Question 5 of 7: Trip Distribution by the Gravity Model

Question 5: 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. A two-zone system with balanced productions and attractions and a symmetric travel-time matrix.

Find. The intra-zonal and inter-zonal trip totals under an inverse-time friction factor and again under an inverse-square-time friction factor, and an explanation of what the change does to the distribution.

Approach. Apply the singly-constrained gravity model, which allocates each zone's productions among destinations in proportion to the product of destination attractiveness and friction factor, then check the resulting attractions against their targets and comment on the sensitivity to the friction exponent.

1. State the model. With no socio-economic adjustment factors, the gravity model distributes the productions of zone $i$ as $$T_{ij} = P_i \frac{A_j F_{ij}}{\displaystyle\sum_{k} A_k F_{ik}}$$ Every row sums to $P_i$ by construction, so trip productions are satisfied exactly; the attractions are matched only approximately unless the model is iterated.
2. Evaluate the friction factors for part (a). With $F_{ij} = 1/t_{ij}$, $$F_{11} = F_{22} = \tfrac{1}{4} = 0.25, \qquad F_{12} = F_{21} = \tfrac{1}{10} = 0.10$$ The intra-zonal friction factor is two and a half times the inter-zonal one, which is the model's numerical statement that a four-minute trip is more attractive than a ten-minute one.
3. Distribute the productions of zone 1. The two competing products are $A_1F_{11} = 320(0.25) = 80$ and $A_2F_{12} = 180(0.10) = 18$, summing to 98. Hence $$T_{11} = 270 \times \frac{80}{98} = 220.4, \qquad T_{12} = 270 \times \frac{18}{98} = 49.6$$ The two add to 270, as required.
4. Distribute the productions of zone 2. Here $A_1F_{21} = 320(0.10) = 32$ and $A_2F_{22} = 180(0.25) = 45$, summing to 77, so $$T_{21} = 230 \times \frac{32}{77} = 95.6, \qquad T_{22} = 230 \times \frac{45}{77} = 134.4$$ Collecting the two categories the question asks for, $$\boxed{\text{intra-zonal} = T_{11} + T_{22} = 220.4 + 134.4 = 354.8 \ \text{trips}}$$ $$\boxed{\text{inter-zonal} = T_{12} + T_{21} = 49.6 + 95.6 = 145.2 \ \text{trips}}$$
5. Check the attraction balance. Summing down the columns gives 316.0 trips attracted to zone 1 against a target of 320, and 184.0 to zone 2 against a target of 180 — a shortfall of 1.3 per cent on zone 1 and an excess of 2.2 per cent on zone 2, which is a normal first-pass result for a singly-constrained model. If the attractions must be matched, the standard remedy is to scale them, $A_j^{\,\prime} = A_j (A_j / C_j)$ where $C_j$ is the computed column total, and redistribute. One iteration gives adjusted attractions of 324.1 and 176.1, and the redistribution $$T_{11} = 221.8, \quad T_{12} = 48.2, \quad T_{21} = 97.5, \quad T_{22} = 132.5$$ whose column totals of 319.3 and 180.7 are within 0.4 per cent of target. The intra-zonal and inter-zonal totals move by barely one trip, so the answers above stand.

6. Repeat with the inverse-square friction factor. With $F_{ij} = 1/t_{ij}^{2}$, $$F_{11} = F_{22} = \tfrac{1}{16} = 0.0625, \qquad F_{12} = F_{21} = \tfrac{1}{100} = 0.01$$ so the intra-zonal factor is now 6.25 times the inter-zonal one rather than 2.5 times — the ratio has been squared, exactly as the exponent implies.
7. Redistribute both rows. For zone 1, $A_1F_{11} = 320(0.0625) = 20$ and $A_2F_{12} = 180(0.01) = 1.8$, summing to 21.8, giving $$T_{11} = 270 \times \frac{20}{21.8} = 247.7, \qquad T_{12} = 270 \times \frac{1.8}{21.8} = 22.3$$ For zone 2, $A_1F_{21} = 3.2$ and $A_2F_{22} = 11.25$, summing to 14.45, giving $$T_{21} = 230 \times \frac{3.2}{14.45} = 50.9, \qquad T_{22} = 230 \times \frac{11.25}{14.45} = 179.1$$ Collecting the categories, $$\boxed{\text{intra-zonal} = 247.7 + 179.1 = 426.8 \ \text{trips}}$$ $$\boxed{\text{inter-zonal} = 22.3 + 50.9 = 73.2 \ \text{trips}}$$ The column totals are now 298.6 against a target of 320 and 201.4 against 180, a deviation of about 6.7 and 11.9 per cent respectively — considerably worse than in part (a), because the stronger distance decay forces trips toward the origin zone irrespective of where the attractions are.

(c) Effect of the friction factor

Squaring the friction factor moves the distribution decisively toward short trips. Intra-zonal travel rises from 354.8 to 426.8 trips, that is from 71.0 to 85.4 per cent of all travel, while inter-zonal travel falls from 145.2 to 73.2 trips, a reduction of very nearly one half. Nothing about the zones changed — the productions, attractions and travel times are identical in both parts — so the entire shift is attributable to the impedance function alone.

The mechanism is straightforward. The friction factor expresses how strongly travellers resist distance or time, and only the ratio of friction factors across the destinations available to a zone influences the split. Raising the exponent from one to two squares that ratio: the intra-zonal advantage of 2.5 to 1 becomes 6.25 to 1, so each origin retains a correspondingly larger share of its own productions. A higher exponent therefore represents a population less willing to travel far, or a network on which longer trips are relatively more onerous. A lower exponent flattens the friction ratio toward unity, and in the limit as the exponent tends to zero the model degenerates into a purely attraction-proportional allocation in which travel time is irrelevant and the trip matrix depends only on the size of each destination.

Two practical consequences follow. First, the friction exponent is the single most influential parameter in the distribution step, so it must be calibrated against observed trip-length frequency distributions rather than assumed; comparing the modelled and surveyed average trip length is the usual test, and the exponent is adjusted until they agree. Second, the deteriorating attraction balance in part (b) shows that the stronger the impedance, the harder the singly-constrained model has to be worked to reproduce the attraction targets, and the more important it becomes to iterate the attraction adjustment or to move to a doubly-constrained formulation. In a Canadian regional model, where intra-zonal trips in large suburban zones are already a substantial share of all travel, an over-stated exponent will inflate them further and will systematically understate the demand on the inter-zonal network that the model was built to test.