HW #3 (1) (2)
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School
Arizona State University *
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Course
432
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
Pages
3
Uploaded by SuperHumanPower871
1)
Defects: 0.04
Inspect 50 units per hour
Hourly rate including fringe benefits is $9
Inspection cost eliminated then $10 in final testing (repair costs)
a)
No inspection would be 0.04 x 50 x 10 = 20. Hourly rate with inspection is $9. So
no,
they should not eliminate the inspection position because no inspection costs
more than inspection.
b)
Cost to inspect each item = hourly rate / number of inspections per hour
9 / 50 = 0.18
It costs $0.18 to inspect each unit.
c)
Cost of no inspection - cost of inspection
20 - 9 = 11.
There is a benefit of saving $11 per hour with the current inspection
process. They would be spending $20 without inspection vs $9 with.
2a)
Cpk = min[(X − LTL)/3σ,(UTL − X)/3σ]
UTL : 1 + 0.01 = 1.01
LTL : 1 - 0.01 = 0.99
Min [(1.002 - 0.99) / 3(0.003) , (1.01-1.002) / 3(0.003)]
Min (1.33, 0.89)
Cpk = 0.89
2b)
Because Cpk value is 0.89 which is less than 1, the process is
not capable
of
producing high quality output consistently.
7)
a&b)
AQL: 0.03
LTPD: .10
Consumer’s risk: .10
Producer’s risk: .05
LTPD / AQL = .10 / 0.03 = 3.33
According to the in class powerpoint on slide 27 on acceptance sampling, using the
chart. C = 5. nAQL = 2.613 and n = 261.3 → 262.
If the sample size is 262 then the allowable defects would be 5.
9)
AQL = 0.15
LTPD = 0.40
LTPD / AQL = 0.40 / 0.15 = 2.667
According to the same chart used in question 7, C = 8.
nAQL = 4.695 and n = 469.5 → 466
a)
A sampling plan would be to sample 466 LSI circuit chips per lot.
b)
You should randomly sample 466 LSI circuit chips per and and if 8 or more defects are
found in the given lot, then the entire lot needs to be rejected.
13) a
) UTL = 4 + 0.003 = 4.003
LTL = 4 - 0.003 = 3.997
Cpk index : = min [ 4.003 - 4.001 / (3 x 0.002) , 4.001 - 3.997 / ( 3 x 0.002)
Cpk min ( 0.33 , 0.67)
Cpk is 0.33
b) 0.33 is less than the 1.33 acceptable range and doesn’t meet the 6-sigma
requirement of at least 2.0 , so they should not use it.
17)
Cpk = min[(X − LTL)/3σ,(UTL − X)/3σ]
UTL = 100 + 10 = 110
LTL = 100 - 10 = 90
a)
Cpk = min[(100 - 90) / 3(4) , (110 - 100) / 3(4)]
= min (0.83 , 0.83)
Cpk = 0.83
(Because Cpk = 0.83 the process is not capable of producing units that would be
satisfactory)
b)
Cpk = min [(92 - 90) / 3(4) , (110 - 92) / 3(4)]
= min (0.167 , 1.5)
Cpk = 0.1667
(Also would not be capable of producing units that would be satisfactory)
c)
The process after the shift significantly decreases from 0.83 to 0.167 meaning the
efficiency decreases and is even less capable of producing high quality output.
-
Then find Z scores associated with UTL and LTL.
-
90 - 92 = -2 / 4 = -.05 (LTL)
-
110 - 92 = 18 / 4 = 4.5 (UTL)
-
Using excel function to calculate LTL -0.5 is 0.3085409
-
Using excel function to calculate UTL 4.5 is 0.9999966.
-
Measurement of parts above UTL to find the percentage of defective parts
-
USL = 1 - 0.9999966 = 0.0000034
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