working hours per year

seabubble

New Member
Hi,

I am a bit confused over the following that I am trying to work out.

If there is 1 user and he works 10 hours a day and 250 days a year (available working hours) his hours are 10 * 25,000 hours, which it shows in the cell.

When formating the cell using custom [H]:MM:SS it returns 60,000 hours

I am trying to calculate how many hours are affected and show it as hours, as I have to add up rows of time.

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=10*250/24

Format as [h]:mm:ss

Keep in mind time values are stored as fractions of a day, and that there are 24 hours in a single day.

Thanks Mark. For a sanity check does this look right?

down
18/01/2005 16:05
up
18/01/2005 16:06
user time down
79:51:40
users
46

Cheers
[/img]

seabubble said:
Thanks Mark. For a sanity check does this look right?

down
18/01/2005 16:05
up
18/01/2005 16:06
user time down
79:51:40
users
46

Cheers
[/img]

I'm afraid I don't follow. It appears that the downtime was 1 min and affected 46 users. So, I don't follow where 79 hrs 51 min 40 sec comes from. Should it just be 46 min?

If I do one user I get
18/01/2005 16:05 18/01/2005 16:06 0:01:00 0:01:00 1:44:10 1

sorry
down
18/01/2005 16:05
up
18/01/2005 16:06
user time down
1:44:10
users
1

Sorry, without cell references and the formula that you're using I'm lost.

all the formulas do is

10*250/24 that gives me the available working hours in a year 2500 that is formated into [H]:mm:ss

I then times this by users then times by downtime

seabubble said:
all the formulas do is

10*250/24 that gives me the available working hours in a year 2500 that is formated into [H]:mm:ss

I then times this by users then times by downtime

Remember =10*250/24 (formatted as [h]:mm:ss) gives you the downtime as a fraction of a day. If you want to multiply downtime hours by the number of affected users then you'd need to use hours (not a fractional days). Use...

=10*250

Don't confuse a cell's formatting with the actual value stored within.

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