Thanks for helping. I have a similar problem and am hoping to follow the work to understand how to solve it.
A company currently using an inspection process in its material receiving the part is trying to install an overall cost reduction program. One possible reduction is the elimination of one inspection position. This position tests material that has a defective content of the average of 0.04. By inspecting all items, the inspector is able to remove all defects. The inspector can inspect 50 units per hour. The hourly rate including fringe benefits for this position is $90.00. If the inspection is eliminated, defects will go into product assembly and will have to be replaced later at a cost of $100.00 each when they are detected in final product testing.
Answer – Defective content of the average D= 0.04
Inspecting rate R = 50units/hour
Cost of Inspector C = $90/hour
Repair cost RC= $100
Cost per hour = D * (R) * RC
= 0.04 * (50) * 100
= $200
If the inspection takes place
Cost per hour = $90
Thus, it is cheaper to inspect in this case. Inspection should take place
= $90/50
= $1.8
Benefit per hour = Cost of no inspection – cost of inspection
= ($200 - $90)
= $110
Benefit per unit = average cost of unit – cost of inspection
= ((0.04)*($100 - $1.8)
= $2.2
Thanks for helping. I have a similar problem and am hoping to follow the work to...
A company currently using an inspection process in its material receiving the part is trying to install an overall cost reduction program. One possible reduction is the elimination of one inspection position. This position tests material that has a defective content of the average of 0.04. By inspecting all items, the inspector is able to remove all defects. The inspector can inspect 50 units per hour. The hourly rate including fringe benefits for this position is $90.00. If the inspection...
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