This document contains 4 practice problems related to reliability and maintenance. Problem 1 calculates the failure rate (FR) of computer chips tested over 500 hours. Problem 2 uses this failure rate to estimate failures per month for 300 chips in a mainframe. Problem 3 calculates the reliability of a system as the product of component reliabilities. Problem 4 determines the expected daily breakdown cost given probabilities of 0, 1, 2, or 3 breakdowns occurring.

1. Practice Problems: Chapter 17, Maintenance and Reliability
Problem 1:
California Instruments, Inc., produces 3,000 computer chips per day. Three hundred are tested for a
period of 500 operating hours each. During the test, six failed: two after 50 hours, two at 100 hours,
one at 300 hours, and one at 400 hours.
Find FR(%) and FR(N).
Problem 2:
If 300 of these chips are used in building a mainframe computer, how many failures of the
computer can be expected per month?
Problem 3:
Find the reliability of this system:
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2. Problem 4:
Given the probabilities below, calculate the expected breakdown cost.
Number of Breakdowns
Daily Frequency
0
3
1
2
2
2
3
3
Assume a cost of $10 per breakdown.
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3. ANSWERS
Problem 1:
FR(%) = failures per number tested = 6/300 = 0.02 = 2%
FR(N) = failures per operating time:
Total time = 300 * 500 = 150,000 hours
Downtime = 2(450) + 2(400) + 1(200) + 1(100) = 2,000 hours
Operating time = Total time – Downtime = 150,000 – 2,000 = 148,000
Therefore: FR(N) = 6/148,000 = 0.0000405 failures/hour
MTBF = 1/FR(N) = 24,691 hours
Problem 2:
Converting the units of FR(N) to months:
FR(N) = 0.0000405 * 24 hours/day * 30 days/month = 0.029 failures/month
FR(N) for the 300 units:
FR(N) = 0.029 failures/month * 300 units = 8.75 failures/month
MTBF for the mainframe:
MTBF = 1/FR(N) = 1/8.75 = 0.11 month = 0.11 * 30 = 3.4 days
Calculation for MTBF assumes that failure of any one chip brings down entire system.
Problem 3:
R = [0.95 + 0.92(1 − 0.95)] * [0.98] * [0.90 + 0.90(1 − 0.90)]
= 0.996 * 0.98 * 0.99 = 96.6%
Problem 4:
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4. Number of Breakdowns
Daily Frequency
Probability
0
3
0.3
1
2
0.2
2
2
0.2
3
3
0.3
Expected number of breakdowns = (0)(0.3) + (1)(0.2) + (2)(0.2) + (3)(0.3)
= 0 + 0.2 + 0.4 + 0.9
= 1.5 breakdowns/day
Expected breakdown cost = Expected number of breakdowns * Cost per breakdown
= 1.5 * $10
= $15/day
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