Module 8 : Specialized Traffic Studies
Lecture 41 : Parking Studies
1 2 3 4 5 6 7
 

License plate method of survey

This results in the most accurate and realistic data. In this case of survey, every parking stall is monitored at a continuous interval of 15 minutes or so and the license plate number is noted down. This will give the data regarding the duration for which a particular vehicle was using the parking bay. This will help in calculating the fare because fare is estimated based on the duration for which the vehicle was parked. If the time interval is shorter, then there are less chances of missing short-term parkers. But this method is very labor intensive.

Numerical Example

The parking survey data collected from a parking lot by license plate method is s shown in the table 1 below. Find the average occupancy, average turn over, parking load, parking capacity and efficiency of the parking lot.
Table 1: Licence plate parking survey data
Bay Time
  0-15 15-30 30-45 45-60
1 1456 9813 - 5678
2 1945 1945 1945 1945
3 3473 5463 5463 5463
4 3741 3741 9758 4825
5 1884 1884 - 7594
6 - 7357 - 7893
7 - 4895 4895 4895
8 8932 8932 8932 -
9 7653 7653 8998 4821
10 7321 - 2789 2789
11 1213 1213 3212 4778
12 5678 6678 7778 8888

Solution

See the following table for solution 2.
Table 2: Licence plate parking survey solution
Bay Time Time  
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
  15 30 45 60 15 30 45 60 Turn over
1 1456 9813 - 5678 1 1 0 1 3
2 1945 1945 1945 1945 1 1 1 1 1
3 3473 5463 5463 5463 1 1 1 1 2
4 3741 3741 9758 4825 1 1 1 1 3
5 1884 1884 - 7594 1 1 0 1 2
6 - 7357 - 7893 0 1 0 1 2
7 - 4895 4895 4895 0 1 1 1 1
8 8932 8932 8932 - 1 1 1 0 1
9 7653 7653 8998 4821 1 1 1 1 3
10 7321 - 2789 2789 1 0 1 1 2
11 1213 1213 3212 4778 1 1 1 1 3
12 5678 6678 7778 8888 1 1 1 1 4
  Accumulation       10 11 9 11  
  Occupancy       0.83 0.92 0.75 0.92 2.25

Columns 1 to 5 is the input data. The parking status in every bay is coded first. If a vehicle occupies that bay for that time interval, then it has a code 1. This is shown in columns 6, 7, 8 and 9 of the table corresponding to the time intervals 15, 30, 45 and 60 seconds.
  • Turn over is computed as the number of vehicles present in that bay for that particular hour. For the first bay, it is counted as 3. Similarly, for the second bay, one vehicle is present throughout that hour and hence turnout is 1 itself. This is being tabulated in column 10 of the table. Average turn over = $ \frac{\mathrm{Sum~of~ turn-over}}{{\mathrm{Total~ number~ of~ bays}}}$ = 2.25
  • Accumulation for a time interval is the total of number of vehicles in the bays 1 to 12 for that time interval. Accumulation for first time interval of 15 minutes = 1+1+1+1+1+0+0+1+1+1+1+1 = 10
  • Parking volume = Sum of the turn over in all the bays = 27 vehicles
  • Average duration is the average time for which the parking lot was used by the vehicles. It can be calculated as sum of the accumulation for each time interval $ \times$ time interval divided by the parking volume = $ \frac{(10+11+9+11)\times 15}{27}$ = 22.78 minutes/vehicle.
  • Occupancy for that time interval is accumulation in that particular interval divided by total number of bays. For first time interval of 15 minutes, occupancy = (10$ \times$100)/12 = 83% Average occupancy is found out as the average of total number of vehicles occupying the bay for each time interval. It is expressed in percentage. Average occupancy = $ \frac{0.83+0.92+0.75+0.92}{4} \times 100$ = 85.42%.
  • Parking capacity = number of bays $ \times$ number of hours = 12$ \times$ 1 = 12 vehicle hours
  • Parking load = total number of vehicles accumulated at the end of each time interval $ \times$ time = $ \frac{(10+11+9+11)\times 15}{60}$ = 10.25 vehicle hours
  • Efficiency = $ \frac{\mathrm{Parking~load}}{\mathrm{Total~number~of~bays}}$ = $ \frac{10.25}{12}\time 100$ = 85.42%.