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Answer:
Answer 1 
Part A

Generalised cost is the sum of monetary and non-monetary cost of a journey. It is used for the purpose of judgement. Monetary cost includes public transport fare journey, fuel cost, wear and tear and any other parking charge.

Non-Monetary Cost involved time which is spent on undertaking the journey. Time is converted into monetary value figure using time value figure, which varies as per traveller income and trip purpose.

The generalised is equivalent to the goods price in demand and supply theory. It is seen that travellers are willing to spent time on some parts of journey rather than spent on others. The same is categorised into following:

  1. Walk from Origin
  2. Wait for the vehicle
  3. Ride in the vehicle
  4. Walk towards the destination

All travellers “dislike” all time spent in travelling, at the same time they dislike walking and also waiting parts of journey more than as compared to in-vehicle journey time, and are willing to pay for such more. This ultimately leads to higher time value for those journey parts than main in-vehicle journey part.

An alternative approach for the same is to apply weighting to time spent on each different part of journey which quantifies the level of dislike a traveller has for time spent on that bit of journey as compared to time spent on vehicles.  

Part B

The model has 5 iterations

Base Model

 

 

Attraction

 

 

Base

 

H1

H2

S

I

Total P

Target P

Production

H1

192

144

1123

524

1983

2382

H2

138

108

476

1021

1743

1986

S

21

26

79

74

200

232

I

25

24

11

16

76

84

 

Total A

376

302

1689

1635

4002

 

 

Target A

415

465

1940

1864

 

4684

 

Increase

39

163

251

229

 

 

Iteration 1

 

 

Attraction

 

 

 

1.P

 

H1

H2

S

I

Total P

Target P

GF

Production

H1

230.6324

172.9743

1348.959

629.4342

2382

2382

1.201210287

H2

157.2392

123.0568

542.3614

1163.343

1986

1986

1.139414802

S

24.36

30.16

91.64

85.84

232

232

1.16

I

27.63158

26.52632

12.15789

17.68421

84

84

1.105263158

 

Total A

439.8632

352.7174

1995.118

1896.301

4684

 

 

 

Target A

415

465

1940

1864

 

4684

 

 

Increase

24.8632

-112.283

55.11849

32.30091

 

 

 

 

Error

5.99%

-24.15%

2.84%

1.73%

 

 

 

 

 

Too Big

Too Big

Too Big

Too Big

 

 

 

Iteration 2

 

 

Attraction

 

 

 

 

 

2A

 

H1

H2

S

I

Total P

Target P

Error

 

Ratio

Production

H1

211.9149

221.7219

1289.888

597.392

2320.916

2382

2.56%

Too Big

1.026319

H2

152.3138

166.2914

546.7377

1164.002

2029.345

1986

-2.18%

Too Big

0.978641

S

23.17819

40.03311

90.74008

84.36453

238.3159

232

-2.72%

Too Big

0.973498

I

27.59309

36.95364

12.6347

18.24098

95.4224

84

-13.60%

Too Big

0.880296

 

Total A

415

465

1940

1864

4684

 

 

 

 

 

Target A

415

465

1940

1864

 

4684

 

 

 

 

Increase

0

0

0

0

 

 

 

 

 

 

GF

1.103723

1.539735

1.148609

1.140061

 

 

 

 

 

Iteration 3

 

 

Attraction

 

 

 

 

A,P

 

H1

H2

S

I

Total P

Target P

Increase Required

Proportion

Production

H1

217.4922

227.5573

1323.836

613.1147

2382

2382

0

1

H2

149.0605

162.7395

535.0598

1139.14

1986

1986

0

1

S

22.56392

38.97214

88.33526

82.12867

232

232

0

1

I

24.29009

32.53016

11.12228

16.05747

84

84

0

1

 

Total A

413.4068

461.7991

1958.353

1850.441

4684

 

 

 

 

Target A

415

465

1940

1864

 

4684

 

 

 

Increase Required

1.593239

3.200868

-18.3531

13.559

 

 

 

 

 

Error

0.38%

0.69%

-0.95%

0.73%

 

 

 

 

 

Proportion

1.003854

1.006931

0.990628

1.007327

 

 

 

 

Iteration 4 

 

 

Attraction

 

 

 

 

 

A,P

 

H1

H2

S

I

Total P

Target P

Increase Required

Proportion

Error

Production

H1

218.3304

229.1346

1311.429

617.6072

2376.501

2382

5.498567848

1.002314

0.23%

H2

149.635

163.8675

530.0454

1147.487

1991.035

1986

-5.035025552

0.997471

-0.25%

S

22.65088

39.24227

87.50741

82.73047

232.131

232

-0.131028367

0.999436

-0.06%

I

24.38371

32.75564

11.01804

16.17513

84.33251

84

-0.332513929

0.996057

-0.40%

 

Total A

415

465

1940

1864

4684

 

 

 

 

 

Target A

415

465

1940

1864

 

4684

 

 

 

 

Increase Required

0

0

0

0

 

 

 

 

 

 

Error

0.00%

0.00%

0.00%

0.00%

 

 

 

 

 

 

Proportion

1

1.539735

1.148609

1.140061

 

 

 

 

 

Iteration 5

 

 

Attraction

 

 

 

 

 

A,P

 

H1

H2

S

I

Total P

Target P

Increase Required

Proportion

Error

Production

H1

219

230

1314

619

2382

2382

0

1

0.00%

H2

149

163

529

1145

1986

1986

0

1

0.00%

S

23

39

87

83

232

232

0

1

0.00%

I

24

33

11

16

84

84

0

1

0.00%

 

Total A

415

465

1942

1862

4684

 

 

 

 

 

Target A

415

465

1940

1864

 

4684

 

 

 

 

Increase Required

-0.01782

0.035545

-1.60104

1.583323

 

 

 

 

 

 

Error

0.00%

0.01%

-0.08%

0.09%

 

 

 

 

 

 

Proportion

0.999957

1.539735

1.148609

1.140061

 

 

 

 

 

Final Solution 

 

 

Attraction

 

 

A,P

 

H1

H2

S

I

Total P

Target P

Production

H1

219

230

1318

620

2388

2382

H2

149

163

527

1142

1981

1986

S

23

39

87

83

232

232

I

24

32

11

16

84

84

 

Total A

415

465

1943

1861

4684

 

 

Target A

415

465

1940

1864

 

4684

Ans C
Base Data

 

 

Attraction

 

 

 

H1

H2

S

I

Total P

Production

H1

30

4

150

40

224

H2

39

17

51

106

213

S

80

61

119

183

443

I

60

111

335

271

777

 

Total A

209

193

655

600

 

Additive Data

Total

 

 

 

 

 

 

 

 

Attraction

 

 

 

H1

H2

S

I

Total P

Production

H1

249

234

1468

660

2612

H2

188

180

578

1248

2194

S

103

100

206

266

675

I

84

143

346

287

861

 

Total A

624

658

2598

2461

 

Question 2
Ans A

 

 

 

 

e^(-0.2*c)

Proportion flow

Number

Path

O1-D1

18

 

0.027323722

24.64%

42

 

01-02-03-D1

18

Model 1

0.027323722

24.64%

42

 

01-D2-D3-D1

23

Model 1

0.010051836

9.07%

15

 

01-02-03-D1

26

Model 2

0.005516564

4.98%

8

 

01-02-03-D1

24

Model 3

0.008229747

7.42%

13

 

O1-O3-D1

22

 

0.01227734

11.07%

19

 

01-D2-D3-D1

24

Model 2

0.008229747

7.42%

13

 

01-D2-D3-D1

28

Model 3

0.003697864

3.34%

6

 

O1-D3-D1

24

 

0.008229747

7.42%

13

Ans B

 

Distance

 

 

North Route

7

 

 

Flow Vehicle Per hour

Speed

Time Taken

Incremental Time

0

50

8.4

0

100

50

8.4

8.4

200

50

8.4

0

300

42

10.00

1.6

400

34

12.4

2.4

500

26

16.2

3.8

600

18

23.3

7.2

700

10

42.0

18.7

800

10

42.0

0.0

900

10

42.0

0.0

1000

10

42.0

0.0

 

Increment

Qa

Ca

Time Taken

Incremental Time

0

0

50.00

8.4

0

100

0

50.00

8.4

8.4

200

0

50.00

8.4

0

300

100

42

10.00

1.6

400

100

34

12.4

2.4

500

100

26

16.2

3.8

600

100

18

23.3

7.2

700

100

10

42.0

18.7

800

100

10

42.0

0.0

900

100

10

42.0

0.0

1000

100

10

42.0

0.0

Ans C

North Route

 

Slope

-0.08

Intercept

66

Va=50

if Qa=<200

Va=74- 0.08Qa

if 200=<Qa=<700

Va=10

if Qa>=700

South Route

 

Slope

-0.09167

Intercept

92.5

Va=65

if Qa=<300

Va=91.5- 0.09167Qa

if 300=<Qa=<900

Va=10

if Qa>=900

At Equilibrium;

 

Ca=Cb

 

Ta=Tb

 

Ta=7/Va

 

Tb=6/Vb

 

7/Va=6/Vb

 

Vb=6/7Va

 

Assuming that solution points lay on the slope of the speed-flow curves, substitute

 

Va=74- 0.08Qa

 

Vb=91.5- 0.09167Qb

 

91.5-0.09167Qb=6/7*(74-0.08Qa)

 

Qb=0.748Qa-300.78

 

Total flow = 1500;

 

Qb=1500-Qa

 

Qa = 1030.163

1030

Qb=470

470

Alternative Model

North Route

Distance

 

 

 

7

 

 

Flow Vehicle Per hour

Speed

Time Taken

Incremental Time

0

50

8.4

0

100

50

8.4

8.4

200

50

8.4

0

300

42

10

1.6

400

34

12.4

2.4

500

26

16.2

3.8

600

18

23.3

7.2

700

10

42

18.7

800

10

42

0

900

10

42

0

1000

10

42

0

 

Distance

 

 

South Route

6

 

 

Flow Vehicle Per hour

Speed

Time Taken

Incremental Time

0

65

5.538462

0

100

65

5.54

5.5

200

65

5.54

0.0

300

65

5.54

0.0

400

55.83

6.45

0.9

500

46.67

7.71

1.3

600

37.50

9.60

1.9

700

28.33

12.71

3.1

800

19.17

18.78

6.1

900

10

36.00

17.2

1000

10

36.00

0.0

 

Equilibrium

 

 

 

Combination

 

 

 

North

South

Average time

MC

500

1000

29.38

 

600

900

30.93

 

700

800

29.62

 

800

700

28.33

Optimal Combination

900

600

29.04

 

1000

500

30.57

 

Question 3
Ans A 

Cost

Small Electric Car

Large Electric Car

Small Diesel Car

Large Diesel Car

Purchase (P)

25

35

14

23

Charger (c)

2

2

0

0

Annual Maintenance (m)

1

2

1

2

Weekly fuel (f)

0

0

45

55

Weekly Electricity (e)

20

30

0

0

Value

-1.208

-1.848

-2.406

-3.042

Probability

50.28%

26.51%

15.17%

8.03%

Ans B

Electric

1

Small

Step 1

Logit Model

Diesel

Electric

2

Large

Diesel

Step 2:

Consider the lower lever decision (Electric & Diesel)

Vf - Small Electric

-1.208

Vf- Small Diesel

-2.406

Pf - Small Electric

76.82%

Pf- Small Diesel

23.18%

Vf - Large Electric

-1.848

Vf- Large Diesel

-3.042

Pf - Large Electric

76.75%

Pf- large Diesel

23.25%

I Small

-0.944254226

V~small =

0

If lamda is 0.75

VSmall =

-0.70819067

I Large

-1.583325476

V~Large =

0

If lamda is 0.75

VSmall =

-1.187494107

Step 4:

Consider the higher level decision (small or large)

elements common to utility are ignored

V Small =

-0.70819

V large =

-1.18749

P small =

0.617583

Plarge =

0.382417

Step 5

Calculate final probability for each mode

Psmall electric =

47.44%

Psmall diesel =

14.32%

Plarge electric =

29.35%

Plarge diesel =

8.89%

Ans C

Microscopic modelling refers to break of process into small parts and then analyzing in a microscopic view. This modelling aims to reduce the turnaround time through efficient combination of activities which shall reduce air traffic delay or delay in construction of airport. This model helps to create scenario analysis through range of subgroups and selection of ideal combination. For instance, combination of cleaning activity in the plane.

Four stage demand modelling comprise 4 stages namely trip generation, trip distribution, modal split and assignment. The last first phase involves estimation of generalised cost and When assignment has been completed this will produce better estimates which can then be used to revise earlier predictions. Further, only when the whole model converges should results be produced. This model helps in better and efficient designing of the airport through cost minimization.

GIS Modelling helps to better understand the geographic area and helps to give the better understanding of the geography and topography of the area for the purpose of effective deign. It helps to create a 3D design model which helps to plan better. 

Question 4 
Answer A

A principal element which is missing from the 4-stage planning model is the time impact on transport planning. A ride-sharing request of travel can be created by people at any time through mobile app and be matched automatically after a few times. The waiting time of same can range from minutes to hours, it may also affect departure time of people, which is missing from four stage model.

Few literatures have considered departure time into account as a principal element when designing models or any schemes. It has been pointed out by Liu and Li that traffic congestion would change over a period of time and they designed the commuters. It has also been found out by Yin et al. that the effect of reducing of ride-sharing on UPV and traffic congestion was not same during morning and evening time, which also affect people choice of departure time. It is therefore departure time choice is important in transport planning, which ascertain the road condition and also the congestion level faced by travellers, which affect the traffic assignment. 

Therefore, ride sharing effects the transport system in many forms as it becomes important to be considered as a new transport model. 

Answer B

Joint tours are a situation where either or both the activities and trips are same for different people involved.

If all the activities and trips are same between all the people than the same is a case of fully joint tour, but if only some of the trips or involved activities are same between different people involved than it is considered to be a partial joint tour.

Fully joint tour example: Husband and wife going together for a holiday vacation

Partial joint tour example: Husband and wife living together, wife going for shopping and husband goes to office and after they meet up again at the cinema afterwards.

Answer C 

User equilibrium is a method most widely used trip assignment method for auto trips. As per UE condition no driver has the opportunity to reduce the travel time by shifting to another route. It means travel time on each path will be same between any Origin-Destination.

Total system travel time can be found very much minimum under System Optimum condition. In this particular research, UE and SO methods for relevant assignments related to trip are compared. 

The total system travel time can be found minimum in case of system Optimum condition. In this type of research, UE and SO method for trip assignments are made comparable. Firstly, the methodologies are compared, and then link flows and time of travel under both given conditions are make compared. As system Optimum is not a natural process. Arterial highway network of Sioux falls, South Dakota comprised of twenty-four nodes and seventy-six links with thirteen origins and destinations is also analysed for both UE and SO conditions. The analyses of network are done through GAMS optimization software. It is found that travel time is less under SO condition as compared to SE condition.

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[Accessed 25 February 2024].

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