HW 5
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Auburn University *
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Industrial Engineering
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Dec 6, 2023
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4-30 The specifications for the thickness of nonferrous washers are 1.0±0.04 mm.
From process data, the distribution of the washer thickness is estimated to be
normal with a mean of 0.98mm and a standard deviation of 0.02mm. The unit
cost of rework is $0.10, and the unit cost of scrap is $0.15. For a daily production
of 10,000 items:
(a) What proportion of the washers is conforming? What is the expected total
daily cost of rework and scrap?
Mean=0.98, standard deviation=0.02, Specification limits are 1.0+/-0.04 mm
Z1 = (0.96-0.98)/0.02 = -1.00
Z2 = (1.04-0.98)/0.02 = 3.00
The proportion of conforming washers = 1-(0.1587+0.0013) = 0.84
Daily cost of scrap = 10000 x 0.1587 x 0.15 = $238.05
Daily cost of rework = 10000 x 0.0013 x 0.10 = $1.30
Daily cost of rework & scrap = 238.05+1.30 = $239.35
(b) In its study of constant improvement, the manufacturer changes the mean
setting of the machine to 1.0mm. If the standard deviation is the same as before,
what is the expected total daily cost of rework and scrap?
Mean=1.0, Standard deviation=0.02
Z1 = (0.96-1.0)/0.02 = -2.00
Z2 = (1.04-1.0)/0.02 = 2.00
Proportion of rework = 0.0228
Proportion of scrap = 0.0228
Total daily cost of scrap & rework = 10000 x 0.0228 x (0.15 x 0.10) = $57
(c) The manufacturer is trying to improve the process and reduces its standard
deviation to 0.015 mm. If the process mean is maintained at 1.0mm, what is the
percent decrease in the expected total daily cost of rework and scrap compared
to that of part (a)?
Mean=1.0, Standard deviation=0.015
Z1 = (0.96-1.0)/0.015 = -2.67
Z2 = (1.04-1.0)/0.015 = 2.67
Proportion of scrap = 0.0038
Proportion of rework = 0.0038.
Total daily cost of scrap & rework = 10000 x 0.0038 x (0.15+0.10) = $9.50
Percentage decrease in total daily cost of rework & scrap compared op part a. =
(239.35-9.50)/239.35 = 0.9603 = 96.03%.
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