Probabilistic Fault Isolation in Embedded Systems Using Prior Knowledge of the System
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(183) # ? m1 5 ' ? 2 H X1 = 0 X2 = 1 X3 = 1 X4 = 0 X5 = 0 + H S11 = X S21 = 0 S31 = 0 S41 = X S51 = 0 8 L H m1 : p(X1 = 0 | S11 = X ) p(X2 = 1 | S21 = 0) p(X3 = 1 | S31 = 0) p(X4 = 0 | S41 = X ) p(X5 = 0 | S51 = 0) H m2 : p(X1 = 0 | S12 = 0) p(X2 = 1 | S22 = X ) p(X3 = 1 | S32 = X ) p(X4 = 0 | S42 = X ) p(X5 = 0 | S52 = 0) m3 : p(X1 = 0 | S13 = X ) p(X2 = 1 | S23 = X ) p(X3 = 1 | S33 = 0) p(X4 = 0 | S43 = 0) p(X5 = 0 | S53 = X ) m4 : p(X1 = 0 | S14 = 0) p(X2 = 1 | S24 = 0) p(X3 = 1 | S34 = X ) p(X4 = 0 | S44 = 0) p(X5 = 0 | S54 = X ). 9 = 4% B8;C m1 H p(X = x | Sm1 ) = p(X1 = 0 | S11 = X ) · p(X2 = 1 | S21 = 0)· p(X3 = 1 | S31 = 0) · p(X4 = 0 | S41 = X ) · p(X5 = 0 | S51 = 0) = (0.3) · (0.01) · (0.01) · (0.3) · (0.99) = 8.9 · 10−6 m1 1 ⇒ αm = 8.9 · 10−6 · 32 ≈ 2.85 · 10−4 x = p(X = x | Sm1 ) · A. H m2 2 αm = 0.144 · 32 ≈ 4.61 x = p(X = x | Sm2 ) · A m3 αx = p(X = x | Sm3 ) · Am3 = 6.24 · 10−4 · 32 ≈ 0.020 m4 4 αm = 2.06 · 10−3 · 32 ≈ 0.066 x = p(X = x | Sm4 ) · A. 1
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(211) x3 0H αm X(x3 =0) p(X(x3 = 0) | M = m, I) = = Am ⎧ 0.3 · 0.01 · 0.99 · 0.3 · 0.99 = 8.821 · 10−4 ⎪ ⎪ ⎨ 0.99 · 0.7 · 0.3 · 0.3 · 0.99 = 0.062 = ⎪ 0.3 · 0.7 · 0.99 · 0.99 · 0.3 = 0.062 ⎪ ⎩ 0.99 · 0.01 · 0.3 · 0.99 · 0.3 = 8.821 · 10−4 πX(x3 =0) =. . M M M M. = m1 = m2 = m3 = m4. p(X(x3 = 0) | M = m, I)p(M = m | I) =. m. =. αm X(x3 =0) m. =. Am. p(M = m | I) =. 1 (8.821 · 10−4 + 0.062 + 0.062 + 8.821 · 10−4 ) = 0.0313 4. ? H. i p(M = mi | I) αm X (x3 = 0) = πX(x3 =0) Ami ⎧ 0.25·8.821·10−4 ⎪ = 0.007 M = m1 ⎪ 0.0313 ⎪ ⎨ 0.25·0.062 0.0313 = 0.493 M = m2 = 0.25·0.062 ⎪ ⎪ 0.0313 = 0.493 M = m3 ⎪ ⎩ 0.25·8.821·10−4 = 0.007 M = m4 0.0313. p(M = mi | X(x3 = 0), I) =.
(212) x3 1H. αm X(x3 =1). p(X(x3 = 1) | M = m, I) = = Am ⎧ 0.3 · 0.01 · 0.01 · 0.3 · 0.99 = 8.910 · 10−6 ⎪ ⎪ ⎨ 0.99 · 0.7 · 0.7 · 0.3 · 0.99 = 0.144 = ⎪ 0.3 · 0.7 · 0.01 · 0.99 · 0.3 = 6.237 · 10−4 ⎪ ⎩ 0.99 · 0.01 · 0.7 · 0.99 · 0.3 = 2.058 · 10−3 πX(x3 =1) =. . M M M M. = m1 = m2 = m3 = m4. p(X(x3 = 1) | M = m, I)p(M = m | I) =. m. =. αm X(x3 =1) m. =. Am. p(M = m | I) =. 1 (8.910 · 10−6 + 0.144 + 6.237 · 10−4 + 2.058 · 10−3 ) = 0.0367 4. ? H. i p(M = mi | I) αm X (x3 = 0) πX(x3 =1) Ami ⎧ 0.25·8.910·10−6 ⎪ ≈ 6 · 10−5 ⎪ 0.0367 ⎪ ⎨ 0.25·0.144 0.0367 = 0.982 = −4 0.25·6.237·10 ⎪ = 0.004 ⎪ 0.0367 ⎪ ⎩ 0.25·2.058·10−3 = 0.014 0.0367. p(M = mi | X(x3 = 1), I) =. +-. = M M M M. = m1 = m2 = m3 = m4.
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(241) 2 # . x1 x2 x3 x4 x5. m1 X 0 0 X 0. 88H m2 0 X X X 0. m3 X X 0 0 X. m4 0 0 X 0 X. m1,2 X X X X 0. m1,3 X X 0 X X. m1,4 X 0 X X X. m2,3 X X X X X. m2,4 0 X X X X. m3,4 X X X 0 X. 1 ? 4% B85:C 88.
(242) p(M = mi | D, I) = psingle (M = mi | D, I)+ pdouble (M = mi,j | D, I) +. , i = j. B85:C. j. psingle pdouble 2 . $%!& 1 B/ C. ( 88 X = (1, 1, 1, 1, 0) + D
(243) & p(M = m | 1 I) = 1r = 10 % 0.01 0.3 ? p(M = m | X, D, I) = { D} = p(M = m | X, I) =. αm X p(M = m | I) Am πX. = 4% B8;C p(M = m | I) Aα πX M = m1 H m. m. p1 (M = m1 | X, I) =. 1 αm 0.1 X p(M = m1 | X, I) = 1.52 · 10−4 = 4.85 · 10−5 m 1 A πX 0.0317. = H M M M M M. = m1 : = m2 : = m3 : = m4 : = m1,2 :. 1.52 · 10−4 1.07 · 10−2 4.63 · 10−5 6.62 · 10−7 0.749. M M M M M. = m1,3 = m1,4 = m2,3 = m2,4 = m3,4. : : : : :. 3.24 · 10−3 3.24 · 10−3 0.227 3.24 · 10−3 3.24 · 10−3. 1 ? 4% B85:C & M = m1
(244)
(245) H p(M = m1 | X, I) =psingle (M = m1 | X, I) + pdouble (M = m1,2 | X, I)+ +pdouble (M = m1,3 | X, I) + pdouble (M = m1,4 | X, I) = =0.756 = 75.6%. 89.
(246) H p(M = m2 | X, I) = 0.990 = 99.0% p(M = m3 | X, I) = 0.234 = 23.4% p(M = m4 | X, I) = 0.010 = 1.0%. ? & 8+ 1 0.9 0.8. Probability. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0. 1. 2 3 BehaviorM ode. 4. & 8+H ? . # ' 2 1 2 # ptriple (M = mi,j,k |D, I) ? . /0 -
(247)
(248) - + * . 7 2 G 2 G G 2 G %
(249) 4
(250) 4% 859 8:.
(251) 1 # ' 895 #
(252) mi,j B4% B85:CC . 8;.
(253) .
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