Question

A manufacturer of disk drives for notebook computers wants a MTBF of at least 50,000 hours. Recent test results for 15 units were one failure at 12,000 hours, one at 24,000 ?hours, and one more at 48,000 hours. The remaining units were still running at 60,000 hours. ?

a) Percent of failures? = __%?(enter your response as a percentage rounded to one decimal? place).

b) Number of failuers per unit-hour? __

?c) The mean time between failures? (MTBF) is the expected time between a repair and the next failure of a? component, machine,? process, or product. MTBF is the reciprocal of? FR(N):

Answer 1

Given,

Total products= 20

No of product failures = 5

Product failure rate (FR)= (5/20) *100 = 25%

MTBF = mean time between failures = 1/FR = 1/0.25 = 4

a) Percentage of failure = (no of disk drive failed / no of disk drives tested) * 100

            = ((1+1+1) / 15) *100 = (3/15) * 100 = 100/5 = 20.0 %

b) Total Time = 60,000 * 15 = 900,000

Non operating time = (60,000 – 12,000) for 1^st failure + (60,000-24,000) for second failure + (60,000-48,000) for 3^rd failure = 48000+36000+12000= 96,000

Operating time = 900,000 – 96,000 = 804,000

No of failure rate per unit hour (FR(N)) = 3/804000 = 1/268000

c) MTBF = 1/ FR(N) = 268,000