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
AC(n) = i=1 Li + Rn
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 f(z) dz = 1 tp 40000 −tp
=
− 40000 40000
R(tp ) = 1 − F (tp) = 1 −
According to the model, 0
C(tp ) = CpR(tp) + Cf (1 − R(tp))
t R(t ) + M (t )(1 R(t )) =
40000−tp p p p − p
t
+ 200 × 40000 p
= 100 ×4000040000
tp tp = 100(80000 + 2tp)
2
− R
tp × 40000 + 0 zf (z) dz 80000t −
p pt
C(tp ) = 0.0143 , tp = 10000
0.01 , tp = 20000
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
, t ∈ [0..15000]
25000
f(t) = 2
25000
, t ∈[15000..25000]
0 , elsewhere
Secondly,
Z tp ( , tp ∈
zf (z) dz = t2
p [0..15000]
M(tp)×(1−R(tp )) = 50000
15000 2 tp z
0 R
50000 +2 1 25000 dz
5000 , tp ∈[15000..20000]
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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 )) = 4500 , = 15000 , R(t p ) = 0.4 , tp = 15000
tp
11500 , tp = 20000 0 , tp = 20000
Using the suggested model C(tp ) = CpR(tp )+Cf (1−R(tp ))
for the given values
tpR(tp)+M (tp)(1−R(tp))
of Cf , Cp yields
0.093 , tp = 5000
0.067 , tp = 10000
C(tp ) = 0.063 , tp = 15000
0.078 , tp = 20000
Therefore 15000 km is the optimal preventive replacement age.
2 , tp ∈ [0..2]
10
Problem 4 Similarly to Problem 3 f(tp ) = 101 , tp ∈ [2..8]
0.6 , tp = 2 0 , elsewhere
0.4 , tp = 4
0.2 , tp = 6
From the graph R(tp) =
0 , tp = 8
(R t
Z p 2×z
dz , ∈ [0..2]
tp
t
M(tp ) × (1 − R(tp)) = zf (z) dz = 0 2 10
2×z tp z
p
=
R
R 10 dz + 2 10 dz , tp ∈ [2..8]
0 0 (t 2
p
10 , tp ∈ [0..2]
= tp2
+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
R(t ) + M (t ) 0.7813 , t = 6 Month
D(tp ) = (1 R(t )) =
tp× p p × − p p
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]
( , t ∈[0..20000]
1
20000
f(t) = 0 , elsewhere
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Similarly to the previous problems, Z tp
tp2
1 , tp < 0
20000−tp
R(tp ) = 20000 , tp ∈ [0..20000] , M (tp)×(1−R(tp)) = zf (z) dz =
40000 0
0 , tp > 20000
Substitution of the given values of Dp and Df into the proposed equation gives:
20000−tp tp
0.00103 , tp = 5000
3× 20000
+ 9 × 20000 120000 + 12 ×tp 0.0008 , tp = 10000
D(tp ) = 20000−tp 2
= 40000 ×tp −t2 = 0.0008 , tp = 15000
tp
tp × 20000
+ 40000
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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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
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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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Next 2 1
σ2 = η2 × [Γ(1 + ) − Γ2(1 + )]
β 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 Cf = 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.
µ 20000 Cf
Problem 8 Using = 1000
= 20 and
Cp
= 2 and the Glasser’s graph
σ
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
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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(tp ) =
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
Cf 150 µ 200 = 20
Cp = 100 = 1.5, σ = 10
From the graph Z = −2.3, ρ = 0.83 (17% of expected savings)
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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
AC(n) = i=1 Li + Rn
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 f(z) dz = 1 tp 40000 −tp
=
− 40000 40000
R(tp ) = 1 − F (tp) = 1 −
According to the model, 0
C(tp ) = CpR(tp) + Cf (1 − R(tp))
t R(t ) + M (t )(1 R(t )) =
40000−tp p p p − p
t
+ 200 × 40000 p
= 100 ×4000040000
tp tp = 100(80000 + 2tp)
2
− R
tp × 40000 + 0 zf (z) dz 80000t −
p pt
C(tp ) = 0.0143 , tp = 10000
0.01 , tp = 20000
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
, t ∈ [0..15000]
25000
f(t) = 2
25000
, t ∈[15000..25000]
0 , elsewhere
Secondly,
Z tp ( , tp ∈
zf (z) dz = t2
p [0..15000]
M(tp)×(1−R(tp )) = 50000
15000 2 tp z
0 R
50000 +2 1 25000 dz
5000 , tp ∈[15000..20000]
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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 )) = 4500 , = 15000 , R(t p ) = 0.4 , tp = 15000
tp
11500 , tp = 20000 0 , tp = 20000
Using the suggested model C(tp ) = CpR(tp )+Cf (1−R(tp ))
for the given values
tpR(tp)+M (tp)(1−R(tp))
of Cf , Cp yields
0.093 , tp = 5000
0.067 , tp = 10000
C(tp ) = 0.063 , tp = 15000
0.078 , tp = 20000
Therefore 15000 km is the optimal preventive replacement age.
2 , tp ∈ [0..2]
10
Problem 4 Similarly to Problem 3 f(tp ) = 101 , tp ∈ [2..8]
0.6 , tp = 2 0 , elsewhere
0.4 , tp = 4
0.2 , tp = 6
From the graph R(tp) =
0 , tp = 8
(R t
Z p 2×z
dz , ∈ [0..2]
tp
t
M(tp ) × (1 − R(tp)) = zf (z) dz = 0 2 10
2×z tp z
p
=
R
R 10 dz + 2 10 dz , tp ∈ [2..8]
0 0 (t 2
p
10 , tp ∈ [0..2]
= tp2
+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
R(t ) + M (t ) 0.7813 , t = 6 Month
D(tp ) = (1 R(t )) =
tp× p p × − p p
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]
( , t ∈[0..20000]
1
20000
f(t) = 0 , elsewhere
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Similarly to the previous problems, Z tp
tp2
1 , tp < 0
20000−tp
R(tp ) = 20000 , tp ∈ [0..20000] , M (tp)×(1−R(tp)) = zf (z) dz =
40000 0
0 , tp > 20000
Substitution of the given values of Dp and Df into the proposed equation gives:
20000−tp tp
0.00103 , tp = 5000
3× 20000
+ 9 × 20000 120000 + 12 ×tp 0.0008 , tp = 10000
D(tp ) = 20000−tp 2
= 40000 ×tp −t2 = 0.0008 , tp = 15000
tp
tp × 20000
+ 40000
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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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
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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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Next 2 1
σ2 = η2 × [Γ(1 + ) − Γ2(1 + )]
β 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 Cf = 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.
µ 20000 Cf
Problem 8 Using = 1000
= 20 and
Cp
= 2 and the Glasser’s graph
σ
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
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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(tp ) =
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
Cf 150 µ 200 = 20
Cp = 100 = 1.5, σ = 10
From the graph Z = −2.3, ρ = 0.83 (17% of expected savings)
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