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4 items are filed under this topic.
How much money do I make per sq ft? 2017-03-28
From jackie:
If I make $12.00 per hour and I cover 750 sq ft in one hour, how much money do I make per sq ft?
Answered by Penny Nom.
How much labor should the firm employ? 2006-10-28
From Christy:
A dressmaking firm has a production function of Q=L-L(squared)/800. Q is the number of dresses per week and L is the number of labor hours per week. Additional cost of hiring an extra hour of labor is $20. The fixed selling price is P=$40. How much labor should the firm employ? What is the resulting output and profit? I am having a difficult time with this, HELP!
Answered by Stephen La Rocque.
What is the total cost of a driveway ... 2006-08-29
From Ismael:
What is the total cost of a driveway 15' wide, 40' long and 4" thick if the concrete costs $60.00 per cubic yard and the labor costs $1.25 per square foot?
Answered by Stephen La Rocque.
Labour efficiency 2005-08-23
From Rob:

The problem, on the surface, seems very simple and yet has created some controversy among a group of accountants. The problem itself has to do with labour efficiency rates and only involves two variables; standard working hours, and actual working hours. The difficulty lies in deriving an efficiency % from these two numbers.

Standard working hours or the targeted number of labour hours required to produce one widget, which I will represent as "s". Actual working hour or the actual number of labour hours require to produce one widget, which I will represent as "a". Labour efficiency I will represent with "E". The prevailing calculation with which I have a problem with is this:

s/a=E or if s=3000, and a=4000 then 3000/4000=75%

What bothers me about the calculation is that the standard hours get represented as a percentage of the actual hours and in my opinion changes the focus of the calculation from standard or target, where it should be, to the actual hours. I cannot define why, but this just seems inherently wrong to me.
The calculation that I use:

(1+((s-a)/s))=E or if s=3000, and a=4000 then (1+((3000-4000)/3000))=66.67%

My calculation is like a %change from standard calculation. However, there is something that also concerns me about my calculation.

If you substitute 100 for a and 50 for s, then you come to a quandary, because if you plug those numbers into the second equation the result is of course zero % efficient which doesn't sit right with me either. If you plug them into the first calculation you get 50% efficiency which doesn't really seem to work either, because you require 100% more hours to do the same work in this case. ???

Is the first calculation correct? Am I missing something altogether? Are both calculations off base?

Answered by Harley Weston.



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