Chapter 04S – Reliability
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CHAPTER 04S
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 point-in-time reliability and the use of exponential and normal distributions in
determining the mean time between failures.
Students seem to have some difficulty with the exponential distribution, especially if they have not had
it in their statistics courses. The coverage of the exponential distribution can be omitted without loss of
continuity. The normal distribution should be included because it paves the way for later use in
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 conceivably can have a low reliability
3. Redundancy refers to backup parts or systems built into a product (or service). Their purpose
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Solutions
1. a. Probability that both components will function, i.e., the system will function (Use Rule 1).
= .9 x .9 = .81
b. Probability that Backup for Component 1 & Switch both will function (Use Rule 1):
Probability that Component 2 or (Backup for Component 2 & Switch) will function
(Use Rule 2):
c. Probability that Backup for Component 1 & Switch both will function (Use Rule 1):
= .90 x .99 = .891
Probability that Backup for Component 2 & Switch both will function (Use Rule 1):
= .90 x .99 = .891
Probability that Component 1 or (Backup for Component 1 & Switch) will function (Use
= .9891 x .9891 = .9783
.9
.9
.9
.9
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2. Probability that all four parts will function, i.e., the product will function (Use Rule 1):
= .96 x .96 x .99 x .99 = .9033
3. Given:
Probability of all three components must be equal.
Probability that all three components will function, i.e., the system will function =.92
4. Given: C = (10P) 2 = cost per component
Cost for 2 Components = 2 x (10P) 2 = 173
2 x 100P2 = 173
5. a. Reliability of system = Probability that all three components will function, i.e., the New
Computer will function (Use Rule 1):
= .97 x .97 x .99 = .9315
b. Given: Probability that Backup Computer will function (same as probability of New
Computer) = .9315
Reliability of system = Probability that New Computer will function or Backup Computer
will function (Use Rule 2):
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c. One backup component will be added. The backup will have a reliability of .92.
Probability that Component 1 or Backup will function (Use Rule 2):
= .98 + [(1 – .98) x .92] = .98 + .0184 = .9984
Reliability of robot = Probability that all four major components will function (Use Rule
Probability that Component 3 or Backup will function (Use Rule 2):
= .94 + [(1 – .94) x .92] = .94 + .0552 = .9952
Reliability of robot = Probability that all four major components will function (Use Rule
1):
= .98 x .95 x .9952 x .90 = .8339
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7. a. Line 1 reliability = Probability that all three machines will function (Use Rule 1):
= .99 x .96 x .93 = .8839
Plan 1: Identical backup line (Line 2)
Given: Reliability of Line 2 = .8839
Reliability of system = Probability that Line 1 or Line 2 will function (Use Rule 2):
= .8839 + [(1 – .8839) x .8839] = .8839 + .1026 = .9865
Plan 2: Backups for each machine. Note: In this scenario, if one machine goes down, we
use the Backup for the machine.
b. In Plan 1, the system will fail if any one original + any one backup fail.
In Plan 2, the system will fail only if a component + its backup fail.
c. Other factors that might enter this decision of which plan to select include space for a line
.99
.96
.93
.99
.96
.93
.99
.96
.93
.99
.96
.93
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8. a. Refer back to Problem 7 – Part a – Plan 1.
Given:
Original Plan 1 Reliability = .9865
Revised Plan 1 (Switch now needed for Line 2)
Probability that Line 1 will function = .8839
Reliability of Switch = .98
b. Refer back to Problem 7 – Part a – Plan 2.
Given:
Original Plan 2 Reliability = .9934
Revised Plan 2 (Switch now needed for each Backup machine)
Given:
Probability that Machine A will function = .99
Probability that Backup for Machine A will function = .99
Reliability of Switch = .98
Chapter 04S – Reliability
Probability that Backup for Machine B & Switch will function (Use Rule 1):
Probability that Machine C will function = .93
Probability that Backup for Machine C will function = .93
Reliability of Switch = .98
Probability that Backup for Machine C & Switch will function (Use Rule 1):
= .93 x .98 = .9114
Probability that Machine C or (Backup for Machine C & Switch) will function (Use Rule
9. Given:
x = reliability for each of five major components
10. Given:
Repeat Problem 9, but now one component will have Backup with reliability = x
System reliability = .98
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11. Not completed in time means no team completes in time:
12. a. Given: MTBF = 30 months
P(no failure before T) = e–T/MTBF
P (no failure before 39):
T/MTBF = 39/30 = 1.3
Using Table 4S.1:
P (no failure before 60):
T/MTBF = 60/30 = 2.0
Using Table 4S.1:
e-2.0 = .1353
b. Given: MTBF = 30 month
P(failure before 33) = 1 – .3329 = .6671
P(failure before 15):
T/MTBF = 15/30 = 0.5
Using Table 4S.1:
e-0.5 = .6065
P(failure before 6) = 1 – .8187 = .1813
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c. Given: MTBF = 30 months
Probability (no failure before T) = e–T/MTBF
Substituting:
T/30 = .70
T = .70 x 30 = 21 months
(2) Given: P(no failure before T) = .15 (1.00 – .85)
.15 = e–T/MTBF
T = 1.90 x 30 = 57 months
(3) Given: P(no failure before T) = .05 (1.00 – .95)
.05 = e–T/MTBF
Re-arranging: