research article peak power demand and energy consumption...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 940936 11 pageshttpdxdoiorg1011552013940936

Research ArticlePeak Power Demand and Energy Consumption ReductionStrategies for Trains under Moving Block Signalling System

Qing Gu1 Tao Tang1 Fang Cao1 Hamid Reza Karimi2 and Yongduan Song13

1 State Key Laboratory of Rail Traffic Control and Safety Beijing Jiaotong University Beijing 100044 China2Department of Engineering Faculty of Engineering and Science University of Agder 4898 Grimstad Norway3 School of Automation Chongqing University Chongqing 400044 China

Correspondence should be addressed to Yongduan Song ydsongbjtueducn

Received 16 July 2013 Accepted 24 September 2013

Academic Editor Xinjie Zhang

Copyright copy 2013 Qing Gu et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In themoving block signalling (MBS) systemwhere the tracking target point of the following train ismoving forwardwith its leadingtrain overload of the substations occurs when a dense queue of trains starts (or restarts) in very close distance interval This is thepeak power demand problem Several methods have been attempted in the literature to deal with this problem through changingtrainrsquos operation strategies However most existing approaches reduce the service quality In this paper two novel approachesmdashldquoService Headway Brakingrdquo (SHB) and ldquoExtending Stopping Distance Intervalrdquo (ESDI)mdashare proposed according to available andunavailable extra station dwell times respectively In these twomethods the restarting times of the trains are staggered and tractionperiods are reduced which lead to the reduction of peak power demand and energy consumption Energy efficient control switchingpoints are seen as the decision parametersNonlinear programmingmethod is used tomodel the process Simulation results indicatethat compared with ARL peak power demands are reduced by 40 and 20 by applying SHB and ESDI without any arrival timedelay respectively At the same time energy consumptions are also reduced by 77and 50by applying SHBandESDI respectively

1 Introduction

Moving block signalling (MBS) [1] was proposed a few dec-ades ago to reduce headway among successive trains in a trackline Theoretically two successive trains are separated by ashort distance which is equivalent to the braking distanceof the following train from its current speed as well as asafety margin This distance can be changed with the limitfor the given operating speed and train characteristics suchas train length and braking rate In MBS when a leadingtrain stops for a long time the following trains will stop atthe tail of the leading train When this leading train restartsthe following trains will start almost simultaneously It couldcause synchronization of the peak demand of trains andincreases the total peak power demand significantly whichis highly energy consumed and may lead to overload of thenearby substations How to reduce peak power demand iscalled ldquoPeak Demand Reductionrdquo (PDR) problem

This problem could be solved by improving the infras-tructure or the train operation strategies Energy storage

system (ESS) is a very important component in modernrailway power supply system Advanced control and man-ufacturing technology improve the stability capacity andweight of ESS [2ndash5] which makes it suitable to be equippedin the substations or in trains It could store the regenerativebraking energy and assist the main power source duringtrainrsquos acceleration period However it is costly to equipESS since the whole system needs a large number of themfor example in urban railway system nearly each of thestations or trains needs an ESS At the same time the costof maintenance for ESS is also huge especially when thefrequency of accelerationbraking is high (which means thehigh frequency of charge and discharge and leads to life timereduction) Furthermore it is inconvenient to be adopted byexisting lines since the reconstruction project is costly andcomplex Therefore changing train operation strategies is amore convenient and economical way to reduce peak powerdemand

There are two kinds of traditional PDR techniques basedon changing operation strategies one is called starting time

2 Mathematical Problems in Engineering

delay (STD) which introduces a starting time delay to eachof the following trains Under this category there are twospecific techniques called single STD and grade STD Thedifference between them is the introduced starting timedelay to each of the following trains which are the same insingle STD technique but different in grade STD (showinga deceasing trend) The other one is called acceleration ratelimit (ARL) which means the acceleration of the followingtrains is limited to a certain extent (or different extents)Under this category there are two specific techniques calledsingle ARL and grade ARL The difference between them isthe limited acceleration rate to each of the following trainswhich are the same in ARL technique but different in gradeARL (showing a deceasing trend) In addition to the abovetechniques there is also a PDR technique called coordinatedPDR It is a combination of the STD and ARL techniqueswith feeding the regenerated power of decelerating trains toaccelerating trains in the same queue by coordinating themovement of queued trains

Takeuchi and his colleagues discussed these techniques in[6ndash8] Simulation results show that the gradedARL techniquehas the best performance in reducing peak power demandamong single STD grade STD single ARL grade ARL andcoordinated PDR techniques In the traditional techniquestime delay is introduced and quality of service is degradedHo and Wong [9] use an expert system to help the operatorsfor decision making and it focuses on the balance betweentime delay and peak power demand In [10] by braking andpowering the trains simultaneously Albrecht proposed a wayto reduce the peak energy consumption and maximize theregenerative energy Chen and his colleagues [11] proposeda method to minimize the peak energy consumption byadjusting the train dwell times at each station in MRT sys-tems Kim et al [12] developed amixed integer programmingmodel to minimize the peak power energy demand thatoccurs when trains are running simultaneously and the basicprinciple of the method is the same with STD All of themethods proposed in [6ndash12] are implemented after the trainsrsquorestarting by adjusting the timetable or driving strategy

Although the existing techniques can reduce peak powerdemand at different degrees they increase the travel timebetween the successive stations and decrease the servicequality And the energy consumption is increased since thereare more traction periods

In this paper in order to reduce peak power demandand energy consumption we first analyze the reasons ofthe formation of the peak power demand and then twonovel approaches are proposed based on the main reasonswith considering the energy efficient driving strategies Oneis for available extra station dwell time named ServiceHeadway Braking (SHB) and the other one is for unavailableextra station dwell time named Extending Stopping DistanceInterval (ESDI) Both of them are real-time adjustmentmethods and can be implemented before the leading trainrsquosrestarting Therefore there is no need to change timetableConsidering energy saving driving strategy could be seen asa kind of driving mode switching process [13ndash16] a nonlin-ear programming model is contributed and the simulationresults show that compared with the best traditional PDR

techniques both of these twomethods could reduce the peakpower demand and energy consumption without any arrivaltime delay increasing And when the available extra stationdwell time is the same as the unavailable extra station dwelltime SHB performs better than ESDI since more peak powerand energy consumption could be reduced

2 Tracking Dynamics and Peak PowerDemand in MBS

21 Tracking Model in MBS Under MBS the tracking targetpoint of the following train moves forward continuously asthe leading train travels The instantaneous distance 119871

119911(119905) of

two successive trains is calculated as

119871119911(119905) = 119878leading (119905) minus 119878following (119905) (1)

where 119878leading(119905) is the position of the leading trainrsquos head and119878following(119905) is the position of the following trainrsquos head

The distance between two successive trainsmust be largerthan the safetymargin at anymoment even if the leading traincomes to a sudden halt so we have

119871119911(119905) ge 119871 safe + 119871 119905 +

119881following(119905)2

2119887 (2)

where 119871119905is the length of the train 119871 safe is the length of

safety margin 119881following(119905) is the instantaneous speed of thefollowing train and 119887 is deceleration rate

Based on (1) and (2) the relation between the leadingtrain and following train should satisfy

119878leading (119905) ge 119871 safe + 119871 119905 + 119878following (119905) +119881following(119905)

2

2119887 (3)

which implies that the instantaneous speed and position ofthe following train should satisfy

119881following (119905)

le radic2 times 119887 times (119878leading (119905) minus 119878following (119905) minus 119871 safe minus 119871 119905)

(4)

119878following (119905)

le 119878leading (119905) minus 119871 safe minus 119871 119905 minus119881following(119905)

2

2119887

(5)

22 Reasons of the Formation of Peak Power Demand Thereason of the formation of the peak power demand isthe restarting of the dense queue and the reasons for theformation of the dense queue are listed as follows

(1) Features of moving block signaling system (Twotrains will start simultaneously if the distance intervalbetween them is 119871 safe + 119871 119905)

(2) Extra dwell time in station

Mathematical Problems in Engineering 3

Sm S3 S2

V

S9984002 S

S9984003

Station A

middot middot middot

Target speed

S1

Figure 1 Formation of dense queue

There are two kinds of extra station dwell time oneis available and the other is unavailable In daily railwayoperation there may be some exceptions such as a passengermay be caught in the door of the train or a short-term surgein passenger flow (ie passenger flow increases sharply after afootball match) In these circumstances adjusting the wholetimetable is not convenient because the circumstances onlyexit in a short period In this case the operator will arrangethe train to stop a little longer and this extra station dwell timeis available However if a train is broken in a station and weare not sure how long we need to fix it in this situation theextra station dwell time is unavailable

Based on the analysis above it is known that peak powerdemand could be reduced by avoiding the dense queueIn order to achieve this goal we first analyze the relationbetween extra station dwell time and the number of delayedtrains

23 Station Delay Propagation Generally speaking eachtrain has a required dwell time at a station If a train stopslonger than the required dwell time we call the extra time asdelay time In this section we focus on the relation betweenthe delay time and the number of delayed trains In MBS thedelay timemay impact the following trains and cause a densequeue Figure 1 shows the formation of the dense queue

As it is shown in Figure 1 there are 119898 trains in the trackTrain 1 stops at station A The position of station A is 119878

1 For

each train the dwell time is 119879dwell The target speeds of thetrains are the same and constant According to the normalcondition each train arrives at station A and stops for 119879dwelland then starts to run

When train 1 starts the positions of following trainsare 119878119894 119894 = 2 3 119898 The tracking time (the running

time of the 119894th train running from 119878119894to 119878119894minus1

) between twosuccessive trains isΔ119905tracking it is also the scheduled departuretime interval However the following trains may become adense queue if train A does not run immediately after 119879dwellDefining119879delay is the delay time of the leading train after119879dwelland 119899 is the number of the delayed trains caused by 119879delayBased on (5) train 119894 which is delayed should stop at point 1198781015840

119894

in other words 1198781015840119894is the stop position of train 119894 in the dense

queue

1198781015840

119894= 119878119894minus1minus 119871 safe minus 119871 119905 (6)

Let Δ119905119894be the running time in which train 119894 arrived at 1198781015840

119894

Based on (2) we have 1198781minus1198781015840

2= 119871119905+119871 safe 1198781minus119878

1015840

3= 2(119871

119905+119871 safe)

1198781minus 1198781015840

4= 3(119871

119905+ 119871 safe) 1198781 minus 119878

1015840

119899= (119899 minus 1)(119871

119905+ 119871 safe) and

Δ1199052= Δ119905tracking minus119879dwell minus ((119871 119905 +119871 safe)V) Δ1199053 = 2timesΔ119905tracking minus

119879dwell minus 2 times ((119871 119905 + 119871 safe)V) Δ119905119899 = (119899 minus 1) times Δ119905tracking minus119879dwell minus (119899 minus 1) times ((119871 119905 + 119871 safe)V)

Letting 119879delay ge Δ119905119899 we have

119879delay

ge (119899 minus 1) Δ119905tracking minus 119879dwell minus (119899 minus 1) times119871119905+ 119871 safeV

(7)

then the number of delayed trains should satisfy

119899 le119879delay + 119879dwell

Δ119905tracking minus ((119871 119905 + 119871 safe) V)+ 1 (8)

24 Peak Power Demand Calculation The power demand ofthe 119894th train 119875

119894is calculated by

119875119894= 119865119894(V) times V

119894 (9)

where 119865119894(V) is the traction force of the 119894th train and V

119894is the

speed of the 119894th train after restartingThe peak power demand of the 119894th train 119875

119894minuspeak iscalculated by

119875119894minuspeak = max 119875

119894 (10)

where 119865119894(119905) is the traction force of the 119894th train and V

119894(119905) is the

highest speed of the 119894th train after restartingThe total peak power demand of all delayed following

trains 119875total(119905) is calculated by

119875total =119899

sum119894=1

119875119894minuspeak (11)

where 119899 is the number of delayed following trains

25 Energy Efficient Control Switching Points Because thereare more traction periods traditional PRD techniques causeenergy consumption increasing Energy conservation is theresearch interest in many fields [17 18] especially in railtransport [13ndash16]Therefore it would be better to reduce peakpower demand and energy consumption at the same timeTrain operation is a switching process the operation modesare switching among traction braking and coasting (with notraction and braking force) There are lots of researches onswitching system [19ndash21] For optimal control of switchingsystem we could apply the methodologies proposed in [2223] However the computation complexity will be a bigproblemwhenwe adopt the technologies above In this paperwe first apply Pontryagin maximum principle to find thediscrete optimal control modes and then treat the controlswitching speed as the decision parameters to obtain areference trajectory

For electric traction systems the motion equations oftrain have the following forms

119889V (119909)

119889119909=119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V)

V (119909)

119889119905 (119909)

119889119909=1

V (119909)

(12)


where 119909 is the position of the train V(119909) is the speed of thetrain at position119909 119905(119909) is the time at which the train is locatedat the position 119909 119906

119891is the relative traction force 119891(V) is

the specific maximum traction force per mass unit 119906119887is the

relative braking force 119887(V) is the specific maximum brakingforce per mass unit and 119903(V) is the specific basic resistance

Note that the basic resistance 119903(V) is usually given byDavis equation

119903 (V) = 1198861V2 + 119886

2V + 1198863 (13)

where 1198861 1198862 and 119886

3are regression coefficients obtained by

fitting test data to the Davis equation 1198863accounts for air

resistance 1198861 1198862account for mass andmechanical resistance

And 119903(V) has following characteristics

1199031015840

(V) gt 0 11990310158401015840

(V) ge 0 (14)

The energy efficient operation problem is modeled as follows

min 119869 = int119883

0

119906119891119891 (V) 119889119909

st 0 le 119906119891le 1 0 le 119906

119887le 1

V le 119881 (119909)

(15)

where 119869 is the specific mechanical work of the traction force119881(119909) is the speed limit The boundary conditions are

119909 (0) = 1199090 119909 (119879) = 119909

119879

V (0) = V0 V (119879) = V

119879

(16)

By applying Pontryagin maximum principle to solve theproblem as specified in (15) with constraints (16) the optimalsolution should maximize the Hamiltonian function

119867 =1199011

V(119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V))

+1199012

Vminus 119906119891119891 (V)

(17)

where 1199011should satisfy the differential equation

1198891199011

119889119909= minus120597119867

120597V (18)

It is easy to prove that the Hamiltonian reaches themaximum with respect to 119906

119891and 119906

119887 And there are five

energy efficient control modes as follows

(i) Full power (119906119891= 1 119906

119887= 0) if (119901

1V) lt 0

(ii) Partial power (119906119891isin (0 1) 119906

119887= 0) if (119901

1V) = 0

(iii) No power and no braking (119906119891= 0 119906

119887= 0) if 0 lt

(1199011V) lt 1

(iv) Partial braking (119906119891= 0 119906

119887isin (0 1)) if (119901

1V) = 1

(v) Full braking (119906119891= 0 119906

119887= 1) if (119901

1V) gt 1

They could be seen as four possible driving phases which are

(i) acceleration with full power(ii) speed holding with partial power or braking(iii) coasting with no power and braking(iv) braking with full braking

Based on the above analysis it is seen that the energysaving strategy relies on these optimal controls Train energysaving driving process is the process of switching from anoptimal control to another and the switching sequence isacceleration speed holding coasting and braking

Then the optimal control switching speeds and runningphases duration are treated as decision parameters Theenergy and power can be expressed by these decision param-eters The power demand is used as the objective function ofa nonlinear programming problem The constraints includerunning time and distance In actual train operation maxi-mum power and braking cannot be applied by consideringthe ride comfort Instead a service accelerationbraking rateis applied Therefore by solving the nonlinear programmingmodel a reference trajectory leading to less power and energyconsumption can be obtained

3 Peak Demand Reduction Techniques

31 Service Headway Braking Strategy Based on the analysisabove we know the restarting of the dense queue in asmall area leads to peak power demand and both of thetwo traditional PDR techniques are carried out after theformation of the dense queue In this section we propose anovel operation strategy to reduce the peak power by avoidingthe formation of a dense queue based on the available extrastation dwell time In the following parts we use reaccelerateto indicate train accelerates from a nonzero initial speed andrestart to indicate train starts from 0ms

311 SHB Strategy Analysis Figure 3 shows the new oper-ation strategy the leading train stops at station A andthe position is 119878

1 The black thick solid line is calculated

by (2) and we call it braking curve in this paper Undernormal circumstance the following train should brake whenit touches the brake curve However if the leading train startsto run when the following train touches braking curve thefollowing train will not brake but reaccelerate Therefore inthe new strategy if the following train touches the brakingcurve when the leading train starts to run and the speedat this moment is not 0ms the moment in which eachtrain reaches the peak power demand can be staggeredMeanwhile since the following trains brake at first they willnot stop too to cause a dense queue By these two reasonsthe total peak power demand can be reduced Based on theanalysis above and considered the energy saving operationprocess a new operation strategy is proposed as follows

As shown in Figure 2(a) the target speed of the trains inthe track is V

1 the leading train has an extra dwell time which

is 119879delay If 119879delay is long enough to cause a dense queue with 119899trains then after 119879dwell let train 119894 (119894 = 2 3 119899) brake withservice braking deceleration 119887 to V

1198942(the position at this time


1

2

Si3 Si5S25S23Si1 Si2 S21 S22

3 i3

i2

i41 1

2223 24

i5

S

Station A

S1Si4

S24

25

(a)

2

S33 S35S31 S32

3 3

32

i533

Station A34

S34

1

S1 S

(b)Figure 2 (a) Following trainsrsquo SHB driving curves after train 1 stops for 119879dwell (b) The position of train 3 when train 2 starts to move

0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Time (s)

Spee

d (m

s)

Figure 3 V-T profile of the following train starts form the brakingprofile

is 1198781198942) if V1198942= 0 then it stops at 119878

1198942for Δ119905

1198942 then accelerates

to V1198943with service acceleration a (the position at this time is

1198781198943) and then runs with V

1198944 (V1198944= V1198943) for Δ119905

1198944to 1198781198944 after

that it reduces traction force and keeps the train moving witha constant deceleration 1198871015840 to V

1198945(the position at this time is

1198781198945 V1198945and 119878

1198945are a pair of points in the braking curve) At

this time train i-1 restarts therefore train 119894starts to track traini-1 according to the moving block tracking distance intervaluntil arrives at station A

Figure 2(b) shows the position of train 3 when train 2starts to move from station A It is observed that train 3will reaccelerate from the braking curve (119878

35) to arrive at

station A (1198781) we use 119879

119905119905to represent this time period and

all of the delayed following trains need this time period toarrive at station A after their leading trainrsquos restarting In thiscircumstance for train 2 the running time between 119878

21and

11987825is 119879delay for train 3 the running time between 119878

31and 11987835

is 119879delay +119879dwell +119879119905119905 for train 119894 the running time between1198781198941and 1198781198945is 119879delay + (119894 minus 2) times (119879dwell + 119879119905119905)

In order to obtain 119879119905119905 a simulation is done as follows

119871119905= 140 (m) 119871 safe = 50 (m) The leading train stops at

station A where the position is 400 (m) 119878leading(0) = 400 (m)Let the following train start from the point in braking curvethe speed and position of the following train is calculated by(4) or (5) the results are shown in Table 1

FromTable 1 we can see that the arrival times of followingtrains seem constant when they reaccelerate from the brakingcurve In order to show the trend of this tracing processFigure 3 gives the V-T profile when the reaccelerating speedand position are 8ms and 178 (m)

Table 1 Running time of following train

119878leading(0)

(m)119881following(0)

(ms)119878following(0)

(m)

Running time of thefollowing train arrives at

station A(s)

400 16 82 4026400 12 138 3982400 8 178 395400 4 202 3942400 0 210 3938

Based on the analysis above the whole strategy is con-sisting of 6 steps braking waiting traction speed holdingcoasting and finally tracking the front train The indexes ofthem are 1 2 6

It is worth noting that this operation strategy is energyefficient because the duration of some steps may be 0 basedon an appropriate target function For example if we get V

1198942=

V1198943= V1198944gt 0 based on a certain target function there will be

no traction process before reaccelerating and the train willbrake and coast until reaccelerating

In this new strategy all the following trains brake at firsttherefore they will not stop too to cause a dense queue atthe same time the other trains are coasting when one trainreaccelerates so the reaccelerating times of the followingtrains are staggered Therefore the peak power demand isavoided

312 Model of SHB In this section we formulate the math-ematical model of the operation process For train i define119905119894119895(119895 = 1 2 5) as the starting time of each step Δ119878

119894119895(119895 =

1 2 5) and Δ119879119894119895(119895 = 1 2 5) are the running distance

and time of each step respectively In order to stagger thereaccelerating times of the following trains we should take anappropriate value to V

1198945 Because if V

1198945= 0 the successive two

trains will reaccelerate simultaneously again and peak powerdemand cannot be reduced On the contrary if V

1198945is close to

V1 the effect of energy saving will be reduced Therefore we

recommend V1gt V1198945ge V12 At the same time for train 119894 the

total running time and distance should satisfy the followingequations

5

sum119895=1

Δ119879119894119895= 119879delay + (119894 minus 2) times (119879dwell + 119879119905119905) (19)

5

sum119895=1

Δ119878119894119895= 1198781minus 1198781198941minus 119871 safe minus 119871 119905 (20)


Since the traction phase is highly energy consumed andpower supported we minimize the peak power Thereforebased on the analysis above the problem could be seen as aconstrained nonlinear programming problem as follows

min119891 =119899

sum119894=1

119901119894(V1198943) (21)

st V1198942minus V1198943le 0

V1198945minus V1198944le 0

minus V119894119895 minusΔ119879119894119895le 0

V1

2le V1198945lt V1

5

sum119895=1

Δ119879119894119895= 119879delay + (119894 minus 2) times (119879dwell + 119879119905119905)

5

sum119895=1

Δ119878119894119895= 1198781minus 1198781198941minus 119871 safe minus 119871 119905

(22)

where5

sum119895=1

Δ119879119894119895=V1198942minus V1

119887+ Δ1198791198942+V1198943minus V1198942

119886

+ Δ1198791198944+V1198945minus V1198944

1198871015840

5

sum119895=1

Δ119878119894119895+ 119871119887=V21198942minus V21

2 times 119887+V21198943minus V21198942

2 times 119886

+ Δ1198791198944times V1198944+V21198945minus V21198944

2 times 1198871015840+

V21198945

2 times |119887|

(23)

where 120572 120573 are penalty factors 120572 120573 gt 0 119886 119887 are serviceacceleration and deceleration of the train 1198871015840 is a very smalldeceleration 119894 is the index of the delayed following trains119894 = 2 119899 j is the index of operation steps 119895 = 1 2 6 V

119894119895

is the starting speed of each step for train 119894 1198781is the position

of the leading train (the leading train) 1198781198941is the position of

the 119894th following train Δ119878119894119895is the running distance of the 119895th

step for train 119894 Δ119879119894119895is the duration of of the 119895th step for train

119894 119871119894119887is the braking distance if train 119894 brakes from V

1198945 119879dwell

is the scheduled station dwell time 119879119905119905is the tracking time in

which train 119894 tracks train i-1 from 1198781198945to 1198781

During the running process of successive following trainsthe time interval among them is119879dwell+119879119905119905When train i startsto move from 119878

1198942 the distance between train 119894 and train i-1 is

shortest at this time the speed of train i is larger than V12

and the distance interval between them is at least (119879dwell +119879119905119905) times V12 according to the general condition in mass transit

system 119879dwell = 10 s 119871 safe = 50m 119871119905= 140m MAX(V

1) =

20ms 119887 = 1ms2 it is easy to know that (119879dwell+119879119905119905)timesV12 gt119871 safe+119871 119905+(V12)

2

2119887 this is satisfied by (3)Therefore it is noneed to consider the distance constraint shown in (3) betweensuccessive following trains in (22) again

Station A

1

2 2 4 5 6

13

SS2ΔS1 ΔS2

ΔS3 ΔS4 ΔS5ΔS6S1

Figure 4 Operation strategy of ESDI

32 Extending Stopping Distance Interval Strategy The pre-vious section shows SHB strategy it could reduce the peakpower demand if the delay time is available In this sectionwe will show a strategy which could reduce the peak powerdemand by decreasing the density of the dense queue Byadopting this strategy the peak power demand could bereduced if the extra station dwell time is unavailable Becausein the same 119879delay the longer the stopping distance interval ofthe following trains the lower densities of the stopped trainqueueTherefore when they restart the peak power demandsare reduced

321 ESDI Strategy Analysis According to the analysis inSection 311 it can be seen that no matter the train reaccel-erates from which point in the braking curve 119879

119905119905is nearly

the same Based on 119879119905119905 we can extend the stopping distance

interval of the following trains to reduce the density of thedense queue The new operation strategy is proposed asfollows

Figure 4 shows the new operation strategyThe operationprocess of the following trains when the leading train dosedoes not move after 119879dwell could be divided into 6 stepsspeed holding braking traction speed holding coastingand braking to stop Define 119905

119894(119894 = 1 2 6) as the starting

time of each step V1= V2 and V

3= 0 V

4= V5 define the

running time and distance of each step as Δ119879119894and Δ119878

119894(119894 =

1 2 6) respectively When the leading train has an extradwell time each of following trainswill run according to these6 steps Define 119872119868 as the traveling distance during 119879

119905119905(MI

is also stopping distance interval) for the following trainsDuring this process the running time of steps 3 to 6 shouldbe equal to 119879

119905119905

In this new strategy the stopping distance interval of thefollowing trains is increased so the density of the queue isdecreased and the peak power demand is reduced

322 Model of ESDI In this section we formulate themathematic model of the operation process According tothe last section it is known that we know that the stoppingintervals of the trains are extended in the new operationstrategy therefore peak power demand can be reduced whenthe trains restart Define MI as the stopping interval of twosuccessive trains we have

119872119868 =

6

sum119894=3

Δ119878119894

=V24

2 times 119886+ V4times Δ1198794+(V26minus V25)

2 times 1198871015840+

V26

2 times |119887|

(24)


In order to grantee the feasibility of the new operationstrategy if leading train stops at the station after 119879dwell thedistance between the first following and the leading trainshould satisfy the following equation

119872119868 le 119878tracking minus 119879dwell times V1 (25)

where 119878tracking is the second trainrsquos position when the leadingtrain stops at station and it is calculated by

119878tracking = (119879traking minusV1

|119887|) times V1+

V21

2 times |119887| (26)

Since the traction phase is highly energy consumed andpower supported we minimize the peak power Based on theanalysis above we use nonlinear programming to model theproblem as follows

min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0

119872119868 le 119878tracking minus 119879dwell times V1

(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)

where 120572 120573 are penalty factors 120572 gt 0 119886 119887 are serviceacceleration and deceleration of the train V

119894is the starting

speed of step 119894 1198781is the position of the leading train (the

leading train) 1198782is the position of the following train 119878tracking

is the second trainrsquos position when the leading train stops atstation 119879dwell is the scheduled station dwell time

4 Simulation and Discussion

In this section a simulation is used to test and verify the newstrategies The length of train (119871

119905) is 140m safety margin

(119871 safe) is 50m service tracking headway is 120 seconds dwelltime (119879dwell) is 10 seconds target speed (V

1) is 16ms service

acceleration rate (119886) is 1ms2 service braking decelerationrate (b) is 1ms2 and position of station A is 3710m (119878

1=

3710)In order to choose 1198871015840 we analyze the practical data of

coasting phase from Dalian Fast Track [24] Because thespeed in coasting phase declines very slowly so the profile ofspeed time could be seen as a straight line and the slope of theline is the deceleration rate of coasting phase We use least-square procedure to fit the speed-time date sectional and theresults are shown in Table 2

From Table 2 it is clear that the higher the coastingstarting speed the greater the deceleration In order to supplya small traction force to keep the train moving in a constantdeceleration 1198871015840 we choose 1198871015840 = 001ms2 as appropriate

Table 2 Measured value of coasting data

Speed range (kmh) Slop (ms2) Average error (m)37ndash31 minus00147 0022741-40 minus00125 0036143-42 minus00147 0036454ndash48 minus00213 0023660-59 minus00237 0029662-61 minus00210 0030179ndash75 minus00315 00237

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3

Figure 5 V-T profile without PDR technique

41 Service Headway Braking Strategy If119879delay is 250 secondsaccording to (8) 3 trains will be delayed (including theleading train) Because we choose Vref = 8 therefore 119879119905119905 =385 is selected from Table 1 Based on (21)-(22) we have

V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)

Figures 9 and 10 and Table 3 show the simulation resultsIn order to show the peak power demand without any

PDR technique and compare it with the performance of tra-ditional PDR technique Figures 5ndash8 are given According toFigures 5 and 6 we can see that the two following trains arestarting simultaneously from 250 s the peak power demandis 251 kwtThe arrival times of train 2 and train 3 are 28938 sand 33876 s respectively

Figures 7 and 8 show the performance of graded ARLtechnique The acceleration of train 2 and train 3 is 05ms2and 03ms2 By applying different acceleration the peakpower demand is reduced to 2207 kwt However the timedelay is increased The arrival times of train 2 and train 3are 29778 s and 40036 s respectively That means the arrivaltimes of the two trains are 84 s and 616 s later than their timeswithout PDR techniques respectively

The performance of applying SHB technique is shownfrom Figures 9 and 10 As can be seen train 2 has a waiting


Table 3 Comparison of the graded ARL and SHB techniques

Arrival time oftrain 2 (s)

Arrival timeof train 3 (s)

Peak powerdemand (kwt)

Energyconsumption (kwsdoth)

Stopping time before arrival (s)Train 2 Train 3

Non-PDR 28938 33876 251 159137 130 4375GradedARL 29778 40036 2207 463402 130 4375

SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)

Figure 6 Peak power demand profile without PDR technique

0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3

Figure 7 V-T profile with graded ARL technique

time 6566 s at the position of 2048m and then restarting at8166 s At this time train 3 is coasting therefore there is nopeak power in this time The two trains restart with differentspeeds in different time points which can be seen form thered and blue circles in Figure 9Thatmeans the reacceleratingtimes of the two trains are staggered Furthermore thetwo trains reaccelerate with nonzero initial speeds whichmeans the reacceleration periods are reduced Therefore thepeak power demand is reduced to 1342 kwt According toFigure 9 the two trains have no speed holding phases and

260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)

Figure 8 Peak power demand profile with graded ARL technique

train 3 has no traction phases before step 6 Therefore theenergy consumption is reduced The arrival times of train 2and train 3 are 2895 s and 33916 s respectively

Table 3 shows the result data of the simulation GradedARL technique can reduce the peak power demand howeverthe arrival times of the two trains are delayed significantlySHB technique has great advantages According to Table 3the delay times of the two trains are very short at the sametime the energy consumption is also reduced to a low leveleven less than value without any PDR technique In additionthe stopping times before arrival of the two trains are reducedto 6566 and 0 respectively

42 Extending StoppingDistance Interval Strategy Accordingto Section 312 we choose 119879

119905119905= 40 s when the leading train

stops at the station and dose not start to run after119879dwell basedon (27) we have

V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)

If the extra dwell time is 250 seconds in graded ARL 3trains will be delayed (including the leading train) accordingto (8) And the length of the dense queue is 380m The per-formance of graded ARL technique can be seen in Figure 5

In ESDI based on (24) the stopping distance intervalfor the following trains (MI) is 38368m Based on (8) 2trains (besides the leading train)will be delayedHowever thelength of the dense queue is 76736m In the same time periodand distance area with graded ARL technique we calculatethe peak power demand and the performance is shown inFigures 11 and 12


0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3

Restarting time of train 2Restarting time of train 3

Train 2Train 3

Restarting position of train 2Restarting position of train 3

Figure 9 V-S and V-T profiles with SHB technique

Table 4 Comparison the PDR techniques







SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)

Figure 10 Peak power demand profile with SHB technique

As we see using ESDI technique the arrival times ofthe following two trains are almost the same as their timeswith no PDR technique Without any PDR technique thestopping distance interval is 190m so the dense queue causedby the extra station time (250 s) is 380m The peak powerdemand after 250 swithin 380m is 251 kwt In ESDI strategystopping distance interval is extended to 38386m which istwo times longer than the distance in ARL That is to say the

restarting position of train 2 belongs to the nearby substationThus the power demands by train 2 and train 3 are notafforded by the same substation Therefore the peak powerdemand after 250 s within 380m is reduced to 1818 kwtComparedwith non-PDR technique the peak power demandis reduced by 18 In ESDI train 2 has a stopping time15398 s at the position of 29446m Train 3 has a stoppingtime 5796 s at the position of 332832m Comparedwith non-PDR technique the stopping time before arrival is increasedby 18

According to above analysis we can see that the newstrategy reduces peak power demand by sacrificing thestopping time before arrival However the advantages ofESDI are still obvious because it is more efficient on energysaving it can reduce energy consumption by 50 comparedwith graded ARL technique

From Table 4 it is seen that although the arrival timesof trains when applying SHB and the arrival times of trainswhen applying ESDI are the same SHB reduces more peakpower demand and energy consumption which means SHBperforms than ESDI when the available and unavailable extrastation dwell times are the same

5 Conclusion

Peak power demand reduction strategies are discussed inthis paper The reasons of peak power demand problem areanalyzed deeply and two main reasons are given


0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3Restarting time of the two trains

Figure 11 V-S and V-T profiles with ESDI technique

250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)

Figure 12 Peak power demand profile with ESDI technique

Based on the reasons and according to different sit-uations two new peak demand reduction techniques areproposed One is Service Headway Braking (SHB) strategywhich is used to reduce peak power demand when theextra station dwell time is available The other is ExtendingStopping Distance Interval (ESDI) strategy which is usedto reduce peak power demand when the extra station dwelltime is unavailable Nonlinear programming approach isintroduced to model the operation strategy The simulationresults show that compared with the best traditional PDRtechniques SHB can reduce 40 of peak power demand and77 of energy consumption without increasing the arrivaltime delay ESDI can reduce 20 of peak power demand and50 of energy consumption without increasing the arrivaltime delay Therefore SHB has a better performance thanESDI when the available and unavailable extra station dwelltimes are the sameThe basic principle of the two strategies isto avoid the formation of the dense queue Therefore both of

the two strategies are implemented before the formulation ofthe dense queue

Conflict of Interests

The authors declare that there is no conflict of interests

Acknowledgments

This work was supported by the National High Technol-ogy Research and Development Program of China (no2011AA110502) and the National Natural Science Foundationof China (no 61134001)

References

[1] L V Pearson Moving block railway signalling [PhD thesis]Loughborough University of Technology Loughborough UK1973

[2] T Chaiyatham and I Ngamroo ldquoOptimal fuzzy gain schedulingof PID controller of superconducting magnetic energy storagefor power system stabilizationrdquo International Journal of Innova-tive Computing Information and Control vol 9 no 2 pp 651ndash666 2013

[3] D Iannuzzi E Pagano and P Tricoli ldquoThe use of energy storagesystems for supporting the voltage needs of urban and suburbanrailway contact linesrdquo Energies vol 6 pp 1802ndash1820 2013

[4] H Lee S Jung Y Cho H Jung H Kim and G Jang ldquoPeakpower reduction and energy efficiency improvement with thesuperconducting flywheel energy storage in electric railwaysystemrdquo Physica C vol 494 pp 246ndash249 2013

[5] Y Arai H Seino K Yoshizawa and K Nagashima ldquoDevelop-ment of superconductingmagnetic bearing with superconduct-ing coil and bulk superconductor for flywheel energy storagesystemrdquo Physica C vol 494 pp 250ndash254 2013

[6] H Takeuchi C J Goodman and S Sone ldquoPeak demand reduc-tion techniques when starting under Moving Block Signallingrdquo


in Proceedings of the International Conference on Developmentsin Mass Transit Systems pp 280ndash285 April 1998

[7] H Takeuchi and C J Goodman ldquoSimulation study of peakdemand reduction strategies when starting undermoving blocksignallingrdquo in Proceedings of the 5th International Conferenceon Computer Aided Design Manufacture and Operation in theRailway and other Advanced Mass Transit Systems pp 187ndash196August 1996

[8] H Takeuchi C J Goodman and S Sone ldquoMoving block sign-alling dynamics performance measures and re-starting queuedelectric trainsrdquo IEE Proceedings vol 150 no 4 pp 483ndash4922003

[9] T K Ho and K K Wong ldquoPeak power demand reductionunder moving block signalling using an expert systemrdquo IEEProceedings vol 150 no 4 pp 471ndash482 2003

[10] T Albrecht ldquoReducing power peaks and energy consumption inrail transit systems by simultaneous train running time controlrdquoAdvances in Transport vol 15 pp 885ndash894 2004

[11] J-F Chen R-L Lin and Y-C Liu ldquoOptimization of an MRTtrain schedule reducing maximum traction power by usinggenetic algorithmsrdquo IEEE Transactions on Power Systems vol20 no 3 pp 1366ndash1372 2005

[12] K M Kim S-M Oh and M Han ldquoA mathematical approachfor reducing the maximum traction energy the case of KoreanMRT trainsrdquo inProceedings of the InternationalMultiConferenceof Engineers andComputer Scientists (IMECS rsquo10) pp 2169ndash2173March 2010

[13] K K Wong and T K Ho ldquoDynamic coast control of trainmovement with genetic algorithmrdquo International Journal ofSystems Science vol 35 no 13-14 pp 835ndash846 2004

[14] P G Howlett P J Pudney and X Vu ldquoLocal energy minimiza-tion in optimal train controlrdquo Automatica vol 45 no 11 pp2692ndash2698 2009

[15] R Liu and I M Golovitcher ldquoEnergy-efficient operation of railvehiclesrdquo Transportation Research A vol 37 no 10 pp 917ndash9322003

[16] E Khmelnitsky ldquoOn an optimal control problem of trainoperationrdquo IEEE Transactions on Automatic Control vol 45 no7 pp 1257ndash1266 2000

[17] Y Zhang S F Jin and W Y Yue ldquoAn enhanced energysaving strategy for an active DRX in LTE wireless networksrdquoInternational Journal of Innovative Computing Information andControl vol 9 no 11 pp 4373ndash4387 2013

[18] S Ali and D Kim ldquoEnergy conservation and comfort man-agement in building environmentrdquo International Journal ofInnovative Computing Information and Control vol 9 no 6 pp2229ndash2244 2013

[19] Z Wu M Cui P Shi and H Karimi ldquoStability of stochasticnonlinear systems with controlled state-dependent switchingrdquoIEEETransactions onAutomatic Control vol 58 no 8 pp 1904ndash1918 2013

[20] R Wang P Shi Z Wu and Y Sun ldquoStabilization of switcheddelay systems with polytopic uncertainties under asynchronousswitchingrdquo Journal of the Franklin Institute vol 350 no 8 pp2028ndash2043 2013

[21] R Wang Z-G Wu and P Shi ldquoDynamic output feedback con-trol for a class of switched delay systems under asynchronousswitchingrdquo Information Sciences vol 225 pp 72ndash80 2013

[22] X Xu and P J Antsaklis ldquoOptimal control of switched systemsbased on parameterization of the switching instantsrdquo IEEETransactions on Automatic Control vol 49 no 1 pp 2ndash16 2004

[23] S C Bengea and R A DeCarlo ldquoOptimal control of switchingsystemsrdquo Automatica vol 41 no 1 pp 11ndash27 2005

[24] Z Y Yu ldquoTrain operation experimental report in Dalian fasttrackrdquo Tech Rep Beijing Jiaotong University

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


delay (STD) which introduces a starting time delay to eachof the following trains Under this category there are twospecific techniques called single STD and grade STD Thedifference between them is the introduced starting timedelay to each of the following trains which are the same insingle STD technique but different in grade STD (showinga deceasing trend) The other one is called acceleration ratelimit (ARL) which means the acceleration of the followingtrains is limited to a certain extent (or different extents)Under this category there are two specific techniques calledsingle ARL and grade ARL The difference between them isthe limited acceleration rate to each of the following trainswhich are the same in ARL technique but different in gradeARL (showing a deceasing trend) In addition to the abovetechniques there is also a PDR technique called coordinatedPDR It is a combination of the STD and ARL techniqueswith feeding the regenerated power of decelerating trains toaccelerating trains in the same queue by coordinating themovement of queued trains

Takeuchi and his colleagues discussed these techniques in[6ndash8] Simulation results show that the gradedARL techniquehas the best performance in reducing peak power demandamong single STD grade STD single ARL grade ARL andcoordinated PDR techniques In the traditional techniquestime delay is introduced and quality of service is degradedHo and Wong [9] use an expert system to help the operatorsfor decision making and it focuses on the balance betweentime delay and peak power demand In [10] by braking andpowering the trains simultaneously Albrecht proposed a wayto reduce the peak energy consumption and maximize theregenerative energy Chen and his colleagues [11] proposeda method to minimize the peak energy consumption byadjusting the train dwell times at each station in MRT sys-tems Kim et al [12] developed amixed integer programmingmodel to minimize the peak power energy demand thatoccurs when trains are running simultaneously and the basicprinciple of the method is the same with STD All of themethods proposed in [6ndash12] are implemented after the trainsrsquorestarting by adjusting the timetable or driving strategy

Although the existing techniques can reduce peak powerdemand at different degrees they increase the travel timebetween the successive stations and decrease the servicequality And the energy consumption is increased since thereare more traction periods

In this paper in order to reduce peak power demandand energy consumption we first analyze the reasons ofthe formation of the peak power demand and then twonovel approaches are proposed based on the main reasonswith considering the energy efficient driving strategies Oneis for available extra station dwell time named ServiceHeadway Braking (SHB) and the other one is for unavailableextra station dwell time named Extending Stopping DistanceInterval (ESDI) Both of them are real-time adjustmentmethods and can be implemented before the leading trainrsquosrestarting Therefore there is no need to change timetableConsidering energy saving driving strategy could be seen asa kind of driving mode switching process [13ndash16] a nonlin-ear programming model is contributed and the simulationresults show that compared with the best traditional PDR

techniques both of these twomethods could reduce the peakpower demand and energy consumption without any arrivaltime delay increasing And when the available extra stationdwell time is the same as the unavailable extra station dwelltime SHB performs better than ESDI since more peak powerand energy consumption could be reduced

2 Tracking Dynamics and Peak PowerDemand in MBS

21 Tracking Model in MBS Under MBS the tracking targetpoint of the following train moves forward continuously asthe leading train travels The instantaneous distance 119871

119911(119905) of

two successive trains is calculated as

119871119911(119905) = 119878leading (119905) minus 119878following (119905) (1)

where 119878leading(119905) is the position of the leading trainrsquos head and119878following(119905) is the position of the following trainrsquos head

The distance between two successive trainsmust be largerthan the safetymargin at anymoment even if the leading traincomes to a sudden halt so we have

119871119911(119905) ge 119871 safe + 119871 119905 +

119881following(119905)2

2119887 (2)

where 119871119905is the length of the train 119871 safe is the length of

safety margin 119881following(119905) is the instantaneous speed of thefollowing train and 119887 is deceleration rate

Based on (1) and (2) the relation between the leadingtrain and following train should satisfy

119878leading (119905) ge 119871 safe + 119871 119905 + 119878following (119905) +119881following(119905)

2

2119887 (3)

which implies that the instantaneous speed and position ofthe following train should satisfy

119881following (119905)

le radic2 times 119887 times (119878leading (119905) minus 119878following (119905) minus 119871 safe minus 119871 119905)

(4)

119878following (119905)

le 119878leading (119905) minus 119871 safe minus 119871 119905 minus119881following(119905)

2

2119887

(5)

22 Reasons of the Formation of Peak Power Demand Thereason of the formation of the peak power demand isthe restarting of the dense queue and the reasons for theformation of the dense queue are listed as follows

(1) Features of moving block signaling system (Twotrains will start simultaneously if the distance intervalbetween them is 119871 safe + 119871 119905)

(2) Extra dwell time in station


Sm S3 S2

V

S9984002 S

S9984003

Station A


Target speed

S1






1 For





119894


queue

1198781015840



119894


2= 119871119905+119871 safe 1198781minus119878

1015840

3= 2(119871

119905+119871 safe)

1198781minus 1198781015840

4= 3(119871

119905+ 119871 safe) 1198781 minus 119878

1015840

119899= (119899 minus 1)(119871

119905+ 119871 safe) and




119879delay


(7)


119899 le119879delay + 119879dwell




119875119894= 119865119894(V) times V

119894 (9)


119894is the



119875119894minuspeak = max 119875

119894 (10)


119894(119905) is the



119875total =119899

sum119894=1

119875119894minuspeak (11)




119889V (119909)

119889119909=119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V)

V (119909)

119889119905 (119909)

119889119909=1

V (119909)

(12)







119903 (V) = 1198861V2 + 119886

2V + 1198863 (13)

where 1198861 1198862 and 119886





1199031015840

(V) gt 0 11990310158401015840

(V) ge 0 (14)


min 119869 = int119883

0

119906119891119891 (V) 119889119909

st 0 le 119906119891le 1 0 le 119906

119887le 1

V le 119881 (119909)

(15)


119909 (0) = 1199090 119909 (119879) = 119909

119879

V (0) = V0 V (119879) = V

119879

(16)


119867 =1199011

V(119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V))

+1199012

Vminus 119906119891119891 (V)

(17)


1198891199011

119889119909= minus120597119867

120597V (18)


119891and 119906



(i) Full power (119906119891= 1 119906

119887= 0) if (119901

1V) lt 0


119887= 0) if (119901

1V) = 0


119887= 0) if 0 lt

(1199011V) lt 1


119887isin (0 1)) if (119901

1V) = 1

(v) Full braking (119906119891= 0 119906

119887= 1) if (119901

1V) gt 1















1

2


3 i3

i2

i41 1

2223 24

i5

S

Station A

S1Si4

S24

25

(a)

2

S33 S35S31 S32

3 3

32

i533

Station A34

S34

1

S1 S


0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Time (s)

Spee

d (m

s)



1198942for Δ119905




1198944 (V1198944= V1198943) for Δ119905

1198944to 1198781198944 after



1198781198945 V1198945and 119878




35) to arrive at

station A (1198781) we use 119879



21and


31and 11987835







119878leading(0)



(m)


station A(s)

400 16 82 4026400 12 138 3982400 8 178 395400 4 202 3942400 0 210 3938



1198942=





119894119895(119895 =






1198945is close to




5

sum119895=1


5

sum119895=1




min119891 =119899

sum119894=1

119901119894(V1198943) (21)

st V1198942minus V1198943le 0

V1198945minus V1198944le 0

minus V119894119895 minusΔ119879119894119895le 0

V1

2le V1198945lt V1

5

sum119895=1


5

sum119895=1


(22)

where5

sum119895=1

Δ119879119894119895=V1198942minus V1

119887+ Δ1198791198942+V1198943minus V1198942

119886

+ Δ1198791198944+V1198945minus V1198944

1198871015840

5

sum119895=1

Δ119878119894119895+ 119871119887=V21198942minus V21

2 times 119887+V21198943minus V21198942

2 times 119886

+ Δ1198791198944times V1198944+V21198945minus V21198944

2 times 1198871015840+

V21198945

2 times |119887|

(23)


119894119895






1198945 119879dwell








1) =


2


Station A

1

2 2 4 5 6

13

SS2ΔS1 ΔS2





119905119905is nearly




119894(119894 = 1 2 6) as the starting


3= 0 V

4= V5 define the


119894(119894 =


119905119905(MI


119905119905



119872119868 =

6

sum119894=3

Δ119878119894

=V24


2 times 1198871015840+

V26

2 times |119887|

(24)






|119887|) times V1+

V21

2 times |119887| (26)


min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0


(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)










1) is 16ms service


1=






0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)













SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Sm S3 S2

V

S9984002 S

S9984003

Station A


Target speed

S1






1 For





119894


queue

1198781015840



119894


2= 119871119905+119871 safe 1198781minus119878

1015840

3= 2(119871

119905+119871 safe)

1198781minus 1198781015840

4= 3(119871

119905+ 119871 safe) 1198781 minus 119878

1015840

119899= (119899 minus 1)(119871

119905+ 119871 safe) and




119879delay


(7)


119899 le119879delay + 119879dwell




119875119894= 119865119894(V) times V

119894 (9)


119894is the



119875119894minuspeak = max 119875

119894 (10)


119894(119905) is the



119875total =119899

sum119894=1

119875119894minuspeak (11)




119889V (119909)

119889119909=119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V)

V (119909)

119889119905 (119909)

119889119909=1

V (119909)

(12)







119903 (V) = 1198861V2 + 119886

2V + 1198863 (13)

where 1198861 1198862 and 119886





1199031015840

(V) gt 0 11990310158401015840

(V) ge 0 (14)


min 119869 = int119883

0

119906119891119891 (V) 119889119909

st 0 le 119906119891le 1 0 le 119906

119887le 1

V le 119881 (119909)

(15)


119909 (0) = 1199090 119909 (119879) = 119909

119879

V (0) = V0 V (119879) = V

119879

(16)


119867 =1199011

V(119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V))

+1199012

Vminus 119906119891119891 (V)

(17)


1198891199011

119889119909= minus120597119867

120597V (18)


119891and 119906



(i) Full power (119906119891= 1 119906

119887= 0) if (119901

1V) lt 0


119887= 0) if (119901

1V) = 0


119887= 0) if 0 lt

(1199011V) lt 1


119887isin (0 1)) if (119901

1V) = 1

(v) Full braking (119906119891= 0 119906

119887= 1) if (119901

1V) gt 1















1

2


3 i3

i2

i41 1

2223 24

i5

S

Station A

S1Si4

S24

25

(a)

2

S33 S35S31 S32

3 3

32

i533

Station A34

S34

1

S1 S


0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Time (s)

Spee

d (m

s)



1198942for Δ119905




1198944 (V1198944= V1198943) for Δ119905

1198944to 1198781198944 after



1198781198945 V1198945and 119878




35) to arrive at

station A (1198781) we use 119879



21and


31and 11987835







119878leading(0)



(m)


station A(s)

400 16 82 4026400 12 138 3982400 8 178 395400 4 202 3942400 0 210 3938



1198942=





119894119895(119895 =






1198945is close to




5

sum119895=1


5

sum119895=1




min119891 =119899

sum119894=1

119901119894(V1198943) (21)

st V1198942minus V1198943le 0

V1198945minus V1198944le 0

minus V119894119895 minusΔ119879119894119895le 0

V1

2le V1198945lt V1

5

sum119895=1


5

sum119895=1


(22)

where5

sum119895=1

Δ119879119894119895=V1198942minus V1

119887+ Δ1198791198942+V1198943minus V1198942

119886

+ Δ1198791198944+V1198945minus V1198944

1198871015840

5

sum119895=1

Δ119878119894119895+ 119871119887=V21198942minus V21

2 times 119887+V21198943minus V21198942

2 times 119886

+ Δ1198791198944times V1198944+V21198945minus V21198944

2 times 1198871015840+

V21198945

2 times |119887|

(23)


119894119895






1198945 119879dwell








1) =


2


Station A

1

2 2 4 5 6

13

SS2ΔS1 ΔS2





119905119905is nearly




119894(119894 = 1 2 6) as the starting


3= 0 V

4= V5 define the


119894(119894 =


119905119905(MI


119905119905



119872119868 =

6

sum119894=3

Δ119878119894

=V24


2 times 1198871015840+

V26

2 times |119887|

(24)






|119887|) times V1+

V21

2 times |119887| (26)


min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0


(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)










1) is 16ms service


1=






0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)













SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















119903 (V) = 1198861V2 + 119886

2V + 1198863 (13)

where 1198861 1198862 and 119886





1199031015840

(V) gt 0 11990310158401015840

(V) ge 0 (14)


min 119869 = int119883

0

119906119891119891 (V) 119889119909

st 0 le 119906119891le 1 0 le 119906

119887le 1

V le 119881 (119909)

(15)


119909 (0) = 1199090 119909 (119879) = 119909

119879

V (0) = V0 V (119879) = V

119879

(16)


119867 =1199011

V(119906119891119891 (V) minus 119906

119887119887 (V) minus 119903 (V))

+1199012

Vminus 119906119891119891 (V)

(17)


1198891199011

119889119909= minus120597119867

120597V (18)


119891and 119906



(i) Full power (119906119891= 1 119906

119887= 0) if (119901

1V) lt 0


119887= 0) if (119901

1V) = 0


119887= 0) if 0 lt

(1199011V) lt 1


119887isin (0 1)) if (119901

1V) = 1

(v) Full braking (119906119891= 0 119906

119887= 1) if (119901

1V) gt 1















1

2


3 i3

i2

i41 1

2223 24

i5

S

Station A

S1Si4

S24

25

(a)

2

S33 S35S31 S32

3 3

32

i533

Station A34

S34

1

S1 S


0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Time (s)

Spee

d (m

s)



1198942for Δ119905




1198944 (V1198944= V1198943) for Δ119905

1198944to 1198781198944 after



1198781198945 V1198945and 119878




35) to arrive at

station A (1198781) we use 119879



21and


31and 11987835







119878leading(0)



(m)


station A(s)

400 16 82 4026400 12 138 3982400 8 178 395400 4 202 3942400 0 210 3938



1198942=





119894119895(119895 =






1198945is close to




5

sum119895=1


5

sum119895=1




min119891 =119899

sum119894=1

119901119894(V1198943) (21)

st V1198942minus V1198943le 0

V1198945minus V1198944le 0

minus V119894119895 minusΔ119879119894119895le 0

V1

2le V1198945lt V1

5

sum119895=1


5

sum119895=1


(22)

where5

sum119895=1

Δ119879119894119895=V1198942minus V1

119887+ Δ1198791198942+V1198943minus V1198942

119886

+ Δ1198791198944+V1198945minus V1198944

1198871015840

5

sum119895=1

Δ119878119894119895+ 119871119887=V21198942minus V21

2 times 119887+V21198943minus V21198942

2 times 119886

+ Δ1198791198944times V1198944+V21198945minus V21198944

2 times 1198871015840+

V21198945

2 times |119887|

(23)


119894119895






1198945 119879dwell








1) =


2


Station A

1

2 2 4 5 6

13

SS2ΔS1 ΔS2





119905119905is nearly




119894(119894 = 1 2 6) as the starting


3= 0 V

4= V5 define the


119894(119894 =


119905119905(MI


119905119905



119872119868 =

6

sum119894=3

Δ119878119894

=V24


2 times 1198871015840+

V26

2 times |119887|

(24)






|119887|) times V1+

V21

2 times |119887| (26)


min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0


(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)










1) is 16ms service


1=






0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)













SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1

2


3 i3

i2

i41 1

2223 24

i5

S

Station A

S1Si4

S24

25

(a)

2

S33 S35S31 S32

3 3

32

i533

Station A34

S34

1

S1 S


0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Time (s)

Spee

d (m

s)



1198942for Δ119905




1198944 (V1198944= V1198943) for Δ119905

1198944to 1198781198944 after



1198781198945 V1198945and 119878




35) to arrive at

station A (1198781) we use 119879



21and


31and 11987835







119878leading(0)



(m)


station A(s)

400 16 82 4026400 12 138 3982400 8 178 395400 4 202 3942400 0 210 3938



1198942=





119894119895(119895 =






1198945is close to




5

sum119895=1


5

sum119895=1




min119891 =119899

sum119894=1

119901119894(V1198943) (21)

st V1198942minus V1198943le 0

V1198945minus V1198944le 0

minus V119894119895 minusΔ119879119894119895le 0

V1

2le V1198945lt V1

5

sum119895=1


5

sum119895=1


(22)

where5

sum119895=1

Δ119879119894119895=V1198942minus V1

119887+ Δ1198791198942+V1198943minus V1198942

119886

+ Δ1198791198944+V1198945minus V1198944

1198871015840

5

sum119895=1

Δ119878119894119895+ 119871119887=V21198942minus V21

2 times 119887+V21198943minus V21198942

2 times 119886

+ Δ1198791198944times V1198944+V21198945minus V21198944

2 times 1198871015840+

V21198945

2 times |119887|

(23)


119894119895






1198945 119879dwell








1) =


2


Station A

1

2 2 4 5 6

13

SS2ΔS1 ΔS2





119905119905is nearly




119894(119894 = 1 2 6) as the starting


3= 0 V

4= V5 define the


119894(119894 =


119905119905(MI


119905119905



119872119868 =

6

sum119894=3

Δ119878119894

=V24


2 times 1198871015840+

V26

2 times |119887|

(24)






|119887|) times V1+

V21

2 times |119887| (26)


min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0


(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)










1) is 16ms service


1=






0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)













SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











min119891 =119899

sum119894=1

119901119894(V1198943) (21)

st V1198942minus V1198943le 0

V1198945minus V1198944le 0

minus V119894119895 minusΔ119879119894119895le 0

V1

2le V1198945lt V1

5

sum119895=1


5

sum119895=1


(22)

where5

sum119895=1

Δ119879119894119895=V1198942minus V1

119887+ Δ1198791198942+V1198943minus V1198942

119886

+ Δ1198791198944+V1198945minus V1198944

1198871015840

5

sum119895=1

Δ119878119894119895+ 119871119887=V21198942minus V21

2 times 119887+V21198943minus V21198942

2 times 119886

+ Δ1198791198944times V1198944+V21198945minus V21198944

2 times 1198871015840+

V21198945

2 times |119887|

(23)


119894119895






1198945 119879dwell








1) =


2


Station A

1

2 2 4 5 6

13

SS2ΔS1 ΔS2





119905119905is nearly




119894(119894 = 1 2 6) as the starting


3= 0 V

4= V5 define the


119894(119894 =


119905119905(MI


119905119905



119872119868 =

6

sum119894=3

Δ119878119894

=V24


2 times 1198871015840+

V26

2 times |119887|

(24)






|119887|) times V1+

V21

2 times |119887| (26)


min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0


(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)










1) is 16ms service


1=






0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)













SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














|119887|) times V1+

V21

2 times |119887| (26)


min119891 = 119901 (V4)

st V3minus V2le 0

V6minus V5le 0


(119879dwell + Δ1198791) times V1 +V21

2 times |119887|+

6

sum119894=3

Δ119878119894= 1198782minus 1198781

6

sum119894=3

Δ119879119894minus 119879119905119905= 0

(27)










1) is 16ms service


1=






0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



V22= 0 V

23= 957 V

25= 8 Δ119905

24= 0

V32= 13043 V

33= 13043

V35= 1008 Δ119905

34= 0

(28)













SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















SHB 2895 33916 1342 105016 6566 0

250 260 270 280 290 300 310 320 330 3400

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)


0 50 100 150 200 250 300 350 400 4500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3



260 280 300 320 340 360 380 400 4200

5

10

15

20

25

Time (s)

P(t

)m

(kW

t)







V4= V5= 16 V

6= 1592 Δ119879

4= 0 (29)




0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3


Train 2Train 3










SHB 2895 33916 1342 105016 6566 0ESDI 290 340 1818 2286 15398 5796

250 260 270 280 290 300 310 320 330 3400

2

4

6

8

10

12

14

Time (s)

P(t

)m

(kW

t)






5 Conclusion



0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 500 1000 1500 2000 2500 3000 3500 40000

2

4

6

8

10

12

14

16

Distance (m)

Spee

d (m

s)

0 50 100 150 200 250 300 3500

2

4

6

8

10

12

14

16

Time (s)

Spee

d (m

s)

Train 2Train 3




250 260 270 280 290 300 310 320 330 34002468

101214161820

Time (s)

P(t

)m

(Wt)






Acknowledgments


References


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra




































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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