If our company has historical scrap of 6%, a. what is the amount of steel needed (it rusts with a shrinkage rate of 1%) for our finished product of 8 cubic feet if waste is known to be 0.75 cubic feet? b. what is our efficiency?
A) Scrap available = 6%
It's rusts with shrinkage rate of 1%
=> Net scrap available= 6×(1 - 1/100) = 5.94%
Total steel manufactured = finished product+ waste
Ts= 8+0.75=8.75 cubic feet out of which 5.94% is supplied by scrap
=> Amount of steel needed=8.75×(1-5.94/100)=8.23 cubic feet
(Answer)
B) Efficiency (n) = output×100/input = OP×100/IP
Output=OP= 8 cubic feet
Input=IP= steel needed+ stored scrap; steel needed=8.23 cubic feet
5.94% scrap=8.23×5.94/100= 0.48886 cubic feet
=> net stored scrap= 0.48886×6/5.94= 0.4938 cubic feet
Input=IP= 8.23+0.4938= 8.7238 cubic feet
Efficiency (n) = 8×100/8.7238 = 91.703%
(Answer)
If our company has historical scrap of 6%, a. what is the amount of steel needed...
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