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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
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 O n e can use the model from section 2.5 (see 2.5.2). In this problem
Cap = 100, Cu = 200,
∫ tp tp
F (z) ds = 1—
= 40000 − top
R (top) = 1 − F (top) =
0 40000 40000
1−
According to the model,
P R (top) + Cu (1 − R (top)) =
C (t)p =
Taper (top) + M (top) (1 − R (top))
100 × 40000 −tp + 200 × t p 100(80000 + 2t)p
= 40000 ∫ tp 40000
=
80000tp − t2
40000−t p
T× + of (z) ds
p 40000 0 p
0.0143, top = 10000
0.01 , top = 20000
C (top) =
0.0093, top = 30000
0.01 , top = 40000
Calculations above indicate that the optimal age is 30000 km.
Problem 3 F i r s t l y , one can find f (t). Since the area below the
probability density curve is equal to 1, the area of each rectangle on the 5
Figure 2.40 is 1.
It follows then, that
1
25000 , t ∈ [0...15000]
F (t) 2 , t ∈ [15000...25000]
= 250
00
0 , elsewhere
Secondly,
∫ (t2
Tp p , top ∈ [0...15000]
M (top)×(1−R(top)) = zf (z) dz = 50000
150002
∫ tp
0 50000
+2 z
15000 25000 dz , tp ∈ [15000..20000]
,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
Using the suggested model C(tp) = CpR(t p)+Cf (1−R(t p)) 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 (t p) = 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]
∫02 2×z ∫t p
M (tp) × (1 − R(tp)) = zf (z) dz = 10
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 = 6 Month
tp × R(tp) + M (tp) × (1 − R(tp))
0.8824 , t p = 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
f (t) = 20000
0 , elsewhere
@
@SSeeisismmicicisisoolalatitoionn
, 5
Similarly to the previous problems,
1 , tp < 0 ∫ tp
t2
, tp ∈ [0..20000] , M (tp)×(1−R(tp)) =
20000−tp p
R(tp) = 20000
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
p p
D(tp) = 20000
20000−tp
20000
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
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 O n e can use the model from section 2.5 (see 2.5.2). In this problem
Cap = 100, Cu = 200,
∫ tp tp
F (z) ds = 1—
= 40000 − top
R (top) = 1 − F (top) =
0 40000 40000
1−
According to the model,
P R (top) + Cu (1 − R (top)) =
C (t)p =
Taper (top) + M (top) (1 − R (top))
100 × 40000 −tp + 200 × t p 100(80000 + 2t)p
= 40000 ∫ tp 40000
=
80000tp − t2
40000−t p
T× + of (z) ds
p 40000 0 p
0.0143, top = 10000
0.01 , top = 20000
C (top) =
0.0093, top = 30000
0.01 , top = 40000
Calculations above indicate that the optimal age is 30000 km.
Problem 3 F i r s t l y , one can find f (t). Since the area below the
probability density curve is equal to 1, the area of each rectangle on the 5
Figure 2.40 is 1.
It follows then, that
1
25000 , t ∈ [0...15000]
F (t) 2 , t ∈ [15000...25000]
= 250
00
0 , elsewhere
Secondly,
∫ (t2
Tp p , top ∈ [0...15000]
M (top)×(1−R(top)) = zf (z) dz = 50000
150002
∫ tp
0 50000
+2 z
15000 25000 dz , tp ∈ [15000..20000]
,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
Using the suggested model C(tp) = CpR(t p)+Cf (1−R(t p)) 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 (t p) = 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]
∫02 2×z ∫t p
M (tp) × (1 − R(tp)) = zf (z) dz = 10
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 = 6 Month
tp × R(tp) + M (tp) × (1 − R(tp))
0.8824 , t p = 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
f (t) = 20000
0 , elsewhere
@
@SSeeisismmicicisisoolalatitoionn
, 5
Similarly to the previous problems,
1 , tp < 0 ∫ tp
t2
, tp ∈ [0..20000] , M (tp)×(1−R(tp)) =
20000−tp p
R(tp) = 20000
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
p p
D(tp) = 20000
20000−tp
20000
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
