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Stern  
#1 Posted : 26 February 2015 13:44:07(UTC)
Rank: Super forum user
Stern

Hi all, Having one of those days where I just can't get my head round something I know will turn out to be simple! It's to do with the "Average number of persons employed during the year" figure used in the standard AIR calculation. We're a construction company and operate multiple sites up and down the country. Currently this is how I will work out the average number of people employed during our Jan - Dec 12 month reference period but something doesn't seem right...? Firstly we calculate the average number of people that were on each job each day: 1. Total number of "sign ins" to site (eg 5 guys working for 20 days would be 100 "sign ins") 2. Divide this number by the number of days on site to give the average number on site each day. To calculate the average across the whole business: 1. Add up all the answers from step 2 above. 2. Divide by the total number of projects worked in the year to show a company wide average. Is this the best way of doing things or would a better (more accurate) way of doing things be to simply: 1. Count up all the people inducted onto site 2. Divide by the number of jobs worked. Or is there another way which i'm completely missing!? Any help would be greatly appreciated
toe  
#2 Posted : 27 February 2015 00:03:58(UTC)
Rank: Super forum user
toe

IM not sure if this helps or not, but I get this information from our payroll people, they simply know who has been paid which can be calculated into number of employees employed throughout the year.
chris42  
#3 Posted : 27 February 2015 08:59:43(UTC)
Rank: Super forum user
chris42

Depends option two is the simplest, however if you also intend to work out AFR, then option 1 gives you the total number of "sign ins" which would allow you to calc hours worked. Don't option 1 and option 2 give you the same answer ? Chris
Animax01  
#4 Posted : 27 February 2015 14:54:30(UTC)
Rank: Super forum user
Animax01

I would definitely speak with Payroll, this is how I do it as it can take onboard any overtime worked too.
achrn  
#5 Posted : 02 March 2015 08:46:10(UTC)
Rank: Super forum user
achrn

Chris42 wrote:
Don't option 1 and option 2 give you the same answer ?
No. They give different answers, and both are wrong anyway. Suppose two sites. One works for a year with 20 people on site every day. The other works 3 days with two differnet people on site each day. The first process gives you answers of 20 and 2 respectively. Add the answers up = 22, divide by number of sites = 11. You can't possibly have an average number of people employed of 11 if there were more than 20 employed each individual day. Anyhow, the second option proposed doesn't work either. It gives you the average number of different people on each site, which doesn't seem a terribly useful figure. Apart from anything else, if that first site was the same 20 people every day it would give you an answer of 13 (wrong again), but if there was turnover and 60 different people worked on the site in the period, it would give you 33 (also wrong). You need to add up teh sign-ins across all sites and divide by the number of working days in a year (whatever your organisation regards as 'standard'). So if you have a 250 day working year (5 days a week, less bank holidays, roughly): site 1 will have 5000 person-days worked site 2 will have 6 person-days average employment = 5006/250 = 20.02
chris42  
#6 Posted : 02 March 2015 09:50:23(UTC)
Rank: Super forum user
chris42

You are correct achrn The number I worked out I had not taken into account that the say 20 people could be twenty different people on different days (so more inductions), otherwise you do get the same all be it wrong number (11). But in your example would not the last part be 5006/253= 19.8 Chris
achrn  
#7 Posted : 02 March 2015 10:04:22(UTC)
Rank: Super forum user
achrn

Chris42 wrote:
But in your example would not the last part be 5006/253= 19.8
No. I'm not sure how you get the 253 being used in the denominator. You know the answer cannot be less than 20 because in the scenario set out there were at least 20 people at work every (working) day, and the average of a set of numbers cannot be less than the lowest number. Obviously, that's easy to spot in this fake scenario, but the real world will not be so obliging. You could look at each day individually and add them up. In the scenario given there are 247 days with 20 people at work, and 3 days with 22 people at work. (247 * 20 + 3 * 22) / 250 = 20.02. But the actual calculation depends on lots of assumptions (how many days make one of your standard years, how you treat part-days / overtime / paid leave / unpaid leave etc.) There are different ways you could treat all of those things, and you ought to have a standard approach across the organisation for all the metrics you report. So the best advice probably remains to ask payroll to just tell you the answer.
chris42  
#8 Posted : 02 March 2015 10:14:36(UTC)
Rank: Super forum user
chris42

achrn wrote:
Chris42 wrote:
But in your example would not the last part be 5006/253= 19.8
No. I'm not sure how you get the 253 being used in the denominator. You know the answer cannot be less than 20 because in the scenario set out there were at least 20 people at work every (working) day, and the average of a set of numbers cannot be less than the lowest number. .quote] But its an average, so would you not add the 3 days at one site to the 250 days at the other? Agree though wages / accounts /personnel is the place to go. Chris
Stern  
#9 Posted : 03 March 2015 13:34:32(UTC)
Rank: Super forum user
Stern

Hi all, Thanks very much for the responses so far. Just to clarify, there is no issue in obtaining figures of work hours etc, i am just really struggling with how those numbers can then be "crunched" to give me the necessary "Average number employed during 12 month perdiod" number needed for the AIR calculation. I can still see two different ways of doing this which both seem theoretically correct. I we take an example year where a company did three jobs that year Job 1: Duration - 100 days, no of sign ins - 2000, average staff on site each day 20. Job 2: Duration - 50 days, no of sign ins - 200, average staff on site each day 4. Job 3: Duration - 200 days, no of sign ins - 2000, average staff on site each day 10. **Average Staff calculation option 1** Uses number of man days in relation to number of calendar work days in year. Total "sign ins" on all jobs (aka "man days"): 2000 + 2000 + 200 = 4200 Number of "calendar" work days in a year: 250 (guessing!) So.... 4200 man days divided by 250 calendar days = 16.8 on average on site each day. **Average Staff calculation option 2** Takes each job's individual daily average staff number and calculates an overall yearly average. Total of average staff on each job 20+4+10 = 34 Number of jobs in year = 3 So.... 34 (total of averages) divided by 3 jobs = 11.3 on average on site each day This is still driving me insane as both look like legitimate methods but both give different answers. Help!!
achrn  
#10 Posted : 03 March 2015 15:50:37(UTC)
Rank: Super forum user
achrn

Stern wrote:
This is still driving me insane as both look like legitimate methods but both give different answers.
Option 1 is correct. Option 2 is wrong. It isn't the average number employed each day during the year. Use this example: Job 1 - 250 days, 2500 sign-ins (ie, 10 people working every working day) Job 2 - 2 days, 3 sign-ins (ie, almost negligible) Job 3 - 2 days, 2 sign-ins (ie, even smaller) Obviously, the average employed in the year is very slightly over 10, because every single day you had at least 10 people working. Method 1: (2500+3+2)/250 = 10.02 Method 2: (10+1.5+1)/3 = 4.17 Method 2 is obviously wrong. The average of a set of numbers cannot be lower than the lowest number. Look at it on a day-by day basis. To save my keyboard do just four-weeks worth of data: Job 1 - 20 days, 200 sign-ins Job 2 - 2 days, 3 sign-ins Job 3 - 2 days, 2 sign-ins The number employed each day in the period week 1: 10, 10, 10, 10, 10 week 2: 10, 10, 12, 11, 10 week 3: 10, 10, 10, 10, 10 week 4: 10, 10, 11, 11, 10 What is the average? Method 2 says the average of all those 10s, 11s and 12 is 4.17. It's obviously not right that the average of a load of 10s, 11s and 12s is 4, isn't it?
Stern  
#11 Posted : 03 March 2015 16:30:33(UTC)
Rank: Super forum user
Stern

achrn wrote:
Stern wrote:
This is still driving me insane as both look like legitimate methods but both give different answers.
Option 1 is correct. Option 2 is wrong. It isn't the average number employed each day during the year. Use this example: Job 1 - 250 days, 2500 sign-ins (ie, 10 people working every working day) Job 2 - 2 days, 3 sign-ins (ie, almost negligible) Job 3 - 2 days, 2 sign-ins (ie, even smaller) Obviously, the average employed in the year is very slightly over 10, because every single day you had at least 10 people working. Method 1: (2500+3+2)/250 = 10.02 Method 2: (10+1.5+1)/3 = 4.17 Method 2 is obviously wrong. The average of a set of numbers cannot be lower than the lowest number. Look at it on a day-by day basis. To save my keyboard do just four-weeks worth of data: Job 1 - 20 days, 200 sign-ins Job 2 - 2 days, 3 sign-ins Job 3 - 2 days, 2 sign-ins The number employed each day in the period week 1: 10, 10, 10, 10, 10 week 2: 10, 10, 12, 11, 10 week 3: 10, 10, 10, 10, 10 week 4: 10, 10, 11, 11, 10 What is the average? Method 2 says the average of all those 10s, 11s and 12 is 4.17. It's obviously not right that the average of a load of 10s, 11s and 12s is 4, isn't it?
SOLD! Thanks achrn, you're a life saver, that makes complete sense now. I knew it would be something simple but I was just hitting a brick wall with it! Thanks for your help (and everybody else too!) Right, back to sanity...
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