Explanation Given: The annual household income is $ 150 , 000 . The retail employment of neighborhood is 220 . Formula Used: Write the expression for the first route travel time. t 1 = 2 + 3 ( x 1 c 1 ) ..................(I) Here, t 1 is the first route travel time, x 1 is the route flow in thousands of vehicles per hour and c 1 is the first route capacity in thousands of vehicles per hour. Write the expression for the second route travel time. t 2 = 4 + 2 ( x 2 c 2 ) ..................(II) Here, t 2 is the second route travel time, x 2 is the route flow in thousands of vehicles per hour and c 2 is the second route capacity in thousands of vehicles per hour. Write the expression for derivative of system optimum function. S ( x ) = t 1 x 1 + t 2 x 2 ..................(III) Here, S ( x ) is the system optimum function. Write the expression for relation between x 1 and x 2 . x 1 + x 2 = 3.5 ..................(IV) Write the expression for first total travel route time. t 1total = 1000 t 1 x 1 ..................(V) Here, t 1total is the first total travel route time. Write the expression for first total travel route time. t 2total = 1000 t 2 x 2 ..................(VI) Here, t 2total is the second total travel route time. Write the expression for total travel time need to be saved before construction. t total = t 1total + t 2total ..................(VII) Here, t total is the total travel time need to be saved before construction. Write the expression for relation between x 1 and x 2 after construction. x 1 + x 2 = q ..................(VIII) Here, q is the traffic flow. Write the expression for reduction is peak hour demand. Δ q = q 2 − q 1 ..................(IX) Here, Δ q is the reduction is peak hour demand, q 1 is the peak hour demand before construction and q 2 is the peak hour demand after construction. Calculation: Substitute 2.5 for c 1 in equation (I). t 1 = 2 + 3 ( x 1 2.5 ) = 2 + 1.2 x 1 ..................(X) Substitute 4 for c 2 in equation (II). t 2 = 4 + 2 ( x 2 4 ) = 4 + 0.5 x 2 ..................(XI) Substitute 2 + 1.2 x 1 for t 1 , 4 + 0.5 x 2 for t 2 3.5 − x 1 for x 2 in equation (III). S ( x ) = ( 2 + 1.2 x 1 ) x 1 + ( 4 + 0.5 x 2 ) ( x 2 ) = ( 2 + 1.2 x 1 ) x 1 + ( 4 + 0.5 ( 3.5 − x 1 ) ) ( 3.5 − x 1 ) = 2 x 1 + 1.2 x 1 2 + 20.125 − 5.75 x 1 − 1.75 x + 0.5 x 1 2 = 1.7 x 1 2 − 5.5 x 1 + 20.125 ..................(XII) Differentiate equation (XIII) with respect to x 1 . d d x 1 ( S ( x ) ) = d d x 1 ( 1.7 x 1 2 − 5.5 x 1 + 20.125 ) = 2 ( 1.7 x 1 ) − 5.5 = 3.4 x 1 − 5.5 ..................(XIII) Substitute 0 for d d x 1 ( S ( x ) ) in equation (XIII). 0 = 3.4 x 1 − 5.5 3.4 x 1 = 5.5 x 1 = 5.5 3.4 = 1.62 Substitute 1.62 for x 1 in equation (IV). 1.62 + x 2 = 3.5 x 2 = 3.5 − 1.62 = 1.88 Substitute 2 + 1.2 x 1 for t 1 and 1.62 for x 1 in equation (V). t 1total = 1000 ( 2 + 1.2 x 1 ) x 1 = 1000 ( 2 + 1.2 ( 1.62 ) ) ( 1.62 ) = 1000 ( 3.944 ) ( 1.62 ) = 6389.28 veh-min Substitute 4 + 0.5 x 2 for t 2 and 1

The reduction in peak hour demand.

PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.

7th Edition

ISBN: 9781119610526

Author: Mannering

Publisher: WILEY

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Chapter 8, Problem 17P

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The reduction in peak hour demand.

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