P: n/a

Access 2000 question
Hi any help or advise required if possible
i am trying to create a formula for a field. here is what i have
currently
rem:
IIf(Time()<[Hours]![StartTime],MinToTime(([Hours]*60)),MinToTime((([Hours]*60))([Hours]![NonProductive]*60)(IIf(Time()>"09:00:00"
And
Time()<="10:27:00",(((Time()[Hours]![StartTime])*24*60)*[Personnel]),IIf(Time()>"10:27:00"
And
Time()<="13:30:00",(((Time()[Hours]![StartTime])*24*6012)*[Personnel]),IIf(Time()>"13:30:00"
And
Time()<="15:27:00",(((Time()[Hours]![StartTime])*24*6042)*[Personnel]),(((((Time()[Hours]![StartTime])*24)*60)54)*[Personnel])))))))
here is it simplified.
i have a table which holds the start/end time for each day, the number
of hours for each day and the number of nonproductive hours for the
day.
Our day starts at 09:00 and finishes at 18:00  Time diff 9 hours.
during this time period there is 1 hours worth of breaks making a
working day of 8 hours per person. for this excise lets say there is
only one person. Now that persons day is made up of 2 hours
nonproductive work and 6 hours of productive work.
now what i am trying to achieve is to spread a 6 hour productive
working day over a 9 hour shift, now the breaks are not a problem, but
incorporating the nonproductive time over the course of the day, is
doing my head in, i cannot just throw it in at the start or end of the
day i need to account for it thruout the day
any help would be appreciated  
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 sm*******@burbidge.co.uk wrote: Access 2000 question
Hi any help or advise required if possible
i am trying to create a formula for a field. here is what i have currently
rem: IIf(Time()<[Hours]![StartTime],MinToTime(([Hours]*60)),MinToTime((([Hours]*60))([Hours]![NonProductive]*60)(IIf(Time()>"09:00:00" And Time()<="10:27:00",(((Time()[Hours]![StartTime])*24*60)*[Personnel]),IIf(Time()>"10:27:00" And Time()<="13:30:00",(((Time()[Hours]![StartTime])*24*6012)*[Personnel]),IIf(Time()>"13:30:00" And Time()<="15:27:00",(((Time()[Hours]![StartTime])*24*6042)*[Personnel]),(((((Time()[Hours]![StartTime])*24)*60)54)*[Personnel])))))))
here is it simplified.
i have a table which holds the start/end time for each day, the number of hours for each day and the number of nonproductive hours for the day.
Our day starts at 09:00 and finishes at 18:00  Time diff 9 hours. during this time period there is 1 hours worth of breaks making a working day of 8 hours per person. for this excise lets say there is only one person. Now that persons day is made up of 2 hours nonproductive work and 6 hours of productive work.
now what i am trying to achieve is to spread a 6 hour productive working day over a 9 hour shift, now the breaks are not a problem, but incorporating the nonproductive time over the course of the day, is doing my head in, i cannot just throw it in at the start or end of the day i need to account for it thruout the day
any help would be appreciated
Perhaps this will help get you started: http://groups.google.com/group/comp....ed54b162a1aa3c
James A. Fortune CD********@FortuneJames.com
Interesting site: http://www.doityourself.com  
P: n/a

Thanks for your help, however my boss finally got it to work, here is
the formula
IIf(Time()<[Hours]![StartTime],MinToTime([Hours]*60[NonProductive]*60),+MinToTime(+[Hours]*60[NonProductive]*60DateDiff("n",[StartTime],Time())*[Personnel]+IIf(Time()>"10:15:00",IIf(Time()<"10:27:00",DateD iff("n","10:15:00",Time()),12),0)*[Personnel]+IIf(Time()>"13:00:00",IIf(Time()<"13:30:00",DateD iff("n","13:00:00",Time()),30),0)*[Personnel]+IIf(Time()>"15:15:00",IIf(Time()<"15:27:00",DateD iff("n","15:15:00",Time()),12),0)*[Personnel]))   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 1301
 replies: 2
 date asked: Feb 14 '06
