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If our company has historical scrap of 6%, a. what is the amount of steel needed...

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?

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Answer #1

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)

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