Question

(iii) [14 marks] Consider a simple network between a single OD pair (from zone A to zone B), with travel time functions Travel time functions: Link 1 Link 3 Link 2 t1 = 1 + 0.5x1 t2 = 3 + x2 t3 = 2 + x3 t4 = 3 + 2x4 Link 4 Figure 2 Transportation Network In which xị, i = 1,2,3,4 are the link flows. The OD demand from A to B is D = 15 units. Solve for the user equilibrium link flow, link travel time, path flow, path travel time and OD travel time with equivalent mathematical formulation. Define your path set clearly.

Answer 1

a Travel time for all the links is given by travel time funution ti=1+0.5x, t2 = 3+X2 tz = 2 +X3 ty = 3+2x4 and total Horos from Lis 215 1 to 4 Assume X, X ₂ X z 244 20 which makes t = 1, tq=3 t3=2 & tu 3 Since the shortest path is link all flows are assigned to making x, 15 & X2=0 User 12 equilliheum Subs hituting travel time in ean min = 2(x) = f ( to.sxdx, & $ (3 722 ) dore, +L*6+x)des + ]*C3+24) dow] = 4 + 0.5x + 3x2 + x2 + 202 + 3 + 324 st : x + X2 + x3 + xy = 15 X +X2 =15

substitute X2=15-21 in the above formulation, on : 2 (z) = X1+0.5X? + 3 (15-36) + (15-3,3% = 2x + 0.25%, +45-3x, +225-2x15x, + x, ? ! 2 =xito. 25x1 +45-3x, + 112.5-158, +0.51, 2 20- 15x, ? - 17x, +157.5 = 11.33 x X2 = 15 - 30 X2 = 15-11.33 x = 3.67 2 2 x3 =15-xu z(x) = 203 + x ² + 3x4 + 20 = 2(15-44) + (15-24) ² + 3x4 + 224 2 30-224 +225 - 2x15 ou + xy + 3x4 + 4x4 30-1234 + x + 1125 - 45*4 +0.5x4 + 3x4 + 4 = 1425 - 1554 +5.504 xu = = 136. X3 = 15-xy X = 15-(-1.36) 22 = 15+1.36 = 16:36

ti = 1 +0.5x = 1+0.5X11:33 = 60 t2 = 3+82 = 3+3067 = 6-67 tz = 2+X3 = 27 16.36 = 18.36 tu = 3+224 = 3 +2×(1-36) = 5.72