SUPPLEMENT TO CHAPTER 4
RELIABILITY
Teaching Notes
1.
2.
3.
The Main topics of this chapter are
Quantifying Reliability
Role of Redundancy
Availability
Reliability is a measure of the ability of a product, part, or system to perform its intended function under a
prescribed set of conditions.
Quantitative methods include the use of probabilities (addition, multiplication, complements) in
determining reliability and the use of Exponential and Normal distributions in determining the mean time
between failures (used in availability).
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, i.e. not failing.
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.
Solutions
1.
a. P(operate) = .92 = .81
b.
.9
.9
.9
.9
[.90 + .10(.90)] [.90 + .10(.90)] = .9801
c. [.90 + .99(.10)(.90)]2 = .9783
2.
.96 x .96 x .99 x .99 = .9033
3.
X3 = .92
x = .9726
Instructor’s Manual, Chapter 4 Supplement
53
Solutions (continued)
4.
C = (10P) 2 per component
2 (10P) 2 = 173
100P2 = 86.5
P2 = .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 1st: [.98 + (1 - .98) x .98] x .95 x .94 x .90 = .8034
If 2nd: .98 x [.95 + (1 - .95) x .95] x .94 x .90 = .8270
If 3rd: .98 x .95 x [.94 + (1 – .94) x .94] x .90 = .8349
If 4th: .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)]
The fourth component should be backed up.
c. The one with a reliability of .90 since it poses the greatest risk of failure. The system
reliability will then be .86814.
7.
a.
#1: Pline = .99 x .96 x .93 = .8839
P(line works) + P(line fails) x P(backup works)
= .8839 + [(1 - .8839) x (.8839)] = .9865
#2:
.99
.96
.93
.99
.96
.93
.99
.96
.93
.99
.96
.93
P: .99 + [(1 - .99) x .99]
.96 + [(1 - .96) x .96]
= .9999
= .9984
Overall: .9999 x .9984 x .9951 = .9934
Plan 2 is better (.9934 > .9865)
.93 + [(1 - .93) x .93]
= .9951
b. In #1 the system will fail if any one original and any one backup fail.
In #2 the system will fail only if a component and its backup fail.
c. Space for a line versus space for individual backups, ease of shifting to backups when
needed, possible cost differences.
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Operations Management, 2/ce
8.
a.
Solutions (continued)
8.
a. Rsystem = .8839 + [(1 - .8839) x .98 x .8839]
= .9845
The decrease in reliability is .9865 - .9845 = .002
a. Rsystem = (.99 + [(1 - .99) x .98 x .99])(.96 + [(1 - .96) x .98 x .96])(.93 + [(1 - .93) x .
98 x .93]) = .9997 x .9976 x .9938 = .9911
The decrease in reliability is .9934 - .9911 = .0023
9.
x = reliability
x5 = .98
x = .996
10.
[x + (1 – x)x](x4)
= x5 + (1 –x)x5 = 2x5 – x6 = x5 (2 - x) = .98
x = .995 [by trial and error]
11.
Not completed in time means no team completes in time:
P(not team #1) x P(not team #2) x P(not team #3)
= (1 – .9) x (1 – .8) x (1 – .7) = .1 (.2) (.3) = .006
12.
a.
(1)
(2)
(3)
T
39
48
60
b. (1)
(2)
(3)
33
15
6
e–T/MTBF
.2725
.2019
.1353
T/MTBF
1.3
1.6
2.0
1.1
.5
.2
[from Table 4S-1]
1 – .3392 = .6608
1 – .6065 = .3935
1 – .8187 = .1813
1– (e–T/MTBF)
c.
T
(1)
(2)
(3)
(4)
13.
50%
85%
95%
99%
approx .7
1.9
3.0
4.6
21 mo.
57
90
138
MTBF = 30 months
a. T = 30 months
30
T / MTBF =
= 1 .0
30
1 – e–T/MTBF = 1 – .3679 = .6321
b. 1 – e–T/MTBF = .10, so
e–T/MTBF = .90. Hence,
T/MTBF = .10
T = .10(30 months) = 3 months.
Instructor’s Manual, Chapter 4 Supplement
55
Solutions (continued)
14.
MTBF = 5,000 hours
a. T = 6,000
T / MTBF =
6,000
=1.2
5,000
F (T)
e–T/MTBF = .3012
b. T = 1,000
T / MTBF =
1,000
= .2
5,000
.1813
–T/MTBF
1–e
= 1 – .8187 = .1813
c. P(1,000 ≤ x ≤ 6,000) = .8187
–.3012
.5175
15.
MTBF = 6 years
T
a.
>9
b.
<12
c.
9<T<12
d.
>21
0
.5175
1,000
.3012
6,000
hours
.8647
T/MTBF
e–T/MTBF
F (T)
1.5
.2231
2.0
1 – .1353 = .8647
.8647 – (1 - .2231) =.0878
3.5
.0302
.2231
.0302
.0878
16.
µ = 41 mo.
σ = 4 mo.
0
9
12
21
years
38 − 41
= −.75. Probability = .2266 (From App. B Table B)
4
a.
< 38 : z =
b.
40 < T < 45 : z =
40 − 41
= −.25. Probability = .5 - .4013 = .0987 (From App. B Table B)
4
45 − 41
=1.00. Probability = .3413 (From App. B Table A)
4
Total probability = .4400
z 45 =
c. µ ± 2 months is µ ± .5σ = 2(.1915) = .3830 (App. B Table A)
.1915
c.
.0987
b.
.3413
.1915
a.
41
.2266
38
56
41
40 41
45
39 41 43
Operations Management, 2/ce
Solutions (continued)
17.
µ = 6 years
.9987
σ = .5 years
a. (1)
(2)
5−6
= −2.00
.5
1 − .0228 = .9772
(Appendix B, Table B)
> 5 yr : z =
.9772
-2
0
6−6
= 0.00
.5
= .5000
7.5 − 6
= +3.00
(3)
.5
= .9987( Appendix B, Table B)
.5000
21.
22.
-4
0
< 4 yr : z =
2%
20.
0
4−6
= −4.00
.5
Therefore, approximately zero.
a. 2%: Find 2% in App. A Table B:
z is -2.055.
µ + zσ = 6 - 2.055(.5) = 4.97 yr.
b. 5%: Find 5% in App. A Table B:
z is -1.645.
µ + zσ = 6 - 1.645(.5) = 5.18 yr.
b.
19.
3
> 6 yr : z =
7.5 yr : z =
18.
0
5%
-2.055
0
4.97
6
MTBF
MTBF + MTR
MTBF
Availability =
MTBF + MTR
142
Availability A =
= .953
142 + 7
65
Availability B =
= .970
65 + 2
Availability =
Current Availability =
z-scale
yr-scale
-1.645
0
z-scale
5.18
6
yr-scale
40
300
= .93 b.
= .98
40 + 3
300 + 6
50
a.
= .962
50 + 2
a.
100
= .962
100 + 4
Instructor’s Manual, Chapter 4 Supplement
57
a. Increase in MTBF = (.05)(100) = 5 hrs.
New MTBF = 100 + 5 = 105 hrs.
105
105
Availability =
=
= .9633
105 + 4 109
b. Reduction in MTR = (.1)(4 hrs.) = .4 hrs.
New MTR = 4 hrs - .4 hrs = 3.6 hrs.
100
Availability =
= .9653
100 + 3.6
Since .9653 > .9633, designer should choose to reduce MTR, especially because it costs less.
23.
a.
X −µ
4 − 4.7
=
= −2.33
σ
.3
P(Z ≤ −2.33) = .0099 (from Appendix B, Table B)
Z =
We would expect approximately 1% of the batteries to fail before the warranty period ends.
b. Since 54 months = 4.5 years, X = 4.5.
X −µ
4.5 − 4.7
=
= −.67
σ
.3
P(Z ≤ −.67) = .2514 (from Appendix B, Table B)
Z =
We would expect approximately 25% of the batteries to fail before the warranty period ends.
Therefore, for each individual battery, the company would have to charge 25 – 1 = 24% of
(price of the battery + $5) more.
c. In addition to price of the battery, the company should consider:
1. Possible lost future sales of this type of battery as well as lost sales of other products
manufactured and sold by the company due to a high volume of replaced batteries;
2. Possible loss of good will, reputation, and poor image in the market due to higher failure
rate;
3. The capacity to handle the additional load of battery production and battery exchanges
due to failures;
4. The amount of additional business generated as a result of adding the premium battery.
(In other words the company must consider the trade-off between the additional business
generated from the premium battery vs. the cannibalization of the current base and the
existing batteries.)
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Operations Management, 2/ce
a. Increase in MTBF = (.05)(100) = 5 hrs.
New MTBF = 100 + 5 = 105 hrs.
105
105
Availability =
=
= .9633
105 + 4 109
b. Reduction in MTR = (.1)(4 hrs.) = .4 hrs.
New MTR = 4 hrs - .4 hrs = 3.6 hrs.
100
Availability =
= .9653
100 + 3.6
Since .9653 > .9633, designer should choose to reduce MTR, especially because it costs less.
23.
a.
X −µ
4 − 4.7
=
= −2.33
σ
.3
P(Z ≤ −2.33) = .0099 (from Appendix B, Table B)
Z =
We would expect approximately 1% of the batteries to fail before the warranty period ends.
b. Since 54 months = 4.5 years, X = 4.5.
X −µ
4.5 − 4.7
=
= −.67
σ
.3
P(Z ≤ −.67) = .2514 (from Appendix B, Table B)
Z =
We would expect approximately 25% of the batteries to fail before the warranty period ends.
Therefore, for each individual battery, the company would have to charge 25 – 1 = 24% of
(price of the battery + $5) more.
c. In addition to price of the battery, the company should consider:
1. Possible lost future sales of this type of battery as well as lost sales of other products
manufactured and sold by the company due to a high volume of replaced batteries;
2. Possible loss of good will, reputation, and poor image in the market due to higher failure
rate;
3. The capacity to handle the additional load of battery production and battery exchanges
due to failures;
4. The amount of additional business generated as a result of adding the premium battery.
(In other words the company must consider the trade-off between the additional business
generated from the premium battery vs. the cannibalization of the current base and the
existing batteries.)
58
Operations Management, 2/ce