HOMEWORK ASSIGNMENT COST OF QUALITY A PROCESS IS STABLE AT A 1.5% NONcONFORMITY
ID: 3204221 • Letter: H
Question
HOMEWORK ASSIGNMENT COST OF QUALITY A PROCESS IS STABLE AT A 1.5% NONcONFORMITY RATE WHERE THE PLANT PRODUCES 200,000 UNITS PER MONTH. THE FINAL INSPECTION CAPTURES 1 OUT OF 10 NONCONFORMITIES WITH ONLY 1 OUT OF EVERY 25 NONcONFORMITIES RETURNED FOR WARRANTY RESPONSE. USING THE INFORMATION CALCULATE THE TOTAL COST OF QUALITY: (1) FINALINSPECTION PER PART COSTS $5IUNIT (2) COST OF REJECTION AT FINAL INSPECTION $50 (3) WARRANTY OF COST OF CUSTOMER RETURN IS 250 PER UNIT (4) COST OF WARRANTY FOR UNRETURNED UNITS IS 100 PER UNIT (5) TESTING EQUIPMENT COST $100,000IYEAR (6) TRAINING COST TO USE TESTING EQUIPMENT IS 20,000 PER YEAR A. CALCULATE THE TOTAL COST 0F QUALITY FOR THIS SITUATION SHOW ALL WORK. B. DIVIDE THE COSTS INTO THE FOUR MAIN CATEGORIES: prevention, appraisal internal failure and external failure SHOW HOW THE COSTS FALL INTO THE VARIOUS CATEGORIES.Explanation / Answer
Given plant production = 2,00,000
non confirmity rate = 1.5%
i.e no of non conforming products produced = 0.015*200000 = 3000 defects / month
Since final inspection captures 1 out of 10 non confirmities, total non confirming items captured for inspection = 300
Since 1 out of every 25 return for warranty respeonse, no of total pieces returned due to this reason = 120
Costs :
a) Inspection all parts = 3000 * $5 = $15000
b) Cost of bad parts (defect parts that are rejected) = 300 * $50 = $ 15000
c) Total warranty of cost of customer return = 120 * $ 250 = 30000
d) total cost of unterned bad = (300-120)* $ 100 = 18000
e) Testing equipment cost = $ 100000 / year = 100000/12 /month
f) Training cost = $ 20,000 /year = 20,000 /12 /month
Total cost = $88,000
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Prevention costs : a), f)
Appraisal costs : e)
Internal failure costs: b)
external failure costs: c), d)
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