1
Solution 1.1.
c
1
c
2
c
3
Solution 1.2.
x
1
x
2
x
3
ξ(x)
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
with x = {x
1
, x
2
, x
3
}
Solution 1.3.
I
ξ
(1) =
1
4
[(0 0) + (1 0) + (1 0) + (1 0)] =
3
4
I
ξ
(2) = I
ξ
(3) =
1
4
[(0 0) + (0 0) + (1 0) + (1 1)] =
1
4
Solution 1.4. The path vectors are (101), (110) and (111).
We note that (111) is not a minimal path vector:
(110) < (111) but ξ(110) ξ(111)
(101) < (111) but ξ(101) ξ(111)
The minimal path vectors are (110) and (101) which correspond to the min-
imal path sets P
1
= {c
1
, c
2
} and P
1
= {c
1
, c
3
}, respectively.
Solution 1.5. Minimal path vectors lead to parallel arrangement of series struc-
tures.
The l = 2 minimal paths lead to ξ(x) = 1 (1 x
1
x
2
)(1 x
1
x
3
) and to an
arrangement of l = 2 parallel groups of components.
1
c
1
c
2
c
1
c
3
Solution 1.6. The cut vectors are (000), (001), (010), (011) and (100).
We note that:
(001) > (000) but ξ(001) ξ(000), so 000 is not a minimal cut vector.
(011) > (001) but ξ(011) ξ(001), so 001 is not a minimal cut vector.
(011) > (010) but ξ(011) ξ(010), so 010 is not a minimal cut vector.
The minimal cut vectors are (011) and (100) which correspond to the minimal
cut sets C
1
= {c
1
} and C
2
= {c
2
, c
3
}, respectively.
Solution 1.7. Minimal cut vectors lead to series arrangement of parallel struc-
tures.
The k = 2 minimal cut vectors lead to ξ(x) = x
1
(x
2
+x
3
) and to an arrange-
ment of k = 2 serial groups of components. The original diagram is already in
the desired form.
Solution 1.8.
1
23
4
5
6
7
λ
1
λ
2
λ
3
λ
1
λ
3
λ
1
λ
2
2
Solution 1.9.
(λ
1
+ λ
2
+ λ
3
) λ
3
λ
2
λ
1
0 0 0
0 (λ
1
+ λ
2
) 0 0 λ
1
λ
2
0
0 0 (λ
1
+ λ
3
) 0 0 λ
3
λ
1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Solution 1.10. The reliability is given by the sum of the probabilities of being
in states 7, 3 and 5 (which correspond to operational states) : R = P
7
+P
5
+P
3
.
Then, we must solve the equations system below.
P
7
(λ
1
+ λ
2
+ λ
3
) =
d
dt
P
7
(1)
P
7
λ
3
P
3
(λ
2
+ λ
3
) =
d
dt
P
3
(2)
P
7
λ
2
P
5
(λ
1
+ λ
3
) =
d
dt
P
5
(3)
(4)
P
7
(t) = e
(λ
1
+λ
2
+λ
3
)t
(5)
P
3
(t) =
λ
3
λ
1
+ λ
2
+ λ
3
e
(λ
1
+λ
2
+λ
3
)t
+ e
(λ
2
+λ
3
)t
(6)
P
5
(t) =
λ
2
λ
1
+ λ
2
+ λ
3
e
(λ
1
+λ
2
+λ
3
)t
+ e
(λ
1
+λ
3
)t
(7)
(8)
2
Solution 2.1.
R = P {ξ(X) = 1}
= P
(
n
Y
i=1
X
i
= 1
)
=
n
Y
i=1
P {X
i
= 1}
=
n
Y
i=1
R
i
3
Solution 2.2.
I
R(q)
(i) =
For the numerical example,
I
R
(1) = 0.4 × 0.8 = 0.32
I
R
(2) = 0.75 × 0.8 = 0.60
I
R
(3) = 0.75 × 0.4 = 0.30
3
Solution 3.1.
R = E{ξ(X)}
= E
(
1
n
Y
i=1
(1 X
i
)
)
= 1 E
(
n
Y
i=1
(1 X
i
)
)
= 1
n
Y
i=1
E {1 X
i
}
= 1
n
Y
i=1
(1 R
i
)
Solution 3.2.
I
R(q)
(i) =
For the numerical example,
I
R
(1) = (1 0.4) × (1 0.8) = 0.12
I
R
(2) = (1 0.75) × (1 0.8) = 0.05
I
R
(3) = (1 0.75) × (1 0.4) = 0.15
4
4
Solution 4.1.
x
1
x
2
x
3
x
4
ξ(x)
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 1
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
R =
P
P {Path vectors}
The state function based on path vector is:
ξ(x) = (1 x
1
) x
2
(1 x
3
) x
4
+ (1 x
1
) x
2
x
3
(1 x
4
) + (1 x
1
) x
2
x
3
x
4
+ x
1
Expanding this expression, 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 4.2. It’s easy to see that the minimal path vectors are (1000), (0110)
and (0101). The corresponding path sets are {c
1
}, {c
2
, c
3
} and {c
2
, c
4
}.
c
2
c
2
c
3
c
4
c
1
Note that the corresponding state function is:
ξ(x) = 1 (1 x
1
)(1 x
2
x
3
)(1 x
2
x
4
)
5
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
}.
c
1
c
2
c
3
c
4
c
1
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
Expanding this expression, we get:
ξ(x) = x
2
1
x
2
x
3
x
4
+ x
2
1
x
2
x
3
+ x
2
1
x
2
x
4
x
2
1
x
2
+ x
2
1
x
3
x
4
x
2
1
x
3
x
2
1
x
4
+ x
2
1
+ 2x
1
x
2
x
3
x
4
2x
1
x
2
x
3
2x
1
x
2
x
4
+x
1
x
2
x
1
x
3
x
4
+ x
1
x
3
+ x
1
x
4
x
2
x
3
x
4
+ x
2
x
3
+ x
2
x
4
Then, the reliability is given by
R = 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
1
R
3
R
4
R
1
R
3
R
1
R
4
+ R
1
+ 2R
1
R
2
R
3
R
4
2R
1
R
2
R
3
2R
1
R
2
R
4
+R
1
R
2
R
1
R
3
R
4
+ R
1
R
3
+ R
1
R
4
R
2
R
3
R
4
+ R
2
R
3
+ R
2
R
4
= 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 4.5. Suppose that component c
2
is a key component. When R
2
= 1,
the system is equivalent to a parallel system with three components: c
1
, c
3
, c
4
.
When R
2
= 0 the system behaves as having only component c
1
. Then:
R = [1 (1 R
1
)(1 R
3
)(1 R
4
)]R
2
+ R
1
(1 R
2
)
Expanding this expression, we get:
R = 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
7