d. Suppose this part is in production and 100,000 of them are needed. If the cos
ID: 2075055 • Letter: D
Question
d. Suppose this part is in production and 100,000 of them are needed. If the cost of aluminum is $0.85 l in how much money would be wasted? Assume your measurements are the nominal size of the part. Tolerances are ±0.005. on oversized parts use extra material. The distributio per dimension. How many extra parts are needed to be made to reach the 100,000 acceptable parts? (show method and calculations) all dimensions. Assume all under tolerance parts are rejected (completely wasted), n of error is normal with a standard deviation of 0.001Explanation / Answer
Tolearnces Actual Dim STDDEV Dim 0.005 -0.005 Mean Differe Critera 0.5803 0.0001527 0.5853 0.5753 0.5801473 Accepted 0.52 0.004 0.525 0.515 0.516 Accepted 1.13333333 0.007488881 1.138333 1.128333 1.125844449 Not Accepted 2.365 0.002291288 2.37 2.36 2.362708712 Accepted 0.252 0.003464102 0.257 0.247 0.248535898 Accepted 0.254433 0.003385154 0.259433 0.249433 0.251047846 Accepted 1.7465 0.005766281 1.7515 1.7415 1.740733719 Not Accepted 2.0256666 0.066455499 2.030667 2.020667 1.959211101 Not Accepted Volume 2.038 in3 Cost of Aluminum 0.85 $/in3 No. Items Rejected is 3 No,.s Density of aluminum 0.1 ib/in3 Weight of one Assemble piece. 0.2038 IB Cost of Assemble Piece 1.7323 Dollar (Volume * Price) If 100000 piece is to be make than total assebmle piece will be 100000/8 12500 Assemble Pieces To Assemble ,3 Non accepted part to be Produce again with in Tolerance So, for 12500 Pieces 12500/3 4167 No.s of each Part to be Made volume Cost Total Price for 4167 No.s Volume of 1st 1.133*0.580*.252*2 0.33119856 inn3 0.281519 in $ 1173.089 Volume of 2nd Part 1.7465*.2525*2*.196 0.17286857 inch3 0.146938 in $ 612.2918 Volume of 3 rd Part .2025*.254*.196*2 0.02016252 inch3 0.017138 in $ 71.41464 Consider Thk 0.196 inch Total Extra Cost will be 1856.795 dollar to Complete 100000 Pieces.
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