verify use of TIME Function, Find Quantity Level compare to time-d

[nastech]

(070709) verify use of TIME Function, Find Quantity Level compare to time-day.

If I can get feedback on correct use of the following, thanks.
Hello, have asked this one for awhile, got answer: try WORKDAY()
was / is the wrong answer. Have been trying to find the acceptable value of
a quantity, at any given time of day, compared to a set total quantity.

If the following is correct for a "relative position" equation, in percent:
=(last/from)/(to-from)*100

then trying to find the acceptable volume quantity level, for a percentage
position of the time-of-day (9:30am-4pm / 1600 hrs), might be:

=CT9>(((($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*100)%*CU9)

where CT9 is the Last Quantity, $DC$3 is a time stamp (with: Cntrl+Shift+;
),
and CU9 is the Total Quantity being compared to.

with being told "it cant be done" / Time Function didn't just jump out at
you in help... / otherwise I ask you, is this correct?

 

[Toppers]

I think this is all you need, which returns TRUE/FALSE

=CT9>(($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*CU9

I don't see why you need the 100% if you are simply trying to establish
whether or not the rate of production is above/below that expected in
relation to elapsed time vs total (shift) time.

HTH

 

[nastech]

Thank you much, working on this long time, no responses.. if you have
corrections for the following, else is for others..

What you wrote seems like the right answer, will test if does the same
thing;
- as part of another OR'd "Volume" formula, where todays volume was not yet
factored. (where this example should still work)
=CT9>(($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*CU9

- for a 3 level cond. format volume levels, lo-mid-hi, does the following
sound correct? Measure is against CU9, using fixed / absolute cells:

cond. format good to remote this to 1 absolute cells?: (3 levels this to L5
/ L6 / L7
=>(($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*M5 / M6 / M7

cond. format will say: CT9>$L$5 or $L$6 or $L$7

-----

 

[nastech]

typo, get rid of the > (greater than) sign at front:

=>(($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*M5 / M6 / M7

=(($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*M5 / M6 / M7

 

[nastech]

thanks for your answer, I have found 1 problem with my formula; where max
time is supposed to be 1600 or 4pm, if I download data after 4 pm (but last
volume occurs at 4pm), I get an inflated value because of $DC$3 time stamp
is later than 4pm. not sure if just because of the changes made for the
conditional formatting, don't think so, this is what I am doing:

fixed cell $L$4: =(($DC$3-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*$M$4
and the volume column cond. format top condition (blue):

fixed cell $M$4: merely has minimum level to meet for that condition
cond. format, CT9: CT9>=$L$4

all of this may not have been neccessary to tell, but a time later than 1600
in $DC$3 inflates the minimum level.

QUESTION: Is the best answer to OR(CT9>=$L$4,CT9>=$M$4)

thanks
-----

 

[nastech]

is this correct:

=((IF(NOW()<TIME(16,0,0),$DC$3,TIME(16,0,0))-TIME(9,30,0))/(TIME(16,0,0)-TIME(9,30,0)))*$M$4

 

[nastech]

Found answer, previous is wrong, but included shorter way of doing (what
correction is);

=((IF($DC$3<TIME(16,0,0),$DC$3,"16:00")-"9:30")/("16:00"-"9:30"))*$M$4

 

[Toppers]

... better ?

=((MIN($DC$3,TIME(16,0,0))-"9:30")/("16:00"-"9:30"))*$M$4

 

[nastech]

Thanks.. will use less space.. you believe I've been trying to figure this
out / probably first asked ~2 years ago.... will make sheet accurate.

 

[nastech]

thanks for your help, the following reduction seems to work

=((MIN($DC$3,"16:00")-"9:30")/("16:00"-"9:30"))*$M$5
=((MIN($DC$3,TIME(16,0,0))-"9:30")/("16:00"-"9:30"))*$M$5
=((IF($DC$3<TIME(16,0,0),$DC$3,"16:00")-"9:30")/("16:00"-"9:30"))*$M$4