A transit line runs on a l = 10 km corridor and has stops at every intersection,
ID: 1709892 • Letter: A
Question
A transit line runs on a l = 10 km corridor and has stops at every intersection, S = 400 m apart. The maximum transit vehicle speed is upsilon_max = 30 km/hr, but for every stop passengers experience an in-vehicle delay of t_s = 20 sec. Note that at their origin and destination stops passengers experience only half the stop delay. All passengers work at offices located at the start of the transit line and they travel to destinations that are uniformly distributed on the corridor. Each passenger gets off at the station closest to their destination, regardless of which direction they need to walk. Passengers walk at speed upsilon_w = 4.8 km/hr. Assume that passengers arrive at the transit stops just on time to board the transit vehicle. (a) What is the door-to-door travel time for the average passenger? (b) How would the door-to-door travel time for the average passenger change if the length of the corridor was such that the average trip length is now 6km? (c) What is the optimal spacing that minimizes the door-to-door travel time for the average passenger from part (a)? What is the minimum door-to-door travel time for the average passenger? (d) What spacing would you choose to ensure an integer number of stops and why?Explanation / Answer
Please find the solution in the attached file
For 6 km transition
optimal condition
Spacing of 500 m
Delay time
passenger delay
Distance from origin
door to door time(in Sec)
Distance from 0rigin for 6 km
time(sec)
Distance
delay time of station
delay time for passenger at each station(sec)
door to door time(sec)
Origin
0
20
10
0
0
0
20
10
1
20
10
400
143
400
143
500
20
10
80
2
20
10
800
211
800
211
1000
20
10
160
3
20
10
1200
279
1200
279
1500
20
10
240
4
20
10
1600
347
1600
347
2000
20
10
320
5
20
10
2000
415
2000
415
2500
20
10
400
6
20
10
2400
483
2400
483
3000
20
10
480
7
20
10
2800
551
2800
551
3500
20
10
560
8
20
10
3200
619
3200
619
4000
20
10
640
9
20
10
3600
687
3600
687
4500
20
10
720
10
20
10
4000
755
4000
755
5000
20
10
800
11
20
10
4400
823
4400
823
5500
20
10
880
12
20
10
4800
891
4800
891
6000
20
10
960
13
20
10
5200
959
5200
959
6500
20
10
1040
14
20
10
5600
1027
5600
1027
7000
20
10
1120
15
20
10
6000
1095
6000
1095
7500
20
10
1200
16
20
10
6400
1163
Average time
619
8000
20
10
1280
17
20
10
6800
1231
8500
20
10
1360
18
20
10
7200
1299
9000
20
10
1440
19
20
10
7600
1367
9500
20
10
1520
20
20
10
8000
1435
10000
20
10
1600
21
20
10
8400
1503
Average time
840
22
20
10
8800
1571
23
20
10
9200
1639
24
20
10
9600
1707
Destination
25
20
10
10000
1775
Average time
959
For origin and last station door to door time is calculated as the sum of walkway time +delay of last station+delay of remaining station+delay of origin station+journey time between station
Let us assume from each stop passenger walkway to their home is 100 m with a speed of 4.8 km/hr which gives walk way time as ((100/4800)*3600)=75sec
Since for optimal condition the walway distance should be kept zero and the transit interval should be kept 500 m so that the passenger can directly reach to their door and this spacing is the best and provide integer in number and there is optimal delay.
For 6 km transition
optimal condition
Spacing of 500 m
Delay time
passenger delay
Distance from origin
door to door time(in Sec)
Distance from 0rigin for 6 km
time(sec)
Distance
delay time of station
delay time for passenger at each station(sec)
door to door time(sec)
Origin
0
20
10
0
0
0
20
10
1
20
10
400
143
400
143
500
20
10
80
2
20
10
800
211
800
211
1000
20
10
160
3
20
10
1200
279
1200
279
1500
20
10
240
4
20
10
1600
347
1600
347
2000
20
10
320
5
20
10
2000
415
2000
415
2500
20
10
400
6
20
10
2400
483
2400
483
3000
20
10
480
7
20
10
2800
551
2800
551
3500
20
10
560
8
20
10
3200
619
3200
619
4000
20
10
640
9
20
10
3600
687
3600
687
4500
20
10
720
10
20
10
4000
755
4000
755
5000
20
10
800
11
20
10
4400
823
4400
823
5500
20
10
880
12
20
10
4800
891
4800
891
6000
20
10
960
13
20
10
5200
959
5200
959
6500
20
10
1040
14
20
10
5600
1027
5600
1027
7000
20
10
1120
15
20
10
6000
1095
6000
1095
7500
20
10
1200
16
20
10
6400
1163
Average time
619
8000
20
10
1280
17
20
10
6800
1231
8500
20
10
1360
18
20
10
7200
1299
9000
20
10
1440
19
20
10
7600
1367
9500
20
10
1520
20
20
10
8000
1435
10000
20
10
1600
21
20
10
8400
1503
Average time
840
22
20
10
8800
1571
23
20
10
9200
1639
24
20
10
9600
1707
Destination
25
20
10
10000
1775
Average time
959
For origin and last station door to door time is calculated as the sum of walkway time +delay of last station+delay of remaining station+delay of origin station+journey time between station
Let us assume from each stop passenger walkway to their home is 100 m with a speed of 4.8 km/hr which gives walk way time as ((100/4800)*3600)=75sec
Since for optimal condition the walway distance should be kept zero and the transit interval should be kept 500 m so that the passenger can directly reach to their door and this spacing is the best and provide integer in number and there is optimal delay.
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