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# Vacation Accrual

Hi, I'm moving away from a job where I was a contractor. I worked and got paid 110 hours every 4 weeks. I also, over the years, always got 4 weeks off, so I got 110 hours off every year.

I've will have worked 33 pay periods since Jan 1, 2016, when I leave at the end of the week. How can I calculate how much vacation time I've accrued?

I'm not sure if anyone here can answer, as it's basically an accounting question, but maybe, so here it is. Thanks.
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mel200
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1 Solution

Assuming your vacation year is Jan 1 to Dec 31, you get

(33 / number of pay period in a year)  x 4 weeks less any vacation you took this year. That should suffice nicely and how I do it for clients.
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Author Commented:
Thanks, John, So that is 33 divided by 13 (52 weeks divided by 4) = 2.5x4=10. But I'm not sure what that 10 represents.
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Excel VBA DeveloperCommented:
^John's formula is correct.

Since you stated you're on pay period 33, that agrees with a weekly pay cycle (August 7-13 is week 32 or 33, depending on the fiscal calendar).

No points, but that formula calculates to 69.81 hours total, less any vacation you've taken this year.
[(33/52)*110] = 69.81 (then subtract vacation used from that total).
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It would be 33 / 52 x 4  = 2.54 weeks. (you have an extra 4)   2.54 weeks  / 4 allowed x 110 hours = about 70 hours.
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Author Commented:
Thanks, all!!
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You are very welcome and I was happy to help
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Author Commented:
Hi- wait, one thing. The reason I had the extra four is this. You had 33x52 weeks in a year, But my pay period is for 4 weeks.

It does correspond to a weekly pay period, yes, but every 4 weeks, not every week. It seems off to me. I've worked 7.5 out of 12 months, so 62%  of my work year. 62% of 110 hours is 68.2 hours, I think I might use that.
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You can do that. I said roughly 70 hours and you got 68 (I just approximate) so that is good.
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Author Commented:
Thanks, John. I'm ok at designing websites, but arithmetic has always made my head hurt. :)
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Yes, I understand. I do accounting consulting so my head is always full of numbers.
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