SUPPLEMENT TO CHAPTER 4
RELIABILITY
Teaching Notes
The main topics of this chapter are:
1.
Quantifying Reliability
2.
Availability
3.
Improving Reliability
Reliability is a measure of the ability of a product, part, or system to perform its intended function under a
prescribed set of conditions. There are three important aspects of reliability: (1) reliability as a
probability; (2) definition of failure; and (3) prescribed set of operating conditions.
Quantitative methods include the use of probabilities (addition, products, complements) in determining
pointintime reliability and the use of exponential and normal distributions in determining the mean time
between failures.
Students seem to have some difficulty with exponential distribution, especially if they have not had it in
their statistics courses.
The coverage of exponential distribution can be omitted without loss of
continuity.
The normal distribution should be included because it paves the way for later use of inventory
management and quality control sampling theory.
Answers to Discussion and Review Questions
1.
Reliability
is a measure of the ability of a product or service to perform its
intended function under a prescribed set of conditions.
2.
If a product is composed of a large number of parts, it can conceivably have a
low reliability because its reliability is a function of the products of the individual
reliabilities.
For example, if a product has 20 parts, each with a reliability of .99, and all
must operate, the overall product reliability will only be about .99
20
= .818.
3.
Redundancy refers to backup parts or systems built into a product (or service).
Their purpose is to increase reliability by taking over in the event that a primary part or
system fails.
Instructor’s Manual, Chapter 4 Supplement
69
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1.
a.
P(operate) = .9
2
= .81
b.
[.90 + .10(.90)] [.90 + .10(.90)] = .9801
c.
[.90 + .99(.10)(.90)]
2
= .9783
2.
.96 x .96 x .99 x .99 = .9033
3.
X
3
= .92x = .9726
4.
C = (10P)
2
per component
2 (10P)
2
= 173
100P
2
= 86.5
P
2
= .865
P = .93
5.
a.
97 x .97 x .99 = .9315
b.
.9315 + (1 – .9315) x .9315 = .9953
[i.e., P(work) + P(not work) x P(backup works)]
c.
.9315 + [(1 – .9315) x .98 x .9315] = .994
[i.e., P(work) + [P(not work) x P(switch works) x P(backup works)]
6.
a.
.98 x .95 x .94 x .90 = .7876
b.
If 1
st
: [.98 + (1 – .98) x .98] x .95 x .94 x .90 = .8034
If 2
nd
: .98 x [.95 = (1 – .95) x .95] x .94 x .90 = .8270
If 3
rd
: .98 x .95 x [.94 + (1 – .94) x .94] x .90 = .8349
If 4
th
: .98 x .95 x .94 x [.90 + (1 – .90) x .90] = .8664
[i.e., for any case, P(all other work) x P(that one fails) x P(backup works)]
c.
The one with a reliability of .90 since it poses the greatest risk of failure. The system
reliability will then be .86814.
Operations Management, 9/e
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 Spring '11
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