Expanding this function, we get:
ξ(x) = x
1
x
2
2
x
3
x
4
− x
1
x
2
x
3
− x
1
x
2
x
4
+ x
1
− x
2
2
x
3
x
4
+ x
2
x
3
+ x
2
x
4
Then the reliability is:
R = E{ξ(X)} = R
1
R
2
R
3
R
4
−R
1
R
2
R
3
−R
1
R
2
R
4
+R
1
−R
2
R
3
R
4
+R
2
R
3
+R
2
R
4
Solution 4.3. The cut vectors are
(0000) (0001) (0010) (0011) (0100)
R = 1 −
P
P {Cut vectors}
The state function based on cut vectors is:
ξ(x) = 1 − [(1 − x
1
) (1 − x
2
) (1 − x
3
) (1 − x
4
) + (1 − x
1
) (1 − x
2
) (1 − x
3
) x
4
+ (1 − x
1
) (1 − x
2
) x
3
(1 − x
4
) + (1 − x
1
) (1 − x
2
) x
3
x
4
+ (1 − x
1
) x
2
(1 − x
3
) (1 − x
4
)
Expanding this function, we get:
ξ(x) = x
1
x
2
x
3
x
4
− x
1
x
2
x
3
− x
1
x
2
x
4
+ x
1
− x
2
x
3
x
4
+ x
2
x
3
+ x
2
x
4
Then the reliability is:
R = E{ξ(X)} = R
1
R
2
R
3
R
4
−R
1
R
2
R
3
−R
1
R
2
R
4
+R
1
−R
2
R
3
R
4
+R
2
R
3
+R
2
R
4
Solution 4.4. We note that:
• (0001) > (0000) but ξ(0001) ≯ ξ(0000), so 0000 is not a minimal cut
vector.
• (0011) > (0010) but ξ(0011) ≯ ξ(0010), so 0010 is not a minimal cut
vector.
• (0011) > (0001) but ξ(0011) ≯ ξ(0001), so 0001 is not a minimal cut
vector.
The minimal cut vectors are (0011) and (0100). The corresponding cut sets
are {c
1
, c
2
}, {c
1
, c
3
, c
4
} and {c
2
, c
4
}.
Note that the corresponding state function is:
ξ(x) = (1 − (1 − x
1
)(1 − x
2
)) (1 − (1 − x
1
)(1 − x
3
)(1 − x
4
))
6