5. Suppose that the results of a peak-hour trip generation analysis for a five-zone small urban area is as follows: Trip Productions: Pi=11,000 P-93,300 Ps=12,900 P-203,500 Ps-90,700 person-trips Trip Attractions: A,=135,800 A;-57,500 A,-80,900 A,-72,000 As-85,200 person-trips Given the following friction factor matrix (based on travel time and cost factors) as follows: Fy Matrix To zone ilj 3 4 5 1 0.78 0.44 0.55 0.28 0.28 From Zone 2 0.44 1.30 0.40 0.70 0.33 3 0.55 0.40 1.30 0.44 0.80 0.28 0.70 0.44 1.80 0.44 0.28 0.33 0.80 0.44 1.80

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5. Suppose that the results of a peak-hour trip generation analysis for a five-zone small urban area is as follows:

Trip Productions: 
- \(P_1 = 11,000\) 
- \(P_2 = 93,300\) 
- \(P_3 = 12,900\) 
- \(P_4 = 203,500\) 
- \(P_5 = 90,700\) person-trips

Trip Attractions:
- \(A_1 = 135,800\) 
- \(A_2 = 57,500\) 
- \(A_3 = 89,900\) 
- \(A_4 = 72,000\) 
- \(A_5 = 85,200\) person-trips

Given the following friction factor matrix (based on travel time and cost factors) as follows:

\[
F_{ij} \text{ Matrix}
\]

\[
\begin{array}{c|ccccc}
 & \text{To zone} & 1 & 2 & 3 & 4 & 5 \\
\hline
\text{From Zone} & 1 & 0.78 & 0.44 & 0.55 & 0.28 & 0.28 \\
 & 2 & 0.44 & 1.30 & 0.40 & 0.70 & 0.33 \\
 & 3 & 0.79 & 0.55 & 0.44 & 0.30 & 0.60 \\
 & 4 & 0.28 & 0.70 & 0.44 & 1.80 & 0.44 \\
 & 5 & 0.28 & 0.33 & 0.80 & 0.44 & 1.80 \\
\end{array}
\]

Calculate the expected peak-hour trip distribution from zone 2 to zone 5 (\(T_{25} = ?\)).
Transcribed Image Text:5. Suppose that the results of a peak-hour trip generation analysis for a five-zone small urban area is as follows: Trip Productions: - \(P_1 = 11,000\) - \(P_2 = 93,300\) - \(P_3 = 12,900\) - \(P_4 = 203,500\) - \(P_5 = 90,700\) person-trips Trip Attractions: - \(A_1 = 135,800\) - \(A_2 = 57,500\) - \(A_3 = 89,900\) - \(A_4 = 72,000\) - \(A_5 = 85,200\) person-trips Given the following friction factor matrix (based on travel time and cost factors) as follows: \[ F_{ij} \text{ Matrix} \] \[ \begin{array}{c|ccccc} & \text{To zone} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{From Zone} & 1 & 0.78 & 0.44 & 0.55 & 0.28 & 0.28 \\ & 2 & 0.44 & 1.30 & 0.40 & 0.70 & 0.33 \\ & 3 & 0.79 & 0.55 & 0.44 & 0.30 & 0.60 \\ & 4 & 0.28 & 0.70 & 0.44 & 1.80 & 0.44 \\ & 5 & 0.28 & 0.33 & 0.80 & 0.44 & 1.80 \\ \end{array} \] Calculate the expected peak-hour trip distribution from zone 2 to zone 5 (\(T_{25} = ?\)).
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