STUDY NOTES & REFERENCE GUIDE
Covers All 5 Chapters
SOLUTIONS + LECTURE SLIDES
CourseSity ‐ Stuvia | Page 1 of 204
, STUDY NOTES & REFERENCE GUIDE
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
Σn
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,
∫ tp tp
R (tp) = 1 − F (tp) = 1 − f (z) dz = 1 = 40000 − tp
—
0 40000 40000
According to the model,
CpR(tp) + Cf (1 − R(tp))
C(t p) = =
tpR(tp) p+ M (tp)(1 − R(tp))
100 × 40000−tp + 200 × t 100(80000 + 2tp )
40000
= 4000 =
0
40000−tp ∫ tp
t × + zf (z) 80000tp − t2
dz
p 40000 0 p
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
2500 , t ∈ [0..15000]
f (t) = 0 , t ∈ [15000..25000]
2
2500
0
0 , elsewhere
Secondly,
∫ ( t2
tp p , tp ∈ [0..15000]
M (tp)×(1−R(tp)) = zf (z) dz = 50000 ∫ tp
0
150002
5000
+2 15000
z
dz , tp ∈ [15000..20000]
0 25000
@
@SSeeisismmicicisisoolalatitoionn
CourseSity ‐ Stuvia | Page 2 of 204
, STUDY NOTES & REFERENCE GUIDE
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 (t p ) × (1 − R(tp )) = 4500 , tp = 15000 , R( t p
) = 0. 4 , tp = 15000
11500 , tp = 20000 0 , tp = 20000
Using the suggested model C(tp) = CpR(tp)+Cf (1−R(tp)) for the given values
tp R(tp)+M (tp)(1−R(tp))
of Cf , Cp yields
0.093 , tp = 5000
C(tp ) = 0.067 , tp = 10000
0.063 , tp = 15000
0.078 , tp = 20000
Therefore 15000 km is the optimal preventive replacement age.
2
10 , tp ∈ [0..2]
Problem 4 Similarly to Problem 3 f (tp) 1
10 , tp ∈ [2..8]
=
0 , elsewhere
0.6 , tp = 2
0. 4 , t p = 4
From the graph R(tp) =
0. 2 , t p = 6
0 , tp = 8
(∫ t
∫ tp
p 2×z
dz ∫ , t ∈ [0..2]
∫02 p
dz , t ∈ [2..8] =
t
M (tp) × (1 − R(tp)) = zf (z) dz = 10
2×z
dz + p z
0 0 10 2 10 p
( tp2 , tp ∈ [0..2]
10
= 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 ) = = 0.7813
tp × R(tp) + M (tp) × (1 − R(tp)) , tp = 6 Month
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) = 2000
0
0 , elsewhere
@
@SSeeisismmicicisisoolalatitoionn
CourseSity ‐ Stuvia | Page 3 of 204
, STUDY NOTES & REFERENCE GUIDE
5
Similarly to the previous problems,
1 , tp < 0 ∫ tp
t2
R(tp) = 20000−t , tp ∈ [0..20000] , M (tp)×(1−R(tp)) zf (z) dz = p
p
20000 0 40000
0 , tp =>
20000
Substitution of the given values of Dp and Df into the proposed equation gives:
0.00103 , tp = 5000
3× 20000−tp
+9× tp 120000 + 12 × t 0.0008 , t = 10000
20000 p p
D(tp) = 20000
20000−t t2p = 40000 × tp − t 2 = 0.0008 , t = 15000
p
tp × 2000 + 4000
p
p
0 0 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
@
@SSeeisismmicicisisoolalatitoionn
CourseSity ‐ Stuvia | Page 4 of 204
Covers All 5 Chapters
SOLUTIONS + LECTURE SLIDES
CourseSity ‐ Stuvia | Page 1 of 204
, STUDY NOTES & REFERENCE GUIDE
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
Σn
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,
∫ tp tp
R (tp) = 1 − F (tp) = 1 − f (z) dz = 1 = 40000 − tp
—
0 40000 40000
According to the model,
CpR(tp) + Cf (1 − R(tp))
C(t p) = =
tpR(tp) p+ M (tp)(1 − R(tp))
100 × 40000−tp + 200 × t 100(80000 + 2tp )
40000
= 4000 =
0
40000−tp ∫ tp
t × + zf (z) 80000tp − t2
dz
p 40000 0 p
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
2500 , t ∈ [0..15000]
f (t) = 0 , t ∈ [15000..25000]
2
2500
0
0 , elsewhere
Secondly,
∫ ( t2
tp p , tp ∈ [0..15000]
M (tp)×(1−R(tp)) = zf (z) dz = 50000 ∫ tp
0
150002
5000
+2 15000
z
dz , tp ∈ [15000..20000]
0 25000
@
@SSeeisismmicicisisoolalatitoionn
CourseSity ‐ Stuvia | Page 2 of 204
, STUDY NOTES & REFERENCE GUIDE
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 (t p ) × (1 − R(tp )) = 4500 , tp = 15000 , R( t p
) = 0. 4 , tp = 15000
11500 , tp = 20000 0 , tp = 20000
Using the suggested model C(tp) = CpR(tp)+Cf (1−R(tp)) for the given values
tp R(tp)+M (tp)(1−R(tp))
of Cf , Cp yields
0.093 , tp = 5000
C(tp ) = 0.067 , tp = 10000
0.063 , tp = 15000
0.078 , tp = 20000
Therefore 15000 km is the optimal preventive replacement age.
2
10 , tp ∈ [0..2]
Problem 4 Similarly to Problem 3 f (tp) 1
10 , tp ∈ [2..8]
=
0 , elsewhere
0.6 , tp = 2
0. 4 , t p = 4
From the graph R(tp) =
0. 2 , t p = 6
0 , tp = 8
(∫ t
∫ tp
p 2×z
dz ∫ , t ∈ [0..2]
∫02 p
dz , t ∈ [2..8] =
t
M (tp) × (1 − R(tp)) = zf (z) dz = 10
2×z
dz + p z
0 0 10 2 10 p
( tp2 , tp ∈ [0..2]
10
= 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 ) = = 0.7813
tp × R(tp) + M (tp) × (1 − R(tp)) , tp = 6 Month
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) = 2000
0
0 , elsewhere
@
@SSeeisismmicicisisoolalatitoionn
CourseSity ‐ Stuvia | Page 3 of 204
, STUDY NOTES & REFERENCE GUIDE
5
Similarly to the previous problems,
1 , tp < 0 ∫ tp
t2
R(tp) = 20000−t , tp ∈ [0..20000] , M (tp)×(1−R(tp)) zf (z) dz = p
p
20000 0 40000
0 , tp =>
20000
Substitution of the given values of Dp and Df into the proposed equation gives:
0.00103 , tp = 5000
3× 20000−tp
+9× tp 120000 + 12 × t 0.0008 , t = 10000
20000 p p
D(tp) = 20000
20000−t t2p = 40000 × tp − t 2 = 0.0008 , t = 15000
p
tp × 2000 + 4000
p
p
0 0 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
@
@SSeeisismmicicisisoolalatitoionn
CourseSity ‐ Stuvia | Page 4 of 204
