1. Simulation data collection The evaluation of the efficiency of this building ’ s elevators Ryu Sook Jea Kim Sung Min Lee Yoo Chol Ma Yaowen

2. Data type Collect time period:5:30---6:30 Place: our building Elevator1 (only one for example) and elevator2 Number of people waiting outside at each floor Number of people go out of each floor

3. Some factors 1.elevator1 is stopped only at 1,4,5,6,7 floor, so there is no people waiting at 2,3,4 floor,and no people go out of 2,3,4 floor. 2.Because of the 1st factor, most people at one of 4,5,6,7 floor would go up or down of these four floor (or between the each 2 of these 4 floors) on foot, they do not use elevator, so we ignore the stop during each two floor of elevator in this case. (only from 1 to 5—7 or from 7---5 to1). 3. Usually, two elevators don’t come at the same time, so during one period, there are two different queues for two elevator, that means if some persons come to wait the elevator, they would waiting only one elevator coming, during which time, queue number of the other elevator is 0. 4.Because we collect data during the dinner time, all most all people move between 1st and 5,6,7 floor, so maybe it is more suitable for simulate operation of a elevator( operation of two elevators) during one special time 5.If person waiting to go down find elevator coming with “up moving” to sever other people in the lift to a upper floor, he will be one of the 2 queues according to the first coming time of these two lifts.

4. Collecting data: check two processes : find the number of people enter the elevator and go out at different destination during up and down processes at different time. Number out at floor

5. Distribution of number of go out at 4st floor for up process

6. Distribution Summary Distribution:Poisson Expression:POIS(0.0465) Square Error:0.000006 Chi Square Test Number of intervals= 1 Degrees of freedom = -1 Test Statistic = 9.84e-005 Corresponding p-value< 0.005 Data Summary Number of Data Points= 43 Min Data Value = 0 Max Data Value = 1 Sample Mean = 0.0465 Sample Std Dev = 0.213 Histogram Summary Histogram Range = -0.5 to 1.5 Number of Intervals= 2

7. Distribution of number of go out at 5st floor for up process

8. Distribution Summary Distribution:Lognormal Expression:-0.5 + LOGN(0.644, 0.316) Square Error:0.011530 Chi Square Test Number of intervals= 2 Degrees of freedom = -1 Test Statistic = 1.15 Corresponding p-value< 0.005 Data Summary Number of Data Points= 43 Min Data Value = 0 Max Data Value = 3 Sample Mean = 0.186 Sample Std Dev = 0.627 Histogram Summary Histogram Range = -0.5 to 3.5 Number of Intervals= 4

9. Distribution of number of go out at 6st floor for up process

10. Distribution Summary Distribution:Beta Expression:-0.5 + 7 * BETA(0.419, 2.48) Square Error:0.010587 Chi Square Test Number of intervals= 3 Degrees of freedom = 0 Test Statistic = 3.31 Corresponding p-value< 0.005 Data Summary Number of Data Points= 41 Min Data Value = 0 Max Data Value = 6 Sample Mean = 0.512 Sample Std Dev = 1.25 Histogram Summary Histogram Range = -0.5 to 6.5 Number of Intervals= 7

11. Distribution of number of go out at 7st floor for up process

12. Distribution Summary Distribution:Beta Expression:-0.5 + 7 * BETA(0.48, 3.33) Square Error:0.008462 Chi Square Test Number of intervals= 3 Degrees of freedom = 0 Test Statistic = 2.87 Corresponding p-value< 0.005 Data Summary Number of Data Points= 42 Min Data Value = 0 Max Data Value = 6 Sample Mean = 0.381 Sample Std Dev = 1.06 Histogram Summary Histogram Range = -0.5 to 6.5 Number of Intervals= 7

13. Distribution of number of go out at 1st floor for down process

14. Distribution Summary Distribution:Beta Expression:-0.5 + 8 * BETA(0.47, 1.73) Square Error:0.009244 Chi Square Test Number of intervals= 4 Degrees of freedom = 1 Test Statistic = 1.7 Corresponding p-value= 0.209 Data Summary Number of Data Points= 43 Min Data Value = 0 Max Data Value = 7 Sample Mean = 1.21 Sample Std Dev = 1.83 Histogram Summary Histogram Range = -0.5 to 7.5 Number of Intervals= 8

15. Data collection for queue of person wait to go up at each floor during one period. All most all queue waiting happened at 1st floor.

16. Distribution of 1st floor queue for up process

17. Distribution Summary Distribution:Lognormal Expression:-0.5 + LOGN(1.16, 1) Square Error:0.004199 Chi Square Test Number of intervals= 3 Degrees of freedom = 0 Test Statistic = 0.836 Corresponding p-value< 0.005 Data Summary Number of Data Points= 61 Min Data Value = 0 Max Data Value = 6 Sample Mean = 0.721 Sample Std Dev = 1.25 Histogram Summary Histogram Range = -0.5 to 6.5 Number of Intervals= 7

18. Data collection for queue of person wait to go down at each floor during one period. All most all queue waiting happened at 5,6,7 floor.

19. Distribution of 4th floor queue for down process

20. Distribution Summary Distribution:Erlang Expression:-0.5 + ERLA(0.0707, 8) Square Error:0.000681 Chi Square Test Number of intervals= 1 Degrees of freedom = -2 Test Statistic = 0.0251 Corresponding p-value< 0.005 Data Summary Number of Data Points= 61 Min Data Value = 0 Max Data Value = 2 Sample Mean = 0.0656 Sample Std Dev = 0.309 Histogram Summary Histogram Range = -0.5 to 2.5 Number of Intervals= 3

21. Distribution of 5th floor queue for down process

22. Distribution Summary Distribution:Erlang Expression:-0.5 + ERLA(0.0707, 8) Square Error:0.000681 Chi Square Test Number of intervals= 1 Degrees of freedom = -2 Test Statistic = 0.0251 Corresponding p-value< 0.005 Data Summary Number of Data Points= 61 Min Data Value = 0 Max Data Value = 2 Sample Mean = 0.0656 Sample Std Dev = 0.309 Histogram Summary Histogram Range = -0.5 to 2.5 Number of Intervals= 3

23. Distribution of 6th floor queue for down process

24. Distribution Summary Distribution:Exponential Expression:-0.5 + EXPO(1.07) Square Error:0.025077 Chi Square Test Number of intervals= 3 Degrees of freedom = 1 Test Statistic = 6.83 Corresponding p-value= 0.00923 Data Summary Number of Data Points= 61 Min Data Value = 0 Max Data Value = 4 Sample Mean = 0.574 Sample Std Dev = 0.974 Histogram Summary Histogram Range = -0.5 to 4.5 Number of Intervals= 5

25. Distribution of 7th floor queue for down process

26. Distribution Summary Distribution:Lognormal Expression:-0.5 + LOGN(0.638, 0.286) Square Error:0.003630 Chi Square Test Number of intervals= 2 Degrees of freedom = -1 Test Statistic = 0.851 Corresponding p-value< 0.005 Data Summary Number of Data Points= 61 Min Data Value = 0 Max Data Value = 2 Sample Mean = 0.164 Sample Std Dev = 0.489 Histogram Summary Histogram Range = -0.5 to 2.5 Number of Intervals= 3

27. By collecting the data and make distribution, we evaluate and make sure what kind of data we should use for the distribution in simulation model.

28. model

29. Sub model

30. Further study Add two elevator data and analysis Finish modeling and add data

31. Thank you!