@LECTSOLUTIONSSTUVIA
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
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) + C f (1 − R(tp))
C(t p) = =
tpR(tp) + M (tp)(1 − R(tp))
40000−t p tp
100 × 40000 + 200 × 40000 100(80000 + 2tp )
= 40000−t p ∫ tp =
t × + zf (z) dz 80000tp − t2
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 1 .5
It follows then, that
1
25000
2
, t ∈ [0..15000]
f (t) = 25000 , t ∈ [15000..25000]
0 , elsewhere
Secondly
, ∫ tp ( t2
p , tp ∈ [0..15000]
M (tp)×(1−R(tp)) = zf (z) dz = 50000
15000 2
∫ tp z
0 + dz , tp ∈ [15000..20000]
250000 15000 25000
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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 (t p) × (1 − R(tp )) = p
4500 , tp = 15000 , R(t ) = 0.4 , tp = 15000
11500 , tp = 20000 0 , tp = 20000
C pR(t p)+C f (1−R(t p))
Using the suggested model C(tp) = for the given values
t p R(t p )+M (tp)(1−R(tp))
of Cf , Cp yields
0.093 , tp = 5000
C(t p ) = 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 , tp = 4
From the graph R(tp) =
0.2 , tp = 6
0 , tp = 8
(∫ t
∫ tp p 2×z
dz , t ∈ [0..2]
∫ 20 2×z10 ∫t p
M (tp) × (1 − R(tp)) = zf (z) dz = dz + p z
dz , t ∈ [2..8] =
0 0 10 2 10 p
( t2p , 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) = 20000
0 , elsewhere
EERDDDFDFDDas
, 5
Similarly to the previous problems,
1 , tp < 0 ∫ tp
20000−tp t2p
R(tp) = 20000 , tp ∈ [0..20000] , M (tp)×(1−R(tp)) = zf (z) dz =
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 20000 p p
D(tp) 20000−tp t2p = 40000 × tp − t2 = 0.0008 , t = 15000
=
tp × 20000
+ 40000
p p
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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6
Using the Γ function table (see Appendix 7) and the formula
2 1
σ2 = η2 × [Γ(1 + ) − Γ2(1 + )]
β β
gives
σ2 = (55000)2 × [1.1765Γ(1.1765) − Γ2(1.5882)] = 3025 × 106 × (1.0883 − 0.796) =
= 885.2 × 106
σ = 29753 km and therefore µ
= 49000
= 1.65. Now we can proceed to Glasser’s
σ 29753
paper analysis (Figure 2)
Figure 2: Problem 6 Glasser’s graph
According to the graph ρ = 0.92, which means 8% of expected improvement,
and Z = −0.6. Furthermore,
tp = µ + Z × σ = 49000 − 0.6 × 29753 = 31148
Problem 7 Firstly, sort the data in increasing order
Hours 80 100 115 130 150 170 200
Days 3.33 4.17 4.79 5.42 6.25 7.08 8.33
EERDDDFDFDDas
, 7
Figure 3: Problem 7 Weibull plot
Figure 4: Problem 7 Glasser’s Block graph
Secondly, it is reasonable to assume that the data belongs to the Weibull
distribution. Then using the median ranks table (see Appendix 8) we get
Weibull graph (Figure 3). From Figure 3 β = 3, µ = 5.7 × 24 =
136.8,
η = 6.5 × 24 = 156.
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8
Next
2 1
σ2 = η2 × [Γ(1 + β ) − Γ2(1 + β )]
and therefore σ2 = (156)2 × [Γ(1.67) − Γ2 (1.33)] = 24336 × 0.1059 = 2576
σ = 50.75. Using the obtained result µ = 136.8 = 2.7 and C f = 10
σ 50.75 Cp
Figure 5: Problem 7 Glasser’s Age Graph
From Figures 4 and 5: Z = −1.5, ρ = 0.36, tp = 136.8 − 1.5 × 50.75 = 60.68
for Block replacement policy and Z = — 1.55, ρ = 0.38, tp = 58.14 for age-based
replacement policy.
Problem 8 Using σ = 20000
1000
= 20 and Cf
Cp
= 2 and the Glasser’s graph
µ
we get the following approximations: Z = — 2.1, ρ = 0.6 (40% of expected
improvement).
tp = µ + Z × σ = 20000 − 2100 = 17900
Problem 9 µ
= 150000
= 15 and Cf
= 10 Using the Glasser’s graph for block
σ 10000 Cp
replacement (Figure 6) we get: ρ = 0.14, Z = −3.3.
(a)
tp = µ + Z × σ = 150000 − 33000 = 117000
(b)
1 − ρ = 0.86 or 86% of expected improvement.
(c)
2000 $
R-o-o-F cost= = 0.0133
150000 km
Optimal policy cost= ρ×R-o-o-F= 0.14 × 0.0133 = 0.0018 $
km
EERDDDFDFDDas
, 9
Figure 6: Problem 9 Glasser’s graph
Problem 10
(a) One of the appropriate models is described in section 2.4.2
Cp + Cf × H(tp)
C(t p ) =
tp
(b) The most convenient way of solving the problem with the provided
information is to use Glasser’s graph for block replacement (Figure 7).
Figure 7: Problem 10 Glasser’s Block graph
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10
Cf 150 µ 200
= = 1.5, = = 20
Cp 100 σ 10
From the graph Z = −2.3, ρ = 0.83 (17% of expected savings)
tp = µ + Z × σ = 200 − 23 = 177
Problem11
Cf 200 + 700 900
= = =3
Cp 200 + 100 300
and
µ 150
= = 10
σ 15
Using the Glasser’s Block replacement graph:
Z = −2.2, ρ = 0.45 (55% of expected improvement) tp = 150 − 33 = 117
Problems 12-16 OREST is a user-friendly piece of software designed espe-
cially for solving optimal replacement decision problems. The following screen
shots are meant to give the reader some idea on how to input data and use it
for analysis. At any point the user can access the help file by pressing ”Help”
button in the top right corner of the window.
Figure 8: OREST screen shot
Figure 8 illustrates the main selection window of the program, where you
can choose one of the available options:
• Data Input (see Figure 9), where one can create/change components
• Trend Analysis
• Weibull Analysis, where statistical analysis can be performed and saved/printed
EERDDDFDFDDas
, 11
• Deterministic Replacement
• Age Preventive Replacement, where replacement decisions are made.
Figure 9: OREST screen shot
The main questions addressed in problems 12-16 are:
• Prepare the data for OREST
• Fit the Weibul model to data, determine the key parameters and perform a
certain analysis
• Given the ratio between preventive and failure replacement costs, find an
optimal replacement time
• Estimate a number of failures during a specified period of time under the
optimal policy
• Given a replacement policy, find associated expected costs
Here is how OREST can help the user to answer the above questions.
OREST deals with sets of data which are expected to contain independent
samples from the Weibull distribution (see Appendix 2). Essentially, in order
to qualify for that, elements of a set have to be independent. For this reason,
for example, cumulative times to failures will not do. However, the set of inter-
failure times is the one we can use (under the assumption of independence).
Fitting the data set to the Weibull model is done automatically upon press-
ing the correspondent button in the ”Weibull Analysis” window. Further analy-
sis can be performed using graphs of p.d.f., C.d.f, Reliability and failure rate
functions.
At time of creating a new component, values of preventive and failure costs
are fixed. Using this data, the optimal replacement policy can be obtained in
”Age Preventive Replacement” window. Note that every time an analysis is
performed the user has an opportunity to print the report.
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
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) + C f (1 − R(tp))
C(t p) = =
tpR(tp) + M (tp)(1 − R(tp))
40000−t p tp
100 × 40000 + 200 × 40000 100(80000 + 2tp )
= 40000−t p ∫ tp =
t × + zf (z) dz 80000tp − t2
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 1 .5
It follows then, that
1
25000
2
, t ∈ [0..15000]
f (t) = 25000 , t ∈ [15000..25000]
0 , elsewhere
Secondly
, ∫ tp ( t2
p , tp ∈ [0..15000]
M (tp)×(1−R(tp)) = zf (z) dz = 50000
15000 2
∫ tp z
0 + dz , tp ∈ [15000..20000]
250000 15000 25000
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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 (t p) × (1 − R(tp )) = p
4500 , tp = 15000 , R(t ) = 0.4 , tp = 15000
11500 , tp = 20000 0 , tp = 20000
C pR(t p)+C f (1−R(t p))
Using the suggested model C(tp) = for the given values
t p R(t p )+M (tp)(1−R(tp))
of Cf , Cp yields
0.093 , tp = 5000
C(t p ) = 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 , tp = 4
From the graph R(tp) =
0.2 , tp = 6
0 , tp = 8
(∫ t
∫ tp p 2×z
dz , t ∈ [0..2]
∫ 20 2×z10 ∫t p
M (tp) × (1 − R(tp)) = zf (z) dz = dz + p z
dz , t ∈ [2..8] =
0 0 10 2 10 p
( t2p , 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) = 20000
0 , elsewhere
EERDDDFDFDDas
, 5
Similarly to the previous problems,
1 , tp < 0 ∫ tp
20000−tp t2p
R(tp) = 20000 , tp ∈ [0..20000] , M (tp)×(1−R(tp)) = zf (z) dz =
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 20000 p p
D(tp) 20000−tp t2p = 40000 × tp − t2 = 0.0008 , t = 15000
=
tp × 20000
+ 40000
p p
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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6
Using the Γ function table (see Appendix 7) and the formula
2 1
σ2 = η2 × [Γ(1 + ) − Γ2(1 + )]
β β
gives
σ2 = (55000)2 × [1.1765Γ(1.1765) − Γ2(1.5882)] = 3025 × 106 × (1.0883 − 0.796) =
= 885.2 × 106
σ = 29753 km and therefore µ
= 49000
= 1.65. Now we can proceed to Glasser’s
σ 29753
paper analysis (Figure 2)
Figure 2: Problem 6 Glasser’s graph
According to the graph ρ = 0.92, which means 8% of expected improvement,
and Z = −0.6. Furthermore,
tp = µ + Z × σ = 49000 − 0.6 × 29753 = 31148
Problem 7 Firstly, sort the data in increasing order
Hours 80 100 115 130 150 170 200
Days 3.33 4.17 4.79 5.42 6.25 7.08 8.33
EERDDDFDFDDas
, 7
Figure 3: Problem 7 Weibull plot
Figure 4: Problem 7 Glasser’s Block graph
Secondly, it is reasonable to assume that the data belongs to the Weibull
distribution. Then using the median ranks table (see Appendix 8) we get
Weibull graph (Figure 3). From Figure 3 β = 3, µ = 5.7 × 24 =
136.8,
η = 6.5 × 24 = 156.
, @LECTSOLUTIONSSTUVIA
8
Next
2 1
σ2 = η2 × [Γ(1 + β ) − Γ2(1 + β )]
and therefore σ2 = (156)2 × [Γ(1.67) − Γ2 (1.33)] = 24336 × 0.1059 = 2576
σ = 50.75. Using the obtained result µ = 136.8 = 2.7 and C f = 10
σ 50.75 Cp
Figure 5: Problem 7 Glasser’s Age Graph
From Figures 4 and 5: Z = −1.5, ρ = 0.36, tp = 136.8 − 1.5 × 50.75 = 60.68
for Block replacement policy and Z = — 1.55, ρ = 0.38, tp = 58.14 for age-based
replacement policy.
Problem 8 Using σ = 20000
1000
= 20 and Cf
Cp
= 2 and the Glasser’s graph
µ
we get the following approximations: Z = — 2.1, ρ = 0.6 (40% of expected
improvement).
tp = µ + Z × σ = 20000 − 2100 = 17900
Problem 9 µ
= 150000
= 15 and Cf
= 10 Using the Glasser’s graph for block
σ 10000 Cp
replacement (Figure 6) we get: ρ = 0.14, Z = −3.3.
(a)
tp = µ + Z × σ = 150000 − 33000 = 117000
(b)
1 − ρ = 0.86 or 86% of expected improvement.
(c)
2000 $
R-o-o-F cost= = 0.0133
150000 km
Optimal policy cost= ρ×R-o-o-F= 0.14 × 0.0133 = 0.0018 $
km
EERDDDFDFDDas
, 9
Figure 6: Problem 9 Glasser’s graph
Problem 10
(a) One of the appropriate models is described in section 2.4.2
Cp + Cf × H(tp)
C(t p ) =
tp
(b) The most convenient way of solving the problem with the provided
information is to use Glasser’s graph for block replacement (Figure 7).
Figure 7: Problem 10 Glasser’s Block graph
, @LECTSOLUTIONSSTUVIA
10
Cf 150 µ 200
= = 1.5, = = 20
Cp 100 σ 10
From the graph Z = −2.3, ρ = 0.83 (17% of expected savings)
tp = µ + Z × σ = 200 − 23 = 177
Problem11
Cf 200 + 700 900
= = =3
Cp 200 + 100 300
and
µ 150
= = 10
σ 15
Using the Glasser’s Block replacement graph:
Z = −2.2, ρ = 0.45 (55% of expected improvement) tp = 150 − 33 = 117
Problems 12-16 OREST is a user-friendly piece of software designed espe-
cially for solving optimal replacement decision problems. The following screen
shots are meant to give the reader some idea on how to input data and use it
for analysis. At any point the user can access the help file by pressing ”Help”
button in the top right corner of the window.
Figure 8: OREST screen shot
Figure 8 illustrates the main selection window of the program, where you
can choose one of the available options:
• Data Input (see Figure 9), where one can create/change components
• Trend Analysis
• Weibull Analysis, where statistical analysis can be performed and saved/printed
EERDDDFDFDDas
, 11
• Deterministic Replacement
• Age Preventive Replacement, where replacement decisions are made.
Figure 9: OREST screen shot
The main questions addressed in problems 12-16 are:
• Prepare the data for OREST
• Fit the Weibul model to data, determine the key parameters and perform a
certain analysis
• Given the ratio between preventive and failure replacement costs, find an
optimal replacement time
• Estimate a number of failures during a specified period of time under the
optimal policy
• Given a replacement policy, find associated expected costs
Here is how OREST can help the user to answer the above questions.
OREST deals with sets of data which are expected to contain independent
samples from the Weibull distribution (see Appendix 2). Essentially, in order
to qualify for that, elements of a set have to be independent. For this reason,
for example, cumulative times to failures will not do. However, the set of inter-
failure times is the one we can use (under the assumption of independence).
Fitting the data set to the Weibull model is done automatically upon press-
ing the correspondent button in the ”Weibull Analysis” window. Further analy-
sis can be performed using graphs of p.d.f., C.d.f, Reliability and failure rate
functions.
At time of creating a new component, values of preventive and failure costs
are fixed. Using this data, the optimal replacement policy can be obtained in
”Age Preventive Replacement” window. Note that every time an analysis is
performed the user has an opportunity to print the report.
