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Guideline – Demonstrating Prior Use v4
Page 25 of 30
Calculation Method 2 – BS EN 61508 1oo1 Calculation
For the same case (with all failures documented).
Using the basis of 960 component years provides:
Dangerous Undetected 1 in 960 years
λ
DU
=
1 960
= 1.04 x 10
−3
per year =
1.04 x10
−3
8760
= 1.2 x 10
−7
per hour
(118 FITS)
See Footnote
9
Dangerous Detected 3 in 960 years
λ
DD
=
3 960
= 3.10 x10
−3
per year =
3.10 x10
−3
8760
= 3.6 x 10
−7
per hour
(356 FITS)
Safe Detected 2 in 960 years
λ
SD
=
2 960
= 2.08 x10
−3
per year =
2.08 x10
−3
8760
= 2.4 x10
−7
per hou
r (237 FITS)
Safe Undetected =
λ
SU
= 0
λ
D
=
118 +356 FITS (4.74 x10
-7
per hour)
Assuming perfect testing with all failures repaired to original condition:
Reference: BS EN 61508-2:2010 Annex C
EFGH IFJK ILFMNJOP = Q
ER
+ Q
ET
+ Q
RR
Q
ER
+ Q
ET
+ Q
RR
+ Q
RT
Safe Fail Fraction = 237 + 0 + 356
237 + 0 + 356 + 118
Safe Fail Fraction = 593 711
Safe Fail Fraction = 0.83
Reference: BS EN 61508-6:2010 B.3.2.2.1 1oo1
From the failure data records the following can be quantified:
T
1
=
Proof Test Interval 1 year = (8760 hours)
MTTR = Mean Time to Restore
Σ
#._ ` #._ ` $ ` a ` a ` +0
"
= 7
hours
Using the formula:
bIR
Fcd
= [(Q
RT
) + (Q
RR
)] N
MH
9
Failures In Time (FIT) is the number of failures that can be expected in 1x10
-9
(E-09) failures per hour.