Solutions + Lecture Slides for Maintenance, Replacement, and Reliability (3rd Edition) by Albert Tsang – 2022 | All 5 Chapters Covered

Covers All 5 Chapters




SOLUTIONS + LECTURE SLIDES

, 3

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,
Cp R(tp ) + Cf (1 − R(tp ))
C(tp ) = =
tp R(tp ) + M (tp )(1 − R(tp ))
40000−tp tp
100 × 40000 + 200 × 40000 100(80000 + 2tp )
= 40000−tp t =
80000tp − t2p
R p
tp × 40000 + 0 zf (z) dz


 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 51 .
It follows then, that

1
 25000 , t ∈ [0..15000]

2
f (t) = 25000 , t ∈ [15000..25000]

0 , elsewhere


Secondly,
t2p
(
Z tp
50000 , tp ∈ [0..15000]
M (tp )×(1−R(tp )) = zf (z) dz = 150002
R tp z
0 50000 +2 15000 25000
dz , tp ∈ [15000..20000]




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,4

To find R(t) for the given values of tp one can use Figure 2.40 (R(t) is the
area under f (z) for z > t).
 

 500 , tp = 5000 
0.8 , tp = 5000

2000 
, tp = 10000 0.6 , tp = 10000
M (tp ) × (1 − R(tp )) = , R(tp ) =


 4500 , tp = 15000 0.4

 , tp = 15000
11500 , tp = 20000 0 , tp = 20000
 

p p fC R(t )+C (1−R(t ))
p
Using the suggested model C(tp ) = tp R(tp )+M (tp )(1−R(tp ))
for the given values
of Cf , Cp yields


 0.093 , tp = 5000

0.067 , t = 10000
p
C(tp ) =
0.063 , tp = 15000


0.078 , tp = 20000


Therefore 15000 km is the optimal preventive replacement age.

2
 10
 , tp ∈ [0..2]
1
Problem 4 Similarly to Problem 3 f (tp ) = , tp ∈ [2..8]
 10
0 , elsewhere


0.6


, tp =2
0.4 , tp =4
From the graph R(tp ) =
0.2
 , tp =6

0 , tp =8


(R t
tp p 2×z
, tp ∈ [0..2]
Z
M (tp ) × (1 − R(tp )) = 10 dz R
zf (z) dz = R02 2×z =
t z
0 0 10
dz + 2 p 10 dz , tp ∈ [2..8]
( t2
p
10 , tp ∈ [0..2]
= t2p +4
20 , tp ∈ [2..8]

After substitutions, the suggested formula gives:


 0.9375 , tp = 2

Tp × R(tp ) + Tf × (1 − R(tp ))  0.7692 , tp = 4 Days
D(tp ) = =
tp × R(tp ) + M (tp ) × (1 − R(tp ))  0.7813 , tp = 6 M onth

0.8824 , tp = 8


Clearly, preventive replacement after 4 months of operation is the most prefer-
able.

Problem 5 For the uniform distribution over [0..20000]
(
1
, t ∈ [0..20000]
f (t) = 20000
0 , elsewhere




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, 5

Similarly to the previous problems,

1
 , tp < 0 Z tp
t2p
20000−tp
R(tp ) = , tp ∈ [0..20000] , M (tp )×(1−R(tp )) = zf (z) dz =
 20000 0 40000
0 , tp > 20000



Substitution of the given values of Dp and Df into the proposed equation gives:


0.00103 , tp = 5000
20000−tp tp 
3× 20000 + 9× 20000 120000 + 12 × tp 0.0008 , tp = 10000
D(tp ) = 2 = =
20000−t
tp × 20000 p +
tp 40000 × tp − t2p 0.0008 , tp = 15000
40000


0.0009 , tp = 20000



Hence, there are two equally preferable replacement ages among the given four.


Problem 6 Weibull paper analysis (Figure 1) gives estimations
µ = 49000 km, η = 55000 km, β = 1.7




Figure 1: Problem 6 Weibull plot




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