"Team Development Two elevator scheduling analysis report"

"Team Development Two elevator scheduling analysis report"

1. Mission Overview:

1.1 Task background: Imagine, Shijiazhuang Railway University basic teaching building elevator configuration is as follows: The building has 18 floors, 4 elevators, many passengers use these elevators daily (passenger weight: Average 70 kg max 120 kg, min 45 kg). Other constant data: elevator speed, open/close time, passenger time to be in/out of the elevator. Can be justified on these assumptions.
Elevator Scheduling design requirements:
(1) Each pair of students will design a set of interfaces and class definitions such an algorithm provider can provide him/her with the "Elevator scheduler" class.
(2) Avoid the "bus" worst-case algorithm. The algorithm takes the elevator as a bus, from the bottom to the top, stopping at each floor, opening the door, letting people in and out, then closing the door and moving on. After reaching the top level, it will go down. The algorithm can meet all the requirements, but it is obviously not the fastest algorithm.
(3) An elevator scheduling tip: When the total weight in the maximum limit of 45 kg, or the number of passengers already in the largest, elevators do not need more external requests to stop.
(4) The elevator scheduler does not know how many passengers are waiting for each layer, and it does not know how many passengers will appear. This is the same as the real-world situation.

1.2 Mission Objective: the basic teaching building since put into use, because of the number of classes, more floors, will appear during recess or rush hour when the elevator congestion is not enough or wait longer phenomenon, so you can consider designing a more reasonable elevator scheduling system, improve the operating procedures of the elevator , so that students can arrive in the classroom on time and minimize waiting time.

1.3 Research object: The average usage of four elevators in different time period.

1.4 Research methods: field visits, observation of the use of four elevators in the basic education, data collection, ask some of the students of the course arrangement (including the course time, number of classes and teacher arrangements, etc.).

1.5 Research results:

Name of Elevator	Stop floor/floor	Weight Limit/kg	Operating speed (m/s)	Peak Period number/person			Opening/Closing Time/s	Passenger access to elevator time/s
				1, 5 before class	Big Recess	4,8 after class
1	8-18 Single Layer	1150	2.6	10	16	9	3	40
2	8-18 Single Layer	1150	2.6	12	16	8	3	40
3	8-18 Double Layer	1150	2.6	11	15	8	3	40
4	8-18 Double Layer	1150	2.6	12	13	7	3	40

3. Application System:

3.1 system Function Description: The system to achieve the overall scheduling of four elevators as the main foundation, in the set of procedures as far as possible to shorten the average elevator transport personnel time. The overall framework is to achieve the following functions: on each floor of the elevator outside the button is pressed up or down, the elevator to accept instructions on the corresponding floor to stop, the internal passengers on which floor, the elevator will be docked.

3.2 Improvement measures: Some of the details of the problem can reduce the total operating time of the elevator to a certain extent.

(1) Each elevator is based on the first floor, so that an elevator can be parked in ( -2,2,4) or at ( -1,3) stop (which can be used for the students of -2,-1,2,3,4 a total of 5 floors), so that elevator second can be parked in (6,8) or at (5,7,9) stop (which can be 5- 9 A total of 5 floors of students use), so that elevator third can be parked in (11,13) or at (10,12,14) stop (which can be used for 10-14 a total of 5 floors of students), so that elevator fourth can be parked in (16,18) or at (15,17) stop (which can be 15- 18 A total of 4 floors of students use) as for the parity of the docking, can be docked according to the parity of the week.

(2) Find the best floor where the elevator stops when no one is riding (assuming there is a total of n floors on the floor, the elevator stops on the X floor, the total number of passengers to the level I is total[i], so that the total number of stairs is ∑{total[i]*| i-x|} (i = 1, 2, ..., N). So to find an integer x to make ∑{total[i]*| i-x|} (i = 1, 2, ..., N) has the lowest value. ）

(3) When the elevator in the elevator in the total weight of the people to reach the limit, only when someone out of the elevator floor stop and no longer in someone on the elevator floor stop;

Developer Team: Li min Liu Zixiang

"Team Development Two elevator scheduling analysis report"