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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