SOLUTIONS MANUAL Fault-Tolerant Systems 2nd Edition by Israel Koren, C. Mani Krishna | Chapter 2-11

, Solutions to
Chapter 2 Exercises

1. The lifetime (measured in years) of a processor is exponentially distributed, with a mean
lifetime of 2 years. You are told that a processor failed sometime in the interval [4, 8]
years. Given this information, what is the conditional probability that it failed before it
was 5 years old?
Solution:
Denote the lifetime of the processor by T . Since E(T ) = 2, λ = 0.5 and the distribution
function of T is F(t) = 1 − e−0.5t . Using the conditional probability formula:

Prob{[T < 5] ∩ [4 ≤ T ≤ 8]}
Prob{T < 5 | 4 ≤ T ≤ 8} =
Prob{4 ≤ T ≤ 8}
Prob{4 ≤ T < 5}
=
Prob{4 ≤ T ≤ 8}
F(5) − F(4)
=
F(8) − F(4)
e−2 − e−2.5
= = 0.455
e−2 − e−4

2. The lifetime of a processor (measured in years) follows the Weibull distribution, with
parameters λ = 0.5 and β = 0.6.
(a) What is the probability that it will fail in its first year of operation?
(b) Suppose it is still functional after t = 6 years of operation. What is the conditional
probability that it will fail in the next year?
(c) Repeat parts (a) and (b) for β = 2.

1

,2 Solutions to Chapter 2 Exercises

(d) Repeat parts (a) and (b) for β = 1.

Solution:
(a) Denote the lifetime of the processor by T . The distribution function of T is F(t) =
0.6
1 − e−0.5t . The probability that T is no greater than one year is F(1) = 0.393.
(b) We use the conditional probability formula:

Prob{6 < T ≤ 7}
Prob{6 < T ≤ 7 | T > 6} =
Prob{T > 6}
F(7) − F(6)
= = 0.045
1 − F(6)

3. To get a feel for the failure rates associated with the Weibull distribution, plot them for
the following parameter values as a function of the time, t:

(a) Fix λ = 1 and plot the failure rate curves for β = 0.5, 1.0, 1.5.
(b) Fix β = 1.5 and plot the failure rate curves for λ = 1, 2, 5.

Solution:
The failure rate for the Weibull distribution is

λ(t) = λβtβ−1


2.5 2.5
5
=1.
β=1.5




2 2 ; β
λ=
2
λ=5;
Failure Rate




Failure Rate




β=1.5
1.5 λ=1; 1.5

λ=1; β=1.0 =1.5
1 1 λ=1; β
λ=1; β=0.5
0.5 0.5


0 0
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
t t

Figure 2.1: The failure rate for the Weibull distribution.


4. Write the expression for the reliability Rsystem (t) of the series/parallel system shown in
Figure 2.2, assuming that each of the five modules has a reliability of R(t).
Solution:
The system can be decomposed into a series system consisting of one unit with the leftmost

Solutions for Fault-Tolerant Systems, 2nd Edition, by Koren and Krishna c 2020 Elsevier Science (USA) – Do not copy

, Solutions to Chapter 2 Exercises 3




Figure 2.2: A 5-module series-parallel system.

4 blocks and the second unit with the rightmost block. If the reliability of the leftmost 4
blocks is RA (t), the system reliability is RA (t)R(t).
Now, we calculate RA (t). This subsystem consists of a parallel arrangement of one unit
consisting of the bottom block and another consisting of the other 3 blocks. If RB (t) is the
reliability of the top 3 blocks, RA (t) = 1 − (1 − RB (t))(1 − R(t)).
Next, we calculate RB (t): this subsystem consists of a series arrangement of one block with
another consisting of two blocks in parallel. Hence, we have

RB (t) = R(t)(1 − (1 − R(t))2 )
Substituting all intermediate results yields

Rsystem = R5 (t) − 3R4 (t) + 2R3 (t) + R2 (t)

5. The lifetime of each of the seven blocks in Figure 2.3 is exponentially distributed with
parameter λ. Derive an expression for the reliability function of the system, Rsystem (t),
and plot it over the range t = [0, 100] for λ = 0.02.




Figure 2.3: A 7-module series-parallel system.

Solution:
As before, we decompose this structure into two substructures connected in series. The
left substructure is four blocks in parallel; the right substructure consists of a series ar-
rangement of two blocks in parallel with one block.
The reliability of this system is thus given by:
h ih i
1 − (1 − R(t))4 1 − (1 − R(t))(1 − R2 (t)) = R7 (t)−5R6 (t)+9R5 (t)−6R4 (t)−2R3 (t)+4R2 (t)

Solutions for Fault-Tolerant Systems, 2nd Edition, by Koren and Krishna c 2020 Elsevier Science (USA) – Do not copy