Question 5 of 7: Singly Constrained Gravity Model

Question 5: Singly Constrained 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 stated productions and attractions for a base year and a target year, an intra-zonal travel distance of 5 km, an inter-zonal distance of 10 km, and an inverse-square friction function.

Find. The base-year and target-year trip matrices, split into intra-zonal and inter-zonal totals, and the factors other than distance that shape trip distribution.

Approach. Evaluate the friction factors, then apply the singly (production) constrained gravity model with a separate denominator for each production zone, and read the intra-zonal and inter-zonal totals off the resulting matrix.

1. Part (a) — evaluate the friction factors. With the inverse-square deterrence function, $$\begin{aligned}F_{ii}&=\frac{1}{5^{2}}=0.04\\ F_{ij}&=\frac{1}{10^{2}}=0.01\end{aligned}$$ Staying inside a zone is therefore made four times as attractive as crossing to the other zone, purely by the deterrence function — and that ratio, not the raw distances, is what drives the answer.
2. Write the singly constrained gravity model. The production-constrained form distributes each zone's productions in proportion to the attractiveness of every destination, discounted by distance: $$T_{ij}=P_i\,\frac{A_j F_{ij}}{\displaystyle\sum_{k} A_k F_{ik}}$$ The denominator is specific to the production zone $i$, so it must be recomputed for each row; using a single common denominator for both rows is the classic error and breaks the production constraint.
3. Distribute the productions of zone 1. The weighted attractions seen from zone 1 are $A_1F_{11}=700\times0.04=28$ and $A_2F_{12}=300\times0.01=3$, so the row denominator is 31: $$\begin{aligned}T_{11}&=450\times\frac{28}{31}=406.45\\ T_{12}&=450\times\frac{3}{31}=43.55\end{aligned}$$
4. Distribute the productions of zone 2. From zone 2 the weighted attractions are $A_1F_{21}=700\times0.01=7$ and $A_2F_{22}=300\times0.04=12$, giving a denominator of 19: $$\begin{aligned}T_{21}&=550\times\frac{7}{19}=202.63\\ T_{22}&=550\times\frac{12}{19}=347.37\end{aligned}$$ Collecting the four cells:

   Part (a) — base-year trip matrix

   From \ To	Zone 1	Zone 2	Row total	Target $P_i$
   Zone 1	406.45	43.55	450.00	450
   Zone 2	202.63	347.37	550.00	550
   Column total	609.08	390.92	1,000.00	1,000

5. Report the intra-zonal and inter-zonal totals. The diagonal cells are the intra-zonal trips and the off-diagonal cells are the inter-zonal trips: $$\boxed{\text{intra-zonal}=406.45+347.37=753.82 \text{ trips}}$$ $$\boxed{\text{inter-zonal}=43.55+202.63=246.18 \text{ trips}}$$ Just over 75 per cent of all travel stays within its own zone, which is what an inverse-square deterrence function does to a system whose internal distance is half the external one.
6. Disclose that the attractions are not reproduced. The row totals match the productions exactly — that is what the production constraint guarantees — but the column totals do not match the given attractions: the model sends 609.08 trips to zone 1 against a stated attraction of 700 (a shortfall of 12.99 per cent) and 390.92 to zone 2 against 300 (a surplus of 30.31 per cent). This is inherent in the singly constrained form. Matching both margins requires a doubly constrained model solved by iterative proportional fitting (Furness balancing), alternately scaling rows and columns until both sets of margins are met. The question supplies one friction formula and no balancing factors, so the single pass is the intended answer, but stating the limitation explicitly is part of a complete response.
7. Part (b) — repeat for the target year. Only the productions and attractions change; the distances and hence the friction factors are unchanged. From zone 1 the weighted attractions become $950\times0.04=38$ and $450\times0.01=4.5$, a denominator of 42.5; from zone 2 they become $950\times0.01=9.5$ and $450\times0.04=18$, a denominator of 27.5. Hence $$\begin{aligned}T_{11}&=600\times\frac{38}{42.5}=536.47\\ T_{12}&=600\times\frac{4.5}{42.5}=63.53\end{aligned}$$ $$\begin{aligned}T_{21}&=800\times\frac{9.5}{27.5}=276.36\\ T_{22}&=800\times\frac{18}{27.5}=523.64\end{aligned}$$

   Part (b) — target-year trip matrix

   From \ To	Zone 1	Zone 2	Row total	Target $P_i$
   Zone 1	536.47	63.53	600.00	600
   Zone 2	276.36	523.64	800.00	800
   Column total	812.83	587.17	1,400.00	1,400

8. Report the forecast split and compare it with the base year. Summing the diagonal and the off-diagonal, $$\boxed{\text{intra-zonal}=536.47+523.64=1{,}060.11 \text{ trips}}$$ $$\boxed{\text{inter-zonal}=63.53+276.36=339.89 \text{ trips}}$$ Total travel grows by 40 per cent, from 1,000 to 1,400 trips, and intra-zonal travel grows almost exactly in step (a factor of 1.406), so the intra-zonal share barely moves, from 75.4 per cent to 75.7 per cent. That is the expected behaviour: with the distances unchanged, the deterrence function is unchanged, and the split responds only to the modest reweighting of the attractions. The mismatch between modelled and stated attractions persists — 812.83 against 950 for zone 1 — so the Furness caveat carries into the forecast.
9. Part (c) — list the factors other than travel distance. Distance is only a proxy for the impedance that actually governs destination choice. In order of practical importance: travel time by the relevant mode, including access, waiting and transfer time, which is a better impedance measure than distance because it reflects congestion and network quality; travel cost — fares, fuel, tolls and especially parking charges at the destination; and the size and quality of the attraction, since a zone with more retail floor area, more employment or a hospital draws trips out of proportion to its bare attraction count. Beyond these, trip purpose matters strongly, because work trips tolerate far greater impedance than shopping trips and each purpose needs its own friction function; socio-economic characteristics of the producing zone (income, car ownership, household size) change how far a household is willing to travel; and land-use mix determines whether a trip can be satisfied locally at all.
10. Complete the list with the network and behavioural factors. Mode availability and service quality shape the impedance a traveller actually experiences, which is why distribution and mode choice are increasingly solved jointly rather than in sequence. Network structure and physical barriers — rivers, rail corridors, limited bridge or interchange capacity, and in Canadian practice the seasonal reliability of a route — can make two zones that are close in kilometres far apart in practice. Time of day and congestion level change impedance within the day, which is the reason peak and off-peak matrices are built separately. Institutional and social factors matter too: school catchments, jurisdictional and municipal boundaries, language and cultural ties, and the growth of telework and e-commerce, which suppress some trip purposes altogether. Finally, intervening opportunities — the number of acceptable destinations encountered before the one under consideration — is itself an alternative theory of distribution and is a real determinant of choice.