The Gravity Model: Derivation and Calibration

The Gravity Model: Derivation and Calibration Philip A. Viton October 28, 2014 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 1 / 66

Introduction We turn now to the Gravity Model of trip distribution. As previously noted, this is most widely used model in current practice. From our perspective, it is based on two important ideas, one new and one old. 1. It is based on a parametrized theoretical model, meaning that the theory depends (unlike the Average Growth Factor model) on constants that need to be estimated. 2. It utilizes the same idea of improvement via iteration as the Average Growth Factor model, though, as we will see, in a slightly different context. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 2 / 66

Derivation of the Gravity Model Physics The gravity model is based on Newton s gravitational theory from physics, interpreted in a transportation context. In physics, the attractive force between two bodies is directly proportional to their masses, and inversely proportional to the square of the distance between them: where: h ij = g m i m j d 2 ij h ij is the attractive force between bodies i and j. m i and m j are the bodies masses. d ij is the distance between them. g is Newton s gravitational (proportionality) constant, approximately 6.7 10 8 in cgs (centimeters/grammes/seconds) units. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 3 / 66

Derivation of the Gravity Model Transportation The derivation of the transportation version of the gravity model is a simple argument from analogy: The Newtonian attractive force term (h ij ) is analogized to directed interzonal trip-making: h ij T ij. The masses are analogized to total trips in and out of zones, so that m i O i and m j A j. The distance term is retained, but in an attempt to be more general (whether there is a serious theoretical basis for this is a question) the power-2 term is permitted to be arbitrary. There is no reason to believe that the Newtonian proportionality term applies here, so we replace it by a different one. The result is the first form of the (transportation) gravity model: T ij = θ O i A j d β ij Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 4 / 66

Geographical Justification Tobler s First Law of Geography states that: Everything is related to everything else, but closer things more so (W. Tobler, Cellular Geography in S. Gale and S. Olssohn, eds, Philosophy in Geography, 1979, pp. 379 386). The gravity model is consistent with the First Law, but is not implied by it (there are other possible functional forms that are equally consistent with the First Law). Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 5 / 66

Travel-Time Factors (I) In the first form of the gravity model, let s write T ij = θ O i A j d β ij 1/d β ij = F ij The F ij are called the travel-time factors. With this notation, the first form of the transportation gravity model becomes T ij = θo i A j F ij Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 6 / 66

Travel-Time Factors (II) There are several reasons for writing the model this way. It looks neater. Even if we have data on (say, centroid-to-centroid) distances, it will be hard to estimate the power-law β using linear techniques. More importantly, this can be considered a way to sneak other influences into the model. If travel between zones i and j depends on, say, the number of jobs present in zone j, we might want to write the denominator of the gravity model say telling us that trips depend on distance, jobs and a lot of other factors. Writing all those influences as F ij provides a way of doing this. Note that the F s are to be considered as unknowns. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 7 / 66

Gravity Model with Conservation of Origins (I) We now show that by requiring the gravity model to satisfy conservation of origins, we can eliminate the proportionality constant θ. With the introduction of the travel-time factors, version 1 of the gravity model is: T ij = θo i A j F ij For fixed i, sum both sides over the destinations j: j T ij = θo i A j F ij j On the left, j T ij is just O i. On the right, we can take terms not involving j outside the summation: O i = θo i A j F ij j Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 8 / 66

Gravity Model with Conservation of Origins (II) Now divide both sides by O i 1 = θ A j F ij j and solve for θ : θ = 1 j A j F ij Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 9 / 66

Gravity Model with Conservation of Origins (III) Now insert this into the gravity model. Since we are interested in T ij (which already uses j) we re-label the summation index (as m) in the expression defining θ. The result is: T ij = 1 m A m F im O i A j F ij Then we have our second form of the transportation gravity model: T ij = O i A j F ij m A m F im Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 10 / 66

Gravity Model with Conservation of Origins (IV) There are two important things to note about this form of the model: 1. It is guaranteed to satisfy conservation of origins exactly. Therefore we will never need to check whether this condition is satisfied or not. Of course, the same does not apply to conservation of attractions: there is no reason to suppose that this condition will be satisfied; and in fact it usually will not be. 2. The travel-time factors F ij all depend on an unknown parameter (the parameter β). This means that the interpretation of these F s is not simply as a growth rate, which was something we could calculate given data on the present and some predictions about the future. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 11 / 66

Gravity Model with Zonal Adjustment Factors As an empirical matter, early users of the gravity model observed that it did not appear to yield very accurate predictions. In order to improve the predictive accuracy, they introduced a set of zonal adjustment factors, to be denoted K ij. When these were incorporated, the transportation gravity model took the form T ij = As to the zonal adjustment factors, note that: O i A j F ij K ij m A m F im K im (1) 1. They have no basis whatever in theory, unlike the travel-time factors. They are introduced simply to improve predictions. 2. The matrix K {K ij } is another Z Z = Z 2 parameters which need to be estimated. It might reasonably be asked whether this will be asking too much of the available data. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 12 / 66

Calibration (I) We turn now to the task of calibrating the gravity model of trip distribution specified as: T ij = O i A j F ij K ij m A m F im K im The standard method for doing this is an iterative method developed by the US Bureau of Public Roads (now the Federal Highway Administration), known as the BPR method. As we have seen, the transportation form of the gravity model requires us to estimate two sets of unknown parameters: the travel time factors (F {F ij }) and the zonal adjustment factors (K {K ij }). Note that both of these are Z Z matrices. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 13 / 66

Calibration (II) The underlying idea of the BPR method is to choose these unknowns in order to make the gravity model work as well as possible for our present-day data. That is, unlike the growth factor model calibrations, the BPR calibration method makes no use of future data. What the BPR method does is find the F s and K s that reproduce T 0, the present-day (observed) trip matrix. Once we have calibrated the model (ie found estimates of the travel-time and zonal adjustment factors, which we will denote by ˆF { ˆF ij }) and ˆK { ˆK ij } respectively), we will assume that these are stable (ie unchanged) as between our current period and the future period for which we wish to distribute the trips. This makes prediction simply a matter of plugging in the future (predicted) data and utilizing our estimated ˆF and ˆK. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 14 / 66

BPR Method : Summary (I) The BPR method involves the following steps: 0. (optional) reduce the dimensionality of the problem by imposing similarity assumptions on the different F -factors. We will assume that pairs of zones with similar interzonal travel times involve the same travel-time factor. Thus we will require additional data: the current interzonal travel times. These sets of similar F -factors define what we will call superzones. For now, set the K ij 1 (ie ignore them). 1. Iterative step: estimate the F s based on observed superzonal total trips. That is, we use the gravity model formulation (which will involve only the F s, since we are ignoring the K s) to reproduce as closely as we wish, the observed superzonal total trips. The result of this step (Step 1) is estimates of the travel-time factors, denoted ˆF. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 15 / 66

BPR Method : Summary (II) 2. At the end of Step 1 we have a candidate trip matrix (an estimate of our target T 0 ). However, though it will satisfy conservation of origins, it will usually not satisfy conservation of attractions. So we perform another iterative step, called row-and column factoring, to approximately balance the trip matrix. This is Step 2. 3. At the end of Step 2 we have the F s and an approximately balanced trip matrix. But it still does not reproduce the original T 0. We reproduce T 0 by using the one set of parameters we have so far ignored, namely the zonal adjustment factors (the K matrix). This is Step 3. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 16 / 66

BPR Method : Example In my view, the simplest way to understand the fairly complicated details of the BPR calibration method is to develop the formal expressions for the calibration steps alongside an actual example. So that is what we will do here. Note : I will post to the course website an abbreviated version of the example, containing just the computations. You may find it helpful to refer to this as you review the details. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 17 / 66

Example Data Present-day (observed) trip-interchange matrix: T 0 = 100 350 100 550 240 150 210 600 60 120 200 380 400 620 510 1530 Present-day(observed) interzonal travel-times matrix: t 0 = 1 6 11 7 3 12 15 13 4 We shall use a convergence criterion of 5% in all iterative steps (α L = 0.95, α H = 1.05). Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 18 / 66

Step 0 (I) We will assume that pairs of zones with similar interzonal travel times involve the same travel-time factor. This will define a set of superzones. We operationalize this by assuming that all pairs of zones i and j satisfying 0 t 0 ij < δ for a given interzonal travel-time difference δ, utilize the same travel-time factor, F 1. And we take all zonal pairs satisfying δ t 0 ij < 2δ to utilize a second travel time factor, F 2. And then 2δ t 0 ij < 3δ are assumed to involve a third travel-time factor F 3. And so on. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 19 / 66

Step 0 (II) In our example we shall take δ = 5 minutes. This defines the following 3 superzones: with superzonal trip totals: Superzone Zonal Pairs 1 {1, 1}, {2, 2}, {3, 3} 2 {1, 2}, {2, 1} 3 {1, 3}, {2, 3}, {3, 1}, {3, 2} O S = [450, 590, 490] Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 20 / 66

Step 0 (III) In our example, we will be calculating 3 distinct travel-time factors, one for each superzone. It is a simple matter to parlay these three into a full set F of factors. Formally, we define a mapping (rule) φ that takes our 3 estimates of the travel time factors (F 1, F 2, F 3 ) and produce an estimate of the full F ij matrix. In our case this rule is: Finally we set φ(f 1, F 2, F 3 ) = K 1 = F 1 F 2 F 3 F 2 F 1 F 3 F 3 F 3 F 1 1 1 1 1 1 1 1 1 1 (This amounts to ignoring the K s for now). Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 21 / 66.

Step 1 (I) In Step 1 we will estimate the distinct superzonal travel-time factors F i, as set up in step 0. In this step we work exclusively with superzonal (aggregated) travel. But in that context, the method is very much like the way we calibrated the growth-factor models: we generate an estimate of T, and we compute error factors by comparing actual superzonal totals (ie those implied by T 0 ) with those implied by the current T. If we have not converged we update the F s (not T as in the average growth factor model) and try again. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 22 / 66

Step 1 (II) An iteration in Step 1 consists of the following tasks: 1. Use the previously computed (or assumed) F s and the gravity model to generate a new T matrix. 2. Compare the total superzonal travel implied by this T matrix with the superzonal totals implied by T 0, via error ratios E i. 3. If convergence is not satisfied, update the F s and try again. 4. The updating rule generates a new set of superzonal F s by multiplying the old F s by the error ratios. In general F k i = F k 1 i E k i (where i indexes the superzones). Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 23 / 66

Step 1, Iteration 1 (I) Initial estimate: F 0 i = (1, 1, 1). Apply φ and get F 0 = 1 1 1 1 1 1 1 1 1 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 24 / 66

Step 1, Iteration 1 (II) Apply the gravity model and find: T 1 = Tij 1 = O0 i A0 j F ij 0 m A 0 mf 0 im = 143.791 222.876 183.333 550 156.863 243.137 200 600 99.3464 153.987 126.667 380 400 620 510 1530 (Of course this is just T 1 ij = O 0 i A0 j m A 0 m = O 0 i A0 j S(T 0 ). Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 25 / 66

Step 1, Iteration 1 (III) Convergence check against the superzone totals (NB: not the origins or attractions): Target 450 590 490 Actual 513.595 379.739 636.667 Error ratio Ei 1 0.876177 1.5537 0.769634 Note that the targets are the actual superzonal trip totals. We see that we have not converged. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 26 / 66

Step 1, Iteration 2 (I) The new superzonal factors Fi 1 error ratios: are the previous F s corrected by the Fi 1 = Fi 0 Ei 1 = [1.0, 1.0, 1.0] [0.876177, 1.5537, 0.769634] = [0.876177, 1.5537, 0.769634] Apply φ to obtain the full set of new travel-time factors: F 1 = 0.876177 1.5537 0.769634 1.5537 0.876177 0.769634 0.769634 0.769634 0.876177 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 27 / 66

Step 1, Iteration 2 (II) Apply the gravity model: T 2 = Tij 2 = O0 i A0 j F ij 1 m A 0 mf 1 im = 112.97 310.507 126.522 550 239.457 209.307 151.236 600 94.9643 147.195 137.841 380 447.392 667.009 415.599 1530 The Appendix, beginning at slide 57, contains the detailed computations behind this result. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 28 / 66

Step 1, Iteration 2 (III) Convergence test: Still no convergence. Target 450 590 490 Actual 460.119 549.964 519.917 Error ratio Ei 2 0.978009 1.0728 0.942458 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 29 / 66

Step 1, Iteration 3 (I) New interzonal F i : Fi 2 = Fi 1 Ei 2 = [0.876177, 1.5537, 0.769634] [0.978009, 1.0728, 0.942458] = [0.856 909, 1. 666 8, 0.725 347] Apply φ : F 2 = 0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 30 / 66

Step 1, Iteration 3 (II) Apply the gravity model: T 3 = 107.966 325.512 116.522 550 255.134 203.306 141.56 600 93.6824 145.208 141.11 380 456.782 674.026 399.191 1530 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 31 / 66

Step 1, Iteration 3 (III) Convergence check: These have all converged. Target 450 590 490 Actual 452.381 580.647 496.972 Error ratio 0.994736 1.01611 0.985971 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 32 / 66

Step 1, Final Result At the end of Step 1 we have our final estimate of the travel-time factors ˆF ˆF ij, which is the F matrix we obtained at the convergent iteration of our step-1 computations. For our example: ˆF = 0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 33 / 66

Step 2 (I) At the end of step 1 we have estimated the F s. We also have, associated with those F s (and under the assumption that the K s are all 1) an estimate of the trip matrix. As compared with T 0 (the trip matrix we are trying to reproduce) the current estimate T satisfies conservation of origins (automatically) but: probably does not satisfy conservation of attractions (even approximately) probably does not reproduce the individual elements of T 0 Step 2 is an attempt to remedy the first problem (conservation of attractions). Note that it is possible that at the end of Step 1 we do satisfy conservation of attractions (approximately). In that case we omit Step 2. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 34 / 66

Step 2 (II) If we need to do step 2 (ie if we don t approximately satisfy conservation of origins at the end of Step 1) then an iteration of Step 2 consists of: 1. A column factoring : this scales each column of the trip matrix to ensure conservation of attractions. But in doing this, we destroy conservation of origins; so 2. We perform a row factoring. This scales the rows of the result in part (1) to restore conservation of origins. But it will destroy conservation of attractions. At the end of a Step-2 iteration we check conservation of attractions. If it is not approximately satisfied, then we perform another iteration; and we continue until conservation of attractions is approximately satisfied. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 35 / 66

Step 2, Setup (I) Continuing our example, based on ˆF fron Step 1 and K 1 we have an estimate of the travel-time matrix we re trying to reproduce, namely the final T matrix from the previous step: T 3 = 107.966 325.512 116.522 550 255.134 203.306 141.56 600 93.6824 145.208 141.11 380 456.782 674.026 399.191 1530 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 36 / 66

Step 2, Setup (II) Is this good enough? We need to check only the columns (conservation of attractions) for convergence: Target 400 620 510 Actual 456.782 674.026 399.191 Error ratio 0.875691 0.919845 1.27758 Note that we check against the actual attractions A 0 we have no more use for the superzones. Clearly not good enough. So we need to do Step 2. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 37 / 66

Step 2, Setup (III) To simplify the notation, let s rename our starting point (T 3 ) to be T 1. At each iteration of step 2 (given that we need to do it at all, as we have to in this example) we must do: 1. a column factoring 2. a row factoring 3. a convergence check on the columns. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 38 / 66

Step 2, Iteration 1, Column Factoring The column factors are the previous error ratios, so we have: 0.875691, 0.919845, 1.27758 We multiply the elements in each column by its column factor giving: T 1a = 94.5446 299.421 148.866 542.832 223.419 187.01 180.854 591.283 82.0368 133.569 180.279 395.885 400 620 510 1530 Example: the (1, 2) element of T 1 (= 325.512) is in column 2, so we use the second column factor, giving 325.512 0.919845 = 299. 421 = T 1a 12. Note that this satisfies conservation of attractions, but no longer satisfies conservation of origins. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 39 / 66

Step 2, Iteration 1, Row Factoring The row factors are the row-wise (originations) error ratios of T 1a : Target 550 600 380 Actual 542.832 591.283 395.885 Error ratio 1. 013 2 1. 014 74 0.959 875 and multiplying each row by its row factor gives: T 1b = 95.793 303.375 150.832 550 226.712 189.767 183.521 600 78.7451 128.209 173.046 380 401.25 621.351 507.398 1530 Example: the (1, 2) element of T 1a (= 299.421) is in row 1, so we use the first row factor, giving 299.421 1. 013 2 = 303. 375 = T 1b 12. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 40 / 66

Step 2, Iteration 1, Convergence Check We need to check only the columns (ie the attractions), since the row factoring assures conservation of origins. We have converged. Target 400 620 510 Actual 401.25 621.351 507.398 Error ratio 0.996884 0.997825 1.00513 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 41 / 66

Step 3 (I) We now have an approximately balanced T matrix (ie T 1b ) derived on the hypothesis that K 1, namely T 1b = 95.793 303.375 150.832 550 226.712 189.767 183.521 600 78.7451 128.209 173.046 380 401.25 621.351 507.398 1530 This satisfies conservation of origins, satisfies conservation of attractions to within our convergence criterion, but still does not reproduce the individual elements of T 0. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 42 / 66

Step 3 (II) In order to reproduce the individual elements of T 0 we use the zonal adjustment factors, which we have ignored up to now. The zonal adjustment (K ) factors are now computed as ˆK ij = Tij 0 Tij 1b (ie on the last (convergent) T matrix from step 2) and where means element-by-element division. Then ˆK = = 100 350 100 240 150 210 60 120 200 1.04392 1.15369 0.662989 1.05861 0.790443 1.14429 0.761952 0.93597 1.15576 95.793 303.375 150.832 226.712 189.767 183.521 78.7451 128.209 173.046 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 43 / 66

BPR Method: Example Results Summary For our calibration of the gravity model using the BPR method we have found: ˆF = 0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909 (from the final result of Step 1); and ˆK = 1.04392 1.15369 0.662989 1.05861 0.790443 1.14429 0.761952 0.93597 1.15576 (from Step 3). Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 44 / 66

Discussion : Step 2 (I) Remember that the motivation for the BPR method is to construct the F and K factors to allow us to exactly reproduce our observed trip matrix T 0. However, a quick calculation shows that we have not succeeded in this: we find, plugging in our calibrated results: T ij = O0 i A0 j ˆF ij ˆK ij m A 0 ˆF = m im ˆK im 109.62 365.25 75.14 550 273.38 162.66 163.96 600 73.24 139.44 167.33 380 456.23 667.35 406.42 1530 which does not reproduce our original observed trip matrix. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 45 / 66

Discussion : Step 2 (II) If you think about this for a minute you ll realize that the problem seems to lie in Step 2, the row-and-column factoring. For one thing, this step has nothing to do with the gravity model at all: it s just an ad-hoc way of arriving at a balanced interim trip matrix. For another, you might reasonably say, why do it at all? If we can exactly reproduce T 0 by first doing Step 1 and then adjusting the results via the zonal adjustment factors in Step 3, won t we automatically have a balanced trip matrix? It seems to me that the answer to this is Yes; suggesting that Step 2 is dispensable, despite the BPR. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 46 / 66

Calibration Without Step 2 (I) At the end of Step 1 we had an interim estimate of T 0 as: T 3 = 107.966 325.512 116.522 550 255.134 203.306 141.56 600 93.6824 145.208 141.11 380 456.782 674.026 399.191 1530 implying that we could take: ˆK = = 100 350 100 240 150 210 60 120 200 0.92622 1.07523 0.858208 0.940682 0.737804 1.48347 0.640461 0.826402 1.41734 107.966 325.512 116.522 255.134 203.306 141.56 93.6824 145.208 141.11 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 47 / 66

Calibration Without Step 2 (II) And a simple calculation shows that if we plug these new estimates ˆF = ˆK = 0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909 0.92622 1.07523 0.858208 0.940682 0.737804 1.48347 0.640461 0.826402 1.41734 into the gravity model, we do indeed exactly reproduce our original trip matrix T 0. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 48 / 66

Prediction (I) In order to predict a future trip matrix remember that this is the object of the whole exercise we need: predictions of the future origins by zone, O i predictions of the future attractions by zone, A j possibly, predictions of the future interzonal travel times, if we believe these have changed Given these, we assume that the ˆF and ˆK matrices are stable over time, and derive our predictions by plugging everything into the gravity model. That is,we take T ij = O i A j ˆF ij ˆK ij m A m ˆF im ˆK im Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 49 / 66

Prediction (II) Note that there is no guarantee that this T will satisfy conservation of attractions One possibility is to perform an artificial balancing step on the results even though this is not mandated by the gravity model. We could do this is via row-and-column factoring (Step 2) on T. (Remember that conservation of origins is guaranteed). So the suggestion is: first compute T according to the gravity model formula, and then if necessary go through Step 2 (row-and-column factoring) to arrive at an approximately balanced T. If we do this, we report T as our final prediction. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 50 / 66

Prediction Example (I) We work with the conventionally-calibrated gravity model, ie the one in which we do Step 2, whose results are shown in slide 44. Suppose that we predict: O = (650, 590, 350) A = (350, 700, 540) So that total trip-making rises from 1530 to 1590. We assume that the travel-time matrix is unchanged. There is an interesting question of what to do when an interzonal travel time changes and falls into a superzonal category not covered by the calibration (eg, if in our example the 15-minute time became 16 minutes). About all we can reasonably do is to include this in the original superzone, even though it really doesn t belong there. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 51 / 66

Prediction Example (II) To start, plug in O and A into the gravity model, using the calibrated ˆF and ˆK matrices. The result is: T = 106.057 455.976 87.9665 650 236.616 181.66 171.724 590 56.2564 138.209 155.535 350 398.93 775.845 415.225 1590 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 52 / 66

Prediction Example (III) Is this good enough? We need to check (only) conservation of attractions, based on our predictions (A ): Target 350 700 540 Actual 398.93 775.845 415.225 Error ratio 0.877347 0.902242 1.3005 all of which are outside our 5% convergence criteria So if we are concerned about balance/consistency, it may be a good idea to do some row-and-column factoring. Let s do this. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 53 / 66

Prediction Example (IV) The column factors are: (0.877347, 0.902242, 1.3005). The column-factored matrix is: 93.049 411.401 114.4 618.85 207.595 163.901 223.327 594.822 49.3563 124.698 202.273 376.327 350 700 540 1590 The row factors are: (1.05033, 0.991893, 0.930042). And the row-factored matrix is: 97.7325 432.109 120.159 650 205.912 162.572 221.516 590 45.9035 115.974 188.122 350 349.548 710.655 529.797 1590 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 54 / 66

Prediction Example (V) Good enough? We check: Target 350 700 540 Actual 349.548 710.655 529.797 Error ratio 1.00129 0.985006 1.01926 all of which are OK, so we take as our approximately-balanced predicted trip interchange matrix: T = 97.7325 432.109 120.159 650 205.912 162.572 221.516 590 45.9035 115.974 188.122 350 349.548 710.655 529.797 1590 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 55 / 66

Appendix Appendix Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 56 / 66

Applying the Gravity Model (I) This appendix does the detailed calculations in which we apply the gravity model to the second iteration of Step 1. We have (from slide 27): F 1 = O 0 i = 550, 600, 380 A 0 j = 400, 620, 510 0.876177 1.5537 0.769634 1.5537 0.876177 0.769634 0.769634 0.769634 0.876177 We will compute the T 2 matrix, as shown in slide 28, by: ij = O0 i A0 j F ij 1 m A 0 mf 1 T 2 im Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 57 / 66

Applying the Gravity Model (II) Before we get started, one preliminary note. The denominator of the expression for T ij is of the form Z A m F im = A m F im m m=1 The important thing to note is that this does not involve j. In other words, we will use the same denominator for all elements of T in the same row: T 11, T 12 and T 13 (row 1) all involve the same denominator; as do T 21, T 22 and T 23 (row 2). For a Z Z problem, you have to compute only Z distinct denominators. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 58 / 66

Applying the Gravity Model (III) For our sample problem, we have a Z = 3 zone region, so all summations run from m = 1 to m = 3. We begin by computing the three denominators, since they will be used repeatedly. i = 1 : 3 A 0 mf 1 3 im = A 0 mf1m 1 m=1 m=1 = A 0 1F 0 11 + A 0 2F 0 12 + A 0 3F 0 13 = (400 0.876177) + (620 1.5537) + (510 0.769634) = 1706. 278 14 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 59 / 66

Applying the Gravity Model (IV) Denominators, continued: i = 2 : 3 A 0 mf 1 3 im = A 0 mf2m 1 m=1 m=1 = A 0 1F 0 21 + A 0 2F 0 22 + A 0 3F 0 23 = (400 1.5537) + (620 0.876177) + (510 0.769634) = 1557. 223 08 i = 3 : 3 A 0 mf 1 3 im = A 0 mf3m 1 m=1 m=1 = A 0 1F 0 31 + A 0 2F 0 32 + A 0 3F 0 33 = (400 0.769634) + (620 0.769634) + (510.876177) = 1231. 876 95 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 60 / 66

Applying the Gravity Model Now we begin computing the individual terms. Row 1: i = 1; j = 1 : T11 2 = O0 1 A0 1 F 11 1 m A 0 mf1m 1 = 550 400 0.876177 1706. 278 14 1. 927 589 4 105 = 1706. 278 14 = 112. 970 i = 1; j = 2 : T12 2 = O0 1 A0 2 F 12 1 m A 0 mf1m 1 = 550 620 1.5537 1706. 278 14 5. 298 117 105 = 1706. 278 14 = 310. 507 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 61 / 66

Applying the Gravity Model (V) Final entry for Row 1: i = 1; j = 3 : T13 2 = O0 1 A0 3 F 13 1 m A 0 mf1m 1 = 550 510 0.876177 1706. 278 14 2. 457 676 49 105 = 1706. 278 14 = 126. 522 Note that the denominators for all the entries in row 1 (as they will be in each row) are the same. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 62 / 66

Applying the Gravity Model (VI) Row 2: i = 2; j = 1 : T21 2 = O0 2 A0 1 F 21 1 m A 0 mf2m 1 600 400 1.5537 = 1557. 223 08 = 239. 457 i = 2; j = 2 : T22 2 = O0 2 A0 2 F 22 1 m A 0 mf2m 1 600 620 0.876177 = 1557. 223 08 = 209. 307 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 63 / 66

Applying the Gravity Model (VII) Final entry for Row 2: Row 3: i = 2; j = 3 : T23 2 = O0 2 A0 3 F 23 1 m A 0 mf2m 1 600 510 0.769634 = 1557. 223 08 = 151. 236 i = 3; j = 1 : T31 2 = O0 3 A0 1 F 31 1 m A 0 mf3m 1 380 400 0.769634 = 1231. 876 95 = 94. 964 3 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 64 / 66

Applying the Gravity Model (VIII) Row 3, concluded: i = 3; j = 2 : T32 2 = O0 3 A0 2 F 32 1 m A 0 mf3m 1 380 620 0.769634 = 1231. 876 95 = 147. 195 i = 3; j = 3 : T33 2 = O0 3 A0 3 F 33 1 m A 0 mf3m 1 380 510 0.876177 = 1231. 876 95 = 137. 841 Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 65 / 66

Applying the Gravity Model (IX) Putting all these together we get: T 2 = T 2 ij = 112.97 310.507 126.522 550 239.457 209.307 151.236 600 94.9643 147.195 137.841 380 447.392 667.009 415.599 1530 which matches the result in slide 28. Philip A. Viton CRP/CE 5700 () Gravity Model October 28, 2014 66 / 66