The Gravity Model: Derivation and Calibration

The Gravity Model: Derivation and Calibration

Philip A. Viton

October 28, 2014

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 di¤erent context.

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:

h_ij =gmimj

d_ij² where:

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.

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: hij 7!^Tij.

The masses are analogized to total trips in and out of zones, so that m_i 7!^{Oi and mj} 7!^Aj.

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 di¤erent one.

The result is the …rst form of the (transportation) gravity model:

T_ij =θO_iA_j d_ij^β

Geographical Justi…cation

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).

Travel-Time Factors (I)

In the …rst form of the gravity model, Tij =θO_iA_j

d_ij^β let’s write

1/d_ij^β =F_ij

The F_ij are called the travel-time factors. With this notation, the …rst form of the transportation gravity model becomes

T_ij =θOiA_jF_ij

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

in‡uences 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 in‡uences as F_ij provides a way of doing this. Note that the F ’s are to be considered as unknowns.

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 =θOiA_jF_ij

For …xed i , sum both sides over the destinations j:

∑

T_ij =

∑

j

θOiA_jF_ij

On the left,∑jTij is just Oi. On the right, we can take terms not involving j outside the summation:

Oi =θOi

∑

j

AjFij

Gravity Model with Conservation of Origins (II)

Now divide both sides by O_i

1= θ

∑

j

A_jF_ij

and solve for θ :

θ = ¹

∑jA_jF_ij

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 de…ning θ. The result is:

T_ij = ¹

∑mA_mF_imO_iA_jF_ij

Then we have our second form of the transportation gravity model:

T_ij = ^OiAjFij

∑mA_mF_im

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 satis…ed or not.

Of course, the same does not apply to conservation of attractions:

there is no reason to suppose that this condition will be satis…ed; 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.

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 = ^OiAjFijKij

∑mA_mF_imK_im (1)

As to the zonal adjustment factors, note that:

1. They have no basis whatever in theory, unlike the travel-time factors.

They are introduced simply to improve predictions.

2. The matrix K f^Kijgis another Z Z =Z² parameters which need to be estimated. It might reasonably be asked whether this will be asking too much of the available data.

Calibration (I)

We turn now to the task of calibrating the gravity model of trip distribution speci…ed as:

Tij = ^OiAjFijKij

∑mA_mF_imK_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^Fijg) and the zonal adjustment factors (K f^Kijg)^. Note that both of these are Z Z matrices.

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 …nd the F ’s and K ’s that reproduce T⁰, 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 Fˆ f^Fˆijg)^{and ˆK} f^Kˆijgrespectively), 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 .

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 di¤erent 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 de…ne what we will call superzones.

For now, set the Kij 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 .

BPR Method : Summary (II)

2. At the end of Step 1 we have a candidate trip matrix (an “estimate”

of our target T⁰). 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 T⁰. We reproduce T⁰ by using the one set of parameters we have so far ignored, namely the zonal adjustment factors (the K matrix). This is Step 3.

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 …nd it helpful to refer to this as you review the details.

Example Data

Present-day (observed) trip-interchange matrix:

T⁰ =

100 350 100 550 240 150 210 600 60 120 200 380 400 620 510 1530 Present-day(observed) interzonal travel-times matrix:

t⁰ = 2 4

1 6 11

7 3 12

15 13 4 3 5

We shall use a convergence criterion of 5% in all iterative steps (αL =0.95, αH =1.05).

Step 0 (I)

We will assume that pairs of zones with similar interzonal travel times involve the same travel-time factor. This will de…ne a set of “superzones”.

We operationalize this by assuming that all pairs of zones i and j satisfying 0 t_ij⁰ < δ

for a given interzonal travel-time di¤erence δ, utilize the same travel-time factor, F1. And we take all zonal pairs satisfying

δ t_ij⁰ <2δ

to utilize a second travel time factor, F₂. And then 2δ t_ij⁰ <3δ

are assumed to involve a third travel-time factor F₃. And so on.

Step 0 (II)

In our example we shall take δ=5 minutes.

This de…nes the following 3 superzones:

Superzone Zonal Pairs

1 f^{1, 1}g^,f^{2, 2}g^,f^{3, 3}g 2 f^{1, 2}g^,f^{2, 1}g

3 f^{1, 3}g^,f^{2, 3}g^,f^{3, 1}g^,f^{3, 2}g with superzonal trip totals:

O_S = [450, 590, 490]

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 de…ne a mapping (rule) φ that takes our 3

estimates of the travel time factors (F₁, F₂, F₃) and produce an estimate of the full F_ij matrix. In our case this rule is:

φ(F₁, F₂, F₃) = 2 4

F1 F2 F3

F₂ F₁ F₃ F₃ F₃ F₁

3 5 .

Finally we set

K 1

= 2 4

1 1 1 1 1 1 1 1 1

3 5

(This amounts to ignoring the K ’s for now).

Step 1 (I)

In Step 1 we will estimate the distinct superzonal travel-time factors Fi, 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 T⁰)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.

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 T⁰, via error ratios Ei.

3. If convergence is not satis…ed, 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_i^k =F_i^{k 1} E_i^k (where i indexes the superzones).

Step 1, Iteration 1 (I)

Initial estimate: F_i⁰ = (1, 1, 1). Apply φ and get

F⁰ = 2 4

1 1 1 1 1 1 1 1 1

3 5

Step 1, Iteration 1 (II)

Apply the gravity model and …nd:

T¹ = T_ij¹ = ^O

0 iA⁰_jF_ij⁰

∑mA⁰_mF_im⁰

=

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_ij¹ =O_i⁰A⁰_j ∑mA⁰_m =O_i⁰A⁰_j S(T⁰).

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 E_i¹ 0.876177 1.5537 0.769634

Note that the “targets” are the actual superzonal trip totals. We see that we have not converged.

Step 1, Iteration 2 (I)

The new superzonal factors F_i¹ are the previous F ’s “corrected” by the error ratios:

F_i¹ = F_i⁰ E_i¹

= [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:

F¹= 2 4

0.876177 1.5537 0.769634 1.5537 0.876177 0.769634 0.769634 0.769634 0.876177

3 5

Step 1, Iteration 2 (II)

Apply the gravity model:

T² = T_ij² = ^O

0 iA⁰_jF_ij¹

∑mA⁰_mF_im¹

=

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.

Step 1, Iteration 2 (III)

Convergence test:

Target 450 590 490

Actual 460.119 549.964 519.917 Error ratio E_i² 0.978009 1.0728 0.942458 Still no convergence.

Step 1, Iteration 3 (I)

New interzonal F_i :

F_i² = F_i¹ E_i²

= [0.876177, 1.5537, 0.769634] [0.978009, 1.0728, 0.942458]

= [0.856 909, 1. 666 8, 0.725 347] Apply φ :

F²= 2 4

0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909

3 5

Step 1, Iteration 3 (II)

Apply the gravity model:

T³ =

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

Step 1, Iteration 3 (III)

Convergence check:

Target 450 590 490

Actual 452.381 580.647 496.972 Error ratio 0.994736 1.01611 0.985971 These have all converged.

Step 1, Final Result

At the end of Step 1 we have our …nal estimate of the travel-time factors Fˆ Fˆ_ij, which is the F matrix we obtained at the convergent iteration of our step-1 computations. For our example:

Fˆ = 2 4

0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909

3 5

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 T⁰ (the trip matrix we are trying to reproduce) the current estimate T satis…es conservation of origins (automatically) but:

probably does not satisfy conservation of attractions (even approximately)

probably does not reproduce the individual elements of T⁰

Step 2 is an attempt to remedy the …rst 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.

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 satis…ed, then we perform another iteration; and we continue until conservation of attractions is approximately satis…ed.

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

…nal T matrix from the previous step:

T³ =

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

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 A⁰ — we have no more use for the superzones.

Clearly not good enough. So we need to do Step 2.

Step 2, Setup (III)

To simplify the notation, let’s rename our starting point (T³)to be T¹. 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.

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 T¹ (=325.512)is in column 2, so we use the second column factor, giving 325.512 0.919845=299. 421=T₁₂^1a. Note that this satis…es conservation of attractions, but no longer satis…es conservation of origins.

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

…rst row factor, giving 299.421 1. 013 2=303. 375=T₁₂^1b.

Step 2, Iteration 1, Convergence Check

We need to check only the columns (ie the attractions), since the row factoring assures conservation of origins.

Target 400 620 510

Actual 401.25 621.351 507.398 Error ratio 0.996884 0.997825 1.00513 We have converged.

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 satis…es conservation of origins, satis…es conservation of attractions to within our convergence criterion, but still does not reproduce the individual elements of T⁰.

Step 3 (II)

In order to reproduce the individual elements of T⁰ we use the zonal adjustment factors, which we have ignored up to now. The zonal adjustment (K) factors are now computed as

Kˆij =T_ij⁰ T_ij^1b

(ie on the last (convergent) T matrix from step 2) and where means element-by-element division. Then

Kˆ = 2 4

100 350 100 240 150 210 60 120 200

3 5

2 4

95.793 303.375 150.832 226.712 189.767 183.521 78.7451 128.209 173.046

3 5

= 2 4

1.04392 1.15369 0.662989 1.05861 0.790443 1.14429 0.761952 0.93597 1.15576

3 5

BPR Method: Example Results Summary

For our calibration of the gravity model using the BPR method we have found:

Fˆ = 2 4

0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909

3 5

(from the …nal result of Step 1); and

Kˆ = 2 4

1.04392 1.15369 0.662989 1.05861 0.790443 1.14429 0.761952 0.93597 1.15576

3 5

(from Step 3).

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 T⁰.

However, a quick calculation shows that we have not succeeded in this: we …nd, plugging in our calibrated results:

T_ij = ^O

i0A⁰_jFˆ_ijKˆ_ij

∑mA⁰_mFˆ_imKˆ_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.

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 T⁰ by …rst 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.

Calibration Without Step 2 (I)

At the end of Step 1 we had an interim estimate of T⁰ as:

T³ =

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:

Kˆ = 2 4

100 350 100 240 150 210 60 120 200

3 5

2 66 4

107.966 325.512 116.522 255.134 203.306 141.56 93.6824 145.208 141.11

3 77 5

= 2 4

0.92622 1.07523 0.858208 0.940682 0.737804 1.48347 0.640461 0.826402 1.41734

3 5

Calibration Without Step 2 (II)

And a simple calculation shows that if we plug these new estimates

Fˆ = 2 4

0.856909 1.6668 0.725347 1.6668 0.856909 0.725347 0.725347 0.725347 0.856909

3 5

Kˆ = 2 4

0.92622 1.07523 0.858208 0.940682 0.737804 1.48347 0.640461 0.826402 1.41734

3 5

into the gravity model, we do indeed exactly reproduce our original trip matrix T⁰.

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^y predictions of the future attractions by zone, A^y_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

y

iA^y_jFˆ_ijKˆ_ij

∑mA^y_mFˆ_imKˆ_im

Prediction (II)

Note that there is no guarantee that this T will satisfy conservation of attractions

One possibility is to perform an arti…cial 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: …rst 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 …nal prediction.

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^y= (650, 590, 350) A^y= (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.

Prediction Example (II)

To start, plug in O^y and A^y 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

Prediction Example (III)

Is this good enough? We need to check (only) conservation of attractions, based on our predictions (A^y):

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.

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

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

Appendix

Appendix

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):

F¹ = 2 4

0.876177 1.5537 0.769634 1.5537 0.876177 0.769634 0.769634 0.769634 0.876177

3 5

O_i⁰ = 550, 600, 380 A⁰_j = 400, 620, 510

We will compute the T² matrix, as shown in slide 28, by:

T_ij² = ^O

0 i A⁰_jF_ij¹

∑mA⁰_mF_im¹

Applying the Gravity Model (II)

Before we get started, one preliminary note.

The denominator of the expression for Tij is of the form

∑

A_mF_im =

∑

A_mF_im

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: T₁₁, T₁₂ and T₁₃ (row 1) all involve the same denominator; as do T₂₁, T₂₂ and T₂₃ (row 2). For a Z Z problem, you have to compute only Z distinct denominators.

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 :

∑

A⁰_mF_im¹ =

∑

A⁰_mF_1m¹

= A⁰₁F₁₁⁰ +A⁰₂F₁₂⁰ +A⁰₃F₁₃⁰

= (400 0.876177) + (620 1.5537) + (510 0.769634)

= 1706. 278 14

Applying the Gravity Model (IV)

Denominators, continued:

i =2 :

∑

A⁰_mF_im¹ =

∑

A⁰_mF_2m¹

= A⁰₁F₂₁⁰ +A⁰₂F₂₂⁰ +A⁰₃F₂₃⁰

= (400 1.5537) + (620 0.876177) + (510 0.769634)

= 1557. 223 08

i =3 :

∑

A⁰_mF_im¹ =

∑

A⁰_mF_3m¹

= A⁰₁F₃₁⁰ +A⁰₂F₃₂⁰ +A⁰₃F₃₃⁰

= (400 0.769634) + (620 0.769634) + (510 .876177)

= 1231. 876 95

Applying the Gravity Model

Now we begin computing the individual terms. Row 1:

i =1; j =1 : T₁₁² = ^O

0 1A⁰₁F₁₁¹

∑mA⁰_mF_1m¹

= ^{550 400 0.876177} 1706. 278 14

= 1. 927 589 4 10⁵ 1706. 278 14

= 112. 970 i =1; j =2 : T₁₂² = ^O

10A⁰₂F₁₂¹

∑mA⁰_mF_1m¹

= ^{550 620 1.5537} 1706. 278 14

= ^{5. 298 117 10}

5

1706. 278 14

= 310. 507

Applying the Gravity Model (V)

Final entry for Row 1:

i =1; j =3 : T₁₃² = ^O

10A⁰₃F₁₃¹

∑mA⁰_mF_1m¹

= ^{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.

Applying the Gravity Model (VI)

Row 2:

i =2; j =1 : T₂₁² = ^O

20A⁰₁F₂₁¹

∑mA⁰_mF_2m¹

= ^{600 400 1.5537} 1557. 223 08

= 239. 457

i =2; j =2 : T₂₂² = ^O

0 2A⁰₂F₂₂¹

∑mA⁰_mF_2m¹

= ^{600 620 0.876177} 1557. 223 08

= 209. 307

Applying the Gravity Model (VII)

Final entry for Row 2:

i =2; j =3 : T₂₃² = ^O

20A⁰₃F₂₃¹

∑mA⁰_mF_2m¹

= ^{600 510 0.769634} 1557. 223 08

= 151. 236 Row 3:

i =3; j =1 : T₃₁² = ^O

30A⁰₁F₃₁¹

∑mA⁰_mF_3m¹

= ^{380 400 0.769634} 1231. 876 95

= 94. 964 3

Applying the Gravity Model (VIII)

Row 3, concluded:

i =3; j =2 : T₃₂² = ^O

0 3A⁰₂F₃₂¹

∑mA⁰_mF_3m¹

= ^{380 620 0.769634} 1231. 876 95

= 147. 195

i =3; j =3 : T₃₃² = ^O

30A⁰₃F₃₃¹

∑mA⁰_mF_3m¹

= ^{380 510 0.876177} 1231. 876 95

= 137. 841

Applying the Gravity Model (IX)

Putting all these together we get:

T²=T_ij² =

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.