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RPRA 4. Availability 1

Engineering Risk Benefit Analysis

1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82,

ESD.72, ESD.721

RPRA 4. Availability

George E. Apostolakis

Massachusetts Institute of Technology

Spring 2007

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RPRA 4. Availability 2

Definitions

Unavailability: q(t) = Pr[down at t]

Availability: a(t) 1-q(t) = Pr[up at t]

q(t) + a(t) = 1

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RPRA 4. Availability 3

Unattended components

q(t) = F(t) = Pr[T < t]

Example: 2-out-of-3 system of exponential components

3t2tss )e1(2)e1(3)t(F)t(Q

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RPRA 4. Availability 4

Continuously monitored repairable

components

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RPRA 4. Availability 5

Continuously monitored repairable

components (2)

Average unavailability:

For the exponential failure distribution:

MTTRMTTF

MTTRq +=

1.0forq:Note,1

1

q

1MTTF

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Example

2

3

1A B

i MTTF (hrs) MTTR (hrs)

1 800 82 600 15

3 600 15

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Example (2)

What is the reliability of the system for one

month assuming that no repair is available?

Step 1: System Logic

Minimal path sets: { Y1, Y2} { Y1, Y3}

Structure function for success:

YS = 1 - (1 - Y1Y2)(1 - Y1Y3) Y

S= Y

1(Y

2+ Y

3- Y

2Y

3)

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Example (3)

Step 2: Reliability of the system in terms of component

reliabilities:

RS = R1(R2 + R3 - R2R3)

Component reliabilities:

R1 = exp(-720/800) = 0.407

R2 = R3 = exp(-720/600) = 0.301

Therefore, RS = 0.208

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RPRA 4. Availability 9

Example (4)

What is the availability of the system assuming that the

repair process starts immediately upon detection of

failure?

Structure function for success

YS = Y1(Y2 + Y3 - Y2 Y3)

Availability of the system in terms of componentavailabilities:

AS

= a1(a

2+ a

3- a

2a

3)

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Example (5)

Component availabilities:

a1 = 1 (8/800) = 0.990

a2 = a3 = 1 (15/600) = 0.975.

Therefore, AS = 0.989.

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Note

Minimal cut sets: X1 and X2X3

Structure function for failure:

XS = 1 - (1 - X1)(1 - X2X3)

XS = X1 + X2X3 - X1X2X3

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Note (2)

The unreliability of the system for one month is:

FS = F1 + F2F3 - F1F2F3

where: F1 = 1 - exp(-720/800) = 0.593

F2 = F3 = 1 - exp(-720/600) = 0.699.

Thus,FS = 0.792 = 1 - 0.208 = 1 - RS

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PROBABILITY OF SYSTEM FAILURE

OR SUCCESS

1. Determine the structure function.

2. Express system (un)reliability or(un)availability as a function of component

(un)reliabilities or (un)availabilities.

3. Determine component (un)reliabilities or

(un)availabilities.