \text{trips}$$ $$T_{12} = 150\left(\frac{27.067}{63.855}\right) = 150(0.42394) = 63.58\ \text{trips}$$ Zone 1 sends 57.6 per cent of its trips to itself even though zone 2 offers twice the attraction, because the friction factor more than offsets the attraction advantage: $2.718 > 2.0$.
4. Distribute the trips produced by zone 2. By symmetry of the method but not of the data, $$\sum_k A_k F_{2k} = 100(0.13534) + 200(0.36788) = 13.534 + 73.576 = 87.109$$ $$T_{21} = 150\left(\frac{13.534}{87.109}\right) = 23.30\ \text{trips},\qquad T_{22} = 150\left(\frac{73.576}{87.109}\right) = 126.70\ \text{trips}$$ Here the attraction advantage and the friction advantage point the same way, so zone 2 retains 84.5 per cent of its own trips.
5. Aggregate the base-year result. $$\boxed{\text{Intra-zonal} = T_{11}+T_{22} = 86.42 + 126.70 = 213.11\ \text{trips (71.0 per cent)}}$$ $$\boxed{\text{Inter-zonal} = T_{12}+T_{21} = 63.58 + 23.30 = 86.89\ \text{trips (29.0 per cent)}}$$ The two sum to 300, which is the total production, as the production constraint requires.
6. (b) Repeat for the target year. The friction factors are unchanged, so only the productions and attractions move. For zone 1, $$\sum_k A_kF_{1k} = 150(0.36788) + 250(0.13534) = 55.182 + 33.834 = 89.016$$ $$T_{11} = 200\left(\frac{55.182}{89.016}\right) = 123.98,\qquad T_{12} = 200\left(\frac{33.834}{89.016}\right) = 76.02$$ and for zone 2, $$\sum_k A_kF_{2k} = 150(0.13534) + 250(0.36788) = 20.300 + 91.970 = 112.270$$ $$T_{21} = 200\left(\frac{20.300}{112.270}\right) = 36.16,\qquad T_{22} = 200\left(\frac{91.970}{112.270}\right) = 163.84$$
7. Aggregate the target-year result. $$\boxed{\text{Intra-zonal} = 123.98 + 163.84 = 287.82\ \text{trips (72.0 per cent)}}$$ $$\boxed{\text{Inter-zonal} = 76.02 + 36.16 = 112.18\ \text{trips (28.0 per cent)}}$$ summing to 400, the target-year total production. Total travel has grown by one third, from 300 to 400 trips, and the intra-zonal share has crept up by one percentage point — from 71.0 to 72.0 per cent — purely because zone 1's attraction grew proportionally faster (50 per cent) than zone 2's (25 per cent), which slightly strengthens the shorter destination for zone 1 producers.
8. (c) Explain the role of the friction factor. The friction factor is the model's representation of the deterrence of distance, and it is the only term that distinguishes one destination from another once attraction is accounted for. Its effect is best shown by removing it. If $\beta = 0$, so that $F_{ij}=1$ for every pair, the model collapses to a pure attraction share: $$T_{ij} = P_i\frac{A_j}{\sum_kA_k}\;\Longrightarrow\;T_{11}=150\left(\tfrac{100}{300}\right)=50,\quad T_{22}=150\left(\tfrac{200}{300}\right)=100$$ giving 150 intra-zonal trips, a share of exactly 50 per cent in the base year and, by the same calculation, 50 per cent in the target year. The friction factor as specified therefore raises the intra-zonal share from 50 per cent to 71 per cent — it is responsible for the entire concentration of travel near its origin. Pushing the other way, a steeper decay of $\beta = 0.2$ would raise the base-year intra-zonal share to 86.2 per cent, and in the limit of very large $\beta$ every trip stays home.
9. Draw out the two planning consequences. First, the friction factor controls trip length, not trip volume: the production constraint fixes the row totals, so a change in $\beta$ redistributes the same trips over shorter or longer distances. It is therefore the single most important parameter to calibrate against an observed trip-length frequency distribution, and calibrating it against total trips instead tells you nothing. Second, because the friction factors here are held constant between the two years, the change from 71.0 to 72.0 per cent is entirely attributable to the changed land-use pattern. That is exactly the comparison a planner wants: if a transportation improvement had shortened the inter-zonal travel time, $F_{\text{inter}}$ would have risen, the inter-zonal share would have grown, and the model would have reproduced the induced longer-distance travel that new capacity typically generates.