Explanation Given: The interstate shipment distance is 200 miles . State Lumber Produced Lumber Consumed 1 5880 980 2 3300 10 , 000 3 9800 8000 Formula used: The formula to calculate the impedance factor is given by Impedance factor = 1 d ......(I) Here, d is the shipment distance. The formula to calculate the number of trips that are produced in zone 1 and attracted to zone 1 is given by T i j = [ P i A j F i j ∑ j = 1 n A j F i j ] ......(II) Here, T i j is the number of trips between each zone, P i is total number of trips produced in zone i , A j is number of trips attracted to zone j and F i j is an impedance factor for zone j . The formula to calculate the adjusted attraction factors for attraction zone in iteration 2 is given by A j k = A j C j ( k − 1 ) × A j ( k − 1 ) ......(III) Here, A j k is the adjusted attraction Factor for attraction zone j , k is the number of iteration, C j k is actual attraction total for zone j , A j is the desired attraction total for attraction zone j and j is the attraction zone. Calculation: Calculate the impedance factor for zone 1 to zone 2 . Substitute 200 miles for d in equation (I). Impedance factor = 1 200 = 0.005 The remaining impedance factor for zone 1, 2 and 3 is shown below. State 1 2 3 1 0.0050 0.00125 0.0010 2 0.00125 0.0050 0.0025 3 0.0010 0.0025 0.0050 Calculate the number of trips that are produced in zone 1 and attracted to zone 1 . Substitute 5 , 880 for P 1 , 980 for A 1 and 0.005 for F 11 in equation (II). T 11 = [ 5 , 880 × 980 × 0.005 ( 980 × 0.005 ) + ( 10 , 000 × 0.00125 ) + ( 8 , 000 × 0.001 ) ] = 28 , 812 25.4 = 1 , 134.3 Calculate the number of trips that are produced in zone 1 and attracted to zone 2 . Substitute 5 , 880 for P 1 , 980 for A 1 , 10 , 000 for A 2 , 0.00125 for F 12 and 0.005 for F 11 in equation (II). T 12 = [ 5 , 880 × 10 , 000 × 0.00125 ( 980 × 0.005 ) + ( 10 , 000 × 0.00125 ) + ( 8 , 000 × 0.001 ) ] = 73 , 500 25.4 = 2 , 894 The remaining number of trips that are produced in zone i and consumed to zone j is shown below, Zone 1 2 3 Given P 1 1 , 134 2 , 894 1 , 852 5 , 880 2 57 2 , 137 927 3 , 330 3 146 3 , 713 5 , 941 9 , 800 Computed A 1 , 337 8 , 924 8 , 720 18 , 980 Given A 980 10 , 000 8 , 000 In this case, the total trips consumed for each zone don't match the attractions that were given in the above table. Thus, adjust the attraction factors. To produce a doubly constrained gravity model, repeat the trip distribution computations using modified attraction values so that the number attracted will be increased or reduced required. Calculate the adjusted attraction factors for the attraction zone 1 , iteration 2 . Substitute 980 for A j , 980 for A j ( k − 1 ) and 1 , 337 for C j ( k − 1 ) in equation (III). A j k = 980 1 , 137 × 980 = 960 , 400 1137 = 844

The estimation of tonnage of lumber that will travel between the three states.

Traffic and Highway Engineering

5th Edition

ISBN: 9781305156241

Author: Garber, Nicholas J.

Publisher: Cengage Learning

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Question

Chapter 12, Problem 16P

To determine