Survey data suggest the trip interchange matrix shown below. Calibrate the friction factors for one iteration, assuming travel times are 2.0 minutes for interzonal trips and 1.0 minute for all intrazonal trips. Assume the friction factor takes the form t-b.
What will be an ideal response?
Based on the trip table, the productions and attractions are:
The initial guess is to assume b = 1. Thus the intrazonal friction factors are
1/1 = 1 and the interzonal friction factors are 1/2 = 0.5
The gravity model is then applied with these initial friction factors. For
example, the trips from zone 1 to zone 1 are found to be approximately 28.
The remaining trips are found in a similar manner.
We now compare predicted and actual results for each trip impedance. In
this particular example, we only have two trip impedances: impedances for
intrazonal trips and impedances for interzonal trips.
? For intrazonal trips (T11, T22, T33) a perfect friction factor would have
predicted 10 + 50 + 90 = 150 trips. The guess with b = 1 meant that the model
showed 28 + 75 + 125 = 228 trips.
? For interzonal trips (T12,T13, T21, T23, T31, and T32) a perfect friction factor
would have predicted 40 + 70 + 20 + 80 + 30 + 60 = 300 trips. The guess with b
= 1 predicted, however, 35 + 56 + 15 + 60 + 16 + 39 = 222 trips
The adjusted friction factors are:
? Intrazonal travel = (1)(150)/228 = 0.658
? Interzonal travel = (0.5)(300)/222 = 0.676
The regression equation will be:
F = t–b
Ln(F) = – b ln (t)
Ln (0.658) = –b ln(1)
Ln (0.676) = –b ln(2)
b = 0.565
Thus the friction factor based on one iteration is t–b. For intrazonal travel (with
travel time = 1.0) the friction factor is thus 1.0. For interzonal travel time with
travel time = 2.0, the friction factor is thus 2.0–0.565 = 0.676.
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