Question 5 of 7: Singly-Constrained Gravity Model for Three Zones

Question 5: Singly-Constrained Gravity Model for Three Zones (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.

Friction factor F_ij = 1/t_ij². Total productions 100 + 160 + 140 = 400 trips, and total attractions 130 + 70 + 200 = 400 trips, so the two margins balance in total.

Find. The full 3 × 3 trip matrix, separating the intra-zonal (diagonal) from the inter-zonal (off-diagonal) trips; and a qualitative factor affecting distribution together with the mechanism for admitting it to the model.

Approach. Apply the singly-constrained gravity model — each production zone distributes its own trips in proportion to the attraction of each destination discounted by the friction factor — computing a separate denominator for each production zone.

1. Part (a) — state the model. The production-constrained gravity model is $$\begin{aligned}T_{ij}&=P_i\,\frac{A_j\,F_{ij}}{\sum_{k} A_k\,F_{ik}} \\ \text{with} \\ F_{ij}&=\frac{1}{t_{ij}^{\,2}}\end{aligned}$$ The denominator is summed over destinations for the given production zone i, so there is one denominator per row. Using a single denominator for the whole matrix is the classic error here, and it breaks the production constraint.
2. Compute the friction factors. Squaring and inverting each travel time gives the symmetric matrix

3. Form the row denominator for each production zone. Weighting each friction factor by the destination's attractions, $$\begin{aligned}\textstyle\sum_k A_k F_{1k} &= 130(0.25)+70(0.04)+200(0.111111) \\ &= 32.500+2.800+22.222=57.522 \\ \textstyle\sum_k A_k F_{2k} &= 130(0.04)+70(0.25)+200(0.0625) \\ &= 5.200+17.500+12.500=35.200 \\ \textstyle\sum_k A_k F_{3k} &= 130(0.111111)+70(0.0625)+200(0.25) \\ &= 14.444+4.375+50.000=68.819\end{aligned}$$
4. Distribute each zone's productions. For zone 1, for instance, $$\begin{aligned}T_{11} &= 100\,\frac{32.500}{57.522}=56.50 \\ T_{12} &= 100\,\frac{2.800}{57.522}=4.87 \\ T_{13} &= 100\,\frac{22.222}{57.522}=38.63\end{aligned}$$ and the same operation on rows 2 and 3 completes the matrix below. Each row sums to its own production total by construction, which is the check that the model has been applied correctly.

Separating the two classes of trip, which is what the question asks for: the intra-zonal trips are the diagonal, 56.50 + 79.55 + 101.72 = 237.76 trips, and the inter-zonal trips are everything else, 400 − 237.76 = 162.24 trips. Intra-zonal travel is therefore 59.4 per cent of all travel in the system. That concentration is entirely the work of the friction factor: every zone's shortest travel time is to itself (2 min), and because the deterrence is inverse-square, a 2-minute internal trip is weighted 6.25 times as heavily as a 5-minute trip to a neighbour. Re-running the model with the deterrence switched off (F_ij = constant, so trips follow attractions alone) drops the intra-zonal share to 32.6 per cent, which quantifies exactly how much of the concentration the friction function is responsible for.

Part (b) — a non-quantitative factor and how the gravity model can carry it

The gravity model as written contains only two kinds of information: the size of each zone, through P_i and A_j, and the cost of reaching it, through t_ij. Many real determinants of where people choose to travel are neither. A good example is the perceived quality and security of the destination environment — whether a commercial district feels safe and pleasant to walk in after dark. Two retail zones with identical floor area and identical travel time will not attract equal numbers of trips if one is a well-lit, animated main street and the other a windswept surface-parking precinct. Equally good examples are language or cultural affinity between a residential zone and a destination (well documented in Montreal and in the Greater Toronto Area), the effect of a psychological or physical barrier such as a river, a rail corridor, a provincial boundary or the Canada–United States border, and habit or historical allegiance — people continue to shop where they have always shopped.

There are three practical ways to admit such a factor. The standard device is a socio-economic adjustment factor or K-factor, entered multiplicatively as

where K_ij is calibrated so that the modelled flow on the affected interchange matches the observed flow; a barrier such as a river crossing typically calibrates to K well below 1, and an affinity pair to K above 1. This is honest and effective but it is a residual, not a behavioural explanation, and because it is fitted to today's observations it should be held constant or decayed toward unity in a forecast rather than treated as a permanent property.

The second route is to quantify the factor by proxy and fold it into the impedance, replacing pure travel time with a generalised cost that includes a time penalty representing the deterrent — for example adding an equivalent two minutes to every interchange that crosses the river. This is preferable to a K-factor because it responds correctly when the network changes (build the bridge, remove the penalty). The third route is to segment the model, running separate gravity models by trip purpose, by language group or by household type, so that a factor which cannot be measured directly is captured as a difference between segments.