Question

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."

Answer 1

`scrape rate  = (38)/(1566) = 2.43 \%`

Step-by-step 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

`T_C = Process\ time + handling\ time + (change\ time\ for\ each part)/(total part)`

`T_c = 2.14 + 0.65 + (1.50)/(25) = 2.85` min

total production hour is calculated as

production hour = 80 - 3.8 = 76.2 hr

NUmber of part produce
`= ( 76.2 * 60)/(2.85)`= 1604 parts

acceeptable parts are = 1604 - 38 = 1566 parts

scrape rate

`scrape rate  = (38)/(1566) = 2.43 \%`