Instructors Manual to the book
“Maintenance, Replacement and Reliability:
Theory and Applications”
Third Edition
by Andrew K.S. Jardine and Albert H.C. Tsang
2021
Notes:
(1) The First Edition manual was prepared by Andrey Pak, University
of Toronto, 2005.
(2) This Third Edition manual has been updated to correct some typos
in the Second Edition, update the web site for software used for
solving some end-of-Chapter problems, and include solutions to
problems in Third Edition’s new Chapter 6.
(3) The solutions to problems in Chapter 6 (The Role of Emerging
Technologies in Physical Asset Management) were prepared by
Professor Sharareh Taghipour of Ryerson University, Toronto, 2021.
1
,2
Introduction
This is an Instructors Manual to the book “Maintenance, Replacement and Reliability:
Theory and Applications”, Third Edition, written by Andrew K.S. Jardine and Albert
H.C. Tsang (2021). It contains solutions to all of the problems in the book, in order of
their appearance. In Chapters 2, 4 and Appendix 2 of the book, use of educational
version of special software packages (OREST, AGE/CON, WeibullSoft) is required.
These can be downloaded from: https://bit.ly/31cZRgb
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Chapter 2: Component Replacement Decisions
Problem 1 The following table contains cumulative losses, total costs and
average monthly costs of operation for n = 1, 2, 3, 4. Here
Pn
Li + Rn
AC(n) = i=1
n
where Li stands for loss in productivity during year i with respect to the first
year’s productivity, Ri stands for replacement cost (constant)
Month Productivity Losses Replacement Total Cost Average Cost
1 10000 0 1200 1200 1200
2 9700 300 1200 1500 750
3 9400 600+300 1200 2100 700
4 8900 1100+600+300 1200 3200 800
Clearly, the optimal replacement time is 3 months since the pump is new.
Problem 2 One can use the model from section 2.5 (see 2.5.2). In this problem
Cp = 100, Cf = 200,
Z tp
tp 40000 − tp
R(tp ) = 1 − F (tp ) = 1 − f (z) dz = 1 − =
0 40000 40000
According to the model,
40000−t t
Cp R(tp ) + Cf (1 − R(tp )) 100 × 40000 p + 200 × 40000
p
C(tp ) = = 40000−t R t
tp R(tp ) + M (tp )(1 − R(tp )) tp × 40000 p + 0 p zf (z) dz
100(80000 + 2tp )
=
80000tp − t2p
0.0143 , tp = 10000
0.01 , tp = 20000
C(tp ) =
0.0093 , tp = 30000
0.01 , tp = 40000
Calculations above indicate that the optimal age is 30000 km.
Problem 3 Firstly, one can find f (t). Since the area below the probability
density curve is equal to 1, the area of each rectangle on the Figure 2.40 is 15 .
It follows then, that
1
25000 , t ∈ [0, 15000]
2
f (t) = 25000 , t ∈ [15000, 25000]
0 , elsewhere
Secondly,
Z (
tp t2p
, tp ∈ [0, 15000]
M (tp )×(1−R(tp )) = zf (z) dz = 50000
150002
R tp z
0 50000 +2 15000 25000
dz , tp ∈ [15000, 20000]
