... better ?

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

"nastech" wrote:

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" wrote:

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" wrote:

(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?