A high-production operation was studied during an 80-hr period. During that time, a total of seven equipment breakdowns occurred for a total lost production time of 3.8 hr, and the operation produced 38 defective products. No setup was performed during the period. The operation cycle consists of a processing time of 2.14 min, a part handling time of 0.65 min, and a tool change is required every 25 parts, which takes 1.50 min. Determine the scrap rate q in % during the period."

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rejkjavik rejkjavik · Answer 1 · 2020-02-14T10:28:41+01:00

Answer:

[tex]scrape rate = \frac{38}{1566} = 2.43 \%[/tex]

Explanation:

Given data;

studying period of production = 80 hr

lost time of production = 3.8 hr

defective product = 38

Processing time = 2.14 min

handling time = 0.65 min

total part = 25 part

cycle time pf unit operation

[tex]T_C = Process\ time + handling\ time + \frac{change\ time\ for\ each part}{total part}[/tex]

[tex]T_c = 2.14 + 0.65 + \frac{1.50}{25} = 2.85[/tex] min

total production hour is calculated as

production hour = 80 - 3.8 = 76.2 hr

NUmber of part produce[tex] = \frac{ 76.2 \times 60}{2.85}[/tex]= 1604 parts

acceeptable parts are = 1604 - 38 = 1566 parts

scrape rate [tex]q = \frac{defective\ product}{acceptable\ product}[/tex]

[tex]scrape rate = \frac{38}{1566} = 2.43 \%[/tex]