LinkBack URL
About LinkBacks
sccottt 
Not really very clear... Productivity in terms of what ?
Perhaps:
H26: ="00:07"/H24
ie which would equate to 8% of expected given your sample results (ie 100% would be around 65 completed items given time worked)
Not really very clear... Productivity in terms of what ?
Perhaps:
H26: ="00:07"/H24
ie which would equate to 8% of expected given your sample results (ie 100% would be around 65 completed items given time worked)
My Recommended Reading:
Volatility
Sumproduct & Arrays
Pivot Intro
Email from XL - VBA & Outlook VBA
Function Dictionary & Function Translations
Dynamic Named Ranges
thanks for the quick reply. productivity as in items completed in the time they have been logged in given that it should only take 7 mins to complete one item. e.g a person is logged in to the system for 1 hr and completes 1 piece of work so 100% of there time is taken on this one piece but it should only take them 7mins to complete so in 1 hour technically they should have completes roughly 8.5 pieces. hopefully this makes sense
i thought it would be average per case / 7 mins * 100 but i keep getting ridiculous figures. appreciate the help thanks
I'm still curious though as to why the earlier example is not correct... ie what is the result you expect to see ?
If you changed time to 00:35 the formula would return 100% ... ie they completed 5 items which equates to 100% of the items expected.
If you subsequently revert time worked to 07:30 you get 7.78% ... ie they completed 5 items out of the expected 64.28571 items - aka 7.78%
There are currently 1 users browsing this thread. (0 members and 1 guests)
Posting Permissions
Register To Reply
Bookmarks