Ammar AlkhaldiReal Reliability
15 March 2016
PART 1: why you shall not use MTBF!!!
How to Measure Reliability
Measurement can help us to answer the followings question:
Are we doing good or bad ?
Is our performance increasing or decreasing ?
Which unit is performing better ? (Benchmarking)
What/How to improve ?
βYou canβt improve what you canβt measureβ
Why are we measuring things ?
1. MTBF is a misleading indicator.
Example: 1000 Units, one unit fail @ 1 Hour, MTBF = 1000 Hours
1 Unit fail @ 1000 hours, MTBF = 1000 Hours
Is it the same ?
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
2. MTBF Canβt be used for benchmarking.
Example:
SYSTEM #2 seems to be performing better
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
SYSTEM January February
SYSTEM #1 150 π»ππ’ππ
6 πΉππππ’πππ
MTBF = 25 690 π»ππ’ππ
15 πΉππππ’πππ
MTBF = 46
SYSTEM #2 540 π»ππ’ππ
18 πΉππππ’πππ
MTBF = 30 300 π»ππ’ππ
6 πΉππππ’πππ
MTBF = 50
2. MTBF Canβt be used for benchmarking.
Example:
But not really.
Any sense ?
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
SYSTEM January February TOTAL
SYSTEM #1 150 π»ππ’ππ
6 πΉππππ’πππ
MTBF = 25 690 π»ππ’ππ
15 πΉππππ’πππ
MTBF = 46 840 π»ππ’ππ
21 πΉππππ’πππ
MTBF = 40
SYSTEM #2 540 π»ππ’ππ
18 πΉππππ’πππ
MTBF = 30 300 π»ππ’ππ
6 πΉππππ’πππ
MTBF = 50 840 π»ππ’ππ
24 πΉππππ’πππ
MTBF = 35
3. MTBF is time independent.
Example: 12 failures over 12 months, MTBF = (365/12) = 30.4
MTBF = 30.4
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12
F1 F2 F3 F4 F5 F7F6 F8 F9 F10 F11 F12
3. MTBF is time independent.
Example: 12 failures over 12 months, MTBF = (365/12) = 30.4
MTBF= 30.4, But the failure rate is increasing?
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12
F1
F2
F3
F4
F5
F7
F6
F8
F9
F10
F11
F12
3. MTBF is time independent.
Example: 12 failures over 12 months, MTBF = (365/12) = 30.4
MTBF= 30.4, But the failure rate is decreasing?
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12
F1
F2
F3
F4
F5
F7
F6
F8
F9
F10
F11
F12
3. MTBF is time independent.
Example: 12 failures over 12 months, MTBF = (365/12) = 30.4
MTBF= 30.4, But the failure rate is decreasing? When to plan PMs ?
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12
F1
F2
F3
F4
F5
F7
F6
F8
F9
F10
F11
F12
4. MTBF considering normal distribution, is your data so ?
Example:
But first, how different distribution can make different result/decision ?
First of all:-How you shall not measure reliability!!!
π΄π»π©π = π½ =πΆππππππππ π―ππππ
# ππππππππ
How to measure reliability ?
Here is the beautiful normal distribution AKA Bell shape.
Where MEAN = MEDIAN = MODE
The normal distribution
MEAN TIME BETWEEN FAILURESSo we are talking about the mean, and our X-axis is time, and Y-Axis is failures
Here is the beautiful normal distribution AKA Bell shape.
Where MEAN = MEDIAN = MODE
IS YOUR DATA FOLLWING THE NORMAL DISTRBUTION ?
The normal distribution
MEAN TIME BETWEEN FAILURESSo we are talking about the mean, and our X-axis is time, and Y-Axis is failures
Here is the beautiful normal distribution AKA Bell shape.
Where MEAN = MEDIAN = MODE
IS YOUR DATA FOLLWING THE NORMAL DISTRBUTION ?
Letβs see
The normal distribution
MEAN TIME BETWEEN FAILURESSo we are talking about the mean, and our X-axis is time, and Y-Axis is failures
Letβs say we are studding the failure of lightbulb, we have a group of 100 bulb, and we are running in the constant failure rate part of the bath curve (Phase 2)
The normal distribution
Letβs say we are studding the failure of lightbulb, we have a group of 100 bulb, and we are running in the constant failure rate part of the bath curve (Phase 2), weβll assume this rate = 1%,
Remember
MTBF = 1/failure rate
MTBF = 1/1% = 100
MTBF = 100
The normal distribution
Letβs say we are studding the failure of lightbulb, we have a group of 100 bulb, and we are running in the constant failure rate part of the bath curve (Phase 2), weβll assume this rate = 1%,
Remember
MTBF = 1/failure rate
MTBF = 1/1% = 100
MTBF = 100
So half of the population should be failed by the @ 100 hours
Letβs try it
The normal distribution
The data points will followings:-
100 β 1% = 99
99 β 1% = 98.01
98.01-1%= 97.02
97.02 β 1% = 96.05
And so onβ¦
@ 100 hours we left with
37 unitsβ¦
But why ? We suppose to get MEAN=50 unit ???
Simply because the failure pattern unfirming an exponential distribution.
For exponential :
MEAN β MEDIAN β MODE
But is everything followings exponential pattern ?
NO
EVERY FAILURE MODE HAVE ITβS UNIQUE DISTRBUTION SHAPE.
The normal distribution
0
20
40
60
80
100
1201
13
25
37
49
61
73
85
97
109
121
133
145
157
169
181
193
205
217
229
241
253
265
Un
its
Time in Hours
@ 100 hours only 37 units survives
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