October 24, 2007 at 7:25 am
Guys hi,
i have a logical problem and i need your help in solving it.. Any suggestions are welcome.
Imagine i have a door which operates from 12:30 to 15:30. After that time access to the door, is unavailable.
Each person that enters or exits the door, does this by a magnetic card.
Each person can enter or exit the door as many times as he likes since he is in the designated hours (after or before this timespan, door is not operating).
my problem is that not all persons follow the pattern, entry - exit. This is because if 2 persons go at the door at the same time the first will use his magnetic keycard, but the second will hold the door and will not use his magnetic keycard. The same goes for the exit. If the same people (or any two people) exit the room at the same time, the first person will use his magnetic key while the second may or may not use his magnetic ey since the door will be open from the first guy.
This results in having a table that many people have only one access and this may be just an exit, or people may have 2,3,4 accesses without following the patern of [1 entry - 1 exit]. They may have 2 consequtive exits, and then just an entry (without exit) or any combination of entries and/or exits you can imagine.
I want to find the total time spent by each guy the room. Have you got any ideas? If i do this by "case" becomes a lot complicated.
You help would be valuable to me...
October 24, 2007 at 7:43 am
I must be in need of more caffeine...
My first reaction was "have a cattle prod installed at the door - when they scan in two entries or exits - ZAP them...". But I suppose that doesn't help with the counting process :hehe:
In the absence of good data - about the only thing you can do is assume something, and because of that, you're going to have a margin of error. How many of those "double scans" were them entering/exiting at the same time as someone else vs scanning and NOT exiting/entering (for example - scan to enter while in the hall, and before you enter, your boss comes by and asks you a question, so you skip entering until the question is answered).
For that matter - how would you know if ANY of them are good? If someone enters legitimately, then leaves with someone, and later walks back in with someone and finally leaves "legitimately" - you got one span when you should have 2.
I'd have to say - your best bet is to find some way to prevent this from a process standpoint. That of course usually entails some form of "nasty consequence" being leveled for non-compliance.
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Your lack of planning does not constitute an emergency on my part...unless you're my manager...or a director and above...or a really loud-spoken end-user..All right - what was my emergency again?
October 24, 2007 at 7:51 am
October 24, 2007 at 8:28 am
Hi,
May I suggest the following practical solution:
1. Educate the staff as to the importance of the first scan and the last scan for the day ie. reporting to work and leaving from work.
2. Note for each person for each day his first and last scan. Calculate from this as the Total Time Presumed to be at work.
3. Your table should add a record upon entry. Upon exit, update that entry. Write an intelligent routine to modify these values and flag each with either: Accurate, Best, Worst. This is from the point of view of the Manager, not teh employee.
4. Report the Presumed Time at work, the best time at work, the worst time at work.
5. The people with the consistent 'bad' figures will show.
6. Use with care.
I hope that helps. Sorry I am in a hurry.
Osama
October 24, 2007 at 8:38 am
Dfalir (10/24/2007)
Come on mate, I am sure you can think something smarter than that. Suggest something serious! lol 🙂
Garbage in, garbage out. If you don't know when or whether someone entered the building or left it, how can you hope to calculate how much time, if any, they spent in it?
John
October 24, 2007 at 8:52 am
You all are correct guys. What you are saying it makes perfect sense, but i though i should give it a try in order to find the least possible error. Anyway i will take your advice and do the following. If there is more than one anomaly in the entry-exit accesses i will charge employees as beeing there the full time span of 3 hours. Next time they will learn to use their cards.
... Lets fire the suckers! lol 🙂
Now seriously, thank you all. I communicated your support to the management. 🙂 And it proved valuable.
Sincere thanks to you all, for your time and effort.
October 24, 2007 at 8:54 am
The way ours works is if you don't scan in you cannot scan out and vice versa.All cards scanned in but not out are reset at the end of the day with a blanket out code at closing time. People tend to learn to scan in so they don't have to wait for someone else to leave so they can leave.
Anyone with a blanket out code gets the cattle prod:w00t:
October 24, 2007 at 9:19 am
I think it would be safe to assume if someone entered once, the exited once, unless they spent the night. Why couldn't there be an update script to set all the null exits to 1 where enter is not null, and vice versa. Then you would still be left guessing how many people gained access without scanning in or out. Good luck.
Greg
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The glass is at one half capacity: nothing more, nothing less.
October 25, 2007 at 12:35 am
October 25, 2007 at 12:46 am
I've been in this situation a few months ago, whereby i need to calculate the time difference between the entry and exit times....:hehe:
I resolved the problem by just adding column to hold the type of the card reader (such as entry reader, exit reader and both) in my master table....:)
--Ramesh
October 25, 2007 at 1:23 am
It's not just a problem of database, program, script,... This is organizational problem. The best you can do is to analyze data and produce report by departments with all irregular entries/exits. Send this to the departments's chiefs and give them a chance to decide what to do whit that people/data. Some additional flag in table (updateable by responsible person) can also be helpful. When some of them doesn't get a salary your data(their behavior) will become perfect.
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