For Problem 12-18, a field study of travel times was conducted which yielded the travel times of 1.5 minutes for intrazonal trips and 4.0 minutes for interzonal factors. Perform the first iteration of friction factor calibration.
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 1/1.5
= 0.67 and the interzonal friction factors are 1/4 = 0.25
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 35.
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 35.0 + 85.9 + 135.7 = 256.6 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, 32.7 + 52.3 + 12.8 + 51.3 + 12.7 + 31.6 = 193.4 trips
The adjusted friction factors are:
? Intrazonal travel = (0.67)(150)/256.6 = 0.392
? Interzonal travel = (0.25)(300)/193.4 = 0.388
The regression equation will be:
F = t–b
Ln(F) = –b ln (t)
Ln (0.392) = –b ln(1.5)
Ln (0.388) = –b ln(4)
b = 2.31 for intrazonal trips
b = 0.68 for interzonal trips
Thus the friction factor based on one iteration is t-b. For intrazonal travel (with
travel time = 1.5) the friction factor is thus (1.5)–2.31 = 0.392. For interzonal travel
time with travel time = 4.0, the friction factor is thus (4.0)–0.68 = 0.388.
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