Given: Study area with four transportation analysis zones, and origin-destination survey results. Provide a trip distribution calculation using the gravity model for two iterations; assume Kij = 1.





What will be an ideal response?


The mathematical formulation for the gravity model as provided as Equation 12.3:



Since Kij = 1, this factor does not affect calculations. The iterative application of

Equation 12.3 is as follows:

Iteration 1

T11 = 3,400 × ((2,800 × 1.51) /

((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97)))

T11 = 3,400 × (4,228 / 16287)

T11 = 883

T12 = 3,400 × ((6,500 × 0.92) /

((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97)))

T12 = 3,400 × (5,980 / 16,287)

T12 = 1,248

T13 = 3,400 × ((2,550 × 0.71) /

((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97)))

T13 = 3,400 × (1,811 / 16,287)

T13 = 378

T14 = 3,400 × ((4,400 × 0.97)/

((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97)))

T14 = 3,400 × (4268 / 16,287)

T14 = 891

T21 = 6,150 × ((2,800 × 0.92) /

((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04)))

T21 = 6,150 × (2,576 / 18,917)

T21 = 837

T22 = 6,150 × ((6,500 × 1.30) /

((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04)))

T22 = 6,150 × (8,450 / 18,917)

T22 = 2747

T23 = 6,150 × ((2,550 × 1.30) /

((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04)))

T23 = 6,150 × (3,315 / 18,917)

T23 = 1078

T24 = 6,150 × ((4,400 × 1.04) /

((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04)))

T24 = 6,150 × (4576 / 18,917)

T24 = 1488

T31 = 3,900 × ((2,800 × 0.71) /

((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92)))

T31 = 3,900 × (1,988 / 17,801)

T31 = 436

T32 = 3,900 × ((6,500 × 1.30) /

((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92)))

T32 = 3,900 × (8,450 / 17,801)

T32 = 1851

T33 = 3,900 × ((2,550 × 1.30) /

((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92)))

T33 = 3,900 × (3,315 / 17,801)

T33 = 726

T34 = 3,900 × ((4,400 × 0.92) /

((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92)))

T34 = 3,900 × (4,048 / 17,801)

T34 = 887

T41 = 2,800 × ((2,800 × 0.97) /

((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51)))

T41 = 2,800 × (2,716 / 18,466)

T41 = 412

T42 = 2,800 × ((6,500 × 1.04) /

((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51)))

T42 = 2,800 × (6,760 / 18,466)

T42 = 1025

T43 = 2,800 × ((2,550 × 0.92) /

((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51)))

T43 = 2,800 × (2,346 / 18,466)

T43 = 356

T44 = 2,800 × ((4,400 × 1.51) /

((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51)))

T44 = 2,800 × (6,644 / 18,466)

T44 = 1007



Next, calculate the adjusted attraction factors using Equation 12.4.



Zone 1

Ajk = (2,800 / 2,568) × 2,800

Ajk = 3,053

Zone 2

Ajk = (6,500 / 6,871) × 6,500

Ajk = 6,149

Zone 3

Ajk = (2,550 / 2,538) × 2,550

Ajk = 2,562

Zone 4

Ajk = (4,400 / 4,273) × 4,400

Ajk = 4,531

Now apply the gravity model formula for Iteration 2 using the above adjusted

attraction factors.

Iteration 2

T11 = 3,400 × ((3,053 × 1.51) /

((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97)))

T11 = 3,400 × (4,610 /16,481)

T11 = 951

T12 = 3,400 × ((6,149 × 0.92) /

((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97)))

T12 = 3,400 × (5,657 /16,481)

T12 = 1167

T13 = 3,400 × ((2,562 × 0.71) /

((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97)))

T13 = 3,400 × (1,819 /16,481)

T13 = 375

T14 = 3,400 × ((4,531 × 0.97) /

((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97)))

T14 = 3,400 × (4,395 /16,481)

T14 = 907

T21 = 6,150 × ((3,053 × 0.92) /

((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04)))

T21 = 6,150 × (2,809 / 18,845)

T21 = 917

T22 = 6,150 × ((6,149 × 1.30) /

((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04)))

T22 = 6,150 × (7,994 / 18,845)

T22 = 2609

T23 = 6,150 × ((2,562 × 1.30) /

((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04)))

T23 = 6,150 × (3,331 / 18,845)

T23 = 1087

T24 = 6,150 × ((4,531 × 1.04) /

((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04)))

T24 = 6,150 × (4,712 / 18,845)

T24 = 1538

T31 = 3,900 × ((3,053 × 0.71) /

((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92)))

T31 = 3,900 × (2,168 / 17,660)

T31 = 479

T32 = 3,900 × ((6,149 × 1.30) /

((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92)))

T32 = 3,900 × (7,994 / 17,660)

T32 = 1765

T33 = 3,900 × ((2,562 × 1.30) /

((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92)))

T33 = 3,900 × (3,331 / 17,660)

T33 = 736

T34 = 3,900 × ((4,531 × 0.92) /

((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92)))

T34 = 3,900 × (4,169 / 17,660)

T34 = 921

T41 = 2,800 × ((3,053 × 0.97) /

((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51)))

T41 = 2,800 × (2,961 / 18,555)

T41 = 447

T42 = 2,800 × ((6,149 × 1.04) /

((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51)))

T42 = 2,800 × (6,395 / 18,555)

T42 = 965

T43 = 2,800 × ((2,562 × 0.92) /

((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51)))

T43 = 2,800 × (2,357 / 18,555)

T43 = 356

T44 = 2,800 × ((4,531 × 1.51) /

((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51)))

T44 = 2,800 × (6,842 / 18,555)

T44 = 1032



Observe that the computed attractions approximately equal the given attractions.

A total convergence would be expected in another iteration.

Trades & Technology

You might also like to view...

Always plan the plotting scale prior to adding text or dimensions to the drawings because the scale affects the display size of each item.

Answer the following statement true (T) or false (F)

Trades & Technology

The first step in pre-job safety planning should be _____.

a. ranking the hazards by risk b. controlling or minimizing the hazards c. evaluating the consequences d. identifying the hazards

Trades & Technology

Range is defined as:

a. the portion of the process controlled by the controller. b. the difference ? between the upper and lower range limits. c. is designed to reduce the difference between the set point and process variable by adjusting the controllers output continuously until the offset is eliminated d. enhances controller output by increasing the output in relationship to the changing process variable.

Trades & Technology

The shim used on a drive pinion gear is used to properly position the gear so that it can correctly mesh with the ________

A) Ring gear B) Side gear C) Differential pinion gears D) Both B and C

Trades & Technology