Get Instant Help From 5000+ Experts For

Writing: Get your essay and assignment written from scratch by PhD expert

Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost

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
 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==700 South Route Slope -0.09167 Intercept 92.5 Va=65 if Qa=<300 Va=91.5- 0.09167Qa if 300==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

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.

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.

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.

Cite This Work

My Assignment Help (2022) CENV6153 Transport Modelling [Online]. Available from: https://myassignmenthelp.com/free-samples/cenv6153-transport-modelling/non-monetary-cost-file-A1D4D68.html
[Accessed 25 February 2024].

My Assignment Help. 'CENV6153 Transport Modelling' (My Assignment Help, 2022) <https://myassignmenthelp.com/free-samples/cenv6153-transport-modelling/non-monetary-cost-file-A1D4D68.html> accessed 25 February 2024.

My Assignment Help. CENV6153 Transport Modelling [Internet]. My Assignment Help. 2022 [cited 25 February 2024]. Available from: https://myassignmenthelp.com/free-samples/cenv6153-transport-modelling/non-monetary-cost-file-A1D4D68.html.

Get instant help from 5000+ experts for

Writing: Get your essay and assignment written from scratch by PhD expert

Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost