Manpower Productivity

[reesecc66]

I have recently stepped into a new job and am having a lot of trouble getting my head wrapped around this. It was planned for a contractor to complete a job with a .70 productivity factor. Their shift will be 10.5 hours per day. This includes a :30 minute lunch, and about :45 minutes of safety/line out time at the beginning of each shift. This brings their total available time in the field to 9.25 hours. For them to accomplish a .70 productivity factor how many hours do they need to be working in the field of the available 9.25?

I feel like this is a simple request but for some reason it's not computing in my head...

 

[IRstuff]

You're doing the math backwards; a 0.7 productivity factor means that it takes the contractor 10 hours to do a 7-hr planned task. Basically, the contractor will need 13.2 hrs of actual work time to complete the task

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[reesecc66]

You're right...the expectation is 7 hours of work in a 10 hour shift. So I guess being productive 70% of the time they are on shift. So time being productive would still be 7 hours even if available is only 9.25. Otherwise my calculation would 9.25*.7=6.475. I think the objective is for them to still produce as though they have 10 hours available instead of the 9.25.

 

[hahor]

... is the contractor paid by the hour or does a unit rate contract apply?

 

[reesecc66]

By the hour.

 

[IRstuff]

Well, the "proper" way to look at it is that a "standard" time from something like RS Means will actually take 43% longer, so if RS Means says that it should take 7 hours to install a sink, it'll actually take 10 hours with the 0.7 productivity factor.

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