M. Kim, P. Schonfeld and E. Kim 1 - docs.trb.orgdocs.trb.org/prp/13-1196.pdfM. Kim, P. Schonfeld and E. Kim 1 ... 41 This model also considers how regenerative braking may affect vertical

M. Kim, P. Schonfeld and E. Kim 1

SIMULATION-BASED RAIL TRANSIT OPTIMIZATION MODEL 1

2 Myungseob (Edward) Kim 3 PhD Candidate 4 Department of Civil & Environmental Engineering 5 University of Maryland 6 College Park, MD 20742, United States 7 Tel: 301-405-3160 8 Fax: 301-405-2585 9 Email: [email protected] 10 11 Paul Schonfeld 12 Professor 13 Department of Civil & Environmental Engineering 14 University of Maryland 15 College Park, MD 20742, United States 16 Tel: 301-405-1954 17 Fax: 301-405-2585 18 Email: [email protected] 19 20 Eungcheol Kim (Corresponding author) 21 Associate Professor 22 Department of Civil & Environmental Engineering 23 University of Incheon 24 Incheon 406-772, South Korea 25 Tel: +82-32-835-8469 26 Fax: +82-32-835-0775 27 Email: [email protected] 28 29 30 Submitted for presentation at the 2013 TRB 92nd Annual Meeting of the Transportation Research 31 Board and for publication in the Transportation Research Record 32

33 Word Count: 3,794 + (4 Tables + 10 Figures) * 250 = 7,294 34

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TRB 2013 Annual Meeting Paper revised from original submittal.


ABSTRACT 38

A bi-level optimization process is proposed for determining the depth of a dipped vertical 39 alignment (DVA) between rail transit stations as well as the cruising speed for each direction. 40 This model also considers how regenerative braking may affect vertical alignment decisions. The 41 optimized depth and directional cruising speeds are jointly obtained in a numerical example. 42 Sensitivity analysis shows that regenerative braking reduces total cost, but does not significantly 43 change the optimized DVA depth. The developed model is more realistic and useful than 44 previous ones because it allows unequal station elevations, jointly optimizes decision variables, 45 and considers regenerative braking. 46

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INTRODUCTION 64

Rail transit can transport many passengers, and can generally provide more reliable travel times 65 than highway traffic. Rail transit operations may also reduce overall emissions. Many researchers 66 have explored ways to improve rail transit through rail alignment optimization (1, 2), joint 67 optimization rail and bus transit systems (3), optimized location of rail transit alignments and 68 stations (4, 5), and phased development of rail transit systems (6). Train and crew scheduling 69 have been optimized with genetic algorithms (7, 8). 70

Several studies considered energy-saving design of rail transit alignments (9~12). Kim 71 and Schonfeld introduced the dipped vertical alignment (DVA) profile (9) as a concept for 72 exploiting gravity to reduce acceleration energy and brake wear. They found that the DVA 73 profile reduces travel time as well as the required energy for operating trains. Yeh (10) extended 74 the work of Kim and Schonfeld (9) by jointly optimizing vertical track alignment and operational 75 characteristics (10). More recently, Kim and Chien analyzed various track alignments such as 76 level, convex, and concave profiles for rail transit operations (11). Although the metro systems in 77 New York City and elsewhere considered elevated lines more than a century ago (13), we could 78 not find any evidence that the dipped vertical alignment concept was applied. 79

One major limitation of previous studies is their assumption that trains operate between 80 stations having the same elevation. Kim et al (12) recently relax this constraint, thus allowing 81 more realistic alignments. They also optimize the cruising speed between stations in order to 82 minimize the total cost (i.e., the supplier cost plus user cost). They find that DVA reduces the 83 cost of operations, but without optimizing the DVA depth. Kim et al (12) also constrain the 84 optimized cruising speed to be equal in both directions. These limitations are removed in our 85 present study. 86

Many electric transit vehicles, particularly recently built ones, use dynamic braking as 87 their principal braking system (13). Dynamic braking can also recover braking energy and return 88 it as electric energy to the power supply system. This braking variant is called regenerative 89 braking. Dynamic braking is based on the fact that motors are inherently two-way electric-90 mechanical energy conversion devices. Normally, electric power is used by the motor to propel 91 the vehicle. However, it is possible to reverse the motor into an electric generator, so that when 92 the vehicle is in motion, the kinetic energy of the vehicle is used to generate electricity. The 93 work involved in generating the electricity decelerates the vehicle (13). Thus, regenerative 94 braking may complement and partially substitute for dipped vertical alignments. 95

In this paper we jointly optimize the depth of the DVA and the cruising speeds for each 96 direction. In rail transit systems the Maximum Authorized Speed (MAS) is dictated by the train 97 control system, which is not analyzed in this paper. We analyze here the Maximum Reachable 98 Speed (MRS) based on kinematics, propulsion, braking and passenger comfort considerations. 99 However, the proposed models can impose speed constraints based on any criteria, including the 100 limitations of control and signal systems. We further analyze the effects of regenerative braking 101 with the DVA profile. A numerical case study is used to verify the proposed framework for 102



jointly optimizing the depth and the cruising speed in each direction. The following sensitivity 103 analysis shows the impact on total cost with and without regenerative braking. 104

DIPPED VERTICAL ALIGNMENT PROFILE 105

Previous studies (10, 12) found that a Dipped Vertical Alignment (DVA) is useful in reducing 106 travel time and energy costs. However, they did not optimize the depth for applicable conditions. 107 The value of depth is related to the length of SB (described in Figure 1), assuming that trains 108 reach their maximum gradient along the vertical alignment. 109

Along the alignment the rate of gradient change r should not exceed 0.05 on sag curves 110 and 0.1 on crest curves (14). Thus sag sections should be twice as long as corresponding crest 111 sections. 112

For the DVA shown in Figure 1, Sections 1 and 7 are station platforms. Sections 2 ~ 6 are 113 parabolic curves. Sections 2 and 6 are 1/3 of S1 and S2, respectively. Sections 3 and 5 are 2/3 of 114 S1 and S2. SB is the section that has zero gradient. The total station spacing is the sum of one 115 platform length plus five (parabolic or straight) sections. 116

117 Previous studies (10~13) also considered the % rate of change in grade (r) per 30.48m 118

(100ft) defined as: 119 120

r = (G1 − G2)/L (1) 121

where G1 and G2 = percent grade of the two intersecting tangents; and L = curve length in 122 30.48m (100ft) stations. 123



124

Figure 1 DVA Profile with Depth and SB (Depth & SB ≠ 0) 125

The generalized equations for the vertical alignment for Figure 1 are provided in Table 1. 126

Table 1 Generalized DVA Equations 127 Section

No. Equation (unit: Yi=ft; X, S1, S2, SB, P l = stations) Boundary (units: stations)

1 Y1 = 𝐸𝑎 0 ≤ X ≤ P l /2

2 Y2 = 𝐸𝑎 −3(d1+d2)

S12(X − 𝑃𝑙

2)2 P l /2 ≤ X ≤ 1/3S1 + P l /2

3 Y3 = 𝐸𝑎 + 3(d1+d2)2S12

(X − S1 −𝑃𝑙2

)2 − (d1 + d2) 1/3S1 + P l /2 ≤ X ≤ S1 + P l /2

4 Y4 =𝐸𝑎 + 3(d1+d2)2S12

(X − S1 −𝑃𝑙2

)2 − (d1 + d2) S1 + P l /2 ≤ X ≤ S1 + SB + P l /2

5 Y5 =𝐸𝑎 + 3d22S22

(X − S1 − 𝑆𝐵 −𝑃𝑙2

)2 − (d1 + d2) S1 + SB + P l /2 ≤ X ≤ S1 + SB + 2/3S2 + P l /2

6 Y6 =𝐸𝑎 −3d2S22

(X − S1 − 𝑆𝐵 − S2 −𝑃𝑙2

)2 − (d1) S1 + SB + 2/3S2 + P l /2≤ X ≤ S1 + SB + S2 + P l /2

7 Y7 = 𝐸𝑏 S1 + SB + S2 + P l /2≤ X ≤ S1 + SB + S2 + P l

128

The maximum value of vertical gradient allowed by WMATA (15) is 4% (i.e., +/- 4%). 129 However, assuming some adhesion and control improvements, we explore gradients within +/- 130 4.5%. A maximum gradient cannot be exceeded anywhere along the vertical alignment. In Figure 131 1, the points of the steepest gradient are at the boundaries between Sections 2 and 3, and between 132 Sections 5 and 6. Thus, 133

𝐺𝑚𝑎𝑥(= 4.5) ≤ 2(𝑑1+𝑑2)𝑆1

(1) 134



𝐺𝑚𝑎𝑥(= 4.5) ≤ 2(𝑑2)𝑆2

(2) 135

From equations (1) and (2), S1 and S2 can be rearranged as 136

𝑆1 = 2(𝑑1+𝑑2)𝐺𝑚𝑎𝑥

(3) 137

𝑆2 = 2(𝑑2)𝐺𝑚𝑎𝑥

(4) 138

Total alignment length, S, is summation of S1, S2, and SB: 139

140 S = 𝑆1 + 𝑆2 + 𝑆𝐵 (5) 141

where SB is the length of the section with 0% in the vertical alignment. 142 Substituting equations (3) and (4) into equation (5), the length of SB is 143 144

𝑆𝐵 = S − 𝑆1 − 𝑆2 = S − �2(𝑑1+𝑑2)𝐺𝑚𝑎𝑥

+ 2(𝑑2)𝐺𝑚𝑎𝑥

� = S − �2𝑑1+4𝑑2𝐺𝑚𝑎𝑥

� (6) 145

In equation (6), S, d1, and Gmax are input values. Therefore, we note that the value of SB 146 and the value of d2 are interdependent. The maximum depth (d2) can be found when SB is zero. 147 The maximum depth (d2) is obtained with equation (7). 148

𝑑2𝑚𝑎𝑥 = 𝐺𝑚𝑎𝑥∙S−2𝑑14

(7) 149

150 Also, if the minimum depth (d2 = 0) is applied, SB has the maximum length given by equation (8): 151

𝑆𝐵𝑚𝑎𝑥 = S − � 2𝑑1𝐺𝑚𝑎𝑥

� (8) 152

153 The DVA profiles with the minimum and maximum depth are shown in Figure 2. This depth is 154 optimized in the following section. 155



(a) DVA Profile with minimum depth (depth = 0)

(b) DVA Profile with maximum depth (below, SB =0)

Figure 2 DVA Profiles with (a) minimum and (b) maximum depth values 156 157 DETERMINISTIC SIMULATION MODEL 158

A deterministic simulation model is used to obtain the travel times and energy consumption 159 between stations. The simulation model is adopted from Kim et al (12), and improved to 160 consider regenerative braking. 161

Simulation Assumptions 162

The assumptions made in this model are as follows: 163



1. Parabolic vertical curves are used in which gradients are limited by train climbing ability, 164 2. Horizontal curves are negligible for this analysis, 165 3. A concentrated mass is used to represent a train in motion; thus simulated trains start 166

from the center of a platform and arrive at the center of the next platform. 167 4. The acceleration rate of trains is limited by whichever of the following three factors 168

yields the lowest rate: (1) available power, (2) adhesion (friction) or (3) passenger 169 comfort. Similarly, the braking rate is limited by (1) power, (2) adhesion or (3) passenger 170 comfort. 171

5. The braking system can provide the maximum comfort-limited deceleration rate. 172 6. The cost of braking (i.e. brake maintenance and replacement) is proportional to the 173

absorbed braking energy and can thus be represented with a cost per unit of braking 174 energy. 175

7. Regenerative braking is available, which means the energy can be recovered while the 176 train is braking. The efficiency (% of recovered energy) of the regenerative braking is a 177 given input. No additional cost for using regenerative braking is assumed. 178

179 Modeling Process 180

During the simulation process, the train’s coordinates are updated along the vertical alignment 181 profile at 0.05 sec intervals. For each time interval, the simulation model computes obtains the 182 required tractive energy or braking energy. Maximum acceleration/deceleration rates of +/- 0.15g 183 are assumed, for the passengers’ comfort and safety. For any given DVA profiles, the simulation 184 model finds the braking point where a train must start decelerating so that it stops precisely at the 185 next station. The overall process of the simulation model is shown in Figure 3. Detailed relations 186 and equations are provided in the following section. 187



START

Vehicle CharacteristicsVertical Alignment

Initial Accel, Cruising Speed

Time = Time + Interval

aavail = (TE-Rt)g / ρWt, TE=min{Ft, Fa}

aapplied = Min.{ aavail, amax, acruising}

Update Information (X, Y, V)Use Kinematic Relations

Calculate Energy, Save to TE(sum)

Apply Feasible Braking Rate

Calculate Energy, Save to BE(sum)


Braking Point ?

Y

N

Print ResultTime, Position(X,Y), Station, Speed, TE, BE

END

Save ResultTime, Position(X,Y)Station, Speed, TE


aavail = (TE-Rt)g / ρWt, TE=min{Ft, Fa}

aapplied = Min.{ aavail, amax, acruising}


Save Distance and Speed Array DistanceAtoB(Time), SpeedAtoB(Time)

Vehicle CharacteristicsVertical Alignment

Initial Acceleration Rate


Apply Max. Acceleration Rate


Save Distance and Speed Array DistanceBtoA(Time), SpeedBtoA(Time)

DistanceBtoA(Time) >= Station Spacing

DistanceAtoB(Time) >= Station Spacing

N

N

Run Simulation A to B Direction Run Simulation B to A Direction

For i = 1: the number of array(SpeedAtoB)

For j = 1: the number of array(SpeedBtoA)

Y

SpeedAtoB(i) >= SpeedBtoA(j)

DistanceAtoB(i) >= DistanceBtoA(j)

Print Braking PointBrakingPoint = DistanceAtoB(i)

Y

Y

Y

Run Simulation A to B Direction

STOP ?


N

188

Figure 3 Simulation Process Adapted from Kim et al (12) 189 190 Train Motion and Energy Consumption 191

Newton’s second law is used to determine vehicle acceleration at any point: 192 193

a = F∙gm

� Flbw

� gft/s2

� mlb� (10) 194

195 where a=acceleration; F=force (unit: lbw); g=the acceleration due to gravity; and m=mass of 196 train. The train’s acceleration is limited by net force, which is the difference between available 197 tractive effort, Eavail and train resistance R. Thus, available acceleration aavail is 198



aavail = (Eavail−R)∙gρ∙Wt

�aavailft/s2

�Eavaillbw

� Rlbw

� gft/s2

� ρ−�Wtlb� (11) 199

where ρ = coefficient for rotating masses (usually 1.04~1.10, (13, 16)); Wt=train mass; R=train 200 resistance. However, the available tractive effort Eavail is limited by both tractive force Ft and 201 adhesive force Fa on driving wheels. Hence it is the minimum of these two forces: 202 203

Eavail = min{Ft, Fa} � Ealbw

� Ftlbw

� Falbw

� (12) 204

Tractive force Ft for a rail vehicle is 205 206

Ft = 375∙η∙PV

� Ftlbw

�η−� PHp� Vmile/h

� (13) 207

where P = power; V=speed; η=transmission efficiency. Since η is typically 0.82 (13), equation 208 (13) simplifies to: 209 210

Ft = 308∙PV

� Ftlbw

� PHp� Vmile/h

� (14) 211

The adhesive force is 212 213

Fa =∝∙ Fn � Falbw

� ∝−� Fnlbw

� (15) 214

where Fn = normal force; ∝=adhesive coefficient. 215 Vuchic (13) provides relations between the adhesive coefficient ∝ and train speed V for 216

dry conditions (in tunnels), which are approximately: 217 218

∝= 0.3 − 0.0015 ∙ 1.609V �∝−� Vmile/h

� (16) 219

Fn = Wtcosθ � Fnlbw

�Wtlb� θradian

� (17) 220

θ = tan−1 G100

� θradian

� G%� (18) 221

where θ = angle from level and G = percent gradient. To avoid a negative value of adhesive force 222 in equation (15), we assume that adhesive coefficient from equation (16) applies for speeds up to 223 62.15 mi/h (100km/h), and beyond that speed the coefficient stays constant at 0.15. 224

For resistance computation, we apply the Modified Davis Equation from Hay (14). Unit 225 resistance in pounds per ton is: 226

227

r = 0.6 + 20w

+ 0.01V + K𝑉2

𝑤𝑛+ 20𝐺 + 0.8𝐷 � w

ton� Vmph

� naxles

� G%� (19) 228



where w represents weight per axle in tons. The car resistance is r multiplied by the car weight in 229 tons: 230

Ri = r × Wt �r,Rcarlbw

� Wtton� (20) 231

The total resistance of a train R is the sum of the individual resistances of every vehicle i in that 232 train: 233

R = ∑ Ri𝐼𝑖 � R

lbw� Rcarlbw

� (21) 234

For the comfort and safety of standing passengers in rail transit systems, the longitudinal 235 acceleration and deceleration rates should be limited (usually to ±0.1~0.15). Therefore, the 236 maximum allowable acceleration or deceleration rate a is 237

238

-0.15 ∙ g ≤ a ≤ 0.15 ∙ g � gft/s2

� aft/s2

� (22) 239

where g = acceleration due to gravity. 240 When a constant cruise speed is sought, the acceleration limit acru is 241 242

acru at i = Vcruise−Vi∆t

�acruise at i ft/s2

�Vcruiseft/s

� Vift/s� ∆tsec� (23) 243

244 Therefore, the applied acceleration rate aapplied is 245 246

aapplied = min{aavail, acru at i , amax } �aappliedft/s2

�aavailft/s2

� amaxft/s2

� acru at i ft/s2

� (24) 247

For braking energy estimation, the available braking force is the minimum of two forces 248 249 Fb = min{Fbc, Fba } (25) 250

where Fbc is the maximum comfort-limited braking force that is applicable with standing 251 passengers. Fbc can be obtained by rearranging equation (11): 252 253

Fbc = R + bmax∙ ρ ∙Wtg

�Fbclbw

� bmaxft/s2

� ρ−�Wtlb� Rlbw

� (26) 254

Fba is an adhesive-limited braking force given in equation (15). 255 To compute energy consumption, the applied tractive effort, TEapplied, can be computed 256

after the train acceleration (or deceleration) rate is determined 257 258

Eapplied = R + aapplied∙ ρ ∙Wt

g �TEapplied

lbw� aapplied

ft/s2� ρ−�Wtlb� (27) 259



If Eapplied is positive, some tractive force is needed and if it is negative, some braking force is 260 needed. The power required for generating the applied tractive effort PTE is 261 262

Papplied = Eapplied∙V308

�Pappliedhp

� TEappliedlbw

� Vmile/h

� (28) 263

The energy consumption ∆eTE within time interval ∆t is 264 265

∆e = Papplied ∙∆t

3600∙ 11.341

� ∆ekwh

� Pappliedhp

� ∆ts� (29) 266

Based on the increments of energy specified in equation (29), we then separately accumulate 1) 267 the total energy required for tractive effort and 2) the total braking energy that must be dissipated. 268

JOINT OPTIMIZATION FRAMEWORK 269

Objective Function 270

The objective function minimizes total cost, which is the sum of both directional costs. The 271 objective function is a function of the depth of DVA, and the cruising speed in each direction. 272 For each directional cost between two stations, the user cost and supplier cost are added. The 273 user cost, Cu, is determined by multiplying by the travel time between stations, the number of 274 passengers per car, the number of cars in a train, and the time value for a passenger. The time 275 value of a passenger is given as an input parameter. Supplier cost, Cs, includes the cost for 276 tractive energy, the cost for braking energy, the cost saving from regenerative braking and the 277 vehicle depreciation cost. The directional cost function is: 278

Ct = Cu + Cs=t · Nc · Np · vuser + Nc · t · vvehicle + Et · vtractive + Eb · vbraking - Erb· vrb (30) 279

where Ct = total cost for one directional operation; Cu = user cost; Cs = supplier cost; t = travel 280 time between stations; Nc = cars per train; Np = passengers per car; vuser = passenger time value in 281 dollars/hour; vvehicle = vehicle depreciation per revenue hour; Et = tractive energy required (kwh); 282 vtractive = unit tractive energy cost; Eb = braking energy required (kwh); vbraking = unit braking 283 energy cost; Erb = energy recovered by regenerative braking (kwh); vrb = unit cost of regenerative 284 braking (=vtractive ). 285 286 Optimization Framework 287

Here we provide a two-stage optimization framework for determining the values of the depth and 288 cruising speeds for each direction. The depth of DVA should be identical for both directions 289 (assuming shared tunnels for the two directions), but the optimized cruising speeds may differ. 290 Moreover, with given depth, finding the optimal cruising speed becomes a one-variable 291 optimization problem. Therefore, we determine cruising speed through a constrained 292 optimization process. 293



To optimize the cruising speed for each direction, golden section search and parabolic 294 interpolation are used. To run the golden section search, we must assign lower and upper bounds 295 of the variable (cruising speed here). Lower bound should exceed zero. For the upper bound, we 296 run the simulation model without speed limitation, and then obtain maximum reachable speed. 297 This maximum possible speed obtained is used as the upper bound for the cruising speed 298 optimization. 299

The objective function provided in equation (30) is nonlinear. Thus, we need a nonlinear 300 optimization algorithm to minimize the total cost. We apply a sequential quadratic programming 301 (SQP) to optimize the DVA depth, which is the upper level decision variable. SQP is a nonlinear 302 optimization solver in MATLAB. Since SQP may get trapped in local optima, we create a global 303 solver which utilizes multiple (k) SQPs. The “k” is the number of solvers that we use for finding 304 solutions. k is an input value. With it, a starting point to run each solver is randomly selected 305 between the lower and upper bound of the depth. The lower bound is zero, with which SB has the 306 maximum length shown in equation (8). The upper bound is provided in equation (7). Each 307 solver starts to find the solutions from the randomly assigned starting point, and then stops if the 308 changes from the previous iteration are below 0.01 for the decision variable (depth) and below 309 0.01 for the total cost. The global solution is found from the least cost solution among the 310 solutions obtained by the k solvers. For each solver, the active-set algorithm is used to find the 311 solution because the nonlinear objective function in this problem cannot provide the hessian 312 matrix. The detailed algorithm description can be found in Mathworks (17). 313



314

315

Figure 4 Optimization Framework 316

NUMERICAL EXAMPLE 317

Input Values 318

The input values used for the case study are shown in Table 2. Station spacing is about 13,200 319 feet, and the elevation difference between the two stations is 60 feet. A train has eight cars, with 320 power and brakes on all axles. The passengers’ time value is assumed to be 10$ per person hour. 321 Vehicle depreciation cost is assumed to be 50$ per vehicle.hr. 322

323



Table 2 Input Values for Numerical Example 324 Spatial Inputs

Parameter Default Value Parameter Default Value Station Spacing (km) 4.0 (13,200 ft) Elevation at Station A (m) 302.8 (1,000 ft) Platform Length (m) 181.7 (600 ft) × 2 stations Elevation at Station B (m) 284.6 (940 ft)

Elevation Difference (m) 18.2 (60 ft) Train Inputs

Parameter Default Value Parameter Default Value Car Weight (tons/car) 50 Number of Cars 8 Car Power (kw/car) 520 Number of Axles per Car 4

Cost and User Inputs Parameter Default Value Parameter Default Value

Tractive Effort Unit Cost $0.3/kwh Passengers/car 50 persons Braking Energy Unit Cost $0.2/kwh Passenger Time Value $10/hr Vehicle Depreciation Cost $50/veh.hr 325 Optimization Results 326

Optimization results are shown in Table 3. The optimized depth is 104.76ft, which is close to the 327 maximum depth (105ft) because DVA can reduce travel time significantly, and reduced travel 328 time decreases user cost. Optimized speeds for the downhill and uphill directions are 84.5 mph 329 and 78.2 mph, respectively. Trains operating downhill reach higher speeds more quickly. 330 Therefore, travel time is lower than uphill. Directional cost is $208.13 per a train operating 331 downhill between two stations, and $222.40 uphill. The total cost for both directions with jointly 332 optimized depth and cruising speeds is $ 430.53 per train operating between two stations. 333

334 Table 3 Optimized Results and Total Cost Break-Down 335

Simulation Results

A to B (Downhill) B to A (Uphill) Sum Travel Time (sec) 148.7 156.3 305.0 Max. Speed (mph) 84.5 78.20 - Optimized Depth 104.76 104.76 -

Max. Gradient (%) 4.5 4.5 - Tractive Effort Required (kwh) 64.62 87.85 152.47

Tractive Effort Cost ($) 19.39 26.36 45.75 Braking Energy Required (kwh) 63.58 45.51 109.09

Braking Energy Required Cost ($) 12.72 9.10 21.82 Re-generative Energy Saving ($) 5.72 4.10 9.82

Vehicle Depreciation Cost ($) 16.52 17.37 33.89 User Cost ($) 165.22 173.67 338.89 Total Cost ($) 208.13 222.40 430.53

336

Figures 5~8 show the downhill simulation profiles. Figure 5 shows the relation between 337 the downhill speed and the elevation profile. In Figure 6, the gradient versus the elevation 338 profiles are shown in which the maximum gradient is less than 4.5%. Figure 7 shows the speed 339



versus the acceleration rate profiles. As, expected, the acceleration rate is zero for the cruising 340 section, in which the speed is constant. In Figure 7, at about 10,700ft from the starting station 341 the deceleration rate decreases. The reason is that gradient at the 10,700ft point changes from 342 positive to negative. This decreases the total resistance, so that the available tractive force 343 increases. Therefore, speed decreases slowly. Figure 8 shows that the total resistance is highly 344 dependent on the gradient profile. To limit this paper’s length, the profile figures in the opposite 345 direction (uphill) are omitted. 346

347

348

Figure 5 Speed vs. Elevation Profile (Downhill Direction) 349 350

351 Figure 6 Gradient vs. Elevation Profile (Downhill Direction) 352

353



354

355

Figure 7 Speed vs. Acceleration Profile (Downhill Direction) 356 357

358 Figure 8 Resistance vs. Gradient Profile (Downhill Direction) 359

360 Sensitivity Analysis 361

The following sensitivity analysis shows that regenerative braking is useful for reducing 362 energy cost, but the optimized depth changes are somewhat limited. We explore an extreme case 363 in which regenerative braking efficiency is 100%. In this extreme case, total cost decreases by 364 5.6% and 7.8% compared to the Base Case and No Regenerative Braking case, respectively. The 365 detailed sensitivity analysis results are shown in Table 4. 366

However, the optimized depths are not significantly reduced even if perfect regenerative 367 braking is assumed. Therefore, we note that although the regenerative braking recovers some 368



fraction of the energy cost, the DVA alignment concept is still useful for saving energy in rail 369 transit operations. 370 371 Table 4 Sensitivity Analysis Results 372 Regenerative Braking

Efficiency (%) Optimized Depth (Ft)

Directional Cost (A to B), ($)

Directional Cost (B to A), ($)

Total Cost ($)

Cost Savings (%)

0% 105.00 213.67 226.53 440.20 2.2% 30% (Base) 104.76 208.13 222.40 430.53 0.0%

45% 103.52 205.22 220.33 425.54 -1.2% 100% 100.44 193.71 212.83 406.54 -5.6%

373

The sensitivity results are illustrated in Figures 9~10. Figure 9 shows that the cost 374 reduction in the downhill direction (i.e., A to B) is greater than uphill. . Figure 10 shows that 375 optimal depth change between no regenerative braking and perfect regenerative braking is about 376 five feet. Moreover, according to Vuchic (13) the efficiency (percent energy recovery) of 377 regenerative braking is generally up to 30%, which is our Base Case. 378

379

380

Figure 9 Directional Costs, and Total Cost Changes in Sensitivity Analysis 381 382

Based on the sensitivity analysis, we note that regenerative braking slightly reduces the 383 optimized DVA depth. However, regenerative braking can further reduce some of the 384 construction cost because the optimized depth is reduced. The construction cost saving due to 385 regenerative braking depends very much on local site conditions which are beyond the scope of 386 this paper. 387

388



389

390

Figure 10 Cost Savings vs. Optimal Depth Profiles by Sensitivity Analysis 391 392

CONCLUSIONS 393

In this study, we provide a joint optimization framework in which the DVA depth and the 394 cruising speeds for both directions are jointly obtained through two-stage optimization. Since the 395 previous studies related to this work did not consider the optimization of the cruising speed in 396 each direction and optimal depth, the model provided here can be used for more realistic 397 operational analysis. Furthermore, this model also assesses regenerative braking, which is not 398 considered in the relevant previous studies. For our numerical cases the analysis indicates that 399 regenerative braking recovers some fraction of energy costs, and thus reduces the total cost. 400 Although the optimized DVA depth change from regenerative braking is slight, we note that the 401 accumulated cost reduction from the long-term analysis may be significant. When travel time is 402 less important, the optimized depth of DVA may be decreased. 403

ACKNOWLDEGEMENTS 404

This work was supported by the University of Incheon (International Cooperative) Research 405 Grant in 2012. 406 407 408 409 410 411 412 413



REFERENCES 414

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Documents

M. Kim, P. Schonfeld and E. Kim 1 - docs.trb.orgdocs.trb.org/prp/13-1196.pdfM. Kim, P. Schonfeld and E. Kim 1 ... 41 This model also considers how regenerative braking may affect vertical