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 Title:
 Hysteresis effect in repair and replacement problems with downtime.
 Creator:
 Brown, Jonathan Franklin, 1946
 Publication Date:
 1973
 Language:
 English
Subjects
 Subjects / Keywords:
 Cost functions ( jstor )
Equipment failures ( jstor ) Infinity ( jstor ) Long run costs ( jstor ) Operating costs ( jstor ) Polynomials ( jstor ) Replacement value ( jstor ) Sufficient conditions ( jstor ) Total costs ( jstor ) Unit costs ( jstor )
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 University of Florida
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 University of Florida
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 Copyright Jonathan Franklin Brown. Permission granted to the University of Florida to digitize, archive and distribute this item for nonprofit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
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 14171914 ( OCLC )
0022915789 ( ALEPH )

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HYSTERESIS EFFECT IN REPAIR AND REPLACEMENT
PROBLEMS WITH DOWNTIME
By
JONATHAN FRANKLIN BROWN
A DISSERTATION PRESENTED TO THE GRADUATE
COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA 1973
To my loving wife, Marci,
and to my parents
ACKNOWLEDGMENTS
The author wishes to express his gratitude to those who helped him in the preparation of this dissertation. He especially thanks Dr. B. D. Sivazlian, chairman of the supervisory committee, for introducing the author to the problem. Dr. Sivazlian's guidance and helpful assistance are greatly appreciated. The author would also like to sincerely thank Dr. J. F. Burns, Dr. T. S. Hodgson, Dr. E. J. Muth, and Dr. R. L. Scheaffer, members of the committee, for their advice and encouragement.
The author also acknowledges Mrs. Karen Walker for her excellent typing.
Finally, thanks go tohis wife, Marci, for her confidence and support.
This research was supported in part by U. S. Army Contract No. AROD3112470G92.
iii
TABLE OF CONTENTS
Page
ACKNOWLEDGMENTS ........................................... iii
LIST OF TABLES........................................ ..... vii
LIST OF FIGURES............................................ ix
ABSTRACT......................... .......................... xi
CHAPTERS:
1. THE PROBLEM ...................................... 1
1.1 Introduction.................................. 1
1.2 Optimization Criteria......................... 3
1.3 Literature Review ............................. 4
a. Problem I minimal repair................. 7
b. Problem II major repair ................. 10
c. Problem III hysteresis.................. 14
d. Downtime for Repair...................... 15
1.4 Functional Equation Approach ................... 15
1.5 Problem Formulation........................... 20
a. The Functional Equation for the
Hysteresis Problem........................ 20
b. The Economically Controlled Linear
Hysteresis Problem........................ 29
1.6 Chapter Outline....... ...... ................ 31
2. AGE REPLACEMENT WITH MAJOR REPAIR AND DOWNTIME..... 33
2.1 Introduction .......................... ........ 33
2.2 A Sufficient Condition for a Unique
Replacement Age X...... ............... ..... .... 33
2.3 The Undiscounted Long Run Expected Cost Per
Unit Time ................................. 44
iv
TABLE OF CONTENTS (Continued)
Page
3. AGE REPLACEMENT WITH PARTIAL REPAIR AND DOWNTIME... 47
3.1 Introduction .................................. 47
3.2 General Solution to FunctionalDifferential
Equation with X(x) = X and a(x) = ax.......... 48
3.3 Special Polynomial Forms of g(x)..... ......... 55
a. g(x) = co ................................. 55
b. g(x) = c1x + c ........................... 56
c. g(x) = c2x2 + l + o .................... 67
4. AGE REPLACEMENT WITH ECONOMICALLY CONTROLLED
PARTIAL REPAIR AND DOWNTIME ........................ 80
4.1 Introduction.................................. 80
4.2 General Solution to FunctionalDifferential
Equation with X(x) = A and p(E) = a.......... 84
4.3 Special Polynomial Forms of q(x).............. 89
a. q(x) = c0 + XE............................ 89
b. q(x) = c1x + c0 + XE ...................... 90
c. q(x) = clx + c0 + AEx ........................ 96
5. THE NUMBER OF OPERATING PERIODS IN [O,X] FOR
THE LINEAR HYSTERESIS PROBLEM ........................ 101
5.1 Introduction................ ..... ...... 101
5.2 Probability Mass Function of N(X) ............. 102
5.3 Expected Value of N(X)........................ 113
6. CONCLUSIONS AND RECOMMENDATIONS .. .................. 117
6.1 Conclusions. .......................... 117
6.2 Recommendations for Future Research ........... 118
V
TABLE OF CONTENTS (Continued)
Page
APPENDICES .................. .... s ...... ................... 121
APPENDIX A ............. .............................. 122
APPENDIX B........................ ................... 125
REFERENCES ... ............................ ... ... 146
ADDITIONAL REFERENCES ...................................... 148
BIOGRAPHICAL SKETCH.......................................... 149
vi
LIST OF TABLES
Table Page
1.1 Significant Papers for Problems I and II.......... 16
3.1 Table of Optimum X and Corresponding fl(O) for g(x) = O1x When a = 0.0........................... 62
3.2 Table of Optimum X and Corresponding fl(0) for g(x) = 10x When a = 0.25 .......................... 63
3.3 Table of Optimum X and Corresponding f1(0) for g(x) = 10x When a = 0.50..........................
3.4 Table of Optimum X and Corresponding f1(0) for g(x) = 10x When a = 0.75................ ....... 65
3.5 Table of Optimum X and Corresponding f1(0) for g(x) = 10x When a = 1.0......................... 66
3.6 Table of Optimum X and Corresponding fl(0) for g(x) = 1.75x2 + 7.5x When a = 0.0.................. 75
3.7 Table of Optimum X and Corresponding fl(0) for g(x) = 1.75x2 + 7.5x When a = 0.25................ 76
3.8 Table of Optimum X and Corresponding fl(O) for g(x) = 1.75x2 + 7.5x When a = 0.50................ 77
3.9 Table of Optimum X and Corresponding fl(0) for g(x) = 1.75x2 + 7.5x When a = 0.75 ................ 78
3.10 Table of Optimum X and Corresponding fl(0) for g(x) = 1.75x2 + 7.5x When a = 1.0................ 79
4.1 Table of Optimum Values for E,Xpolicy for a Convex Curve p(E) as in Figure 4.6................ 94
4.2 Table of Optimum Values for E,Xpolicy for a Concave Curve p(E) as in Figure 4.5........ ...... 95
4.3 Table of Optimum Values for E,Xpolicy for a Linear Curve p(E) as in Figure 4.1................ 97
4.4 Table of Optimum Values for E,Xpolicy for a Convex Curve p(E) as in Figure 4.6................ 99
vii
LIST OF TABLES (Continued)
Table Page
4.5 Table of Optimum Values for E,Xpolicy for a Concave Curve p(E) as in Figure 4.5............. 100
B.1 Table of S(Y,B,A) for B = .10..................... 128
B.2 Table of S(Y,B,A) for B = .20...................... 130
B.3 Table of S(Y,B,A) for B = .30 ..................... 132
B.4 Table of S(Y,B,A) for B w .40 .............. .. 134
B.5 Table of S(Y,B,A) for B = .50..................... 136
B.6 Table of S(Y,B,A) for B = .60 ...................... 138
B.7 Table of S(Y,B,A) for B = .70..................... 140
B.8 Table of S(Y,B,A) for B = .80....................... 142
B.9 Table of S(Y,B,A) for B = .90 .................... 144
viii
LIST OF FIGURES
Figure Page
1.1 The variation of f(x) illustrating the cost curves associated with the alternatives of do not replace (h(x)) and replace (K(x) + h(O)) and the decision
regions of do not replace and replace for f(x).... 19
1.2 Sample function of the service age process S(t) for the minimal repair problem with downtime,
where x is the initial equipment age, and X is
the equipment replacement age ..................... 22
1.3 Sample function of the service age process S(t) for the major repair problem with downtime,
where x is the initial equipment age, and X is
the equipment replacement age..................... 23
1.4 Sample function of the service age process S(t) for the hysteresis problem with downtime, where x is the initial equipment age, X is the equipment replacement age, and a(u) is the hysteresis
age function..... .................................... 24
4.1 Example of a linear curve p(E) .................... 81
4.2 Example of a linear curve p(E).................... 81
4.3 Example of a curve p(E)........................... 82
4.4 Example of a curve p(E) ................ 82
4.5 Example of a concave curve p(E).................... 83
4.6 Example of a convex curve p(E) .................... 83
5.1 Sample function of the service age process S(t) when N(X) = 1 for the linear hysteresis problem
with downtime and constant failure rate function.. 105
5.2 Sample function of the service age process S(t) when N(X) = 2 for the linear hysteresis problem
with downtime and constant failure rate function.. 105
5.3 Sample function of the service age process S(t) when N(X) = 3 for the linear hysteresis problem
with downtime and constant failure rate function.. 106
ix
LIST OF FIGURES (Continued)
Figure Page
5.4 Sample function of the service age process S(t) when N(X) = 4 for the linear hysteresis problem
with downtime and constant failure rate function.. 106 B.1 Graph of Logl0 S(Y,B,A) for B = .10 .............. 129
B.2 Graph of Logl0 S(Y,B,A) for B = .20................ 131
B.3 Graph of Logl0 S(Y,B,A) for B = .30............... 133
B.4 Graph of Logl0 S(Y,B,A) for B .40............... 135
B.5 Graph of Log10 S(Y,B,A) for B = .50............... 137
B.6 Graph of Log10 S(Y,B,A) for B = .60............... 139
B.7 Graph of Logl0 S(Y,B,A) for B = .70............... 141
B.8 Graph of Logl0 S(Y,B,A) for B = .80............... 143
B.9 Graph of Logl0 S(Y,B,A) for B = .90............... 145
Abstract of Dissertation Presented to the
Graduate Council of the University of Florida in Partial
Fulfillment of the Requirements for the Degree of Doctor of Philosophy
HYSTERESIS EFFECT IN REPAIR AND REPLACEMENT PROBLEMS WITH DOWNTIME
By
Jonathan Franklin Brown
December, 1973
Chairman: Dr. B. D. Sivazlian
Major Department: Industrial and Systems Engineering A new class of repair and replacement problems called the
hysteresis problem is investigated. This problem is concerned with the amount of recovery of service life following a repair with
arbitrary time distribution, extreme cases of which are minimal and major repair. Incorporated in the cost structure of the hysteresis problem are variable operating costs, repair costs, replacement costs, and loss of revenue per unit of downtime. A mathematical formulation
for the discounted hysteresis problem is presented using the
functional equation approach. Optimal age replacement policies, which minimize the long run total expected discounted cost, are examined for the linear hysteresis problem.
The extreme case of major repair is investigated, and a
generalization of previously obtained results is determined. For
the linear hysteresis problem, specific cost functions are examined
for an equipment subject to a constant failure rate. In addition,
numerical results for the optimal age replacement policy are obtained. An extension of the linear hysteresis problem incorporating an
xi
expenditure repair function to control the amount of recovery of service life is examined. A new operating policy, which is an extension of the age replacement policy, is investigated numerically for different age recovery functions. A renewal process associated with the linear hysteresis problem is examined, and a probability mass function obtained.
xii
CHAPTER 1
THE PROBLEM
1.1 Introduction
The literature concerned with the determination of optimal replacement policies for continuously aging equipment subject to failure and continuous review has examined only the problems of minimal and major repair. The distinction between these problems is in the 'amount' of recovery of equipment life following a repair. Whereas no recovery of equipment life occurs after a minimal repair, there is a complete recovery of equipment life following a major repair. Recovering equipment life is equivalent to reducing service age, the total operating time of the equipment.
For most of the literature, the costs associated with repair and replacement have been considered invariant with respect to the service age of the equipment. In addition, the majority of the literature has been concerned with the determination of necessary and sufficient conditions for minimizing either total expected discounted cost or expected cost per unit time over an infinite time horizon. Not until the last two years has any attention been focused on continuous variable operating and repair costs. Also, only a few papers in the literature have assumed that repairs are not instantaneous.
Between the extremes of minimal repair and major repair exists the situation of partial repair. Following this repair there is a hysteresis effect, which is a partial reduction of service age. This is in contrast to a complete (no) reduction of service age in the case of a major (minimal) repair. The amount of partial reduction (or
2
hysteresis) can be a function of the length of the repair, the expense incurred in undertaking the repair, and the service age of the equipment at the instant of failure. In addition, by varying the amount of partial reduction of service age, it is possible to recover the extreme problems of minimal and major repair. The problem generated by considering the amount of partial reduction of service age shall be called the hysteresis problem. To date,optimal policies have not been determined for the hysteresis problem.
A special case of the hysteresis problem, called the linear hysteresis problem, occurs when the function defining the partial reduction of service age is linear. If the amount of partial repair in this problem is economically controlled, the resulting problem shall be called the economically controlled linear hysteresis problem.
This dissertation presents the formulation of the hysteresis problem using an optimization approach due to Bellman [5) (also Bellman and Dreyfus [6]), called the functional equation approach. Analytic and quantitative results are obtained for the linear hysteresis problem. For the economically controlled linear hysteresis problem, quantitative results are obtained for which a qualitative analysis is performed. Some stochastic elements of the hysteresis problem are also investigated.
The remainder of this chapter is devoted to several developmental aspects of the hysteresis problem. There is a discussion of the basic optimization criteria considered in the literature and in this dissertation. An extensive literature survey is presented. The topics covered include: minimal repair problem, major repair problem,
3
preventive maintenance hysteresistype problem, and downtime for repair. The functional equation approach is introduced and discussed as a means of providing a background for the formulation of the hysteresis problem. This formulation yields a functionaldifferential equation, the solution to which is determined for the major repair problem and the linear hysteresis problem with equipment failure rate function equal to a constant in Chapters 2 and 3, respectively. The functionaldifferential equation for the economically controlled linear hysteresis problem is also developed. This chapter concludes with an outline of the remaining chapters.
1.2 Optimization Criteria
Two approaches to replacement theory, known as the renewal theoretic approach and the functional equation approach, result in different optimization criteria. For equipment subject to failure, the renewal theoretic approach and the functional equation approach result in the criterion of expected cost per unit time and the criterion of total expected discounted cost, respectively. These criteria apply for either a finite or infinite time span. When equipment is not subject to failure and a discount factor is assumed, the only incentive to replace, if ever, is the increasing cost disadvantage of retaining the present equipment versus replacing it with a new equipment. This is known as the classical equipment replacement problem. For this problem, the functional equation approach yields the criterion of total discounted cost.
For equipment subject to failure, the difference between the optimization criteria is a consequence of the method of aggregating
the relevant costs over a given length of time. In the renewal theoretic approach, the focus is on the expected number of occurrences (or renewals) of two or more events over a given length of time. As a result of this approach, many interesting statistics concerning the stochastic process of service age in the repairreplacement problem are obtained. Since associated with each occurrence of an event is a cost, expressions for the expected cost for a given length of time and the expected cost per unit time are readily obtained. In contrast, the functional equation approach accumulates future expected costs in a cost function. Unless these future costs are discounted to the present (concept of present value), the cost function would grow without bound for an infinite time span. For a finite time span, it is still important to consider discounting since future alternatives are being examined in terms of present day values. As a result, expressions such as total expected discounted cost for either a finite or infinite time span arise.
When equipment is not subject to failure, the element of
uncertainty is removed from the optimization criterion. This results in a criterion of total discounted cost for the functional equation approach. The functional equation approach will be discussed in greater detail in Section 1.4.
1.3 Literature Review
The literature of equipment repair and replacement has
differentiated according to the amount of effort expended on the repair of a failed equipment. The amount of repair can take on the range of possibilities from minimal repair to major repair or overhaul. Minimal
5
repair usually involves the repair or replacement of a minor part of the equipment, sufficient to return the equipment to an operational status as soon as possible. An example of minimal repair is the replacement of a blown fuse in an electrical piece of equipment. A major repair (or overhaul) usually involves the intense testing and repairing and/or replacing of parts of the equipment. The status of the equipment after major repair is considered to be as good as new. Due to the expense of the initial investment and/or the size of the equipment, overhauls are usually limited to equipment which is never replaced. For this reason and because major repairs return equipment to an as good as new state, major repairs are equivalent to replacement. Examples of equipment which are subject to major repair are jet engines, electric generators, and diesel engines. For the remainder of this discussion, it is assumed that only failure induces a repair, whether minimal, partial, or major.
A failed equipment which is returned to an operational status after a minimal repair has the same service age after the repair as prior to the repair. Thus, the equipment failure rate function remains unchanged by minimal repair. When the equipment service age attains a predetermined value, the equipment is replaced with an identical new one. This type of repair and replacement problem shall be referred to as Problem I or the minimal repair problem. After a failed equipment undergoes a major repair, the service age of the equipment is that of a new equipment, which is zero. Thus after major repair, the failure rate function is that of a new piece of equipment. When the equipment attains a predetermined service age, it is replaced with an identical
6
new equipment. This type of repair and replacement problem shall be referred to as Problem II or the major repair problem.
An extension to Problems I and II is a repair and replacement problem which lies between these problems. Repairing a failed equipment beyond a minimal repair but not up to a major repair is often feasible. Minimal repair often entails the partial disassembly of the equipment. Instead of simply repairing or replacing the failed part, it may be economically feasible to repair or replace parts which have not failed, since the initial labor cost for disassembling the equipment has already been paid. After such a repair, the service age of the equipment is less than its service age prior to commencing the repair. The partial reduction of service age can be a function of repair costs, length of downtime, and service age at the instant of failure. In addition, replacements are made with identical new equipment. This type of repair and replacement problem shall be referred to as Problem III or the hysteresis problem. It can even be rationalized that a minimal repair results in some recovery of service age.
In order to determine when to repair or replace an equipment, it is necessary to utilize an inspection policy. This policy is used to determine the state, either operating or failed, and the service age of the equipment. Inspection policies can be continuous or discrete. Continuous inspection policies involve constant monitoring of the equipment. Discrete inspection policies are customarily periodic although some random policies have been examined in the literature.
Equipment aging can be either discrete or continuous. Discrete
7
aging is often associated with a periodic inspection policy in which the inspection determines the state of the equipment, where the state can have a value from 0 to S inclusive. The state 0 denotes a failed equipment and the state S denotes a new equipment. After inspection, the equipment is classified into one of S + 1 states and a decision is made to either (a) replace, or (b) perform preventive maintenance, or (c) do nothing. If the equipment can be classified into S + 1 operational states, transitions from one state to other states are specified by some probability mass function, and periodic inspections take place, then the resulting model is one of Markovian Replacement. Continuous aging is associated with both discrete and continuous inspection, although the latter is more common in replacement theory.
The literature of replacement theory will be surveyed for
Problems I, II, and III. Under the different optimization criteria previously discussed, the authors in replacement theory have considered primarily the following costs and parameters. They are
K = fixed cost incurred for planned replacement;
k = instantaneous repair cost incurred at failure;
i = interest rate;
F(*) = lifetime distribution function of the equipment;
X(*) = failure rate function or hazard function.
The failure rate function X(.) is also referred to in the literature as the failure rate. The last section of the literature review covers downtime for repair.
a. Problem I minimal repair
The earliest studies in the area of classical equipment
8
replacement were concerned with comparing an existing equipment with a potential replacement. In 1949, Terborgh [24] presented procedures for comparing a challenger (new equipment) and identical future challengers with a defender (present equipment). Grant [14] was concerned with equipment obsolescence, increasing maintenance cost, and decreasing output (increasing equipment inefficiency). He presented procedures for determining the time to replacement of an operating equipment when more efficient equipment was available; no discount factor was assumed; and, the operations cost was nondecreasing. In conjunction with the Machinery and Allied Products Institute [17], Terborgh extended his earlier work to incorporate the concepts of capital investment. This resulted in the development of the classical MAPI model.
Bellman [5] (also Bellman and Dreyfus [6]) demonstrated a method of approach to problems of replacement with an identical equipment and with a superior equipment (technologically superior). The method used a dynamic programming formulation to set up a discrete time functional equation. In order to determine a replacement age X for the equipment, Bellman's objective function was the maximization of the long run total discounted return (profit). Bellman's model incorporated the concepts of salvage value, increasing operations cost, decreasing output of the equipment, and discount factor i. Failure of the equipment was not considered, and the only incentive to replace the equipment was the increasing cost disadvantage of retaining the equipment versus replacing it with a new equipment. Descamps [10] solved the continuous time version of Bellman's functional equation for
9
the model of replacement with like equipment. He restructured the objective function in order that the long run total discounted cost was minimized.
Employing elementary renewal theory, Barlow and Hunter [1]
considered the problem of scheduling replacement of an equipment given a failure rate X(u), and the fixed costs k and K for a minimal repair and a replacement, respectively. Minimal repair and replacement were assumed to be instantaneous. Further, the failure rate at the termination of the minimal repair was the same as at the commencement of the repair. Using a criterion of minimizing the long run total expected cost per unit time, Barlow and Hunter determined that a unique optimal equipment replacement age X existed as the solution to
X
ul'(u)du =
0
provided that the failure rate was strictly increasing to infinity. The authors also considered an age replacement problem, which will be discussed in the following section.
In 1973, Sivazlian [21] extended Barlow and Hunter's result for minimal repair, which had assumed instantaneous minimal repair, to include an arbitrary distribution of downtime for repair. Incorporated in the model were a variable repair cost k(u), a variable operations cost c(u), a cost for revenue lost per unit of downtime r(u), a failure rate X(u), and a discount factor i. In addition, a general distribution function of repair times was incorporated. A functional equation approach, similar to that of Descamps [10] and Beckmann [4], was used to minimize the long run total expected discounted cost,
10
Sivazlian determined that a necessary and sufficient condition for the existence of a unique optimal equipment replacement age X, where X is the unique solution to the equation
X
[1 e ]dc(w) = K
0
is that c(w) (w > 0) be strictly increasing to infinity. Now
w
J(w) = i j(x)dx
with
j(x) = i + X(x)[l h T(i)] and
(w) = {c(w) + r(w) [j(w) i] + A(w)k(w)} j(w) i
where hT(i) is the LaplaceStieltjes transform with parameter i of the general distribution function of repair times. Due to the structure of c(w), it was not necessary for the failure rate to be strictly increasing to infinity for a unique optimal equipment replacement age to exist.
b. Problem II major repair
Every major repair problem in the literature has considered two fundamental costs: a fixed cost K for a planned replacement, and a fixed cost k for a major repair or replacement at failure. In addition, it has been assumed that a planned replacement costs less than or equal to a repair or replacement at failure (i.e., 0 < K < k). This is logical
since an unscheduled repair or replacement would involve an unplanned stoppage in the use of the equipment, and the possible activation of a backup piece of equipment which would not be as efficient. The majority of the papers in the area of major repair for a continuously aging piece of equipment subject to failure and continuous inspection have concluded that a sufficientcondition for a unique finite optimal equipment replacement age X to exist is that the failure rate be continuous and strictly increasing to infinity.
Considering only the costs k and K and the distribution function F, Barlow and Hunter [1] (also Barlow and Proschan [2], [31) examine the major repair problem with instantaneous major repair and planned replacement. Following a method of objective function formulation (Morse [18]) using renewal theory, Barlow and Hunter constructed an objective function for the average cost per unit time over a finite time span, which was the weighted sum (by the costs, K and k) of the expected number of planned replacements and the expected number of unplanned replacements (or major repair) per unit time over the time span. Then letting the finite time span approach infinity, Barlow and Hunter minimized the long run expected cost per unit time to obtain X as the solution to
X
K
X(X) F(t)dt F(X) = k K
0
X will be finite and unique if the failure rate X(u) is continuous and strictly increasing to infinity. The optimal age for a planned replacement is infinite if either (1) the failure rate X(u) = X = constant, or (2) the cost of a planned replacement is equivalent to the cost of a
12
major repair (or unplanned replacement). The latter is a consequence of it never being economically advantageous to replace prior to failure.
Specific results for the unique equipment replacement age for Barlow and Hunter's [1] major repair problem were obtained by Glasser [13] for the truncated normal, Gamma, and Weibull distributions. For each distribution, a graph was presented from which the unique equipment replacement age is readily ascertained after the computation of a cost ratio.
Fox [12] investigated the effects of incorporating a discount factor i in Barlow and Hunter's major repair problem. Continuous discounting was assumed, and the total loss was equal to the sum of the discounted losses incurred at each planned replacement and failure, He determined that provided the failure rate (for known distributions) was continuous and strictly increasing to infinity, a unique equipment replacement age X exists which minimizes the long run total expected discounted loss (or cost). Denardo and Fox [7] proved that even if the cost of a replacement at failure exceeded the cost of a planned replacement, it is not optimal to replace an operating equipment during periods of decreasing failure rate.
Beckmann [4] approached the same problem studied by Fox [12] by formulating a functional equation in a manner analogous to that of Bellman [5] and Descamps [10]. His objective function minimized the long run total expected discounted cost. A necessary condition for a unique optimal equipment replacement age X to exist was determined,but was not presented in a closed form expression.
Employing renewal theory, Scheaffer [20] extended Barlow and
13
Hunter's [1] major repair problem to include an increasing equipment operations cost cta (a > 0). A cost for operation related to equipment age is reasonable, and intuitively appealing, since as an equipment ages the output may decrease or the equipment may become more inefficient. By minimizing the long run expected cbst per unit time, Scheaffer determined that a finite equipment replacement age X exists which is the unique solution to
X
[(k K)X(X) + acX1] F(t)dt (k K)F(X)
0
X
c[X(X) + J ta dF(t)] = K
0
provided that the failure rate is continuous and nondecreasing. The incorporation of an increasing equipment operations cost has altered the condition determined by Barlow and Hunter that the failure rate be strictly increasing to infinity in order that the equipment replacement age X will be finite and unique. For the exponential distribution function, it was demonstrated that the unique equipment replacement age is finite provided that the operations cost is strictly increasing to infinity. For a constant or decreasing operations cost, an infinite replacement age is optimal for the exponential distribution function.
The area of Markovian Replacement was pioneered by Derman [8]. Under the assumption of Markov Chains with stationary transition probabilities, Derman developed three theorems, which proved that for an equipment subject to Markovian deterioration with costs associated with the observed state (i.e., determined by inspection) and the decision made (replace or do nothing), the optimal stopping rules belong
14
to a subclass of all possible rules. Further, the optimal stopping rules can be determined by solving linear programming problems. Derman's objective function for his results was the long run total expected cost.
Extensions to Derman's work to obtain formal rules based upon the costs for repair at failure, replacement, and the observed state have been made by Derman [9], Kolesar [16], and Ross [19].
c. Problem III hysteresis
Only one paper, and that in the area of Markovian Replacement, has considered the hysteresis effect. Eppen [11] determined conditions under which a policy for preventive maintenance is optimal for a Markovian deteriorating system with a discrete state space. Following each inspection to determine in which one of the S + 1 states (states are 0,..., S) the system is, a decision is made to either leave the system in the present state or place it in a newer and higher state by performing some maintenance action. Since the system deteriorates from S to 0, it improves in the reverse direction. It is assumed that all maintenance actions are performed instantaneously. The cost of improving the system by z states (z integer) is cz (c > 0). For each state an operating cost L(*) is assigned, such that L(.) is a finite, positive, real valued function defined on 0, 1, ..., S which possesses the properties: (1) L(.) is convex and (2) AL < c2 The deterioration of the system is described by onestepdown transition matrices. A functional equation approach was employed to minimize the cost of operating and maintaining the system. The conditions for the policy of preventive maintenance were based on the onestepdown transition matrices.
15
d. Downtime for Repair
Takacs [22], [23] has investigated the stochastic processes associated with operating and repair times. It is assumed that the succession of operating times between failures is independently and identically distributed, as is the succession of repair times. Moreover, the operating and repair times are mutually independent. The operating and repair statesform disjoint sets, such that the stochastic process of system state (operating or repair) takes on a value in each set in turn. Given the successive sojourn times in the two sets, Takacs determined the long run asymptotic distribution of the sum of the sojourn times spent in a repair state. Special cases for the sojourn times of operating and repair times were discussed.
Weiss [25] formulated an approach for incorporating repair
times (or downtimes) in the major repair problem. He assumed that the distribution of repair times was arbitrary and the only cost incurred in connection with the repair was the fixed cost of the repair (i.e., no cost per unit of downtime). Conditions were determined under which an optimal policy for the major repair problem did not necessarily exist.
Table 1.1 summarizes the significant papers for Problems I and II.
1.4 Functional Equation Approach
In this section, an optimization approach known as the functional equation approach is examined. This approach, which is fundamental to dynamic programming,is utilized in the development of the hysteresis
16
Table 1.1. Significant Papers for Problems I and II
0U 0 C) 0 a CC 4U 04i 0O U A 0 4 41 & b 0 C)
W o M CO o 04 0) C f r P%) f C 4 H Author [*] r r o Descamps [ X No V V Yes C FE Barlow and Hunter [] X No F o No Yes C RT
BeSivazlian [21] X YesNo V V Yes No C FE Barlow and Hunter [1] X No F F No No Yes C RT
Barlow and Hunter [1] X No F F No No Yes C RT Fox [12] X No F F No Yes Yes C RT Beckmann [4] X No F F No Yes Yes C FE Scheaffer [20] X No F F V No No C RT
V variable FE functional equation F fixed RT renewal theory D discrete
C continuous
17
problem in the following section. As applied to replacement theory, the functional equation approach examines, for each value of equipment service age, the costs associated with each of the alternatives of replace and do not replace, and selects the one which is smaller. It is assumed that the 'cost curve associated with the alternative do not replace' intersects at least once the 'cost curve associated with the alternative replace', where the curves are a function of equLipmei service age. It therefore follows that alternating decision regions of do not replace and replace exist. Since it is assumed that the first equipment has an initial age less than or equal to X, then only the age X at which the first region of do not replace terminates is of interest,
The different decision regions including the point(s) of indifference are illustrated by the following application of the functional equation approach. Assume continuous aging. Define for x > 0:
h(x) = total expected discounted cost over an infinite time
horizon for an equipment aged x when following an
optimal policy;
K(x) = replacement cost at age x incorporating a salvage
feature;
f(x) = minimum total expected discounted cost over an infinite
time horizon.
The principle of optimality of dynamic programming states that with respect to the current decision of replace or do not replace, ail future decisions are optimal. Then the expected cost of a decision to replace at age x is K(x) + h(O). Further if no replacement occur at age x, the expected cost for this decision is h(x). Then if XI is th:
18
age at which the equipment is replaced, the following functional equation is obtained.
(x) = min K(x) + h(O) x = X (replace' f(x) = min
h(x) 0 < x < X (do not replace) Several conclusions concerning decision regions and continuity of f(x) at x = X can be drawn from this functional equation. Given an age x, a replace policy is preferred to a do not replace policy if and only if
K(x) + h(O) < h(x)
Similarly, a do not replace policy at age x is preferred to a replace policy if and only if
K(x) + h(O) > h(x)
These two conclusions generate the decision regions of do not replace and replace. The cost curves and the alternating decision regions are illustrated in Figure 1.1. At x = X, the decision maker is indifferent to either a policy of replace or do not replace if and only if
K(x) + h(O) = h(x)
For this last conclusion to be true, it is necessary that the function f(x) be continuous at x = X.
Continuity of the function f(x) at x = X implies that the limit of f(x) as x approaches X from the left and the limit of f(x) as x approaches X from the right are equal. Therefore the cost curves
.19
cd
[1 1 ci 4. (1 A 0'
+ (
L4.J
20
for do not replace (h(x)) and replace (K(x) + h(0)) intersect at age X. The importance of continuity of the function f(x) at x = X is best illustrated by assuming the opposite, that is a discontinuity exists at x = X for f(x). Then the following paradox occurs to the left of the discontinuity (i.e., x < X), the cost curve corresponding to do not replace is lower than the cost curve of replace, which implies that the equipment should not yet be replaced. To the right of the discontinuity (i.e., x > X), the cost curve corresponding to replace is lower than the cost curve of do not replace, which implies that the equipment should already have been replaced. To resolve this paradox, it is necessary to require continuity of the function f(x) at x = X. Then a replacement occurs at age X.
1.5 Problem Formulation
a. The Functional Equation for the Hysteresis Problem
It is assumed that the equipment is continuously monitored or reviewed and that failures are detected instantly. Further, the planning horizon is infinite and all replacements are made with identical equipment. All replacements are assumed to occur instantaaneous!y Following a failure, there is a length of time called the downtime during which the equipment is repaired before being returned to an operational status. A zero downtime corresponds to an instantaneous repair. At the termination of the repair, the service age of the equipment .s governed by the hysteresis age function a(x), where x is the service age of the equipment at the instant of failure and a(x) is the service age at the termination of the repair.
21
The range of values that a(x) can assume lies in the interval [0, x]. When a(x) = x, this corresponds to no reduction of sevice age after a repair, which is the minimal repair problem (Problem I). If a(x) = 0, there is a reduction of service age to zero at the termination of the repair, which is the major repair problem (Problem II). For 0 < a(x) < x, a partial reduction, or hysteresis, of service age occurs, which is the hysteresis problem (Problem III), Figures 1.2, 1.3,and 1.4 are sample functions of the service age process S(t) for Problems I, II, and III with downtime for repair, respectively. Thus, the hysteresis age function covers Problems I, II, and III. The linear hysteresis problem where a(x) = ax for 0 < a < 1 will be investigated in this dissertation.
For a specified value of the replacement age X (X > 0),
the following replacement policy is known as an Xpolicy. Given an equipment with service age u, it is retained if 0 < u < X and replaced instantly if X < u < =, where all replacements are made with identical new (service age zero) equipment in an operating state. Define for u > 0:
K(u) = replacement cost at age u incorporating a salvage
feature (nonrecurring cost);
k(u) = repair cost associated with failure at age u and
incurred instantaneously (i.e., at commencement of
repair);
c(u)du = operations cost of equipment between u and u + du
(recurring cost);
r(u) = loss of revenue per unit of downtime for an equipment
with age u and in a state of repair;
X(u)du = probability that the equipment will fail between
ages u and u + du given that the equipment is in an
operating state at age u;
22
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25
a(u) = hysteresis age function for an equipment aged u, where
a(u) is the age of the equipment at the end of a
repair;
i = interest rate expressed over continuous time, where
exponential discounting is employed;
fl(u) = minimum total expected discounted cost over an infinite
time horizon when the equipment has age u and is in
operation;
f0(u) = minimum total expected discounted cost over an infinite
time horizon when repair on a failed equipment aged u
has just commenced.
It is assumed that the functions K(u), k(u), c(u), r(u), and X(u) are all continuous and K(u) is bounded from above.
A functional equation describing the replacement decision of the Xpolicy in a manner analogous to Beckmann [4] and Descamps [10] is now formulated. Since equipment ages continuously in time while operating and is currently in an operating state with present age of x (0 < x < X), the following functional equation is obtained
K(x) + fl(0) x = X (replace) fl(x) = min c(x)dx + (lidx)[l X(x)dx]fl(x+dx) (1.1) + (lidx) X(x)dx f0(x+dx)
+ o(dx) 0 < x < X (do not replace)
where o(dx) is defined by lim o(dx= 0.
dx dx
The first alternative in the functional equation (1.1) corresponds to a replacement action with an associated fixed cost K(x) and a cost of commencing with a new equipment f1(0). The second alternative corresponds to a retain action, thereby incurring an equipment operations
26
cost and the sum of the discounted costs associated with the equipment in an operating state and age x + dx and the equipment having failed at age x + dx and repair commencing.
For 0 < x < X, no replacement occurs and f (x) satisfies the relation
fl(x) = c(x)dx + (1 idx)[l A(x)dx]fl(x + dx)
+ (1 idx) X(x)dx f0(x + dx) + o(dx) (1.2) Before proceeding to form a Taylor series expansion of f0(x + dx) and fl(x + dx), it is necessary to state some assumptions concerning these functions. It is assumed that the properties of f0(x) and fl(x) are continuity, monotone increasing, and existence of the first derivative. Then performing a Taylor series expansion on f0(x + dx) and fl(x + dx), where the remainder term (after the first two terms in the expansion) is of order o(dx) as dx approaches zero, the following is obtained
fl(x) = c(x)dx + (1 idx)[l X(x)dx][fl(x) + f'(x)dx]
+ (1 idx)X(x)dx[f0(x) + f0(x)dx] + o(dx) (1.3) Simplifying (1.3) yields
fl(x) = c(x)dx + fl(x) + fi(x)dx [i + X(x)]fl(x)dx
+ A(x)f0(x)dx + o(dx) (1.4) Rearranging (1.4), dividing the result by dx and letting dx 0, the following is obtained
f'(x) [i + X(x)lfl(x) + X(x)f0(x) = c(x) (1.5)
27
Before solving equation (1.5), it is necessary to obtain an expression for f0(x) in terms of fl(a(x)) using a derivation due to Sivazlian [21]. Assume that an equipment is in an operating state and has a service age u such that 0 < u < X. If the equipment fails between u and u + du, repair commences instantaneously. The length of downtime, or repair time, T is a random variable whose distribution function HT () is not dependent on the failure characteristics of the equipment. It is assumed that the succession of downtimes and service times forms a sequence of mutually independent random variables (Takacs [23]). At the conclusion of the repair, the service age of the equipment is a(u) (0 < a(u) < u), where a(u) is the hysteresis age function. From the characteristics of a(u), it follows that the service age is reduced due to the completion of a repair whenever a(u) < u.
For an equipment aged u which has just failed, the conditional long run total expected discounted cost is
t
k(u) + r(u)ei6 d + fl(a(u))eit (1.6)
0
given that the repair time has length t and terminates at time t at which time the equipment is returned to an operating state with a service age a(u). Then the unconditional long run total expected discounted cost for an equipment aged u which has just failed is
St
f0(u) = [k(u) + r(u)e d6 + fl (a(u))eit]dHT(t)
0 0
= k(u) dHT(t) + (u) (1 e it)dHT(t)
0 0
28
Sfl(a(u)) J eitdHT(t)
0
=k(u) + r + [f1(a(u)) ] eitdH(t) (1.7)
0
Let
hT(s) = etdHT(t) Re s > 0 (1.8)
0
denote the LaplaceStieltjes transform of the repair time distribution function HT('), which exists provided that HT(t) is sectionally continuous in every finite interval 0 < t < M and of exponential order 8 for t > M. Upon using (1.8) in (1.7), the following is obtained
f0(u) = k(u) + r u) [f(a(u)) u) ]hT(i)
Using the last expression to substitute for f0(x) in (1.5) and rearranging yields
f'(x) [i + X(x)]fl(x) + X(x)hT(i)fl(a(x))
= c(x) k(x)X(x) [1 hT(i)]X(x) (1.9) Let
g(x) = c(x) + k(x)A(x) + r [1 h(i)]X(x)
Substituting the prior expression in (1.9) results in the functionaldifferential equation
f'(x) [i + X(x)]fl(x) + X(x)hi)f(a(x)) = g(x) (1.10)
When a(x) = x (minimal repair) or a(x) = 0 (major repair), the
29
solution to (1.10) can be determined by using an appropriate integrating factor. In Chapter 2, the solution to (1.10) with a(x) = 0 is obtained and investigated. A general analytic solution to the functionaldifferential equation (1.10) does not appear to be available or obtainable (Appendix A). However, it is possible to derive a solution expression in the special case when the hysteresis age function is linear and the failure rate function is equal to a constant. This special case is investigated in Chapter 3. Recall that when the hysteresis age function is linear (i.e., a(x) = ax), the resulting problem is the linear hysteresis problem.
b. The Economically Controlled Linear Hysteresis Problem
Consider the linear hysteresis problem with an added refinement. It is desired to control economically the amount of partial repair and the corresponding fractional reduction of service age at the termination of the repair by spending funds at the start of the repair. This expenditure is in addition to the repair cost k(*) incurred at the commencement of the repair and is a function of the level of expenditure E and the age u at which the repair commences. The function so defined is C(E,u), where
C(E,u) = Eum m = 0,1,2,... (1.11) If no funds are expended, that is E equals zero, then a minimal repair will be performed.
The service age at the termination of the repair is a function of the level of expenditure E and the age u at which the repair commences. If C(E,u) is spent on the repair, then the fraction of
30
service age remaining after the repair is p(E), where 0 < p(E) < 1. The service age at the end of the repair is p(E)u. The new problem generated by considering the new functions C(E,u) and p(E) shall be called the economically controlled linear hysteresis problem.
The value of E acts as a decision variable since it controls both p(E) and C(E,u). Consequently, the previously defined Xpolicy is modified to incorporate E, and becomes a twovariable decision policy. For specified values of the level of expenditure E (E > 0) and the replacement age X (X > 0), the following is known as the E,Xpolicy. An equipment with service age u is retained if 0 < u < X. Further if failure has just occurred, C(E,u) is spent on repair to reduce to p(E)u the service age of the equipment. If X < u < , then the equipment is replaced instantaneously, where all replacements are made with identical new (service age zero) equipment in an operating state. Thus, the E,Xpolicy determines when to replace the equipment and the service age following a repair.
Under the added assumptions of the economically controlled linear hysteresis problem, the conditional long run total expected discounted cost for an equipment aged u (u < X) which has just failed is
C(E,u) + k(u) + r(u)eid + fl((E)u)eit (1.12)
( )de + f(p(E)u)e
0
given that the repair time has length t and terminates at time t at which time the equipment is returned to an operating state with a service age p(E)u. Noting the similarity of (1.12) and (1.6), it follows from examining (1.9) that the unconditional long run total expected discounted cost for an equipment aged u which has just failed is
31
r(u) ru)
f 0(u) = C(E,u) + k(u) + i + (p(E)u) T )
Using the last expression to substitute for f0(x) in (1.5) and rearranging yields
fl(x) [i + X(x)]fl(x) + A(x)hT(i)fl(p(E)x)
= (x) k(x)(x) r) [1 hT(i)]X(x) C(E,x)X(x)
(1.13)
Let
q(x) = c(x) + k(x)A(x) + (x) [1 hT(i)(x) + C(E,x)X(x)
Substituting the prior expression in (1.13) results in the functionaldifferential equation
f'(x) [i + X(x)]fl(x) + X(x)hT(i)fl(p(E)x) = q(x) (1.14) for the economically controlled linear hysteresis problem. For the case when the failure rate function is equal to a constant, the solution to the functionaldifferential equation (1.14) is determined in Chapter 4. The E,Xpolicy is investigated quantitatively in Chapter 4 using this solution.
1.6 Chapter Outline
In Chapter 2, results obtained by Barlow and Hunter, Fox,
Beckmann, and Scheaffer for the major repair problem are extended to incorporate the concept of downtime for repair. Starting with the functionaldifferential equation (1.10) with a(x) = 0, a sufficient
32
condition for a unique finite optimal equipment replacement age X to exist is determined for age replacement with major repair and downtime. Several corollaries to this condition are also presented. An expression is obtained for the undiscounted long run expected cost per unit time.
The linear hysteresis problem with downtime for partial repair and constant failure rate function is investigated in Chapter 3. An nth degree polynomial with all positive coefficients is assumed for the cost function g(x), and the functionaldifferential equation (1.10) is solved for a(x) = ax. Three particular polynomial forms are investigated, and sufficient conditions are derived for a unique finite equipment replacement age X to exist for the linear hysteresis problem.
In Chapter 4 the results of Chapter 3 are applied to study the economically controlled linear hysteresis problem with downtime for partial repair. For several curves p(E), two expenditure repair functions are examined: constant and linear. Quantitative results are obtained instead of analytic results, due to the structure of the solution of the functionaldifferential equation (1.14) for each of the expenditure repair and the curve p(E) functions.
Chapter 5 examines the renewal process associated with the number of operating periods between successive replacements. A probability mass function is determined for the number of operating periods in [0,X]. Using this result, the expected value of the number of operating periods in [0,X] is obtained.
Conclusions and recommendations for future research are presented in Chapter 6.
CHAPTER 2
AGE REPLACEMENT WITH MAJOR REPAIR AND DOWNTIME
2.1 Introduction
Previously obtained results for the major repair problem are extended in this chapter. These results, in the form of conditions for the existence of an optimal age replacement policy, called an Xpolicy, were obtained by Barlow and Hunter [1], Fox [12], Beckmann [4], and Scheaffer [20]. They considered the concepts of fixed repair and replacement costs ([1], [12], [4], [201), discounting ([12], [4]), and variable operating cost ([20]).
Additional concepts are incorporated in the investigation of the major repair problem in this chapter. They are variable repair cost, downtime for major repair following a failure, and loss of revenue per unit of downtime. Using the criterion of long run total expected discounted cost, the major repair problem with downtime is investigated utilizing an approach due to Beckmann [4]. A sufficient condition for a unique finite optimal equipment replacement age X to exist is determined for the major repair problem with downtime. Four corollaries to this condition are presented. In addition, an expression for the undiscounted long run expected cost per unit time is determined from fl(0), the minimum total expected discounted cost over an infinite time horizon when the equipment is as good as new (service age zero) and is in operation.
2.2 A Sufficient Condition for a Unique Replacement Age X
The functionaldifferential equation (1.10) of Chapter 1 is
33
34
recalled. It is
f'(x) [i + X(x)]fl(x) + X(x)h (i)f(a(x)) = g(x) (2.1) where
r(x)
g(x) = c k (x) + i [L hT(i)]X(x) (2.2)
The procedure will be to solve the functionaldifferential equation (2.1) for the major repair problem by using an appropriate integrating factor. The boundary condition from the alternative of replace in (1.1), that is
fl(X) = K(X) + fl(0)
is then used to solve for f (0). For major repair, a(x) = 0, and substituting this expression in (2.1) yields
f'(x) [i + X(x)]fl(x) = g(x) fl(0)hT(i)X(x) (2.3)
u
Now J X(O)d R(u) = e u > 0 where R(u), the reliability function, is the probability that an equipment of age 0 will survive age u. Note that R(u) is continuous and R(0) = 1. Then let
R(u) = eiuR(u) u > 0 (2.4) and therefore
d[R(u)] = [i + X(u)lR(u)du (2.5) Note that since R(0) = 1i, then by (2.4) R(0) = 1.
35
Multiplying both sides of (2.3) by R(x) and integrating yields for fl(x)
fl(x) = fl()hT(i) X(u)R(u)du
R (x) R(x)
x
1 g(u)R(u)du (2.6) R(x) 0
If an optimal Xpolicy exists, continuity of the function fl(x) at x = X is required. It then follows that
fl(X) = K(X) + f1(0) (2.7)
Using (2.7) in (2.6), it follows that at x = X
X
fl(0) fl(0)hT (i)x
K(X) + f1(0) X(u)R(u)du R(X) R(X) 0
X
1 g(u)R(u)du R(X) 0
Solving for f1(0) X
K(X)R(X) i g(u)R(u)du fl(0) = 0 X
R(X) i + hT(i) J X(u)R(u)du
0
Or
X
K(X)R(X) + j g(u)R(u)du fl(0) = 0 X(2.8)
1 R(X) hT(i) f X(u)R(u)du
0
Using (2.5) and (O) = 1 in (2.8), the following relation is obtained
36
X
1 R(X) hT(i) J X(u)R(u)du
x 0 x = d[R(u)] hT(i) j X(u)R(u)du
0 0 X X
= [i + X(u)]R(u)du hT(i) I (u)R(u)du
0 0
X
S {i + X(u)[l hT(i)]}R(u)du
0
Let
v(u) = i + X(u)[l hT(i)] u > 0 (2.9)
This is the expression that Sivazlian [21] defined for j(u) in the minimal repair problem (see 1.3a). Then
X X
1 R(X) hT(i) A(u)R(u)du = I v(u)R(u)du (2.10)
0 0
Substituting (2.10) in (2.8) yields
X
K(X)R(X) + g(u)R(u)du fl(0) = 0
j v(u)R(u) du
0
For the remainder of this chapter, the analysis is restricted to the case where the fixed cost of replacement K(X) is independent of the replacement age X (i.e., no salvage feature incorporated in the replacement cost). Since K(X) = K = constant, then
37
X
KR(X) + J g(u)R(u)du
0
fl(0) = 0 (2.11) I v(u)R(u)du
0
A central theorem for age replacement with major repair and downtime is now presented. Theorem 2.1
If g(u) K(u) Ki (u > 0) is strictly increasing to infinity,
v(u)
then a unique finite optimal equipment replacement age X exists which is the solution to the equation
X
Sv(u) g(X) KX(X) Ki g(u) KX(u) Ki R(u)du = K SV(X) v(u)
0
and if the initial age, x, satisfies x < X then an optimal Xpolicy exists for the major repair problem with arbitrary repair time distribution.
Proof:
If the initial equipment age x is such that x < X, then it is clear from substituting (2.11) into (2.6) that if an optimal replacement age X exists, it minimizes both f1(0) and fl(x) for all x. Differentiating (2.11) with respect to X and equating the result to zero yields
X
[KR'(X) + g(X)R(X)] I v(u)R(u)du
0
v(X)A(X)[KR(X) + 5 g(u)R(u)du] = 0 (2.13)
0
38
where
d[R(X)]
R'(X) XX
Rearranging (2.13) and dividing the result by v(X)R(X) J v(u)R(u)du
0
yields
X
KR(X) + f g(u)R(u)du
0 KR'(X) + g(X)R(X)
X v(X)R(X)
i v (u)R(u)du
0
Applying (2.5) in the last expression, the following is obtained
X
KR(X) + f g(u)R(u)du
0 K[i + X(X)]R(X) + g(X)R(X)
X v(X)R(X)
T v(u)R(u)du
0
g (X) KX(X) Ki (2.14) v (X)
If c(u) = 0, k(u) = k, and hT(i) = 1 (instantaneous repair), then g(u) = kX(u) and v(u) = i. Upon substituting these expressions for g(u) and v(u) in (2.14), Beckmann's [4] result is recovered. Since
X X Ri(X) R(O) = d[R(u)] = [i + (u)]R(u)du
0
from (2.5) and R(0) = 1, then
X
(X) = i [i + X(u)]R(u)du (2.15)
0
Substituting (2.15) in (2.14), the following is obtained
39
X
K + [g(u) KX(u) Ki]R(u)du
0 g(X) KA(X) Ki (2.16)
X v(X)
v(u)R(u)du
0 X
Multiplying both sides of (2.16) by v(X) v (u)R(u)du yields
0
X
Kv(X) + v(X) [g(u) KX(u) Ki]R(u)du
0
X
= [g(x) KX(X) Ki] v(u)R(u)du
0
On rearranging the last expression, it follows that
X
v(u)[g(X) KX(X) Ki]R(u)du
0
X
v(X)[g(u) KA(u) Ki]R(u)du = Kv(X)
0
Or
x
g(X) KX(X) Ki (u)du
v(u) v(X) ()R(u)du
0
S(u) K(u) Ki R(u)du = K
0
Simplifying yields
X
v(u) g(X) KA(X) Ki g(u) KX(u) Ki R(u)du K
() v(X) v(u)
0
which is equation (2.12). Now g(u) KX(u) Ki is strictly increasing to infinity by assumption and v(u (u) for u Thusthe integral to infinity by assumption and v(u)(u) > 0 for u > 0. Thusthe integral
40
on the left side of equation (2.12) is strictly increasing to infinity, which implies that a positive value for the equipment replacement age X exists uniquely.
To show that X is the value for which fl(0) attains its absolute minimum, it is sufficient to establish that X is a relative minimum. Using (2.11), a condition for a relative minimum of f (0) at X implies that dx > 0, the following inequality must hold X+dx X KR(X + dx) + f g(u)R(u)du KR(X) + i g(u)R(u)du
0 0
0 > 0 (2.17) X+dx X
J v(u)R(u)du J v(u)R(u)du
0 0
Noting first that
A A
v(u)R(u)du = ihT(i) I R(u)du + [1 hT(i)][l R(A)] > 0
0 0
for A > 0, inequality (2.17) can be written as
X+dx X [KR(X + dx) + g(u)R(u)du] J v(u)R(u)du
0 0 X X+dx > [KR(X) + g(u)R(u)du] v(u)R(u)du
0 0
Multiplying and transposing yields
X+dx X X X+dx
[ (u)R(u)du v(u(u)du g(u)R(u)du v(u)R(u)du
0 0 0 0
X+dx X
> KR(X) v(u)R(u)du KR(X + dx) f v(u)R(u)du
0 0
41
Simplifying X X X+dx
X+dxx
Sg(u)R(u)d v(u)R(u)du g(u)R(u)du J v(u)R(u)du X X+dx > [KR(X) KR(X + dx)]f v(u)R(u)du KR(X) v(u)R(u)du
0 X X+dx (2.18) Rearranging (2.18) and using f d[R(u)] = R(X + dx) R(X)
X
yields
X+dx X
g(u)R(u)du v(u)(u)du
X X+dx
+ [KR(X) + g(u)R(u)dul J v(u)R(u)du
0 X
X
Dividing both sides of the last expression by f v(u)R(u)du results in the following inequality 0
X+dx X+dx f g(u)R(u)du > K J d[R(u)]
0 X
X
KR(X) + g(u)R(u)du X+dx
+ X 0 v(u)R(u)du (2.19) J v(u)R(u)du
0
Using (2.5) and (2.14) in (2.19) yields
X+dx X+dx
S u)g(u)R(u)du K I [i + X(u)]R(u)du
X X
+ g(X) KX(X) Ki v(u)R(u)d v(x)
X 
42
Transposing and collecting terms X+dx
(X) KA(x) Ki[g(u) KX(u) Ki] v(u) v(X) i (u)du > 0
X
Or
X+dx v(u) g(u) KX(u) Ki g(X) KX(X) Ki R(u)du > 0 (2.20)
1 v(u) v(X)
X
Since g(u) K(u) Ki is strictly increasing to infinity by assumption, then for all points X < u < X + dx (dx > 0),
g(u) KX(u) Ki > g(X) KA(X) Ki
v(u) v(X)
and inequality (2.20) is satisfied. In a similar fashion it can be established that
X
v(u) g(X) KX(X) Ki g(u) KX(u) Ki (u)du > 0
J (X) v(u) Xdx
The following corollaries to Theorem 2.1 are submitted without proof and are direct consequences of the theorem. Corollary 2.1
If g(u) KA(u) (u > 0) is strictly increasing to infinity
v(u)
and X(u) is strictly nondecreasing, then a unique finite optimal equipment replacement age X exists which is the unique solution to equation (2.12), and if the initial age, x, satisfies x < X, then an optimal Xpolicy exists for the major repair problem with arbitrary repair time distribution.
43
Corollary 2.2
If g(u) (u > 0) is strictly increasing to infinity and (u)
v(u)
is strictly decreasing, then a unique finite optimal equipment replacement age X exists which is the unique solution to equation (2.12), and if the initial age, x, satisfies x < X, then an optimal Xpolicy exists for the major repair problem with arbitrary repair time distribution.
Corollary 2.3
If [c(u) + k(u)X(u)] and r(u) are increasing functions to infinity and at least one of them is strictly increasing and if X(u) = A = constant, then a unique finite optimal equipment replacement age X exists which is the unique solution to the equation
X
[g(X) g(u)]R(u)du = K
0
and if the initial age, x, satisfies x < X, then an optimal Xpolicy exists for the major repair problem with arbitrary repair time distribution.
Corollary 2.4
If [c(u) + k(u)X(u) KX(u)] (u > 0) is strictly increasing
to infinity, then a unique finite optimal equipment replacement age X exists which is the unique solution to the equation
X
I {[c(X) + k(x)X(X) K (X)] [c(u) + k(u)X(u) KX(u)]}R(u)du = K
0
and if the initial age, x, satisfies x < X, then an optimal Xpolicy exists for the major repair problem with instantaneous repair.
The function g(u) KX(u) Ki
v(u) in the sufficient
44
condition in Theorem 2.1 for the major repair problem with downtime is similar to a corresponding function in the result obtained by Sivazlian [21] for the minimal repair problem with downtime. It is observed that Sivazlian's expression for the function c(u) is equivalent to g(u) (see Section 1.3a and relations (2.2) and (2.9)). The
v(u)
difference in the functions is a multiple of the fixed cost of replacement K. The functional nature of this difference can be rationalized by recalling that major repair is equivalent to replacement with respect to the resulting equipment service age.
2.3 The Undiscounted Long Run Expected Cost Per Unit Time
In order to determine the undiscounted long run expected cost per unit time, denoted by f, from results already obtained, the wellknown result [4], [26] f = lim [ifl(O)] i O
is used. Utilizing (2.11) then iK (X) + i X g(u)R(u)du
f = lim
iO
iJ0 I v(u)R(u)du
0 X
KR(X) + f g(u)R(u)du = lim 0 (2.21) i v v(u) du
0
It is seen from (2.4) that
lim R(u) = R(u) (2.22) i+0
45
Further, from (2.2)
lim g(u) = lim c(u) + k(u)k(u) + r(u) T (u)
iC i
i+0
= c(u) + k(u)X(u) + r(u)X(u) lim h i 0 i
Since lim hT(i) = 1, 'Hospital's rule is employed. Thus
iO
lim g(u) = c(u) + k(u)X(u) r(u)X(u) lim h'(i)
i+O i 0
Providing E(T) exists, E(T) = lim hT(i) and the following is obtained iO
lim g(u) = c(u) + k(u)X(u) + r(u)X(u)E(T) (2.23)
i0O
Using (2.9)
lim v(u)= li 1 + (u) hT(i)
i 0 i i0 i = i + (u) li i
= 1 + X(u)E(T) (2.24) Substituting appropriately (2.22), (2.23), and (2.24) in (2.21) yields
KR(X) + J [c(u) + k(u)X(u) + r(u)X(u)E(T)]R(u)du
f = 0 (2.25) I [1 + X(u)E(T)]R(u)du
0
46
Barlow and Hunter's [1] result for the major repair problem with zero downtime for repair is recovered if c(u) = 0, k(u) = k = constant, and E(T) = 0 (zero downtime). In addition, from Corollary 2.4, the existence and uniqueness of an optimal solution is assured provided that k > K and X(u) is strictly increasing to infinity. Scheaffer's [20] result is recovered by setting c(u) = cu k(u) = k = constant, and E(T) = 0.
CHAPTER 3
AGE REPLACEMENT WITH PARTIAL REPAIR AND DOWNTIME
3.1 Introduction
This chapter examines the hysteresis effect when an age replacement policy, called an Xpolicy, is being followed. In particular, the linear hysteresis problem with nonzero downtime for partial repair and with constant failure rate is investigated. For the functionaldifferential equation (1.10) with failure rate function, X(x), equal to X, and with hysteresis age function, a(x), equal to ax (0 < a < 1), a general solution is determined. This is the sum of two solutions; namely, a complementary and a particular solution.
The complementary solution is determined by assuming it is of the series form. As a consequence, the function S(x,6,XhT(i),a) is obtained, and is examined analytically and graphically in Appendix B. The properties of S(x,6,AhT(i),a) and its derivatives with respect to x are important in obtaining the results in this and the following chapter. Due to the structure of the cost function g(x), when X(x) equals X, it is assumed that the cost function is a member of the class of nth order polynomials, whose coefficients are all positive, in order to obtain the particular solution. The assumption that all the coefficients of the polynomial are positive is later established as one of a pair of sufficient conditions for a unique finite optimal equipment replacement age x to exist.
Three particular polynomial forms with positive coefficients
are examined in this chapter. They are: constant, linear, and quadratic.
47
48
For linear and quadratic polynomials, a pair of sufficient conditions is determined so that a unique optimal equipment replacement age X exists for the linear hysteresis problem; whereas, for constant polynomials, only one sufficient condition is determined. Numerical examples illustrating the linear and quadratic polynomial forms are discussed.
3.2 General Solution to FunctionalDifferential Equation with
X(x) = X and a(x) = ax
The functionaldifferential equation (1.10) of Chapter 1 is recalled. It is
fl(x) [i + X(x)]fl(x) + XA()hT(i)fl[a(x)] = g(x) (3.1) where
r(x)
g(x) = c(x) + k(x)X(x) + i [1 hT(i)]X(x) (3.2)
If the hysteresis age function, a(x), is equal to ax (where
0 < a < 1) in (3.1), the resulting linear differentialdifference equation has variable coefficients due to the failure rate function X(x). As yet, no solution has been found by the author to this differential equation. However, it is possible to obtain a solution by setting the failure rate function, %(x), equal to X, a constant. Letting a(x) = ax (where 0 < a < 1) in (3.1) and A(x) = X = constant in (3.1) and (3.2) yields
fi(x) (i + X)fl(x) + XhT (i)f(ax) = g(x) (3.3) where
g(x) = c(x) + Xk(x) + Xi(x) [1 hT(i)] (3.4)
i
49
The procedure to obtain the general solution to (3.3) consists of determining a complementary solution, a particular solution, and the constant term of these solutions by evaluating the general solution at x = 0.
It is easily verified (Appendix A) that the general solution to (3.3) is of the form
fl(x) = yl(x) + Y2(x) (3.5) where Yl(x) is a solution to the homogeneous equation
yl(x) (i + X)yl(x) + XhT(i)yl(ax) = 0 (3.6) and y2(x) is a solution to the nonhomogeneous equation
y2(x) (i + X)y2(X) + XhT(i)y2(ax) = g(x) (3.7) Equation (3.6) can be solved by assuming a series solution of the form
yl(x) = a.x (3.8) j=0 3
where a., j=0,1,2,..., are constants to be determined.
Using relation (3.8) in (3.6), the following identity in xj, j=0,1,2,..., is obtained
(j+l)a j+xJ (i + X) e U xj + XhT(i) I a.axj 0
j=0 jj+0 J j=0 j
Collecting terms in xj in the last expression yields
S[(j+l)aj+ (i + )a. + h T(i)aj a]xJ = 0
j=0 j
50
which, providing x # 0, results in the difference equation
(j+l)Cj+1 (i + X)a + h T(i).aJ = 0 j=0,1,2,... (3.9)
The solution to (3.9) obtained recursively is
1 j1
a. = a0 k (i + X XhT(i)a ) j > 1
k=0
where a0 is an arbitrary constant whose value is to be determined. From (3.8) then
y(x) = + 1 + n (i + X XhT(i)a k (3.10) j = k=O
Define e = i + A and
0 xj j1
S(x,e,XhT(i),a) 1 + Y n (e XET(i)a (3.11) j=l k=0
The properties of the function S(x,e,AhT(i)a) are discussed in detail in Appendix B. The pertinent ones for this chapter are
(i) for all nonnegative real values of x, S(x,e,AhT(i),a)
is a convergent power series;
(ii) all derivatives of S(x,6,AhT(i),a) with respect to x
exist and are positive for all nonnegative real values
of x; and
(iii) [S(x,e,hT(i),a) 1] is positive for all positive
real values of x.
Equation (3.10) can be rewritten using (3.11) as
Yl(x) = L0S(x,e,XhT(i),a) (3.12)
51
When a = 1, as in the minimal repair problem, (0 XhT(i))x S(x,6,\hT(i),l) = e
and
(6 XhT(i))x Y1(X) = 0e
Similarly when a = 0, as in the major repair problem, Ox
S(x,e,XhT(i),0) = (6 hT(i)) e + 1 and
Ox
(e 1)
yl(x) = a 0(6 XhT(i)) e + aO
For notational convenience, let
S(x,a) = S(x,8,XhT(i),a) (3.13) where
j j1
S(x,a) = 1 + x Tx Pk(a) (3.14) j= 1 k=0
and
Pk(a) = e XhET(i)a k=0,1,2,... (3.15)
0
where a = 1, for all a. Then substituting (3.13) into (3.12) yields
Yl(x) = a0S(x,a) (3.16)
To determine a particular solution Y2(x), it is necessary to solve the linear differentialdifference equation (3.7). It is observed that the term y2(ax) in equation (3.7) constitutes a backward differencing, since 0 < a < 1. Theeffect of this term in equation (3.7) requires an assumption about the nature of g(x) in order to
52
obtain Y2(x). Recall that g(x) is defined by the relation (3.4). Since c(x), k(x), and r(x) are unspecified, it is assumed that each can be represented by a polynomial. Then their linear combination defined by (3.4) is also a polynomial. Assume that g(x) is a member of the class of nth order polynomials whose coefficients are all required to be positive. Then g(x) is polynomial of the form n nl
g(x) = e x + en x + ... + ex + e n n1 1 0 (3.17)
where
e. > 0 i=O,...,n
1
The assumption e. > 0 for i > 1 is later established as one of a pair sufficient conditions for a unique finite optimal equipment replacement age x to exist. Since there is no physical interpretation for a negative cost in the repair and replacement problems under investigation, it is assumed that e0 > 0. Since g(x) has all positive coefficients, then g(x) is a convex function. The particular solution Y2(x) to (3.7) with g(x) given by (3.17) is
Y2(x) = bnxn + bnlxni + ... + blx + b0 (3.18) where the terms b. are determined by the solution to the difference equation
b j= ,.. ,n
j1 Pjl(a) j Pj_ (a) with
e
Sn
n P (a)
n
Solving (3.19) recursively yields
53
e n1 n
1 (j+1) ... (i)e. P (a) j Pi (a) n i=j+1 k=i+l
7 Pk(a)
k=j
+ (j+1 ... (n)en j=O,...,n2
e ne
b n + n (3.20)
ni P _l(a) P (a)P (a) and
e
b =
n P (a)
n
Using (3.20) in (3.18) yields the particular solution
n n1
e x [en_ P n (a) + nen]x n2 e.
n(x) + + x
2 P Pn(a) P nl(a)P n(a) j= P(a)
1n1 n + (j+l) ... (i)e. T Pk(a)
n (a i=j+i k=i+l
x Pk(a)
k=j k
+ (j+l) ... (n)en (3.21)
which is a solution to (3.7) with the nth order polynomial g(x) given
by (3.17).
Thus, the general solution to equation (3.3) with g(x) given
by (3.17) is
fl(x) = y1(x) + y2(x)
n n1 e nx [e nlPnl (a) + ne ]x n2 e. = 0gS(x,a) + P (a) Pn l(a)P (a) + x P (a) n n1 n j=0
Sn1 n + 1 (j+) ... (i)ei Pk(a)
i P(a) i=J+1 k=i+l
w Pk(a)
k=j
+ (j+l) ... (n)e (3.22)
54
It is necessary to evaluate the arbitrary constant 20 in (3.22). For x = 0 in (3.14) S(O,a) = 1
and evaluating (3.22) at x = 0 yields
1 nl n
0 = f (0) (i!)e 7 P (a) + (n!)e T Pk(a) i=0 k=i+l k=0
Substituting the expression just determined for a0 in (3.22), the following is obtained
n n1 ex [e nP n(a) + nen]xnl fl(x) = f (0)S(x,a) + +
1 1 P (a) Pn (a)Pn (a)
n2 e. 1 nl n
+ x + 1 L (j+l) ... (i)e T Pk(a)
j=0 Pj (a) n ( i=j+ a) k=i+l
1 ni n + (j+) ... (n)e (i)e T Pk(a) ( (nen ni=0 i k=i+l SPk(a)
k=0
+ (n!)en [S(x,a) 1] (3.23)
Assuming that g(x) is an nth order polynomial, this last expression is the general solution to the functionaldifferential equation (3.3). The following section applies (3.23) in the discussion of three particular polynomial forms g(x).
55
3.3 Special Polynomial Forms of g(x)
Three polynomial forms of g(x) are investigated in this section; namely, constant, linear, and quadratic, where every coefficient of each polynomial g(x) is assumed to be positive.
a. (x) = c
When g(x) is a constant, this implies that one of three alternatives exists. Either the instantaneous repair cost, the operations cost, and the loss of revenue per unit of downtime cost are all nonzero constants, or two of the costs are nonzero constants and one is zero, or one of the costs is a nonzero constant and two are zero. Under each of these three possibilities there is never an incentive to replace, which implies that the optimal replacement age is infinity. This conjecture will be established.
Using (3.3) with g(x) = co yields the functionaldifferential equation
fi(x) (i + X)fl(x) + XhT(i)fl(ax) = c0 (3.24) Then applying (3.23), which is the general solution to (3.3), to equation (3.24), the general solution to (3.24) can be written as cO
fl(x) = fl()S(x,a) (a) [S(x,a) 1] (3.25)
If an optimal Xpolicy is followed, continuity of the function fl(x) at x = X is required. It then follows that fl(X) = K(X) + fl(0) (3.26) Applying (3.26) in (3.25), it is required that at x = X
56
c
K(X) + fl(0) = fl(O)S(X,a) P(a) [S(X,a) 1] Solving for fl(0) yields
f (0) ) + (3.27)
1 S(X,a) 1 P0(a)
Now K(X) is continuous in X, and further it is a bounded
function, whereas S(X,a) is a strictly increasing function in X to infinity. Thus, f (0) is minimized with respect to X as X approaches infinity. The following theorem summarizes the result. Theorem 3.1
A sufficient condition for a unique optimal equipment replacement age X to exist for the linear hysteresis problem where a(x) = ax, X(x) = X = constant, and g(x) = c0 is that X be infinite.
b. g(x) = clx + c0
Using (3.3) with g(x) = clx + c0 yields the functionaldifferential equation
f'(x) (i + X)fl(x) + XhT (i)f(ax) = clx c0 (3.28) Applying (3.23), which is the general solution to (3.3), to (3.28), the following is obtained
l1x c1 + C0P1(a)
f(x) = fl()S(x,a) + (a) (a)P(a) [S(x,a) 1] (3.29) Using (3.26) in (3.29), it follows that at x = X clX c1 + 0 Pl(a)
K(X) + f1(0) = fl(O)S(X,a) + [S(X,a) 1] fP(a) P0(a)P1(a)
57
Collecting terms in f (0) and solving for f1(0) yields clX
K(X)
0) = P(a) c1 + 0P 1(a) f (0) = +
S(X,a) 1 P0(a)Pl(a)
For the remainder of the discussion of the linear form of g(x), the analysis is restricted to the situation where the fixed cost of replacement K(X) is independent of the replacement age X. Since K(X) = K = constant, then clX
K
Pl(a) C1 + C0 Pl(a) f (0) = + (3.30)
1 S(X,a) 1 P (a)Pl(a)
The following theorem presents the results obtained for the linear form of g(x).
Theorem 3.2
Sufficient conditions for a unique finite optimal equipment replacement age X to exist for the linear hysteresis problem where a(x) = ax, X(x) = X = constant, K(X) = K = constant, and g(x) = clx + c are that c1 > 0 and
(a) [S(X,a) 1] + K P SX(X,a) = 0 (3.31)
Pl(a) Pl( "a) X where X is the unique solution to (3.31). Proof:
If the initial equipment age x is such that x < X, then it is clear from substituting (3.30) into (3.29) that if an optimal replacement age X exists, it minimizes both f1(0) and fl(x) for all x.
58
Differentiating the expression for f1(0) in (3.30) with respect to
X and equating the result to zero, a sufficient condition for the existence of an optimal equipment replacement age X, say X*, is obtained and is given by the equation in X
1a [S(X*,a) 1] + P 1(a) Sx(X*,a) = 0 (3.32)
PI(a) K (a) X where
Sx(X,a) = dX
Note that (3.32) is identical to (3.31). Dividing all terms in
clSX(X ,a)
(3.32) by (a) and rearranging yields
P 1(a)
X S(X*,a) 1 1KP(a)
X S = (3.33) SX (X ,a) cl
Taking the second derivative of f (0) with respect to X results in
clSX(X,a) clSX(X,a) ClX 2
2 { + [K ]s (Xa)} [S(X,a) 1] d f1(0) Pl1(a) Pl(a) [K Pl(a) ]S(X,a) [S(Xa) 1]
dX2 [S(X,a) 1]4
cl clX 2S(X,a)[S(X,a) ] { 1(a) [S(X,a)l] [K P(a) ]Sx(X,a)} [S(X,a) 1]
(3.34)
where
d2S(X,a)
S (X,a) =
XX da 2
dX
Substituting (3.32) into (3.34) and simplifying
59
2 [K ] (X )
d fl1(0) P1(a) ]SXX(X*,a)
dX2 [S(X*,a) 1]2(3.35) x=x*
Recall that for all positive values of X, the functions S(X,a) 1, SX(X,a), and SXX(X,a) are positive. Returning to equation (3.33), let L(X) equal the lefthand side of the equation that is
L(X) = X S(X,a) 1 SX(X,a)
Differentiating L(X) with respect to X yields
dL(X) Sx(X,a)Sx(X,a) SXX(X,a)[S(X,a) 1]
1
dX 2
dX [S (X,a)] 2
Sxx(X,a)[S(X,a) i] 0 for all
> 0 for all x > 0 [Sx(X,a) ]
Furthermore, it is easily verified that L(0) = 0. Therefore, the function L(X) = X [S(X,a) 1]/SX(X,a) is positive for X > 0 and is strictly increasing to infinity. If c1 > 0 then KP (a)/c1 > 0. It therefore follows from (3.33) that the optimum equipment replacement age X, if it exists, is unique.
The existence of an optimal equipment replacement age X which d2fl(0)
minimizes f1(0) is established by showing that 2 > 0.
dX
Equation (3.32) can be rearranged to yield c1X* cl[S(X*,a) 1] Pl(a) Pl(a)Sx(X*,a) (3.36)
60
Using (3.36) in (3.35) and if c1 > 0, the following is obtained
d2 fl(0) clS X(X*,a)
dX2 P (a)S (X*,a)[S(X*,a) 1 >] X=X
which proves existence. Then the condition c1 0 is one of a pair of sufficient conditions for a unique finite optimal equipment replacement age X to exist.
The following is a numerical example of the linear hysteresis problem with g(x) = clx + co.
Example 1
Consider a continuously operating equipment subject to a constant failure rate and with an expected number of failures of 10 per year. Following each failure, the equipment is instantly repaired. Repair of an equipment results in a recovery of usable service life, where the amount of recovery is proportional to the age of the equipment. Assume that a repair recovers half of the service life used by the equipment. The combined repair and operating cost in $ per year of service for the equipment when service age is u is 10.04 u + 7.2. It is anticipated that when the existing equipment is in need of being replaced, the new equipment will have identical characteristics and a fixed cost of $350 will be incurred. Assuming a fixed interest rate of 10% per year, it is required to determine the service life at which it is more economical to replace the existing equipment rather than to continue operating and repairing it.
From the problem statement, it is known that K = $350,
61
X = .1/year, i = .l$/($year), 6 = .2/year, a = 1/2, hT(i) = 1, AhT(i) = .1/year, P1(1/2) = .15, cl = 10.04, and co = 7.2. Then solving numerically equation (3.31), it is determined that the most economic life at which an equipment should be replaced is 10 years.
The linear cost function g(x) = 10x is examined in Tables 3.1
through 3.5 for specific values of a. These values are 0, 0.25, 0.50, 0.75, and 1. Each table presents the optimum economic life in years and the corresponding value of f1(0) for various combinations of K, 6, and XhT(i).
From Tables 3.1 through 3.5, the following observations can be made concerning the overall behavior of X and f1(0) as the variables K, 6, XhT(i), and a are varied one at a time. As the replacement cost K increases, both X and f1(0) increase. This is logical since increasing the replacement cost would delay replacements and incur (in addition to the increase in K) extra operating and repair costs due to the delay in replacing. As 6 increases, the replacement age X increases and f1(0) decreases. Since 6 = i + X,an increase in e while XhT(i) is fixed corresponds to an increase in the interest rate i. Thus future costs are additionally discounted and their is an economic incentive to retain an equipment longer. As AhT(i) increases, the replacement age X decreases and f1(0) increases. Due to the structure of S(X,a), an exception to the last statement occurs when a = 0, in which case X does not vary and f1(0) increases as XhT(i) increases. With 6 fixed, an increase in Xh T(i) corresponds to an increase in the failure rate A. Since an equipment fails more often, there are additional repair costs incurred and the equipment is replaced sooner. As a increases from 0 to 1, the replacement age X decreases and f (0) increases. Increasing a corresponds to less service life being recovered at the termination of a repair. Thus higher operating and repair costs are incurred and the equipment is replaced more often. For the case of minimal repair, that is a = 1i, the values obtained
62
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44 HN H H
0
0 000 r nLnNLnLnLn n LlU Lf ~O~l DO C'NNh lNN H 000 NNNNN o00000
44
H HHH NNNNN NcNNNN
0
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67
for X and f1(0) for a particular K are a function of 6 hT (i), due to the structure of S(X,a). Thus, the valuesof X and f (0) for 6 = 0.4 and .hT(i) = 0.2 arethe same as for 6 = 0.6 and h T(i) = 0.4.
No evaluation can be made concerning the Xpolicy for various values of a for the linear hysteresis problem from the results displayed in the tables. This is due to the cost function g(x) being the same for each value of a, and therefore a minimal repair at age x costs the same as a major repair. Chapter 4 addresses itself to the problem of determining an optimal a and X.
c. g(x) = c2x2 + C1x + cO
Letting g(x) = c2x2 + clx + co in (3.3) results in the functionaldifferential equation
2
fl(x) (i + X)fl(x) + XhT(i)fl(ax) = c2x C cO (3.37)
Using (3.23) which is the general solution to (3.3), the general solution to (3.37) can be written as
2
c2x [2c2 + clP2(a)]x fl(x) = fl(0)S(x,a) + +
P2(a) Pl(a)P2(a) 2c2 + c1P2(a) + c0P1(a)P2(a)
P0(a)P(a)P2(a) [S(x,a) 1] (3.38) P0(a)Pl(a)P2(a) Applying (3.26) in (3.38), it follows that at x = X c2X2 [2c2 + clP2(a)]X K(X) + f1(0) = fl(0)S(X,a) + P2(a) P 1(a)P2(a) P2(a) Pl(a)P2(a) 2c2 + C1P2(a) + 0 P1(a)P2(a)
P0 (a)Pl(a)P2 (a) Collecting terms in f1(0) and solving for f1(0) yields
68
c2X2 [2c2 + clP2(a)]X
K(X)
P2(a) P1(a)P2(a) 1(0) S(X,a) 1
2c2 + c P2(a) + c Pl(a)P2(a)
PO (a)Pl(a)P2 (a) The analysis of the quadratic form of g(x) is now restricted to the case where the fixed cost of replacement K(X) is independent of the replacement age, that is, K(X) = K = constant. Then c2X [2c2 + clP2(a)]X
K
P2(a) P1(a)P2(a)
() S(X,a) 1
2c2 + c1P2(a) + 0 P1(a)P2(a)
+ (3.39) P 0(a)P1(a)P2(a) For notational convenience let S=P (3.40) P2(a)
2c2 + ClP2(a)
S P(a)P2 (a)
Using (3.40) and (3.41), equation (3.39) can be written as
S X2 + 2c2 + c1P2(a) + 0P 1(a)P2(a)
f(0) = + (3.42)
1 S(X,a) 1 P0(a)Pl(a)P2(a)
The following theorem presents the results obtained for the quadratic form of g(x).
Theorem 3.3
Sufficient conditions for a unique finite optimal equipment replace
69
ment age X to exist for the linear hysteresis problem where a(x) = ax,
(x) = = constant, K(X) = K = constant, and g(x) = c2x2 + C1X + cO are that c. > 0 for i=1,2 and
1
(28X + y)[S(X,a) 1] + (K BX2 yX)SX(X,a) = 0 (3.43) where X is the unique solution to (3.43). Proof:
If the initial equipment age x is such that x < X, then it is clear from substituting (3.42) into (3.38) that if an optimal replacement age X exists, it minimizes both f1(0) and fl(x) for all x. Differentiating the expression for f1(0) in (3.42) with respect to X and equating the result to zero, a sufficientcondition for the existence of an optimal equipment replacement age X, say X is obtained and is given by the equation in X*
(2BX* + y)[S(X*,a) 1] + (K BX*2 yX*)SX(X*,a) = 0 (3.44) Note that (3.44) is identical to (3.43). Dividing (3.44) by ySX(X*,a) and rearranging yields
X*2 + x* S(X*a) 1 K Y Y
Letting 6 = B/y in the last expression, the following is obtained
X* + X (26X* + 1)S( ) 1 K (3.45)
6X SX(X*,a) Y
Taking the second derivative of f (0) with respect to X yields
70
d2 fl() { 2B[S(X,a) 1] (28X + y)SX(X,a)}[S(X,a) 1]2
dX2 [S(X,a) 1]4
{(28X y)Sx(X,a) (K X2 yX)Sxx(X,a)}[S(X,a) ]2
[S(X,a) 114
2SX(X,a)[S(X,a) l]{ (28X + y)[S(X,a) i]}
x
[S(X,a) 1]
2Sx(X,a)[S(X,a) l]{ (K X2 yX)Sx(X,a)} (3.46) [S(X,a) 1]
Substituting (3.44) in (3.46) and simplifying
2 2
d fl(0) 28[S(X,a) 1] (K BX* yX*)S xx(X*,a)
dX2 [S(X*,a) 1]2 (347)
Returning to equation (3.45), let M(X) equal the lefthand side of the equation, that is
M(X) = 6X2 + X (26X + 1) S(Xa)
sx(xa)
Differentiating M(X) with respect to X yields
dM(X) S(Xa) 17
dM(X) 26X +1 26 '( ,)
dX SX(X,a)
S S(X,a)SX(X,a) Sxx(X,a)[S(X,a) 1]
(26X + 1) 2 [Sx(X,a)]2
Simplifying
71
S2 S(X,a) 1 (2X + 1) SX(X,a)[S(X,a) 1]
dX SX(X,a) [Sx(X,a)] 2
S(X,a) 1
= [(26X + 1)Sxx(X,a) 26Sx(X,a)] S(X,a) (3.48)
Now using (3.14) and result (B.3) of Appendix B, S (X,a) and SXX(X,a) can be written as
Sx(X,a) = PO(a) + I i Pk(a) (3.49) j=l j k=0 k
xj j+1
SXx(X,a) P0(a)Pl(a) + 1W i + Pk(a) (3.50) j=1 k=0
Then
(26X + 1)SXX(X,a) 26SX(X,a)
= 26[XSxx(X,a) Sx(X,a)] + S (X,a) i i
= 26 (j1)X T Pk(a) 26PO j=2 J k=0 O j j+l
+ PO(a)Pl(a) + X  Pk(a) (3.51) j=1 k=0
Since 6 = 6/y, then using (3.40) and (3.41) 6 can be expressed as 2Pl (a)
2c2 + clP (a)
Substituting the expression obtained for 6 into (3.51) yields
(26X + 1)SXX(X,a) 26SX(X,a)
2c2Pl(a) j)X
2c + P((a))x j Pk(a)
2 c1P2(a) j2 j k=0
72
CO j j+1
+ 7F Pk (a)
j=1 k=0
+ P0(a)Pl(a) 1 2 2 ClP2(a) 2c2 + cl 2(a) If c. > 0 for i=1,2, then
1
2c
0< < 1 2c2 + C1P2(a) which implies that
(26X + 1)SXX(X,a) 26Sx(X,a) > 0 for all X > 0 (3.52)
Using (3.52) in (3.48), it follows that
dM(X) > 0 for all X > 0
dX
Furthermore, M(0) = 0. Therefore the function M(X) is positive for X > 0 and is strictly increasing to infinity. If c. > 0 for i=1,2, then
kP (a)P2 (a)
> 0
2c2 + c1P2(a)
It therefore follows from (3.45) that the optimum equipment replacement age X, if it exists, is unique.
The existence of an optimal equipment replacement age X, which minimizes fl(0),is established by proving that d f(0)
d2 > 0. Equation (3.44) can be rearranged to yield
SX=X*
2 S(X* a) 1i
K BX*2 yX* = (2X* + y) SX*,a (3.53) S X X*.a)
73
Substituting (3.53) in (3.47) yields
d2f 1(0) 2SX(X*,a) + (2BX* + y)SXX(X*,a)
dX2 S(X*,a)[S(X*,a) 1]
Since 6 = /y, then
d2f1(0) (26X* + 1)SXX(X*,a) 26SX(X*,a)
dX2 X TSX(X*,a)[S(X*,a) 1]
dX2 X=X
Using (3.52) in the last expression, it follows that d2f1(0)
> 0
dX X=x*
This proves existence and concludes the proof of Theorem 3.3. The condition c. > 0 for i=1,2 is one of a pair of sufficient conditions for a unique finite optimal equipment replacement age X to exist.
The following is a numerical example of the linear hysteresis problem with g(x) = c2x2 + cx + co. Example 2
Consider the problem posed in Example 1, with one exception. The combined repair and operating cost in $ per year of service for the equipment when service age is u is 1.73 u2 + 7.62 u + 6.5.
From the problem statement, it is known that K = $350, = .1/year, i = .l$/($year), 0 = .2/year, a = 1/2, hT(i) = 1, h T(i) = .1/year, P1(1/2) = .15, c2 = 1.73, c1 = 7.62, and c0 = 6.5. Then solving numerically equation (3.43), it is determined that the
74
most economic life at which an equipment should be replaced is 6.39 years.
The quadratic cost function g(x) = 1.75 x2 + 7.5 x is examined in Tables 3.6 through 3.10 for specific values of a. These values are 0, 0.25, 0.50, 0.75 and 1. Each table presents the optimum economic life in years and the corresponding value for f (0) for various combinations of K, 6, and XhT(i).
The behavior of X and f1(0) observed from Tables 3.6 through
3.10 for the quadratic cost function is the same as for the linear cost function previously discussed, and therefore will not be repeated here.
It should be noted that the sufficient condition on the coefficients of the polynomial g(x) for a unique finite optimal equipment replacement age X to exist is not a necessary condition. There may exist a polynomial g(x) with some negative coefficients which is convex over an interval which contains a unique optimal age X.
75
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CHAPTER 4
AGE REPLACEMENT WITH ECONOMICALLY CONTROLLED PARTIAL REPAIR AND DOWNTIME
4.1 Introduction
In this chapter, the linear hysteresis problem examined in Chapter 3 is extended to incorporate the problem of economically controlling the amount of partial repair following a failure. This requires the addition of several new concepts. In addition to the repair cost k(u) incurred at failure for an equipment aged u, C(E,u) is spent at the commencement of the repair, where E is the level of expenditure. The expenditure repair function C(E,u) is defined to be equal to Eum, where m is some nonnegative integer. If E equals zero, then no additional funds are expended on the repair, and a minimal repair is performed. For E greater than zero, service age can be recovered at the termination of the repair. If C(E,u) is spent on the repair of an equipment aged u, the fraction of service age remaining after the repair is p(E), where 0 < p(E) < 1. Then the service age at the end of the repair is p(E)u. The function p(E) corresponds to the constant a of the linear hysteresis age function, a(x) equals ax.
Some possible shapes for the curve (function) p(E) are
illustrated in Figures 4.1 through 4.6. In each of the figures, the curve p(E) is strictly nonincreasing in E. The linear curve P(E) in Figure 4.1 is a special case of the curve in Figure 4.2 with L2 = 0 and L1 = L3. Figure 4.2 is an example of an initial expenditure threshold where an amount L2 must be expended before any decrease in p(E) is noticed. A similar type of threshold is observed in Figure 4.3.
80
P (E)
1 I"0 E
L
Figure 4.1. Example of a linear curve p(E). p(E)
1      LE
0 E
L2 L3
Figure 4.2. Example of a linear curve p(E).
82
p (E)
1 '  a
0II
L L5 L6
Figure 4.3. Example of a curve p(E). p (E)
1  LF LL
Figure 4.4. Example of a curve p(E).
83
p (E)
1  0 E
L9
Figure 4.5. Example of a concave curve p(E). p (E)
1 
S10
Figure 4.6. Example of a convex curve p(E).
84
Increasing the level of expenditures beyond L4 and up to L5 does not result in a decrease in p(E). In Figure 4.4 a different threshold occurs. At L7 there is a jump in the value of p(E) without any corresponding change in E. Figures 4.5 and 4.6 are examples of piecewise nonlinear curves p(E) which are concave and convex, respectively.
The value of E acts as a decision variable since it controls both C(E,u) and p(E). Consequently, the age replacement policy of Chapters 2 and 3 is modified to incorporate E, and becomes the twovariable decision policy, the E,Xpolicy.
The new problem generated from extending the linear hysteresis problem so that partial repair is economically controlled is called the economically controlled linear hysteresis problem. This problem with nonzero downtime for repair is examined in this chapter. For the functionaldifferential equation (1.14) with failure rate function,
(x), equal. to and with p(E) equal to a (0 < a < 1), a constant, a general solution is determined. The method of obtaining this solution is identical to the method discussed in the introduction to Chapter 3. For a constant operating cost and a constant expenditure repair function C(E,u), a pair of sufficient conditions is determined for which an E,Xpolicy is an optimal policy for the economically controlled linear hysteresis problem. For several curves of p(E) versus E, the E,Xpolicy is investigated for an equipment subject to a linear operating cost and a constant or linear expenditure repair function C(E,u).
4.2 General Solution to FunctionalDifferential Equation with
X(x) = X and p(E) = a
The functionaldifferential equation (1.14) of Chapter i is
85
recalled. It is
f'(x) [i + (x)]fl(x) + X(x)hT(i)fl(p(E)x) = q(x) (4.1) where
q(x) = c(x) + k(x)A(x) + r [1 hT(i)]X(x) + C(E,x)X(x) (4.2) Letting p(E) = a (a constant to be determined later, where 0 < a < 1) in (4.1) and X(x) = X = constant in (4.1) and (4.2) yields
fl(x) (i + X)fl(x) + XhT(i)fl(ax) = j(x) (4.3) q(x) = c(x) + Xk(x) + Xr(x) [1 hT(i)] + XC(E,x) (4.4)
The procedure to obtain the general solution to (4.3) consists of determining a complementary solution, a particular solution, and the constant term of these solutions by evaluating the general solution at x equals zero. It can be verified (Appendix A) that the general solution to (4.3) is of the form fl(x) = yl(x) + Y2(x) (4.5) where Yl(x) is a solution to the homogeneous equation
yl(x) (i + X)yl(x) + XhT(i)yl(ax) = 0 (4.6) and y2(x) is a solution to the nonhomogeneous equation
y'(x) (i + X)y2(x) + XhT(i)y2(ax) = q(x) (4.7) A complementary solution to the homogeneous equation (4.6) has been previously obtained in Chapter 3 (see (3.6) and (3.8) through (3.16)). It is
86
Yl(X) = o0S(x,a) (4.8) where a0 is an arbitrary constant to be determined.
To determine a particular solution y2(x), it is necessary to
solve the linear differentialdifference equation (4.7). It is observed that the term y2(ax) in equation (4.7) constitutes a backward differencing, since 0 < a < 1. The effect of this term in equation (4.7) requires an assumption about the nature of q(x) in order to obtain y2(ax). Observe from (3.4) and (4.4) that q(x) = g(x) +X C(E,x) (4.9) As in Chapter 3, g(x) is assumed to be a member of the class of nth degree polynomials whose coefficients are all required to be positive. Then, g(x) is a polynomial of the form
(x=en nl
g(x) = e x + ex + ... + elx + e0 n n1 1 0 where
e. > 0 i=0,1,...,n
1
From (1.11),C(E,x) is given by
C(E,x) = Exm m=0,1,2,3...
Assuming that m < n, the expressions for g(x) and C(E,x) are substituted in (4.9) to obtain
n+. + m+l m
q(x) =e x + ... +(e + XE)xm
n m+1 m m1
+ e lx + ... + elx + e0, m=0,1,...,n (4.10) Given the assumption on the polynomial g(x) and since E > 0, q(x) is
87
also a member of the class of nth degreepolynomials whose coefficients are all required to be positive. Therefore, q(x) is a convex function. Let
ei i=0,1,...,m,m+l,...,n
e. = (4.11)
e. + XE i=m
Using (4.11) in (4.10), q(x) can be written as
q(x) = ex + e x + ... + ex + e0
where (4.12)
e. > 0 i=0,1,...,n
i
Then the particular solution Y2(x) to (4.7) with q(x) given by (4.12) is
Y2(X) = dn xn + d n + ... + dx + d (4.13) Applying results obtained in Chapter 3 for the particular solution in order to evaluate the di(i=0,1,...,n) in (4.13), it is determined that the particular solution is
n n1 enx [en P nl(a) + ne]xn
exn n1 n1 n
2( Pn (a) Pn (a)P (a) n2 e.
+ I xj i.
j= 0 P (a)
n1 n + (j+l)...(i)e P (a) n i k P(a) i=j+l k=i+l k=j
+ (j+l)... (n)n] (4.14) which is a solution to (4.7) with the nthdegreepolynomial q(x) given by (4.12).
88
Thus,the general solution to (4.3) with q(x) given by (4.12) is
fl(x) = yl(x) + Y2(x)
Sn n1 enx [en P n(a) + ne ]x = c0S(x,a) + +
0 P (a) P (a)P (a) n ni n
n2 e. 1n n
+ x (a) n1 I (j+l)...(i)e T P (a)
j=O Pj (a n i=j+l k=i+l Pk(a)
k=j
+ (j+l...(n)en] (4.15)
To evaluate the arbitrary constant 00, recall that for x = 0 in
(3.14)
S(O,a) = 1
Evaluating (4.15) at x = 0 to solve for a0 and substituting the value
obtained into (4.15) yields n nl e x [e P (a) + ne ]x n n n1 n fl(x) = f (0)S(x,a) + +
1 1 Pn(a) Pn (a)Pn(a)
n2 e. ni n
+ x + 1 [ (j+)... (i)e T Pk(a)
j=0 {P.(a) n i=j+1 k=i+l T Pk(a)
k=j
n1 n + (j+l).. .(n)n] [ (i!). ki Pk(a) n I Pk(a i=0 k=i+l J T Pk(a)
k=O
+ (n!)en][S(x,a) 1] (4.16)
Thus, if q(x) is an nthdegreepolynomial, then equation (4.16)
is the general solution to the functionaldifferential equation (4.3).
The following section applies (4.16) in the discussion of three particular
polynomials g(x).

Full Text 
BIOGRAPHICAL SKETCH
Jonathan Franklin Brown was born on March 2, 1946, in
San Francisco, California. In June, 1963, he was graduated from
BethesdaChevy Chase High School in Bethesda, Maryland. He attended
Johns Hopkins University, and in June, 1967, received the degree of
Bachelor of Engineering Science with a major in Operations Research.
In September, 1967, he entered the Department of Industrial and
Systems Engineering at the University of Florida, and in June, 1969,
was awarded the degree of Master of Science in Engineering. Since
September, 1969, he has held research and teaching assistantships
while pursuing the degree of Doctor of Philosophy in Operations
Research.
Jonathan Brown is a member of the Operations Research Society
of America, The Institute of Management Sciences, the American
Institute of Industrial Engineers, and Alpha Pi Mu, an honorary
society.
He is married to the former Marcene Leba Needel of Baltimore,
Maryland.
14?
123
^ [yx(x) + y2(x)] A(x)ty1(x) + y2(x)]
+ B(x)[y1(a(x)) + y2(a(x))] = C(x)
which is an alternative way of stating that f^(x) = y^(x) + y2(x) is a
solution. Thus, the general solution to (A.2) is the sum of the
complementary and particular solutions.
A sufficient condition for the existence and uniqueness of
the general solution f^(x) is given by the following theorem.
Theorem A.2
Let A(x), B(x), and C(x) be functions which are continuous in
the interval 0 <_ x < X and a(x) is continuous in the interval 0 <_ x <_ X.
Then there exists a solution f^(x) satisfying the differential equation
(A.2) and also the condition
f1(X) = K(X) + f1(0)
Furthermore, this solution is unique.
Obtaining a complementary solution to (A.l) is not a trivial
task. Consider the special case where the hysteresis age function,
a(x), equals ax. In 1836, Gregory [15] examined solutions to equations
of the form f(x) = kf(qx) + c (q > 0, q i= 1) which he labeled a
qdifference equation. His method of solution for qdifference
equations has had little application to differential qdifference
equations. Efforts to apply his method to obtain a complementary
solution to (A.l) when a(x) = ax have met with no success.
One extremely powerful method of differential equations known
Service Age
Figure 1.3. Sample function of the service age process S(t) for the major
repair problem with downtime, where x is the initial equipment
age, and X is the equipment replacement age.
CHAPTER 3
AGE REPLACEMENT WITH PARTIAL REPAIR AND DOWNTIME
3.1 Introduction
This chapter examines the hysteresis effect when an age
replacement policy, called an Xpolicy, is being followed. In
particular, the linear hysteresis problem with nonzero downtime for
partial repair and with constant failure rate is investigated. For
the functionaldifferential equation (1.10) with failure rate function,
A(x), equal to A, and with hysteresis age function, a(x), equal to ax
(0 ^ a <_ 1), a general solution is determined. This is the sum of
two solutions; namely, a complementary and a particular solution.
The complementary solution is determined by assuming it is
of the series form. As a consequence, the function S(x,0,Ah (i),a)
is obtained, and is examined analytically and graphically in Appendix B.
The properties of S(x,0,Ah^(i),a) and its derivatives with respect
to x are important in obtaining the results in this and the following
chapter. Due to the structure of the cost function g(x), when A(x)
equals A, it is assumed that the cost function is a member of the
class of nth order polynomials, whose coefficients are all positive,
in order to obtain the particular solution. The assumption that all
the coefficients of the polynomial are positive is later established
as one of a pair of sufficient conditions for a unique finite optimal
equipment replacement age x to exist.
Three particular polynomial forms with positive coefficients
are examined in this chapter. They are: constant, linear, and quadratic.
47
91
Solving for f_^(0) yields
fl(0)
K(X) p
~S(X,a)
+
Cp + (Cq + AE)Pp(a)
PQ(a)P1(a)
(4.24)
For the remainder of the discussion of the polynomial (4.21), the
analysis is restricted to the situation where the fixed cost of
replacement K(X) is independent of the replacement age X. Since
K(X) = K = constant, then
f L(0)
Ppia)
S(X,a) 1 +
C1 + ^C0 +
PQ(a)P1(a)
(4.25)
For the polynomial (4.21), neither necessary nor sufficient conditions
for an optimal E,Xpolicy to exist can be obtained. This is due to the
various shapes the curve p(E) can assume (see Figures 4.1 through 4.6)
and the structure of the expression obtained for f^(0) in (4.25).
Although the derivative of S(X,a) with respect to a exists, it is
not possible to obtain a closed form expression for the derivative
of the curve p(E), especially at the end points of the interval over
which the curve p(E) is defined. Through another method of optimization,
it is determined that for some curves p(E) and expenditure repair
functions C(E,u), the value of E of the pair E,X, which minimizes
fp(0), corresponds to an end point of the curve p(E).
Since analytic results for the polynomial (4.21) cannot be
determined, quantitative results are obtained. In Chapter 3 for
an equipment subject to a linear operating cost g(x) (i.e., g(x) =
c^x + Cq), a pair of sufficient conditions are determined.
92
Similar conditions can be determined for the polynomial (4.21). For
this polynomial, the conditions are c^ > 0 and
P1(a)
[S(X,a) 1] +
K 
C1X
Sx(X,a) = 0
(4.26)
which are identical to those determined for the linear operating cost
g(x) By specifying a value of a which corresponds to a value of E
since p(E) = a, it is possible to obtain an optimal X (assuming that
c^ > 0) by utilizing the equation (4.26) in a search procedure. The
search procedure is performed until all possible values of a have a
corresponding optimal X. The function f^(O) is then evaluated for
each of th pairs a, X, and the minimum is obtained.
For the polynomial (4.21), several basic curves p(E) are
examined, some of which are displayed in Figures 4.1 through 4.6.
In order to generate a family of curves p(E) for each basic curve, each
basic curve is normalized with respect to the E axis and then
multiplied by different scaling factors. The scaling factor corresponds
to the value of E for which p(E) = 0. In addition, critical variables
of the economically controlled linear hysteresis problem for the
polynomial (4.21) are varied to determine their effect on the optimal
values of the parameters determined for the E,Xpolicy. These variables
are K, c^5 0 (where 0 = i + X), A, and h^,(i). The following discussion
presents the analysis of the quantitative results obtained.
For the concave curve p(E) displayed in Figure 4.5, the results
indicate that either minimal repair or major repair is performed
following an equipment failure. In addition, as the scaling factor is
ACKNOWLEDGMENTS
The author wishes to express his gratitude to those who helped
him in the preparation of this dissertation. He especially thanks
Dr. B. D. Sivazlian, chairman of the supervisory committee, for
introducing the author to the problem. Dr. Sivazlian's guidance and
helpful assistance are greatly appreciated. The author would also
like to sincerely thank Dr. J. F, Burns, Dr. T. S. Hodgson, Dr. E. J.
Muth, and Dr. R. L. Scheaffer, members of the committee, for their
advice and encouragement.
The author also acknowledges Mrs. Karen Walker for her
excellent typing.
Finally, thanks go to his wife, Marci, for her confidence
and support.
This research was supported in part by U. S. Army Contract
No. AROD3112470G92.
iii
107
P{N(X) = 1} = P{W2 > X, W < X}
= P{W2 > xw < X}P{W1 < X}
= P{aXl + x2 > XXl < X}P{Xl < X}
X
P{W2 > xwi = x1}f (x^)dx^
X
P(axx + x2 >
XVL = x1 }f (x..)dx
1 1 1 Xl 1
X
P(x2 > X ax^f (XjM*!
X
P{N(X) = 1} =
X(Xax.) Xx.
e Xe dx.
X
= e
XX
Xx.(1a)
Xe
dx
or
0
P{N(X) = 1} 
e AX XX(la).
0=17 [1 e
(5.8)
(5.9)
Now
P{N(X) = 2} = P{W3 > X, W2 < X, W1 < X}
= p{w3 > xw2 < x, w < x}p{w2 < xw1 < x}p{w1 < X}
= P{a2Xl + ax2 + x3 > xaXl + X2 < X, < X}
.p{aXl + X2 < XX;L < X}P{Xl < X}
X Xax..
2
= P[X > x a X ax ]f (x )f (x )dx dx.
J J J Z. Xl Xo ^ J Z
0 0
38
where
R'(X) =
d[R(X)]
dX
X
Rearranging (2.13) and dividing the result by v(X)R(X)
yields
v(u)R(u)du
X
KR(X) +
g(u)R(u)du
X
= KR(X) + g(X)R(X)
v(X)R(X)
v(u)R(u)du
Applying (2.5) in the last expression, the following is obtained
0
X
KR(X) +
g(u)R(u)du
X
v(u)R(u)du
J
0
K[i 4 A (X) ]R(X) + g(X)R(X)
v(X)R(X)
g(X) KA(X) Ki
v(X)
(2.14)
If c(u) = 0, k(u) = k, and h^i) = 1 (instantaneous repair), then
g(u) = kA(u) and v(u) = i. Upon substituting these expressions
for g(u) and v(u) in (2.14), Beckmann's [4] result is recovered. Since
X
X
R(X) R(0) =
d[R(u)] = 
'i + A(u)]R(u)du
0
from (2.5) and R(0) = 1, then
X
R(x) = 1 
[i + A(u)]R(u)du
(2.15)
Substituting (2.15) in (2.14), the following is obtained
96
illustrates the change from major repair to minimal repair as the
variables are varied one at a time. It should be noted that only
the critical scaling factors at which the switch in the amount of
partial repair occurs are indicated. The majority of the results
obtained for linear curves as in Figure 4.1 indicate that either minimal
or major repair is performed following a failure. Two exceptions are
illustrated in Table 4.3.
c. q(x) = c^x + Cq + XEx
Let
q(x) = c^x + Cq + XEx
(4.27)
Substituting (4.27) in (4.3) yields the functionaldifferential equation
f(x) (i + A)f1(x) + Ah^iijf^ax) = c^x cQ XEx (4.28)
Applying (4.16), which is the general solution to (4.3), the general
solution to (4.28) can be written as
c^x + XEx c + XE + CqP^(a)
Â£l(x) f1(0)SC*.a) + p^Ta) yaVjU) [S(xa) 1]
(4.29)
Applying (4.19) in (4.29), it follows that at x = X
K(X) + f1(0) = f1(0)S(X,a) +
CjX + XEX
P1(a)
C;L + XE + cQP1(a)
PQ(a)P1(a)
[S(X,a) 1]
Solving for f^(O) yields
K(X) 
f1(0) =
cX + XEX
Px(a)
S(X,a) 1
+
cl + XE + CpP^a)
PQ(a)P1(a)
(4.30)
Letting K(X) = K = constant in (4.30), the following is obtained
HYSTERESIS EFFECT IN REPAIR AND REPLACEMENT
PROBLEMS WITH DOWNTIME
By
JONATHAN FRANKLIN BROWN
A DISSERTATION PRESENTED TO THE GRADUATE
COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL
FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1973
Abstract of Dissertation Presented to the
Graduate Council of the University of Florida in Partial
Fulfillment of the Requirements for the Degree of Doctor of Philosophy
HYSTERESIS EFFECT IN REPAIR AND REPLACEMENT
PROBLEMS WITH DOWNTIME
By
Jonathan Franklin Brown
December, 1973
Chairman: Dr. B. D. Sivazlian
Major Department: Industrial and Systems Engineering
A new class of repair and replacement problems called the
hysteresis problem is investigated. This problem is concerned with
the amount of recovery of service life following a repair with
arbitrary time distribution, extreme cases of which are minimal and
major repair. Incorporated in the cost structure of the hysteresis
problem are variable operating costs, repair costs, replacement costs
and loss of revenue per unit of downtime. A mathematical formulation
for the discounted hysteresis problem is presented using the
functional equation approach. Optimal age replacement policies,
which minimize the long run total expected discounted cost, are
examined for the linear hysteresis problem.
The extreme case of major repair is investigated, and a
generalization of previously obtained results is determined. For
the linear hysteresis problem, specific cost functions are examined
for an equipment subject to a constant failure rate. In addition,
numerical results for the optimal age replacement policy are obtained
An extension of the linear hysteresis problem incorporating an
xi
8
replacement were concerned with comparing an existing equipment with a
potential replacement. In 1949, Terborgh [24] presented procedures
for comparing a challenger (new equipment) and identical future
challengers with a defender (present equipment). Grant [14] was
concerned with equipment obsolescence, increasing maintenance cost,
and decreasing output (increasing equipment inefficiency). He
presented procedures for determining the time to replacement of an
operating equipment when more efficient equipment was available; no
discount factor was assumed; and, the operations cost was nondecreasing.
In conjunction with the Machinery and Allied Products Institute [17],
Terborgh extended his earlier work to incorporate the concepts of
capital investment. This resulted in the development of the classical
MAPI model.
Bellman [5] (also Bellman and Dreyfus [6]) demonstrated a
method of approach to problems of replacement with an identical
equipment and with a superior equipment (technologically superior).
The method used a dynamic programming formulation to set up a discrete
time functional equation. In order to determine a replacement age X
\
for the equipment, Bellman's objective function was the maximization
of the long run total discounted return (profit). Bellmans model
incorporated the concepts of salvage value, increasing operations cost,
decreasing output of the equipment, and discount factor i. Failure of
the equipment was not considered, and the only incentive to replace
the equipment was the increasing cost disadvantage of retaining the
equipment versus replacing it with a new equipment. Descamps [10]
solved the continuous time version of Bellman's functional equation for
TABLE OF CONTENTS
Page
ACKNOWLEDGMENTS iii
LIST OF TABLES vii
LIST OF FIGURES ix
ABSTRACT xi
CHAPTERS:
1. THE PROBLEM 1
1.1 Introduction 1
1.2 Optimization Criteria 3
1.3 Literature Review 4
a. Problem I minimal repair................ 7
b. Problem II major repair.... 10
c. Problem III hysteresis.. 14
d. Downtime for Repair 15
1.4 Functional Equation Approach.................. 15
1.5 Problem Formulation 20
a. The Functional Equation for the
Hysteresis Problem 20
b. The Economically Controlled Linear
Hysteresis Problem 29
1.6 Chapter Outline... 31
2. AGE REPLACEMENT WITH MAJOR REPAIR AND DOWNTIME..... 33
2.1 Introduction. 33
2.2 A Sufficient Condition for a Unique
Replacement Age X.. 33
2.3 The Undiscounted Long Run Expected Cost Per
Unit Time 44
iv
Table 3.4. Table of Optimum X and Corresponding f^(0) for
g(x) = lOx When a = 0.75
K
50 150 250 350
0
>hT(i)
X
f1(0)
X
f1(0)
X
f1(0)
X
f1(0)
0.20
0.10
3.32
278.6
5.98
436.9
7.96
525.1
9.65
585.2
0.40
0.10
3.74
73.4
7.50
94.9
10.81
100.3
14.04
101.9
0.20
3.51
121.4
6.67
169.2
9.28
186.6
11.74
194.2
0.30
3.29
267.8
5.94
404.1
7.94
470.4
9.72
509.4
0.60
0.10
4.27
34.3
9.61
37.9
14.82
38.1
20.06
38.1
0.20
3.99
47.4
8.59
54.9
13.01
55.5
17.46
55.6
0.30
3.73
69.9
7.63
86.0
11.25
88.5
14.91
88.9
0.40
3.49
116.0
6.74
153.8
9.60
163.4
12.44
165.9
0.50
3.27
256.6
5.96
369.6
8.11
413.5
10.16
432.4
0.80
0.10
4.90
19.2
12.14
19.7
19.38
19.7
26.63
19.7
0.20
4.59
24.6
11.04
25.6
17.51
25.6
24.00
25.6
0.30
4.29
32.5
9.95
34.7
15.65
34.8
21.38
34.8
0.40
4.00
45.0
8.89
49.8
13.80
50.0
18.77
50.0
0.50
3.74
66.4
7.88
77.6
11.97
78.4
16.16
78.4
0.60
3.49
110.5
6.92
138.4
10.20
142.4
13.59
142.8
0.70
3.26
245.1
6.07
334.0
8.54
357.2
11.09
362.5
58
Differentiating the expression for f^(0) in (3.30) with respect to
X and equating the result to zero, a sufficient condition for the
existence of an optimal equipment replacement age X, say X is
obtained and is given by the equation in X
Pl(a)
[S(X*,a) 1] +
K 
c^X*
P1(a)
Sx(X*,a) = 0
(3.32)
where
Sx(X,a)
dS(X,a)
dX
Note that (3.32) is identical to (3.31). Dividing all terms in
c Sx(X*,a)
(3.32) by and rearranging yields
JVa')
X
A
S(X*,a) 1
Sx(X*,a)
KP1(a)
(3.33)
Taking the second derivative of f^(0) with respect to X
results in
c1Sv(X,a) c1Sv(X,a)
+
d2fl(0) p'(a)  + qu) tK isxx(x>* [s(xa) 1)2
dX
[S(x,a) ir
C1 C1X
2Sx(X,a)[S(X,a) 1] { [S(X,a)l] [K ]Sx(X,a)}
[S(X,a) 1]
(3.34)
where
S (X,a) = d S(f?a}
^ dX2
Substituting (3.32) into (3.34) and simplifying
89
4.3 Special Polynomial Forms of q(x)
In this section, several polynomial forms of q(x) are investigated
in conjunction with curves p(E) in order to determine optimal values
for the parameters of the E,Xpolicy. It is assumed that all the
coefficients of the polynomial q(x) are positive.
a. q(x) = Cq + AE
Recall that in the linear hysteresis problem with g(x) = c^, there
was never an incentive to replace, regardless of the value of a, and
further it was established that the optimal replacement age was infinite.
Letting q(x) = c^ + AE in (4.3) yields the functional
differential equation
f^(x) (i + A)f^(x) + AhT(i)f1(ax) = cQ AE (4.17)
Applying (4.16) which is the general solution to (4.3), the general
solution to (4.17) can be written as
Cq + ^
f,(x) = f x(0) S (x, a) p ^ [S (x,a) 1]
(4.18)
If an optimal E,Xpolicy is followed, continuity of the
function f^(x) at x = X is required. It then follows that
fp(X) = K(X) + fx(0)
(4.19)
Applying (4.19) in (4.18), it is required that at x = X
c + AE
K(X) + f1(0) = f1(0)S(X,a) (a} [S(X,a) 1]
Solving for f^(0) yields
17
problem in the following section. As applied to replacement, theory,
the functional equation approach examines, for each value of equipment
service age, the costs associated with each of the alternatives of
replace and do not replace, and selects the one which is smaller.
It is assumed that the 'cost curve associated with the alternative do
not replace' intersects at least once the 'cost curve associated with
the alternative replace', where the curves are a function of equipment
service age. It therefore follows that alternating decision regions
of do not replace and replace exist. Since it is assumed that the first
equipment has an initial age less than or equal to X, then only the
age X at which the first region of do not replace terminates is of
interest.
The different decision regions including the point(s) of
indifference are illustrated by the following application of the
functional equation approach. Assume continuous aging. Define
for x >_ 0:
h(x) = total expected discounted cost over an infinite time
horizon for an equipment aged x when following an
optimal policy;
K(x) = replacement cost at age x incorporating a salvage
feature;
f(x) = minimum total expected discounted cost over an infinite
time horizon.
The principle of optimality of dynamic programming states that with
respect to the current decision of replace or do not replace, all
future decisions are optimal. Then the expected cost of a decision
to replace at age x is K(x) + h(0). Further if no replacement occurs
at age x, the expected cost for this decision is h(x). Then if X is the
44
condition in Theorem 2.1 for the major repair problem with downtime
is similar to a corresponding function in the result obtained by
Sivazlian [21] for the minimal repair problem with downtime. It is
observed that Sivazlian's expression for the function c(u) is equivalent
to (see Section 1.3a and relations (2.2) and (2.9)). The
v(u)
difference in the functions is a multiple of the fixed cost of
replacement K. The functional nature of this difference can be
rationalized by recalling that major repair is equivalent to replace
ment with respect to the resulting equipment service age.
2.3 The Undiscounted Long Run Expected Cost Per Unit Time
In order to determine the undiscounted long run expected cost
per unit time, denoted by f, from results already obtained, the well
known result [4], [26]
f = lim [if.(0)]
i+0
is used. Utilizing (2.11) then
f = lim <
i^O
/
X
iKR(X) + i g(u)R(u)du
0
\
>
X
lim \
i>0 !
/
v(u)R(u)du
0 X
KR(X) + g(u)R(u)du
0
(2.21)
X
>
0
It is seen from (2.4) that
lim R(u) = R(u)
i^O
(2.22)
CHAPTER 4
AGE REPLACEMENT WITH ECONOMICALLY CONTROLLED
PARTIAL REPAIR AND DOWNTIME
4.1 Introduction
In this chapter, the linear hysteresis problem examined in
Chapter 3 is extended to incorporate the problem of economically
controlling the amount of partial repair following a failure. This
requires the addition of several new concepts. In addition to the
repair cost k(u) incurred at failure for an equipment aged u, C(E,u)
is spent at the commencement of the repair, where E is the level of
expenditure. The expenditure repair function C(E,u) is defined to be
equal to Eu, where m is some nonnegative integer. If E equals zero,
then no additional funds are expended on the repair, and a minimal
repair is performed. For E greater than zero, service age can be
recovered at the termination of the repair. If C(E,u) is spent on the
repair of an equipment aged u, the fraction of service age remaining
after the repair is p(E), where 0 p(E) <_ 1. Then the service age
at the end of the repair is p(E)u. The function p(E) corresponds to
the constant a of the linear hysteresis age function, a(x) equals ax.
Some possible shapes for the curve (function) p(E) are
illustrated in Figures 4.1 through 4.6. In each of the figures, the
curve p(E) is strictly nonincreasing in E. The linear curve P(E) in
Figure 4.1 is a special case of the curve in Figure 4.2 with L^ = 0 and
L^ = L^. Figure 4.2 is an example of an initial expenditure threshold
where an amount L^ must be expended before any decrease in p(E) is
noticed. A similar type of threshold is observed in Figure 4.3.
80
67
for X and f (0) for a particular K are a function of 6 Ah^(i), due
to the structure of S(X,a). Thus, the values of X and f (0) for 0 = 0.4
and Ah (i) = 0.2 are the same as for 6 = 0.6 and Ah (i) = 0.4.
No evaluation can be made concerning the Xpolicy for various values
of a for the linear hysteresis problem from the results displayed in the
tables. This is due to the cost function g(x) being the same for each
value of a, and therefore a minimal repair at age x costs the same as a
major repair. Chapter 4 addresses itself to the problem of determining
an optimal a. and X.
2
c. g(x) = c2x + c^x + Cq
2
Letting g(x) = c2x + c^x + c^ in (3.3) results in the functional
differential equation
fj(x) (i + A)f^(x) + AhT(i)f1(ax) = c2x2 c^x cQ (3.37)
Using (3.23) which is the general solution to (3.3), the general solution
to (3.37) can be written as
f]_(x)
fl(0)S(xa) +
+
:2c2 + 0^2 (a) ]x
P1(a)P2(a) ~
2c2 + cjP^a) + cq?1 (a)P2 (a)
P0(a)P1(a)P2(a) ~
[S(x,a) 1]
(3.38)
Applying (3.26) in (3.38), it follows that at x = X
K(X) + fx(0) = f1(0)S(X,a) +
c2X
P2(a)
+
[2c2 + c1P2(a)]X
P1(a)P2(1T
2C2 + ClP2(a) + CQPl(a)P2(a)
P0(a)P1(a)P2(a)
[ S (X, a) 1]
Collecting terms in f^(0) and solving for f (0) yields
90
(4.20)
Consider an arbitrary function p(E). By assumption, K(X) is
a continuous and bounded function of X. For each value of a (and
corresponding value of E since p(E) = a), S(X,a) is strictly increasing
in X to infinity. Thus, f^(0) is minimized with respect to X and E
as X > co and for E = 0. The following theorem summarizes the result.
Theorem 4.1
Sufficient conditions for an E,Xpolicy to be an optimal
policy for the economically controlled linear hysteresis problem where
p(E) is an arbitrary function, X(x) = X = constant, and q(x) = Cq + XE
are that E = 0 and X be infinite.
b. q(x) = c^x + cQ + XE
Let
q(x) = c^x + Cq + XE
(4.21)
Substituting (4.21) in (4.3) yields the functionaldifferential equation
f(x) (i + X)f^(x) + XhT(i)f^(ax) = c^x c^ XE
(4.22)
Applying (4.16), which is the general solution to (4.3), the general
solution to (4.22) can be written as
f1(x) = f^(0)S(x,a) +
C1X C1 + ^C0 + lE)P^(a)
tS(x,a) 1] (4.23)
Using (4.19) in (4.23), it follows that at x = X
C;l + (cQ + XE)P1(a)
[S(X,a) 1]
PQ(a)P1(a)
CHAPTER 5
THE NUMBER OE OPERATING PERIODS IN [0,X] FOR
THE LINEAR HYSTERESIS PROBLEM
5.1 Introduction
This chapter investigates a renewal process associated with
the linear hysteresis problem. Between successive replacements of the
equipment, as it ages from 0 to X, there exist alternating operating
and repair periods. Operating periods are terminated by failure, not
replacement. At the end of a repair period, there is a partial
reduction of the service age of the equipment. Thus, the termination
of a repair acts as a point of partial regeneration of the service
age process associated with the linear hysteresis problem. At the
conclusion of the last repair period prior to replacement, the equipment
ages continuously, without failure, until replacement occurs at
service age X. The number of operating periods in [0, X], N(X), forms
a counting or renewal process. The importance of the renewal process
N(X) can be understood in an industrial setting. A plant manager has
a piece of equipment which when it fails, prior to replacement, has a
major repair performed on it. He is contemplating reducing the level
of repair; however, he is not certain what the corresponding effect
will be on the expected number of equipment failures prior to replace
ment .
In this chapter, the probability mass function of N(X) is
determined for the linear hysteresis problem with constant failure rate
function. In addition, the expected value of N(X) is obtained.
101
133
Y
Figure B.3. Graph of Log^Q S(Y,B,A) for B = .30.
11
since an unscheduled repair or replacement would involve an unplanned
stoppage in the use of the equipment, and the possible activation of
a backup piece of equipment which would not be as efficient. The
majority of the papers in the area of major repair for a continuously
aging piece of equipment subject to failure and continuous inspection
have concluded that a sufficient condition for a unique finite optimal
equipment replacement age X to exist is that the failure rate be
continuous and strictly increasing to infinity.
Considering only the costs k and K and the distribution function
Fs Barlow and Hunter [1] (also Barlow and Proschan [2], [3]) examine
the major repair problem with instantaneous major repair and planned
replacement. Following a method of objective function formulation
(Morse [18]) using renewal theory, Barlow and Hunter constructed an
objective function for the average cost per unit time over a finite
time span, which was the weighted sum (by the costs, K and k) of the
expected number of planned replacements and the expected number of
unplanned replacements (or major repair) per unit time over the time
span. Then letting the finite time span approach infinity, Barlow and
Hunter minimized the long run expected cost per unit time to obtain X
as the solution to
X
A(X)
F(t)dt F(X)
X will be finite and unique if the failure rate A(u) is continuous and
strictly increasing to infinity. The optimal age for a planned replace
ment is infinite if either (1) the failure rate A(u) = A = constant,
or (2) the cost of a planned replacement is equivalent to the cost of a
16
Table 1.1. Significant Papers for Problems I and II
Author []
Bellman [5]
Descamps [10]
Barlow and Hunter
Sivazlian [21]
Barlow and Hunter
Fox [12]
Beckmann [4]
Scheaffer [20]
Â¡j
t
cti
P.
CD
Pi
t
e
t
t
t
a
u
t
t
a,
cu
pci
u
o
n
t
a
t
t
pc;
u
o
Ml
t
6
o
o
cn
o
u
4J
t
a)
e
t
CJ
t
t1
Ph
(1)
pc:
t
G
O
t
P.
t
P
t
o
o
t
t
o
t
4J
t
t
t
Pu
O
w>
t
t
4J
t
t
O
o
t
t
Q
;>*
t
t
t
o
t t
4J M
CO
O
t t
t
t &Q
pe: t
t
t co
ii t
t o
oo t
PJ it
t
bO
<
t
t
S
a
t
t
a*
w
[1]
[1]
x
x
x
x
X
X
X
X
No
No
No
Yes
No
No
No
No
V
V
F
V
F
F
F
F
F
V
F
F
F
F
V
V
No
V
No
No
No
V
Yes
Yes
No
Yes
No
Yes
Yes
No
Yes
No
Yes
Yes
Yes
No
D
C
C
C
C
C
C
c
ai
p
o"
t
t
t
CJ
t
H
t
O
H
JJ
P
H
O
CO
FE
FE
RT
FE
RT
RT
FE
RT
V variable FE functional equation
F fixed RT renewal theory
D discrete
C continuous
Table 3.3. Table of Optimum X and Corresponding f (0) for
g(x) = lOx When a = 0.50
K
50 150 250 350
6
XhT(i)
X
f1(0)
X
N .
o
<41
X
o
\1
0W
X
f1(0)
0.20
0.10
3.35
271.5
6.10
416.3
8.19
491.9
10.02
540.9
0.40
0.10
3.79
71.3
7.73
89.9
11.28
93.9
14.78
94.9
0.20
3.60
114.7
7.07
151.8
10.11
162.0
13.08
165.2
0.30
3.41
246.3
6.44
343.5
8.99
377.4
11.44
391.0
0.60
0.10
4.34
33.3
9.93
36.3
15.42
36.4
20.92
36.4
0.20
4.12
44.6
9.20
49.7
14.17
50.0
19.17
50.0
0.30
3.91
63.8
8.47
73.3
12.94
74.0
17.42
74.1
0.40
3.71
102.7
7.77
122.7
11.70
124.8
15.68
125.0
0.50
3.51
220.9
7.08
276.0
10.49
284.4
13.94
285.5
0.80
0.10
4.99
18.7
12.50
19.0
20.00
19.0
27.50
19.0
0.20
4.76
23.1
11.75
23.8
18.75
23.8
25.75
23.8
0.30
4.53
29.6
11.01
30.8
17.50
30.8
24.00
30.8
0.40
4.30
39.6
10.26
41.6
16.25
41.7
22.25
41.7
0.50
4.08
56.5
9.52
60.5
15.00
60.6
20.50
60.6
0.60
3.87
91.1
8.78
99.8
13.75
100.0
18.75
100.0
0.70
3.66
196.0
8.05
221.1
12.51
222.2
17.00
222.2
O'*
e
Table 3.6.
Table of Optimum X and Corresponding f (0) for
^ o
g(x) = 1.75x + 7.5x When a = 0.0
K
50 150 250 350
e
AhT(i)
X
f1(0)
X
^(0)
X
f1(0)
X
f1(0)
0.20
0.10
2.91
267.6
4.71
441.3
5.88
547.3
6.82
625.3
0.40
0.10
3.17
71.3
5.36
101.2
6.87
113.9
8.12
121.0
0.20
3.17
107.0
5.36
151.8
6.87
170.9
8.12
181.4
0.30
3.17
214.0
5.36
303.5
6.87
341.8
8.12
362.9
0.60
0.10
3.46
33.8
6.10
41.5
7.99
43.3
9.57
43.9
0.20
3.46
42.3
6.10
51.8
7.99
54.1
9.57
54.9
0.30
3.46
56.3
6.10
69.1
7.99
72.2
9.57
73.2
0.40
3.46
84.5
6.10
103.7
7.99
108.3
9.57
109.9
0.50
3.46
169.0
6.10
207.3
7.99
216.5
9.57
219.7
0.80
0.10
3.78
19.0
6.89
20.9
9.14
21.2
11.01
21.2
0.20
3.78
22.2
6.89
24.4
9.14
24.7
11.01
24.7
0.30
3.78
26.6
6.89
29.3
9.14
29.6
11.01
29.7
0.40
3.78
33.3
6.89
36.7
9.14
37.0
11.01
37.1
0.50
3.78
44.4
6.89
48.9
9.14
49.4
11.01
49.4
0.60
3.78
66.6
6.89
73.3
9.14
74.0
11.01
74.2
0.70
3.78
133.2
6.89
146.6
9.14
148.1
11.01
148.3
5
repair usually involves the repair or replacement of a minor part of
the equipment, sufficient to return the equipment to an operational
status as soon as possible. An example of minimal repair is the
replacement of a blown fuse in an electrical piece of equipment. A
major repair (or overhaul) usually involves the intense testing and
repairing and/or replacing of parts of the equipment. The status of
the equipment after major repair is considered to be as good as new.
Due to the expense of the initial investment and/or the size of the
equipment, overhauls are usually limited to equipment which is never
replaced. For this reason and because major repairs return equipment
to an as good as new state, major repairs are equivalent to replace
ment. Examples of equipment which are subject to major repair are jet
engines, electric generators, and diesel engines. For the remainder
of this discussion, it is assumed that only failure induces a repair,
whether minimal, partial, or major.
A failed equipment which is returned to an operational status
after a minimal repair has the same service age after the repair as
prior to the repair. Thus, the equipment failure rate function remains
unchanged by minimal repair. When the equipment service age attains
a predetermined value, the equipment is replaced with an identical new
one. This type of repair and replacement problem shall be referred to
as Problem I or the minimal repair problem. After a failed equipment
undergoes a major repair, the service age of the equipment is that of
/ a new equipment, which is zero. Thus after major repair, the failure
rate function is that of a new piece of equipment. When the equipment
attains a predetermined service age, it is replaced with an identical
43
Corollary 2.2
If ~;Uv (u > 0) is strictly increasing to infinity and (u)
v(u)
is strictly decreasing, then a unique finite optimal equipment replace
ment age X exists which is the unique solution to equation (2.12),
and if the initial age, x, satisfies x X, then an optimal Xpolicy
exists for the major repair problem with arbitrary repair time distri
bution.
Corollary 2.3
If [c(u) + k(u)A(u)] and r(u) are increasing functions to
infinity and at least one of them is strictly increasing and if
A(u) = A = constant, then a unique finite optimal equipment replacement
age X exists which is the unique solution to the equation
X
[g(X) g(u)]R(u)du = K
0
and if the initial age, x, satisfies x <_ X, then an optimal Xpolicy
exists for the major repair problem with arbitrary repair time distribution.
Corollary 2,4
If [c(u) + k(u)A(u) KA(u)] (u >. 0) is strictly increasing
to infinity, then a unique finite optimal equipment replacement age X
exists which is the unique solution to the equation
X
{[c(X) + k(x)A(X) K (X)] [c(u) + k(u)A(u) KA(u)]}R(u)du = K
0
and if the initial age, x, satisfies x <_ X, then an optimal Xpolicy exists
for the major repair problem with instantaneous repair.
g(u) KA(u) Ki _
m the sufficient
The function
Table B.2. Table of S(Y,B,
k
y
0.0
0.10
0.20
0.30
o
JC
o
0.50
0,0
1.00C
1 .000
1 .COO
1.CC0
l.COO
1.000
0.2
1.177
1,177
1.176
1.176
1.176
1.175
0.4
1.393
1.392
1.39C
1.389
1.387
1.386
0.6
1.659
1.654
1.650
1.646
1.642
1.638
0.8
1.980
1.973
1.966
1.958
1.951
1.942
1.0
2.375
2.363
2.350
2.337
2.323
2.309
1.2
2.856
2.838
2.813
2.796
2.774
2.750
1.4
3.464
3.417
3.36 7
3. 355
3.320
3.2e3
1.6
4.162
4.123
4.080
4.033
3.9S3
3.928
1.8
5.040
4.985
4.925
4.859
4.766
4.708
2.0
6.111
6.037
5.955
5.863
5.763
5.652
2.2
7.420
7.321
7.211
7.087
6.949
6.797
2.4
9.019
8.889
8.742
3.577
8.392
8. 185
2.6
10.971
10.802
10.611
1C.393
10.148
9.871
2.8
13.356
13.139
12.891
12.607
12.284
11.917
3.0
16.268
15.992
15.673
15.306
14.935
14.404
3.2
19.326
19.475
19.068
19.598
18.054
17.427
3.4
24.171
23.729
23.213
22.613
21.914
21.103
3.6
29.478
20.923
28.273
27.511
26.619
25.577
3.8
35.941
35.267
34.450
33.408
32.355
31.021
4.0
43.879
43.014
41.992
40.782
39.348
37.651
4.2
53.549
52.476
51 201
49.694
47.877
45.725
4.4
65.360
64,031
62.445
60.550
58.201
55.562
4.6
79.707
78.143
76.176
73.814
70.973
67.549
4.8
97.408
95.379
92.943
90.008
06.458
82. 158
5.0
110.930
116.429
113.41R
109.777
105.355
99.969
5.2
145.217
142.139
138.424
133.915
128.416
121.686
5.4
177.324
173.540
168.961
163.387
156.562
140.170
5.6
216.540
211.893
206.256
199.375
190.918
180.473
5.8
264.437
253.734
251.803
243.319
232.857
219.077
6.0
322.940
315.945
307.429
296.980
264.053
267.949
6.2
394.395
335.823
375.368
362.510
346.557
326.606
6.4
481.672
471.171
458.344
442.539
422.870
398.183
6.6
588.272
575.414
559.687
540.271
516.047
485.534
6.8
710.473.
702.733
683.461
659.627
629.817
592.143
7.0
877.501
058.24 1
034.634
005.393
760.737
722.266
7.2
1071.738
1048.177
1019.273
903.416
938.373
881.100
7.4
1308.980
1200.166
1244.784
1200.837
1145.522
1074.992
7.6
1598.745
1563.513
1520.217
1466.375
1398.484
1311.689
7.8
1952.6 70
1909.595
1856.628
1790.680
1707.403
1600.663
8.0
2384.956
2332.302
2267.517
2186.786
2004.661
1953.470
8.2
2912.943
2048.588
2769.363
2670.552
2545.377
2384.223
0.4
3557.933
3479.183
3302.317
3261.405
3100.035
2910.170
8.6
4345.477
4249.371
4130.953
3983.053
3795.194
3552.369
8.8
53C7.523
5190.102
5045.348
4864.418
4634.383
4336.508
9.0
6482.590
6339.105
6162.1Q 5
5940.922
5659.320
5294.035
9.2
7917.016
7742.51 2
7526.293
7255.746
6911.090
6463.313
9.4
9670.813
9456.633
9192.413
8061.645
8439.906
7891.176
9.6
11811.922
11550.270
1 1227.419
10823.059
10307.102
9634.859
9.9
14427.032
14107.434
13712.965
13218.691
12587.590
11764.230
10.0
17621.254
17230.705
16748.801
16144.727
15372.863
14364.630
for B = .20.
0.60
0.70
0.80
0.90
l.CC
1.000
1.000
l.COO
l.COO
l.COO
1.175
i.175
1.174
1.174
1. 174
1.334
1.332
1.381
1.379
1.377
1.634
1.630
1.625
1.621
1.616
1.934
1.925
1.916
1.906
1.896
2.293
2.277
2.260
2.243
2. 226
2.725
2.698
2.670
2.641
2.612
3.244
3.202
3.158
3.112
3. C65
3.869
3.306
3.739
3.669
3. 597
4.622
4.530
4.432
4.329
4.221
5.531
5.400
5.259
5, 109
4.953
6.629
6.44 5
6.246
6.03*
5.812
7.956
7.703
7.427
7.131
6.821
9.561
9.216
8.838
8.431
G.C04
11.503
11.040
10.523
9.975
9.393
13.856
13.239
12.553
11.808
11.023
16.703
15.892
14.980
13.934
12.936
20.167
19.097
17.892
16.572
15.180
24.365
22.969
21.336
19.648
17.814
29.460
27.650
25.588
23.308
20.905
35.650
33.314
30.636
27.664
24.532
43.171
40.169
36.708
32.850
28.789
52.314
48.472
44.015
39.029
33.704
63.433
58.533
52.814
46.394
39.646
76.959
70.729
63.416
55.175
46.525
93.419
85.520
76.196
65.651
54.598
113.454
103.467
91.611
70.153
64.071
137.848
125.250
110.211
93.079
75.189
167.556
151.700
132.665
110.909
68.234
203.745
183.328
159.787
132.213
IC3.544
247.837
222.864
192.558
157.602
121.5 LO
301.566
270.309
232.174
188.141
142.593
367.056
327.993
200.08 l
224.562
167.335
446.891
398.144
338.042
268.195
196.370
544.227
403.476
408.189
320.417
230.442
662.919
587.301
493.118
382.966
270.425
807.668
713.657
595.978
457.915
317.346
934.219
867.464
720.597
547.754
372.409
1199.581
1054.723
871.625
655.479
437.026
1462.312
1282.759
1054.719
784.690
512.856
1782.864
1560.495
1276.752
939.755
6C1.842
2173.985
1890.817
1546.076
1125.874
706.269
2651.260
2311.014
1872.863
1349.362
828.813
3233.706
2013.290
2269.475
1617.804
972.623
3944.542
3425.412
2750.951
1940.343
1141.383
4812.098
4171.484
3335.592
2328.022
1339.425
5871.043
5080.973
4045.677
2794.126
1571.932
7163.648
6109.777
4908.277
3354.700
1844.562
8741.535
7541.746
5956.449
4029.094
2164.615
10667.758
9190.336
7230.352
4840.621
2540.199
13019.320
11200.875
8778.895
5817.543
2980.954
130
70
d2f]_(0) { 23 [ S (X ,a) 1] (23X + Y)Sx(X,a)}[S(X,a) l]2
dX2 [S(X,a) l]4
{(23X y)Sx(X,a) (K gX2 yX)Sxx(X,a)}[S(X,a) l]2
[S(X,a) l]4
2Sx(X,a)[S(X,a) 1]{ (23X + y)[S(X,a) 1]}
[S(X,a) l]4
2Sx(X,a)[S(X,a) 1]{ (K 3X2 yX)Sx(X,a)}
[S(X,a) l]4
Substituting (3.44) in (3.46) and simplifying
(3.46)
df1(0)
dX2
X=X*
2B[S(X*a) 1] (K 3X*2 yX^S^X* ,a)
[S(X*,a) l]2
(3.47)
Returning to equation (3.45), let M(X) equal the lefthand
side of the equation, that is
M(X) = 6X + X (26X + 1)
S(X,a) 1
Sx(X,a)
Differentiating M(X) with respect to X yields
dM(X)
dX
= 26X + 1 26
 (26X + 1)<
S(X,a) 1
Sx(X,a)
Sx(X,a)Sx(X,a) S^d.a) [S(X,a) 1]
[Sx(X,a)]2
Simplifying
APPENDIX B
Consider the power series
_ 00 j j1 ,
S(x,0,Ah (i),a) = 1 + Â£ tv it (0 Ah (i)a )
1 3 k=0
(B.l)
Applying the ratio test for power series of the form I a^x11 it is
determined that
R = lim
jH
1 J1 k
TT It (e Ah (i)a )
3 k0
(j+1)! k^Q (6 XMl)a ^
lim
j+oo
JL
6 AhT(i)aJ
where R is the radius of convergence of the power series. Therefore
for any value of x less than infinity, the power series S(x,0,Ah^(i),a)
converges.
Taking the first derivative of S(x,0,Ah (i),a) with respect to
x, the following is obtained
00 j j
= (0 Ah_(i)) + l ~ TT (9 Ah (i)ak) (B.2)
j1 3 k=0
Further, the nth derivative of S(x,0,Ah^(i),a) with respect to x is
given by
12$
143
0.0
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Y
10
Figure B.8. Graph of Log
S(Y,B,A) for B = .80.
7
aging is often associated with a periodic inspection policy in which
the inspection determines the state of the equipment, where the state
can have a value from 0 to S inclusive. The state 0 denotes a failed
equipment and the state S denotes a new equipment. After inspection,
the equipment is classified into one of S + 1 states and a decision
is made to either (a) replace, or (b) perform preventive maintenance,
or (c) do nothing. If the equipment can be classified into S 4 1
operational states, transitions from one state to other states are
specified by some probability mass function, and periodic inspections
take place, then the resulting model is one of Markovian Replacement.
Continuous aging is associated with both discrete and continuous
inspection, although the latter is more common in replacement theory.
The literature of replacement theory will be surveyed for
Problems I, II, and III. Under the different optimization criteria
previously discussed, the authors in replacement theory have considered
primarily the following costs and parameters. They are
K = fixed cost incurred for planned replacement;
k = instantaneous repair cost incurred at failure;
i = interest rate;
F(") = lifetime distribution function of the equipment;
A(0 = failure rate function or hazard function.
The failure rate function !() is also referred to in the literature
as the failure rate. The last section of the literature review
covers downtime for repair.
a. Problem I minimal repair
The earliest studies in the area of classical equipment
14
to a subclass of all possible rules. Further, the optimal stopping
rules can be determined by solving linear programming problems.
Dertnan's objective function for his results was the long run total
expected cost.
Extensions to Derman's work to obtain formal rules based upon
the costs for repair at failure, replacement, and the observed state
have been made by Derman [9], Kolesar [16], and Ross [19].
c. Problem III hysteresis
Only one paper, and that in the area of Markovian Replacement,
has considered the hysteresis effect. Eppen [11] determined conditions
under which a policy for preventive maintenance is optimal for a
Markovian deteriorating system with a discrete state space. Following
each inspection to determine in which one of the S + 1 states (states
are 0,..., S) the system is, a decision is made to either leave the
system in the present state or place it in a newer and higher state by
performing some maintenance action. Since the system deteriorates
from S to 0, it improves in the reverse direction. It is assumed
that all maintenance actions are performed instantaneously. The cost
of improving the system by z states (z integer) is cz (c > 0). For
each state an operating cost L() is assigned, such that L() is a
finite, positive, real valued function defined on 0, 1, .S which
2
possesses the properties: (1) L() is convex and (2) AL < c .
The deterioration of the system is described by onestepdown transition
matrices. A functional equation approach was employed to minimize
the cost of operating and maintaining the system. The conditions for
the policy of preventive maintenance were based on the onestepdown
transition matrices.
2
hysteresis) can be a function of the length of the repair, the expense
incurred in undertaking the repair, and the service age of the equip
ment at the instant of failure. In addition, by varying the amount of
partial reduction of service age, it is possible to recover the
extreme problems of minimal and major repair. The problem generated
by considering the amount of partial reduction of service age shall
be called the hysteresis problem. To date, optimal policies have not
been determined for the hysteresis problem.
A special case of the hysteresis problem, called the linear
hysteresis problem, occurs when the function defining the partial
reduction of service age is linear. If the amount of partial repair
in this problem is economically controlled, the resulting problem shall
be called the economically controlled linear hysteresis problem.
This dissertation presents the formulation of the hysteresis
problem using an optimization approach due to Bellman [5] (also
Bellman and Dreyfus [6]), called the functional equation approach.
Analytic and quantitative results are obtained for the linear
hysteresis problem. For the economically controlled linear hysteresis
problem, quantitative results are obtained for which a qualitative
analysis is performed. Some stochastic elements of the hysteresis
problem are also investigated.
The remainder of this chapter is devoted to several develop
mental aspects of the hysteresis problem. There is a discussion of
the basic optimization criteria considered in the literature and in
this dissertation. An extensive literature survey is presented. The
topics covered include: minimal repair problem, major repair problem,
68
f 1(0)
K(X) 
c2X2 [2c2 + ClP2(a)]X
P^CaT P1(a)P2(a)
S(X,a) 1
2c2 + ClP2('a'> + CQP1(a)P2(a)
P0(a)P1(a)P2(a)
The analysis of the quadratic form of g(x) is now restricted
to the case where the fixed cost of replacement K(X) is independent
of the replacement age, that is, K(X) = K = constant. Then
K
c2X
f1(0)
P2(a)
:2c2 + c1P2(a)]X
P1(a)P2(a)
S(X,a) 1
+
2c2 + c1P2(a) + cQP1(a)P2(a)
P0(a)P1(a)P2(a) ~
(3.39)
For notational convenience let
P2(a)
(3.40)
Y =
2c2 + c1P2(a)
Y1(a)P2(aT
(3.41)
Using (3.40) and (3.41), equation (3.39) can be written as
2 2c + c P (a) + c P (a)P (a)
f K gX yX 2 12 0 1 2
rilU; S(X,a) 1 PQ(a)P1(a)P2(a)
(3.42)
The following theorem presents the results obtained for the quadratic
form of g(x).
Theorem 3.3
Sufficient conditions for a unique finite optimal equipment replace
31
P(E)
Figure 4.1. Example of a linear curve p(E).
p(E)
Figure 4.2. Example of a linear curve p(E).
Table B.8
Table of
v
o
o
0.10
0.0
1,000
1.00c
0.2
1.044
1.044
0.4
1.09 0
1 .097
0.6
1.164
1.161
0. 8
1.24 5
1.230
1.0
1.3*4
1.332
1.2
1.4 64
1.445
1.4
1.611
1.584
1.6
1.791
1.751
1.8
2.010
1.955
2.0
2.278
2.204
2.2
2.635
2.507
2.4
3.C05
2.875
2.6
3.49 3
3.325
2.3
4.089
3.873
3.0
4.917
4.542
3.2
5.706
5.357
3.4
6.793
6.353
3.6
8.120
7.56
3.0
9.740
9.050
4.0
11.720
10.860
4.2
14.137
13.070
4.4
17.090
15.767
4.6
20.697
19.061
4.8
25.102
23.084
5.0
30.432
27.995
5.2
37.054
33.993
5.4
45.031
41.310
5.6
54.835
50.264
5.8
66.360
61.189
6.0
81.435
74.531
6.2
99.349
90.826
6.*
121.169
U0.728
6.6
147.819
135.035
6.3
180.369
164.722
7.0
220.126
200.980
7.2
263.634
245.265
7. *
327.994
299.35 1
7.6
400.4 36
365.412
7.0
488.916
446 .099
8.0
596.988
544.648
8.2
720.935
665 .015
8.4
890.203
812.031
8.6
1C87.126
991.594
8.8
1327.641
1210.911
9.0
1621.407
1478.703
9.2
1930.216
1905.963
9.4
2418.464
2205.578
9.6
2953.743
2693.670
9.8
3607.534
3289.823
1C.0
4406.051
4017.968
20
0.30
0.40
.COO
1.000
1.C00
.044
1.043
1.043
.095
1.C94
1.092
. 157
1.153
1.149
.231
1.223
1.216
.319
1.307
1.294
.426
1.4CS
1.385
.554
1.523
1.492
7C9
1.665
1.619
. 896
1.834
1.769
. 123
2.037
1.948
.398
2.233
2.161
.732
2.578
2.417
.138
2.936
2.723
.632
3.368
3.090
.232
3.893
3.533
.963
4.529
4.066
.353
5.301
4.710
.938
6.239
5.409
.260
7. 380
6.431
.872
8.763
7.573
.833
10.457
8.959
.237
12.514
10.640
.163
15.021
12.683
.735
18.076
15.166
.093
21.600
18.185
.414
26.341
21.360
.909
31.879
26.333
.830
38.636
31.780
.518
46.879
30.417
. 330
56.939
46.504
.771
69.216
56.363
. 396
84.200
68.383
.917
102.491
83.041
.200
124.620
ICO.920
. 296
152.031
122.731
493
105.364
149.343
.362
226.003
101.016
.826
275,622
221.445
.228
336.215
269.811
.435
410.206
328.849
.940
500.560
400.916
.022
610.904
480.895
.090
745.656
596.304
.941
910.225
727.442
. 940
1111.206
887.558
.404
1356.662
1083.065
.948
1656,440
1321.791
.762
2022.564
1613.302
.172
2469.721
1969.277
.345
3015.854
2403.986
o.
i
i
i
y
i
i
i
i
i
i
2
2
2
3
3
4
4
5
6
B
9
11
14
17
20
25
30
36
44
54
66
80
98
119
146
170
217
265
323
395
482
588
719
877
1071
1 308
1598
1951
2383
2911
3555
S(Y,B,A) for B = .80.
A
0.50 0.60 0.70
0.80
070
1.00
1.000
i.coo
1.0C0
1.000
1.000
1.CC0
1.043
1,042
1.042
1.0*1
1.0*1
1.041
1.091
1.0C9
1.C88
1.086
1.085
l C ? 3
1.146
1.142
1.133
1.134
l.m
1.127
1.203
1.201
1.193
1.186
1.150
1.174
1.260
1.267
1.255
1.243
1.232
1.221
1.36*
1.343
1.323
1.30 4
1.237
1.271
1.460
1.429
1.359
1.371
1.345
1. 323
1.572
1.527
1.483
1 4 4 3
1.400
1.377
l .703
1.639
1.578
1.523
1.47*
1.433
1.357
1.768
1.685
1.610
1.545
1.492
2.033
1.918
i.eos
1.705
1.621
1.553
2.252
2.09L
1.941
1.3 10
1.701
1.616
2.505
2.293
2.096
1.926
1.738
1.682
2.SC6
2.523
2.273
2.054
1.550
1.751
3.164
2.804
2.475
2.196
1.979
1. 322
3.591
3.128
2.707
2.354
2.035
1.396
4.103
3.511
2.973
2.530
2.199
1.974
4.716
3.962
3.292
2.726
2.322
2.054
5.452
4.497
3.639
2.946
2.454
2.138
6.337
5.132
4.053
3.194
2.596
2.226
7.404
5.ee8
4.537
3.472
2.750
2. 316
8.691
6.790
5.101
3.706
2.916
2.411
10.245
7.867
5.762
4.142
3.095
2. 509
12.125
9.157
6.538
4.546
3.290
2.612
14.400
10.703
7.45 2
5.005
3.500
2. 718
17.157
12.560
8.530
5.529
3.729
2. 829
20.500
14.794
9.804
6.129
3.970
2.945
24.556
17.483
11.314
6.815
4.249
3. 065
29.462
20.725
13.105
7.605
4.544
3. 190
35.466
24.638
15.236
8.514
4.866
3.320
42.741
29.365
17.773
9.563
5.218
3.456
51.509
35.08L
20.800
10.777
5.604
3. 597
62.354
41.999
24.417
12.184
6.027
3.743
75.457
50.373
28.744
13.820
6.491
3. 896
91.412
60.532
33.928
15.724
7.002
4.055
110.846
72.047
40.148
17.946
7.564
4.221
134.522
87.792
47.617
20.544
8.183
4.393
163.376
105.937
56.590
23.586
8.860
4. 572
198.547
127.979
67.406
27.155
9.624
4. 759
241.427
154.766
80.427
31.351
10.463
4.953
293.712
187.335
96.120
36.290
11.393
5.155
357.480
226.940
115.075
42.115
12.426
5. 366
435.266
275.143
137.958
48.994
13.576
5. 585
530.158
333.802
165.615
5 7.. 1 3 2
14.857
5. 912
645.935
405.215
199.064
66.770
16.267
6. 050
787.209
492.180
239.545
73.204
17.886
6.7 97
959.608
598.110
288.562
91.783
19.676
6.554
1170.014
727.167
347.952
107.932
21.683
6. 821
1426.820
884.435
419.940
127.159
23.937
7. 099
1740.283
1076.116
507.265
150.078
26.473
7. 309
142
74
most economic life at which an equipment should be replaced is 6.39
years.
2
The quadratic cost function g(x) = 1.75 x + 7,5 x is examined
in Tables 3.6 through 3.10 for specific values of a. These values
are 0, 0.25, 0.50, 0.75 and 1. Each table presents the optimum
economic life in years and the corresponding value for f^(0) for
various combinations of K, 0, and Ah^Ci).
The behavior of X and f^(0) observed from Tables 3.6 through
3.10 for the quadratic cost function is the same as for the linear cost
function previously discussed, and therefore will not be repeated
here.
It should be noted that the sufficient condition on the
coefficients of the polynomial g(x) for a unique finite optimal
equipment replacement age X to exist is not a necessary condition.
There may exist a polynomial g(x) with some negative coefficients which
is convex over an interval which contains a unique optimal age X.
P{N(X) = 4} =
X
AX
(la)(la2)(la3)
Ax^(la^)
Ae dx^
AX(la3)
e
(la)(la2)(la3)
Ax., (la)
Ae dx]
0
Ax(la2)
5 P{N(X) = 2}
(la)(laZ)
Ax(la)
8 (la) PX> 3>
(5.19)
Using (5.8) in (5.19) and integrating the first term on the righthand
side of (5.19) yields
P{N(X) = 4} = ^ r 7 [1 e AX(1 a
(la)(la)(la)(la)
AX(la3)
x 0 P{N(X) = 1}
(la)(la ;(1aJ
AX(la2)
5 P{N(X) = 2}
(la)(la ;
AX(la)
e'"(l^I) piN(x) = 3> (5.20)
By induction or otherwise, the following is obtained
AX rN
P{N(X) = r} = [1 e AX(1 3 ']
7T (la3)
j=l
r1 AX(lar_k)
" I P{N(X) = k} r=2,3 ,...
k=l r K
LIST OF FIGURES
Figure
Page
1.1
The variation of f(x) illustrating the cost curves
associated with the alternatives of do not replace
(h(x)) and replace (K(x) + h(0)) and the decision
regions of do not replace and replace for f(x)....
19
1.2
Sample function of the service age process S(t)
for the minimal repair problem with downtime,
where x is the initial equipment age, and X is
the equipment replacement age
22
1.3
Sample function of the service age process S(t)
for the major repair problem with downtime,
where x is the initial equipment age, and X is
the equipment replacement age
23
1.4
Sample function of the service age process S(t)
for the hysteresis problem with downtime, where
x is the initial equipment age, X is the equip
ment replacement age, and a(u) is the hysteresis
p gp funrti nn
24
4.1
Example of a linear curve p(E)
81
4.2
Example of a linear curve p(E)
81
4.3
Example of a curve p(E)
82
4.4
Example of a curve p(E)
82
4.5
Example of a concave curve p(E)
83
4.6
Example of a convex curve p(E)
83
5.1
Sample function of the service age process S(t)
when N(X) = 1 for the linear hysteresis problem
with downtime and constant failure rate function..
105
5.2
Sample function of the service age process S(t)
when N(X) = 2 for the linear hysteresis problem
with downtime and constant failure rate function..
105
5.3
Sample function of the service age process S(t)
when N(X) = 3 for the linear hysteresis problem
with downtime and constant failure rate function..
106
ix
20
for do not replace (h(x)) and replace (K(x) + h(0)) intersect at age X.
The importance of continuity of the function f(x) at x = X is best
illustrated by assuming the opposite, that is a discontinuity exists
at x = X for f(x). Then the following paradox occurs to the left of
the discontinuity (i.e., x < X), the cost curve corresponding to do
not replace is lower than the cost curve of replace, which implies
that the equipment should not yet be replaced. To the right of the
discontinuity (i.e., x > X), the cost curve corresponding to replace
is lower than the cost curve of do not replace, which implies that the
equipment should already have been replaced. To resolve this paradox,
it is necessary to require continuity of the function f(x) at x = X.
Then a replacement occurs at age X.
1.5 Problem Formulation
a. The Functional Equation for the Hysteresis Problem
It is assumed that the equipment is continuously monitored or
reviewed and that failures are detected instantly. Further, the
planning horizon is infinite and all replacements are made with
identical equipment. All replacements are assumed to occur instantaneously.
Following a failure, there is a length of time called the downtime during
which the equipment is repaired before being returned to an operational
status. A zero downtime corresponds to an instantaneous repair. At
the termination of the repair, the service age of the equipment is
governed by the hysteresis age function a(x), where x is the service
age of the equipment at the instant of failure and a(x) is the service
age at the termination of the repair.
10
Sivazlian determined that a necessary and sufficient condition for the
existence of a unique optimal equipment replacement age X, where X is
the unique solution to the equation
X
[1 e~J(w)]dÂ£(w) = K
0
is that c(w) (w _> 0) be strictly increasing to infinity. Now
with
w
J (w)
j(x)dx
0
j(x) = i + A(x)[1 hT(i)]
and
c(w) = yjyy {c(w) + ryW [j (w) i] + A(w)k(w)}
where h^Ci) is the LaplaceStieltjes transform with parameter i of the
general distribution function of repair times. Due to the structure
of c(w), it was not necessary for the failure rate to be strictly
increasing to infinity for a unique optimal equipment replacement age
to exist.
b. Problem II major repair
Every major repair problem in the literature has considered two
fundamental costs: a fixed cost K for a planned replacement, and a
fixed cost k for a major repair or replacement at failure. In addition,
it has been assumed that a planned replacement costs less than or equal
to a repair or replacement at failure (i.e.5 0 < K <_ k). This is logical
Table B.3.
Table of
S(Y,B,A)
for B
O
CO
II
V
0.0
0.1C
O
rsi
O
0.30
O
o
A
0.50
0.60
0.70
o.eo
0.90
l.CC
0.0
l.COO
1.00C
l.COO
l.COO
i.OOC
l.COO
l.COO
1.000
l.COO
\ .000
1.CC0
0.2
1.15 5
1.155
1.154
1.154
1.153
1.153
1.152
i. 152
1.151
1.151
1.150
0.4
1.344
1.342
1.340
1.338
1.336
1.334
1.332
1.330
1.328
1.325
1.323
0.6
1.575
1.571
1.566
1.561
i.556
1.550
1.545
1.539
1.533
1.529
1.522
0. 3
1.853
1.849
1.839
1.929
1.819
1.808
1.797
1.786
1.774
1.762
1.751
1.0
2.203
2.107
2.171
2.154
2.135
2. 1 16
2.097
2.077
2.056
2.035
2.014
1.2
2.624
2.600
2.574
2.546
2.517
2.486
2.454
2.421
2.336
2.35)
2.315
1 .4
3.139
3.102
3.C63
3.C22
2.977
2.930
2.830
2.82e
2.774
2.719
2.664
1.6
3.767
3.715
3.659
3.598
3.533
3.463
3.359
3.311
3.230
3. 149
3.065
1.8
4.535
4.46 3
4.384
4.293
4.205
4.105
3.993
3.884
3.767
3.646
3.525
2.0
5.472
5.375
5.267
5.149
5.019
4. 379
4.727
4.567
4.399
4.227
4.055
2.2
6.617
6.408
6.343
6. 183
6.006
5.812
5.603
5.369
5.145
4.905
4.665
2.4
8.0 16
7.846
7.654
7.440
7.203
6.941
6.656
6.350
6.027
5.695
5. 366
2.6
9. 725
9.503
9.253
8.971
8.656
8.306
7.922
7.503
7.069
6.619
6.172
2.3
11.811
11.527
11.203
10.835
10.421
9.958
9.448
8.893
8.304
7.697
7. 099
3.0
14.360
13.997
13.530
13. 105
12.566
11.961
11.287
10.552
9.767
8.959
8.165
3.2
17.473
17.013
16.482
15.872
15.176
14.388
13.508
12.540
11.504
10.436
9.393
3.4
21.275
20.695
20.022
19.244
18.351
17.334
16.190
14.925
13.566
1 2 166
10.805
3.6
25.919
25.191
24.342
23.356
22.217
20.911
19.433
17.790
16.019
14.194
12.429
3.5
31.591
30.68 1
29.615
28.370
26.924
25.257
23.357
21.235
18.939
16.571
14.296
4.0
33.519
37.306
36.052
34.487
32.659
30.539
26.109
25.391
2 2.4 1 7
19.362
16.445
4.2
46.93C
45.573
43.909
41.949
39.647
36.962
33.868
30.374
26.566
22.640
19.916
4.4
57.315
55.572
53.5C3
51.054
48.165
44.776
40.850
36.393
31.517
26.492
21.758
4.6
69.939
67.753
65.216
62. 165
58.549
54.287
49.320
43.654
37.433
31.022
25.028
4.8
85.357
02.696
79.518
75.727
71.213
65.866
59.602
52.420
44.506
36.353
28.739
6.0
104.189
100.909
96.982
92.200
86.658
79.967
72.088
63.010
52.970
42.631
33.115
5.2
127. no
123.154
118.307
112.486
105.500
97.146
87.250
75.311
63.106
50.029
38.092
5.4
155.204
150.322
144.349
137. 154
128.487
118.078
105.698
91.295
75.255
59.751
43.ei6
5.6
139.590
183.504
176.151
167.27L
156.536
143.589
128.119
110.036
89.827
69.042
50.400
5.3
231.509
224.030
214.989
204.042
190.764
174.608
155.391
132.730
107.316
81.191
57.974
6.0
202.697
273.525
262.419
248.939
232.539
212.607
188.574
160.225
129.322
95.542
66.636
6.2
345.221
333.980
320.345
303.760
283.526
258.847
228.963
193.553
153.569
112.506
76.707
6.4
421.506
407.818
391.092
370.705
345.767
315.245
278.134
233.971
103.932
132.57C
88.234
6.6
514.863
498 .00 1
477.495
452.454
421.748
384.041
338.015
233.005
220.470
156.313
1C1.494
6.8
628.738
600.150
583.020
552.203
514.505
467.971
410.953
342.519
264.463
184.428
116.746
7.0
767.930
742.684
711.904
674.197
627.752
570.376
499.814
414.780
317.466
217.738
134.290
7.2
937.896
907.003
869.314
823.081
766.021
695.337
608.099
502.549
381.358
257.225
154.470
7.4
1145.482
1107.700
1061.569
1004.907
934.846
047.832
740.075
609.189
458.416
304.062
177.683
7.6
1399.020
1352.830
1296.379
1226.966
1140.939
1033.946
900.952
738.797
551.400
359.647
204.334
7.8
1703.711
1652.231
1583.173
1498.167
1392.709
1261.100
1097.093
896.371
663.657
425.650
235.097
8.0
2086.961
2017.922
1933.456
1829.380
1700.092
1530.392
1336.260
1087.996
799.247
504.066
270.425
8.2
2540.950
2464.568
2361.277
2233.908
2075.449
1876.868
1627.924
1321.007
963.087
597.277
311.052
8.4
3113.229
3010.105
2883.816
2727.965
2533.830
2290.075
1983.659
1604.694
1161.156
700.131
357.307
8.6
3802.442
3676.424
3522 .036
3331.377
3093.611
2794.534
2417.589
1949.841
1400.700
0 40.038
411.576
6.8
4644.215
4490.234
4301.520
4068.351
3777.231
3410.419
2946.949
2369.972
1690.521
997.073
473.426
9.0
5672.395
5484.250
5253.598
4968.426
4612.066
4162.352
3592.792
2831.482
2041.305
1184.115
544.56?
9.2
6920.223
6690.352
6416.473
6067.785
5631.656
5080.477
4380.801
3504.370
2466.036
1407.000
626.405
9.4
8462.090
8 18 1 262
7836.793
7410.504
6876.887
6201.531
5342.379
4262.992
2980.479
1672.748
720.537
9.6
10335.559
5992.480
9571.586
9050.461
8397.664
7570.418
6515.844
5187.148
3603.800
1989.723
828.815
9.8
12623.8 16
12204.711
1 1690.434
11053.461
10255.012
9241.945
794T..969
6313.102
4359.254
2367.994
953.365
1C.0
15413.719
14906.742
14278.402
13499.906
12523.430
11283.102
9695.833
7685.117
5275.207
2719.503
1C 96, 632
21
The range of values that a(x) can assume lies in the interval
[0, x]. When a(x) = x, this corresponds to no reduction of service
age after a repair, which is the minimal repair problem (Problem I).
If a(x) = 0, there is a reduction of service age to zero at the
termination of the repair, which is the major repair problem (Problem II)
For 0 <_ a(x) _< x, a partial reduction, or hysteresis, of service age
occurs, which is the hysteresis problem (Problem III) Figures 1.2,
1.3, and 1.4 are sample functions of the service age process S(t) for
Problems I, II, and III with downtime for repair, respectively. Thus,
the hysteresis age function covers Problems I, II, and III. The
linear hysteresis problem where a(x) = ax for 0 <_ a Â£ 1 will be
investigated in this dissertation.
For a specified value of the replacement age X (X > 0),
the following replacement policy is known as an Xpolicy. Given an
equipment with service age u, it is retained if 0 <_ u < X and replaced
instantly if X <_ u < where all replacements are made with identical
new (service age zero) equipment in an operating state. Define for u > 0
K(u) = replacement cost at age u incorporating a salvage
feature (nonrecurring cost);
k(u) = repair cost associated with failure at age u and
incurred instantaneously (i.e., at commencement of
repair);
c(u)du = operations cost of equipment between u and u + du
(recurring cost);
r(u) = loss of revenue per unit of downtime for an equipment
with age u and in a state of repair;
A(u)du = probability that the equipment will fail between
ages u and u + du given that the equipment is in an
operating state at age u;
10
Figure B.9.
Graph of Log
S(Y,B,A) for B = .90.
Service Age
Figure 1.4. Sample function of the service age process S(t) for the hysteresis
problem with downtime, where x is the initial equipment age, X is
the equipment replacement age, and a(u) is the hysteresis age
function.
CHAPTER 1
THE PROBLEM
1.1 Introduction
The literature concerned with the determination of optimal
replacement policies for continuously aging equipment subject to
failure and continuous review has examined only the problems of minimal
and major repair. The distinction between these problems is in the
'amount' of recovery of equipment life following a repair. Whereas
no recovery of equipment life occurs after a minimal repair, there is
a complete recovery of equipment life following a major repair.
Recovering equipment life is equivalent to reducing service age, the
total operating time of the equipment.
For most of the literature, the costs associated with repair
and replacement have been considered invariant with respect to the
service age of the equipment. In addition, the majority of the
literature has been concerned with the determination of necessary and
sufficient conditions for minimizing either total expected discounted
cost or expected cost per unit time over an infinite time horizon. Not
until the last two years has any attention been focused on continuous
variable operating and repair costs. Also, only a few papers in the
literature have assumed that repairs are not instantaneous.
Between the extremes of minimal repair and major repair exists
the situation of partial repair. Following this repair there is a
hysteresis effect, which is a partial reduction of service age. This
is in contrast to a complete (no) reduction of service age in the case
of a major (minimal) repair. The amount of partial reduction (or
1
Table 4.1.
a
Table of Optimum
Convex Curve p(E)
Values for E
as in Figure
,Xpolicy
4.6.
for
q(x) =
lOx + AE
K
e
X
XhT(i)
Scaling Factor
E
X
a
f]_(0)
100
0.6
0.1
0.1
30
6.75
7.13
0.55
37.44
100
0.6
0.3
0.3
30
11.40
6.50
0.40
78.89
200
0.2
0.1
0.1
35
35.00
8.00
0.00
433.81
200
0.4
0.1
0.1
35
28.52
10.26
0.10
93.53
300
0.2
0.1
0.1
40
40.00
10.37
0.00
477.18
400
0.2
0.1
0.1
40
40.00
12.60
0.00
499.75
400
0.6
0.3
0.3
20
14.60
23.87
0.15
74.66
400
0.6
0.3
0.3
25
14.37
22.67
0.25
r". 1
00
1
r^
1
I
300
0.6
0.3
0.3
30
15.15
16.97
0.30
80.51
300
0.6
0.45
0.3
30
9.75
15.62
0.45
86.30
32
condition for a unique finite optimal equipment replacement age X to
exist is determined for age replacement with major repair and downtime.
Several corollaries to this condition are also presented. An expression
is obtained for the undiscounted long run expected cost per unit time.
The linear hysteresis problem with downtime for partial repair
and constant failure rate function is investigated in Chapter 3. An
nth degree polynomial with all positive coefficients is assumed for the
cost function g(x), and the functionaldifferential equation (1.10)
is solved for a(x) = ax. Three particular polynomial forms are
investigated, and sufficient conditions are derived for a unique
finite equipment replacement age X to exist for the linear hysteresis
problem.
In Chapter 4 the results of Chapter 3 are applied to study the
economically controlled linear hysteresis problem with downtime for
partial repair. For several curves p(E), two expenditure repair functions
are examined: constant and linear. Quantitative results are obtained
instead of analytic results, due to the structure of the solution of
the functionaldifferential equation (1.14) for each of the
expenditure repair and the curve p(E) functions.
Chapter 5 examines the renewal process associated with the
number of operating periods between successive replacements. A
probability mass function is determined for the number of operating
periods in [0,X]. Using this result, the expected value of the
number of operating periods in [0,X] is obtained.
Conclusions and recommendations for future research are
presented in Chapter 6.
This dissertation was submitted to the Dean of the College of Engineering
and to the Graduate Council, and was accepted as partial fulfillment of
the requirements for the degree of Doctor of Philosophy.
December,
1973
Dean, Graduate School
TABLE OF CONTENTS (Continued)
Page
APPENDICES 121
APPENDIX A 122
APPENDIX B 125
REFERENCES 146
ADDITIONAL REFERENCES 148
BIOGRAPHICAL SKETCH 149
vi
Service Age
Figure 1.2. Sample function of the service age process S(t) for the minimal
repair problem with downtime, where x is the initial equipment
age, and X is the equipment replacement age.
87
also a member of the class of nth degreepolynomials whose coefficients
are all required to be positive. Therefore, q(x) is a convex function.
Let
i=0,1,. . ,ml,nril,. . ,n
e. =
i
(A.11)
e. + XE
x
i=m
Using (4.11) in (4.10), q(x) can be written as
~ n 1 .
q(x; = e^x + en_^x 4 . + e^x + e^
where
(4.12)
e_^ > 0 i=0,1,. . ,n
Then the particular solution (x) to (4.7) with q(x) given by (4.12)
is
y(x) = d x11 + d. x11 ^ + . 4 d., x 4 dn
z n n1 1 u
(4.13)
Applying results obtained in Chapter 3 for the particular solution in
order to evaluate the d_^(i=0,l n) in (4.13), it is determined
that the particular solution is
. y 1nlPnl(a) + menlx'1'1
y2(x) p (a) P (a)P (a)
n n1 n
+lx~ v>
tt Pk(a)
k=j
n1
[ l (j+1).
i=j4l
n
(i)e TT P, (a)
k=i4l k
4 (j4l) . (n)e ]
n
(4.14)
which is a solution to (4.7) with the nth degree polynomial q(x) given by
(4.12).
31
fQ(u) = C(E,u) + k(u) + i
r(u)
f1(p(E)u) 
hT(i)
Using the last expression to substitute for f^(x) in (1.5)
and rearranging yields
fj(x) [i + X(x)]f1(x) + X(x)hT(i)f1(p(E)x)
= c (x) k(x) A (x) [1 hT(i)]A(x) C(E,x)A(x)
(1.13)
Let
q(x) = c(x) + k(x)A(x) + [1 hT(i)]A(x) + C(E,x)A(x)
Substituting the prior expression in (1.13) results in the functional
differential equation
f(x) [i + A(x)]f1(x) + A(x)hT(i)f1(p(E)x) = q(x) (1.14)
for the economically controlled linear hysteresis problem. For the
case when the failure rate function is equal to a constant, the solution
to the functionaldifferential equation (1.14) is determined in
Chapter 4. The E,Xpolicy is investigated quantitatively in Chapter 4
using this solution.
1.6 Chapter Outline
In Chapter 2, results obtained by Barlow and Hunter, Fox,
Beckmann, and Scheaffer for the major repair problem are extended to
incorporate the concept of downtime for repair. Starting with the
functionaldifferential equation (1.10) with a(x) = 0, a sufficient
56
K(X) + f1(0) = f1(0)S(X,a) 
P0(a)
[S(X,a) 1]
Solving for f^(0) yields
f (o) = KQQ + J0
rS(X,a) 1 PQ(a)
(3.27)
Now K(X) is continuous in X, and further it is a bounded
function, whereas S(X,a) is a strictly increasing function in X to
infinity. Thus, f (0) is minimized with respect to X as X approaches
infinity. The following theorem summarizes the result.
Theorem 3.1
A sufficient condition for a unique optimal equipment replace
ment age X to exist for the linear hysteresis problem where a(x) = ax,
A(x) = A = constant, and g(x) = c^ is that X be infinite.
b. g(x) = c x + cQ
Using (3.3) with g(x) = c^x + c^ yields the functional
differential equation
f^(x) (i + A)f^(x) 4 AhT(i)f1(ax) = c^x cQ (3.28)
Applying (3.23), which is the general solution to (3.3), to (3.28),
the following is obtained
c x c + c_P^(a)
fl(x) = fl(0)S(x,a) + po^7a) [S(x,a) 1] (3.29)
Using (3.26) in (3.29), it follows that at x = X
c X c + CQP^(a)
K(x) + qco) ijtolsd,.) + p (a)p (a)
[S(X,a) 1J
28
Let
+ f1(a(u))
e lt:dHT(t)
0
k(u) + ^ + [f^aCu)) ]
e lt:dHT(t)
(1.7)
hT(s)
0
e StdHT(t) ,
Re s > 0
(1.8)
denote the LaplaceStieltjes transform of the repair time distribution
function H^()i which exists provided that H^(t) is sectionally continuous
in every finite interval 0 t < M and of exponential order 8 for t > M.
Upon using (1.8) in (1.7), the following is obtained
fQ(u) = k(u) + + [f1(a(u)) ]hT(i)
Using the last expression to substitute for f^(x) in (1.5)
and rearranging yields
f(x) [i + X(x)]f1(x) + A(x)h^(i)f^(a(x))
= c(x) k(x)A(x) [1 hT(i)]A(x) (1.9)
Let
g (x) = c (x) + k(x) A (x) + jX^ [1 hT(i)]A(x)
Substituting the prior expression in (1.9) results in the functional
differential equation
fj_(x) [i + A(x)]f1(x) + A(x)hT(i)f1(a(x)) = g(x)
(1.10)
When a(x) = x (minimal repair) or a(x) = 0 (major repair), the
Table B.9. Table of
0.0
o
o
0.20
0.3C
o
4<*
o
0,0
1.900
1.000
i .c:o
1 .GC0
i.c:c
0,2
1.022
1 .022
1. C 2 2
1.022
1.021
0,4
1 .c w
1 .048
l .C47
1.07
1.C46
3.4
1.092
1 .CSO
1 .072
1. C 76
1.0 7 4
0.3
l. L 23
1.119
l i 15
1 1 1C
1. 1C6
1.0
1.172
l .105
1 .153
l. 1 51
1.14 4
1.2
1.2 32
i 2 22
1.211
1.199
1.108
1.4
1.306
l 290
1.274
1.256
1.239
1 6
1 3 5
1.373
1.350
1.325
1.299
1 3
1.535
1.474
1.441
1.4C6
1.371
2.0
1.639
1.597
1.552
1. 504
1.455
2.2
1.802
1.747
1.607
1.622
1.555
2.4
2. CO 2
1.930
1.950
1. 764
1.675
2.6
2.246
2 .152
2 C 4 7
1.935
1.815
2.8
2.5'* 4
2.423
2.288
2. 142
1.989
3.0
2.909
2.754
2 580
2. 392
2.195
3.2
3.353
3.157
2.536
2.695
2.442
3.4
3.856
3.649
3.369
3.063
2.74 1
3.6
4.560
4.249
3.8 97
3.510
3.101
3.8
5. 370
4.982
4.540
4.053
3.536
4.0
6.3 6C
5.977
5.324
4.713
4.062
4.2
7.569
6.969
6.280
5.516
4 7 C 0
u % u
9.045
8.30 2
7. 46
6.493
5.474
4.6
10.848
9 .929
8.868
7.603
6.413
4.3
13.051
11.917
10.604
9. 133
7.553
5.0
15.741
14.344
12.721
1C.901
8.939
5.2
19.C27
17.307
15.306
13.055
10.625
5.4
23.041
20.926
10.461
15.682
12.676
5.6
27,943
25.345
22.313
18.ce?
15.173
5.8
33.Q30
30.743
27.015
22.796
18.213
6.0
41.243
37.335
32.756
27.565
21.917
6.2
50.175
45.385
39.757
33.386
26.431
6.4
61.0^4
55.2L7
48.326
40.409
31.932
6.6
74.409
67.225
59.779
49.159
38.640
6.0
90.635
31.991
71.543
59.742
46.820
7.
1 10.563
99.004
e7 1 30
72.662
56.797
7.2
1 3 4 J 4 3
121.602
106.166
08.435
68.969
7.4
164.498
140.402
129.413
VC 7.693
83.819
7.6
2C0.7 19
101.038
157.005
131.206
101.939
7. 9
244.960
220.399
192.479
159.913
124.053
8.0
298.994
269.583
234.028
194.978
151.044
8.2
364.993
329 .045
286.547
237.790
133.987
8.4
445.604
401.672
349.715
290.069
224.202
8.6
544.063
490.379
426.065
353.916
273.294
8.8
664.320
598. 725
521.094
431.036
333.229
9. 0
81 1.203
731 .058
636.101
527. 109
406.405
9.2
9 J0.60 8
892.609
776.744
643.401
495.750
9.4
1209.732
1090.105
948.425
785.428
604.944
9.6
1477.373
1331.220
1159.112
958.886
738.055
9,3
1304.268
1 625 ,7 35
1414.21?
1170.734
900.720
1C. D
2203,542
1935.448
1727,C2&
1429.471
1099.356
S(Y,B,A) for B = .90.
A
0.50 0.60 0.70
o.eo
O.90
1.C0
1 000
1 .000
1.000
1.000
1 .000
1.000
1.021
1.021
1.021
1.021
1.020
1.020
1,04 5
1,0^4
1.043
1.042
1 .042
1 C 4 1
1,072
1.070
1.067
J .0 66
l .04s
1. C62
1.102
1.098
1.0*24
1 .090
1.097
l C 8 3
1.137
1.129
1.123
1.116
l. 1 10
l. 105
1.176
1.165
1. 154
1.144
1.135
1. 1 27
1.221
1.205
1. 159
1,17 4
1.141
1.150
1.274
1.249
1.227
1.206
1,103
1.174
1.335
1 300
1. 269
L.240
1.217
1.197
1.406
1.359
1.315
1.277
1.246
1. 221
1.489
1.425
1.367
1.317
1.277
1. 246
1.586
1.501
1.425
1.361
1.310
1.271
1.701
1.590
1.490
1.407
1.344
1. 297
1.337
1.692
1.563
1.459
1.379
1.323
1.997
1.310
1,646
1.513
1.417
I, 350
2.1 P9
1.949
1.739
1,574
1 .456
1.377
2.4 16
2. ILL
1.846
1.540
1.497
1.405
2.689
2.301
1.968
1.713
1.341
1.4 33
3.015
2.525
2.107
1.793
1.596
1.462
3.406
2.790
2,268
1.8 31
1.635
1.4?2
3.076
3.104
2.453
1.979
1.686
1. 522
4,442
3.476
2.56 3
2.088
1 74 C
1.553
5.125
3.920
2.917
2.209
1.797
1.584
5.949
4,443
3.207
2.344
1.557
1.6J6
6.944
5.080
3.54 6
2.494
1.921
1.649
8.149
5.337
3.943
2.666
1.999
L.652
9.609
6.744
4.409
2.850
2.062
l. 716
11.377
7.333
4.958
3.075
2.139
1.751
13.523
9.144
5.606
3.321
2.221
1.7 56
16.129
10.722
6.372
3.601
2.309
1.022
19.292
12.625
7.280
3 .920
2.403
1. P.59
23.137
14.922
8.359
4.204
2.503
1 9?6
27.814
17.698
9.642
4.702
2.61C
1.935
33.503
21.055
11.171
5.182
2,726
l. 974
40.426
25.118
12.996
5.735
7.9 50
2.014
48.856
30.040
15.179
6.373
2.903
2. 0 54
59.122
36.007
17.792
7.112
3.127
2.096
71,629
43.245
20.924
7.970
3.233
2. 1 38
86.969
52.030
24.684
8.968
3.451
2.181
105.445
62.697
29.202
10.120
3.633
2.226
128.091
75.658
34.638
11.49Q
3.831
2.271
155.704
91.411
41.184
13.078
4.045
2.316
189.380
1 10.566
49.074
14.942
4.279
2.363
230.455
133.068
50.593
17.131
4.534
2.411
280.560
162.222
70.097
19.709
4.813
2.460
341.693
196,735
83.975
22,74?
5.117
2. 509
416.2 e 7
238.757
100.769
26.336
5.451
2.560
507,314
289.933
121.090
30.500
5.810
2.612
618.405
352.276
145.592
35.608
6.22 2
2.664
753.993
<<23.233
175.497
41,573
6.667
2.718
109
Or
P{N(X) = 3} = e
AX
X s ,, 3s
Ax(1a )
Xax
Ax (1a )
Ae
Ae l
> 0
"1 2
Ax(la) , ,
, 3 dx dxdx
Ae 1/3
(5.13)
w X , 3, Xax 2s
AX r Ax1(1a ) 1 Ax9(la )
Ae L
(1a)
/ A(Xa x ax)(la)
Ae ^ [1 e 1 ]
0
0
dx^dx^
AX
(1a)
X 3. Xaxi 2.
r Ax(1a ) r x Ax9(la )
Ae
Ae
dx1dx2
, X 2X Xax s
AX(la) Ax_(1a ) r 1 Ax9(la)
~~AX L .4.
Ae Ae
(1a)
[e
dx dx2]
0
0
(5.14)
Substituting (5.10) in (5.14) yields
P{N(X) = 3}
AX
e
(1a)
X
Ae
Ax^(la )
2 ^Xax^
Ax2(la )
A e dx^dx2
AX(la)
P{N(X) = 2}
AX
(1a)
X
(1a)(1a ;
Ax.. (1a3) A (Xax..) (la^)
Ae [1 e ]dxn
aAX(la)
(1 a)
P{N(X) = 2}
^AX rX Ax(1a3)
Ae dx.,
(1a)(1a )
0
127
of B's nine values, the parameters Y and A are varied from 0 to 10
in increments of 0.2 and from 0 to 1 in increments of 0.1, respectively.
The function Log^ S(Y,B,A) is plotted against the parameter Y in
Figures B.l through B.9. In each figure for a particular B, eleven
curves are plotted corresponding to values the parameter A can assume.
Whenever space permits, the curves are labeled in the right margin
with the appropriate value of A. The unlabeled curves follow the rule
that a lower curve has a higher value of the parameter A than a higher
curve.
As can be observed from the tables and figures, S(Y,B,A)
increases as Y increases. However, S(Y,B,A) decreases when either A
or B increases. In terms of the original function S(x,0,Xh (i),a),
the value of S(x,0,Ah^(i),a) increases when either 0 or x increases,
and decreases when either Xh^(i) or a increases.
12
major repair (or unplanned replacement). The latter is a consequence
of it never being economically advantageous to replace prior to failure.
Specific results for the unique equipment replacement age for
Barlow and Hunters [1] major repair problem were obtained by Glasser
[13] for the truncated normal, Gamma, and Weibull distributions. For
each distribution, a graph was presented from which the unique equipment
replacement age is readily ascertained after the computation of a cost
ratio.
Fox [12] investigated the effects of incorporating a discount
factor i in Barlow and Hunter's major repair problem. Continuous
discounting was assumed, and the total loss was equal to the sum of
the discounted losses incurred at each planned replacement and failure.
He determined that provided the failure rate (for known distributions)
was continuous and strictly increasing to infinity, a unique equipment
replacement age X exists which minimizes the long run total expected
discounted loss (or cost). Denardo and Fox [7] proved that even if the
cost of a replacement at failure exceeded the cost of a planned
replacement, it is not optimal to replace an operating equipment
during periods of decreasing failure rate.
Beckmann [4] approached the same problem studied by Fox [12]
by formulating a functional equation in a manner analogous to that of
Bellman [5] and Descamps [10]. His objective function minimized the
long run total expected discounted cost. A necessary condition for a
unique optimal equipment replacement age X to exist was determined, but
was not presented in a closed form expression.
Employing renewal theory, Scheaffer [20] extended Barlow and
27
Before solving equation (1.5), it is necessary to obtain an
expression for f^(x) in terms of f^(a(x)) using a derivation due to
Sivazlian [21]. Assume that an equipment is in an operating state
and has a service age u such that 0 <_ u < X. If the equipment fails
between u and u + du, repair commences instantaneously. The length
of downtime, or repair time, T is a random variable whose distribution
function H () is not dependent on the failure characteristics of
the equipment. It is assumed that the succession of downtimes and
service times forms a sequence of mutually independent random
variables (Takcs [23]). At the conclusion of the repair, the service
age of the equipment is a(u) (0 _<_ a(u) <_ u), where a(u) is the hysteresis
age function. From the characteristics of a(u), it follows that the
service age is reduced due to the completion of a repair whenever
a(u) < u.
For an equipment aged u which has just failed, the conditional
long run total expected discounted cost is
k(u) +
r(u)e ^d0
+ f1(a(u))e
it
(1.6)
given that the repair time has length t and terminates at time t at
which time the equipment is returned to an operating state with a
service age a(u). Then the unconditional long run total expected
discounted cost for an equipment aged u which has just failed is
fo(u) =
[k(u) +
r(u)e i6d0 + f1(a(u))e it;]dHT(t)
0
= k (u)
0
dHT(t) +
(1 e lt)dHfj,(t)
0
105
Figure 5.1. Sample function of the service age process
S(t) when N(X) = 1 for the linear hysteresis
problem with downtime and constant failure
rate function.
Service Age
Figure 5.2. Sample function of the service age process
S(t) when N(X) = 2 for the linear hysteresis
problem with downtime and constant failure
rate function.
85
recalled. It is
f(x) [1 + l(x)]f1(x) + X(x)hT(i)f1(p(E)x) = q(x) (4.1)
where
q(x) = c(x) + k(x)X(x) + [1 hT(i)]A(x) + C(E,x)A(x) (4.2)
Letting p(E) = a (a constant to be determined later, where 0 <_ a <_ 1)
in (4.1) and A(x) = A = constant in (4.1) and (4.2) yields
f^(x) (i + Ajf^x) + AhT(i)f 1(ax) = q(x) (4.3)
q(x) = c(x) + Ak(x) + [1 hT(i)] + AC(E,x) (4.4)
The procedure to obtain the general solution to (4.3) consists
of determining a complementary solution, a particular solution, and the
constant term of these solutions by evaluating the general solution at
x equals zero. It can be verified (Appendix A) that the general
solution to (4.3) is of the form
ff(x) = y2(x) + y2(x) (4.5)
where y^(x) is a solution to the homogeneous equation
y(x) (i + A)y1(x) + AhT(i)y1(ax) = 0 (4.6)
and y2(x) is a solution to the nonhomogeneous equation
y'2 (x) (i + A)y2(x) + AhT(i)y2(ax) = q(x) (4.7)
A complementary solution to the homogeneous equation (4.6) has been
previously obtained in Chapter 3 (see (3.6) and (3.8) through (3.16)).
It is
102
5.2 Probability Mass Function of N(X)
Consider a stochastic process in which the equipment ages
continuously while operating and is subject to failure. Upon failure,
it is repaired and returned to an operating state with a service age
equal to a (0 < a < 1) times the service age at the instant of failure.
When an equipment attains a service age X, it is instantly replaced
with an identical new (service age zero) equipment. The stochastic
process, just described, is the service age process for the linear
hysteresis problem.
Define:
P{N(X) = r} = probability that N(X), the number of operating
periods in [0,X], is r, r=0,l,...
W. = equipment service age at the commencement of the
ith repair, i=l,2,...
Let {x}, i=l,2,..., be a sequence of independently and identically
distributed random variables denoting the successive operating times to
failure of the equipment whose distribution function is the negative
exponential with parameter X.
Since is the equipment service age at the commencement of
the ith repair, then for i=l
\ = Xj_ (5.D
and for i=2
W2 = aXl + X2
= aW]_ + x2
and in general
Wi+1 = aWi + xi+l
(5.2)
CHAPTER 6
CONCLUSIONS AND RECOMMENDATIONS
6.1 Conclusions
This dissertation introduced to replacement theory the concept
of partial repair of a failed equipment and the subsequent partial
reduction (or hysteresis) of service age. By varying the amount of
partial reduction of service age, the problems of minimal repair and
major repair could be recovered. It was assumed that the equipment
ages continuously and is subject to failure, where failures are
detected Instantly. Failure resulted in a partial repair being
performed for a length of time, called the downtime for repair.
Upon attaining a predetermined service age X, the equipment was
replaced instantaneously with an identical new one. The repair and
replacement problem generated under these assumptions was the hysteresis
problem. A cost structure incorporating variable operating costs,
repair costs, replacement costs, and loss of revenue per unit of down
time was assumed.
Using an optimization approach known as the functional equation
approach, an expression was obtained for the minimum total expected
discounted cost over an infinite time horizon. For the special case
where the hysteresis age function recovered all the service life (i.e.,
major repair), a sufficient condition for a unique finite optimal replace
ment age X was determined. This condition was an extension to
prior results obtained for the major repair problem. The significant
contributions to this problem were the incorporation of downtime
for repair and variable repair costs.
117
APPENDICES
Table B.6. Table of S(Y,B,A) for B = .60.
A
Y
0.0
o
o
0.20
C.30
o
i*
o
0.50
0.60
0.70
0.60
C 90
1.00
0.0
1.00C
l.OOC
1.CC0
1.000
1.000
1.0C0
l.cocr
1.000
1.000
1.CG0
1.CC0
0.2
1.039
1 .083
1.053
1.C87
1.036
1.066
1.085
1.085
1.034
1.084
1.033
0.4
1.197
1 194
1.192
1. 190
1.189
1.185
1.193
1.190
1.173
1.176
1. 174
0.6
1.329
1.323
1.318
1.312
1.3C6
1.300
1.294
1.280
1.233
1.277
1.271
0.8
1.4 90
1.400
1.469
1.458
1.446
1.434
1.423
1.11
1.399
1.383
1.377
1.0
1.627
1.670
1.651
1.631
1.611
1.591
1.570
1.550
1.530
1.510
1.492
1.2
1.929
1.900
1.871
1.840
1.308
1.775
1.74L
1.709
1.676
1.64 5
1. 616
1.4
2.222
2.181
2.137
2.090
2.041
1.991
1.940
1.089
1 .840
1.794
1.751
1.6
2.531
2.522
2.459
2.391
2.319
2.246
2.17L
2.097
2.025
1.959
1. 896
1.3
3.020
2.938
2.649
2.753
2.652
2.548
2.4 4 L
2.336
2.234
2. 140
2. 054
2.0
3.556
3.445
3.323
3. 191
3.052
2.906
2.758
2.511
2.4 70
2.340
2. 226
2.2
4.210
4.C62
3.899
3.721
3.531
3.332
3.130
2.929
2.738
2.563
2.411
2.4
5.009
4.ei5
4.599
4.362
4.1C3
3.841
3.568
3.298
3.041
2.810
2.612
2.6
5.935
5.733
5.451
5.140
4.804
4.449
4.086
3.726
3.337
3.084
2.929
2.3
7.178
6.853
6.488
6.083
5.644
5.178
4.699
*.225
3.781
3.309
3. 065
3.0
8.634
8.220
7.751
7.229
6.659
6.052
5.426
4.808
4.231
3.728
3. 320
3.2
10.4 13
9.888
9.290
8.621
7.887
7. 103
6.292
5.491
4.747
4.106
3. 597
3.4
12.536
11.924
11.167
10.315
9.376
8.368
7.323
6.292
5.338
4.527
3.896
3.6
15.239
14.41Q
13.455
12.375
11.181
9.893
8.555
7.234
6.018
4.998
4.221
3.3
18.480
17.443
16.245
14.884
13.371
11.734
10.028
8.344
6.802
5.524
4.572
4.0
22.439
21 148
19.648
17.940
16.032
13.959
11.794
9.655
7.706
6.112
4.953
4.2
27.274
25,670
23.001
21.663
19.265
16.651
13.912
11.206
8.753
6.772
5.366
4.4
33.130
31.193
20.869
26.2C0
23.197
19.910
16.457
13.045
9.965
7.512
5.812
4.6
40.393
37.936
35.054
31.733
27.981
23.060
19.520
15.229
11.373
8.344
6. 297
4.8
49.234
46.171
42.604
30.479
33.804
28.651
23.209
17.926
13.011
9.278
6.821
5.0
59.965
56.227
51.820
46.708
40.894
34.466
27.658
20.921
14.919
10.330
7. 3S9
5.2
73.109
69.509
63.072
56.747
49.531
41.528
33.030
24.613
17.146
11.516
8.004
5.4
89.162
83.507
76.008
68.995
60.056
50.109
39.520
29.24
19.751
12.855
8. 671
5.6
103.77C
101.325
93.581
83.942
72.884
60.542
47.369
34.302
22.001
14.366
9.393
5.8
132.719
124.196
114.061
102.185
80.523
73.232
56.870
40.625
26.381
16.076
10.176
6.0
161.971
151.515
139.069
124.451
107.593
88.674
68.379
48.2C9
30.536
18.013
11.023
6.2
197.699
104.889
169.608
151.631
130.852
107.471
82.328
57.315
35.535
20.208
11.941
6.4
24 1.337
225.645
206.901
184.813
159.224
130.360
99.246
68.261
41.368
22.700
12.936
6.6
294.636
275.422
252.444
225.323
193.839
158.241
119.777
81.431
40.252
25.531
14.013
6.8
359.736
336.219
308.062
274.781
236.076
192.210
144.705
97.292
56.380
28.752
15. 180
7.0
439.250
410.474
375.988
335. 1 72
297.617
233.608
174.987
116.411
66.017
32.422
16.445
7.2
536.368
501.167
458.945
400.912
350.522
284.070
211.739
139.475
77.428
36.606
17.814
7.4
654.989
611.939
560.262
490.956
427.303
345.594
256.531
157.321
90 .966
41.303
19.298
7.6
799.872
747.232
684.001
608.913
521.025
420.617
310.951
200.965
107.049
46.842
20.905
7.0
976.834
912.479
035.129
743.190
635,437
512.118
377.162
241.640
126.176
53.009
22.646
8.0
1192.977
1114.311
1C19.709
907. 171
775.114
623.731
457.744
290.847
148.949
60.246
24.53?
8.2
1456.971
1360.323
1245.142
1107.426
945.642
759.890
655.844
350.414
176.091
68.453
26.576
8.4
1779.416
1661.914
1520.479
1351.991
1153.050
926.0L9
675.305
422.561
208.475
77.876
28.789
0.6
2173.252
2029.665
1856.768
1650.671
1408.071
1120.733
820.OIL
509.991
247.152
88.707
31.187
0.3
2654.263
2478.034
2267.499
2015.447
1718.488
1376.112
998.080
615.994
293.386
101.167
33.784
9.0
3241.014
3027.446
2769.156
2460.948
2097.533
1678.023
1214.091
744.572
340.708
115.525
36.598
9.2
3959.430
3697.522
3381.873
3005.049
2560.396
2046.518
1477.359
900.603
414.960
132,079
39.646
9.4
4035.090
4515.930
4130.207
3669.575
3125.624
2496.310
1798.280
1090.023
494.369
151.189
42.945
9.6
5906.445
5515.551
5044.230
4481.164
3815.874
3045.370
2109.540
1320.066
599.625
173.273
46.526
o.o
7214.020
6736.500
6160.625
5472.414
4650.789
3715.643
2666.627
1599.542
703.975
198.820
50.401
10.0
801 1. 1 17
8227.773
7524.180
6683.102
5688.215
4533.906
3248.446
1939.186
841.353
228.405
54.598
138
42
Transposing and collecting terms
X+dx
[g(u) KA(u) Ki] v(u)
X
Or
g(X) KX(x) Ki
v(X)
R(u)du > 0
X+dx
v(u)
X
g(u) KA(u) Ki g(X) KA(X) Ki
v(u) v(X)
R(u)du > 0 (2.20)
Since ^\U is strictly increasing to infinity
v(u)
by assumption, then for all points X u <_ X + dx (dx > 0) ,
g(u) KA(u) Ki g(X) KA(X) Ki
v(u)
v(X)
and inequality (2.20) is satisfied. In a similar fashion it can be
established that
X
Xdx
v(u)
a (X) KA (X) Ki
v(X)
g(u)
 KA(u) Ki
v (u)
R(u)du > 0
The following corollaries to Theorem 2.1 are submitted without
proof and are direct consequences of the theorem.
Corollary 2,1
If ^ (u > 0) is strictly increasing to infinity
v (u)
and A(u) is strictly nondecreasing, then a unique finite optimal equip
ment replacement age X exists which is the unique solution to equation
(2.12), and if the initial age, x, satisfies x X, then an optimal
Xpolicy exists for the major repair problem with arbitrary repair
time distribution.
CHAPTER 2
AGE REPLACEMENT WITH MAJOR REPAIR AND DOWNTIME
2.1 Introduction
Previously obtained results for the major repair problem are
extended in this chapter. These results, in the form of conditions
for the existence of an optimal age replacement policy, called an
Xpolicy, were obtained by Barlow and Hunter [1], Fox [12], Beckmann
[4], and Scheaffer [20]. They considered the concepts of fixed repair
and replacement costs ([1], [12], [4], [20]), discounting ([12], [4]),
and variable operating cost ([20]).
Additional concepts are incorporated in the investigation of
the major repair problem in this chapter. They are variable repair
cost, downtime for major repair following a failure, and loss of
revenue per unit of downtime. Using the criterion of long run total
expected discounted cost, the major repair problem with downtime is
investigated utilizing an approach due to Beckmann [4]. A sufficient
condition for a unique finite optimal equipment replacement age X to
exist is determined for the major repair problem with downtime. Four
corollaries to this condition are presented. In addition, an expression
for the undiscounted long run expected cost per unit time is determined
from f^(0), the minimum total expected discounted cost over an infinite
time horizon when the equipment is as good as new (service age zero)
and is in operation.
2.2 A Sufficient Condition for a Unique Replacement Age X
The functionaldifferential equation (1.10) of Chapter 1 is
33
Table B.4. Table of S(Y,B,A) for B = .40.
A
Y
o
o
o
o
o
ro
O
0.30
0.40
0.50
0.60
0 TO
0.80
0.90
1 a 0 0
0.0
1.000
1 .COO
1COO
1.000
1.C00
1.000
1.000
1.000
1.000
1.000
1.C00
0.2
1.133
1.132
1.132
1.131
1.131
1.130
1.130
1.129
1.129
l.i2e
1.127
0.4
1.295
1.293
1 291
1.289
1.236
1.283
1.28L
1.279
1.276
1.274
1.271
C. 6
1.493
1.488
1.482
1.476
1.471
1.464
1.458
1.452
l A A 6
1.440
1.433
0.8
1.735
1.725
1.714
1.703
1.691
1.679
1.666
1.654
1.641
1.629
1.616
1.0
2.031
2.013
1.994
1.975
1.954
1.933
1.911
1.809
1.867
1.644
1. 822
1.2
2.392
2.364
2.335
2.303
2.270
2.236
2.200
2.164
2.127
2.090
2.054
1.4
2.833
2.792
2.747
2.7C0
2.650
2.597
2.542
2.486
2.429
2.372
2.316
1.6
3.372
3.313
3.249
3.180
3.106
3.020
2.947
2.363
2.778
2.694
2.612
1 5
4 C 3 0
3.948
3.858
3.761
3.656
3.545
3.428
3.307
3.184
3.062
2. 945
2.0
4.633
4.722
4.600
4.465
4.320
4.165
4.coo
3.830
3.656
3.^85
3.320
2.2
5.815
5.667
5.502
5.321
5.123
4.909
4.683
4.447
4.206
3.969
3. 743
2.a
7.014
6.819
6.601
6.360
6.094
5.806
5.499
5. 176
4.048
4.525
4. 221
2.6
8.4 78
8.226
7.941
7.622
7.270
6.886
6.473
6.040
5.598
5.165
4.759
2.8
10.267
9.942
9.573
9.158
8.696
8.109
7.642
7.065
6.476
5.900
5. 366
3. C
12.451
12.037
11.563
11.027
10.427
9.763
9.044
8.282
7.505
6.746
6.050
3.2
15.119
14.594
13.990
13.303
12.529
11.667
10.729
9.732
8.712
7.722
6. 821
3.4
18.378
17.716
16.951
16.074
15.001
13.972
12.756
11.459
10.131
8.847
7.691
3.6
22.359
21.528
20.563
19.452
10.186
16.764
15.197
13.520
11.801
10.145
8.671
3.8
27.221
26.182
24.970
23.568
21.963
20.149
18.141
15.9S3
13.770
1 1.645
9. 777
4.0
33.159
31.865
30.349
28.587
26.559
24.256
21.694
18.931
16.C93
13.380
11.023
4.2
40.412
30.605
36.914
34.700
32.156
29.243
25.986
22.462
18.838
15.300
12.429
4.4
49.270
47.279
44.928
42.173
38.972
35.301
31.176
26.696
22.085
17.7L4
14.013
4.6
60.C90
57.629
54.712
51.201
47.277
42.664
37.457
31.780
25.931
20.410
15.800
4.8
73.306
70.268
66.657
62.394
57.399
51.617
45.063
37.891
30.491
23.539
17.814
5.0
89.447
85.703
81.241
75.955
69.737
62.510
54.280
45.241
35.905
27.173
20.085
5.2
109.163
104.555
99.C49
92.507
84.702
75.766
65.456
54.091
42.339
31.398
22.646
5.4
133.243
127.578
120.794
112.709
103.130
91.905
79.015
64.756
49.993
36.312
25.534
5.6
162.655
155.697
147.347
137.369
125.510
111.559
95.474
77.619
59.109
42.033
28.789
5.8
190.579
190.040
179.773
167.475
152.813
135.502
115.465
93.144
69.976
48.700
32.460
6.0
242.456
231.984
219.372
204.230
186.125
164.676
139.754
111.896
82.941
56.475
36.598
6.2
296.047
283.212
267.730
249. 105
226.775
200.233
169.270
134.560
98.427
65.549
41.264
6.4
361.604
345.782
326.790
303.897
276.383
243.578
205.181
161.970
116.938
76.148
46.525
6,6
441.454
422.202
398.919
370.002
336.932
296.426
248.855
195.141
139.084
80.539
52.457
6.8
539.104
515.541
487.010
452.500
410.840
360.874
301.996
235.304
165.601
103.035
59.145
7.0
658.375
629.543
594.595
552.266
501.059
439.479
366.682
283.959
197.378
120.007
66.666
7.2
804.053
763.782
725.993
674.097
611.197
535.365
445.440
342.932
235.487
139.894
75.189
7.4
981.985
938.848
886.475
022.877
745.661
652.344
541.355
414.445
281.221
163.212
84.775
7.6
1199.309
1146.565
1C82.478
1004.571
909.829
795.073
650.190
501.201
336.149
190.576
95.583
7.3
1464.752
1400.269
1321.Q69
1226.469
1110.276
969.239
800.541
606.494
402.164
222.710
107.770
8.0
1788.967
1710.143
1614.255
1497.465
1355.029
1181.764
974.015
734.332
401.554
260.473
121.510
8.2
2184.957
2038.617
1971.360
1828.429
1653.083
1441.185
1185.449
889.599
577.089
304.004
137.003
8.4
2660.625
2550.887
2407.522
2232.639
2018.818
1757.799
1443.199
1078.247
692.123
357.150
154.470
8.6.
3259.379
3115.503
2940.243
2726.307
2464.458
2144.270
1757.457
1307.524
830.713
418.704
174.164
8.8
3900.925
3805.123
3590.896
3329.239
3008.657
2616.040
2140.664
1586.262
997.791
491.246
196.370
o.o
4062.184
4647.39 5
4385.563
4065.621
3673.230
3191.962
2608.008
1925.230
1199.303
576.794
221.406
9.2
5938.613
5676.172
5356.199
4964.973
4484.801
3095.075
3170.034
2337.553
1442.400
677.750
249.635
9.4
7253.352
6932.730
6541.715
6063.426
5475.938
4753.465
3873.375
2839.228
1736.079
 796.966
281.462
9.6
8859.180
8467.504
7989.715
7405.047
6606.395
5801.555
4721.641
3449.759
2090.729
937.838
317.347
9.8
1082C.551
10342.063
9758.273
9043.652
8164.699
7081.242
5756.613
4192.910
2519.315
1104.405
357.809
LC.O
13216.180
12631.656
1 1918.402
11045.012
9970,133
8643.805
7019.484
5097.723
3037.483
1301.479
4C3. 423
HT
To my loving wife, Marci,
and to my parents
LIST OF FIGURES (Continued)
Figure Page
5.4 Sample function of the service age process S(t)
when N(X) = 4 for the linear hysteresis problem
with downtime and constant failure rate function.. 106
B.l Graph of Log^g S(Y,B,A) for B = .10 129
B.2 Graph of Log Q S(Y,B,A) for B = .20.... 131
B.3 Graph of Log^g S(Y,B,A) for B = .30 133
B.4 Graph of Log^g S(Y,B,A) for B = .40 135
B.5 Graph of Log^g S(Y,B,A) for B = .50 137
B.6 Graph of Log^g S(Y,B,A) for B = .60 139
B.7 Graph of Log.^ S(Y,B,A) for B = .70 141
B.8 Graph of Log^g S(Y,B,A) for B = .80 143
B.9 Graph of Log^ S(Y,B,A) for B = .90 145
x
Table 3.9. Table of Optimum X and Corresponding f^(0) for
g(x) = 1.75x^ + 7.5x When a = 0.75
K
50 150 250 350
8
AhT(i)
X
f1(o)
X
f1(0)
X
f i. (o)
X
fx(o)
0.20
0.10
2.80
291.9
4.41
508.4
5.45
653.2
6.25
767.4
0.40
0.10
3.04
78.1
5.03
117.3
6.38
136.6
7.49
148.5
0.20
2.91
128.3
4.71
204.3
5.90
246.9
6.87
275.9
0.30
2.78
280.7
4.40
473.9
5.44
595.4
6.26
686.2
0.60
0.10
3.32
37.1
5.75
47.7
7.50
50.7
8.96
51.9
0.20
3.17
50.9
5.41
69.0
7.00
75.3
8.32
78.2
0.30
3.04
74.5
5.06
107.0
6.48
120.7
7.67
127.8
0.40
2.90
122.7
4.73
187.5
5.97
219.5
7.00
238.9
0.50
2.77
269.1
4.40
438.0
5.48
534.9
6.34
601.3
0.80
0.10
3.63
20.8
6.55
23.6
8.69
23.9
10.47
24.0
0.20
3.48
26.6
6.21
31.2
8.21
32.0
9.89
32.2
0.30
3.33
35.1
5.85
43.0
7.71
44.7
9.28
45.1
0.40
3.18
48.3
5.50
62.2
7.19
65.9
8.64
67.1
0.50
3.04
70.8
5.14
96.8
6.66
105.3
7.96
108.8
0.60
2.90
116.9
4.79
170.3
6.12
192.2
7.25
203.0
0.70
2.77
257.2
4.44
400.8
5.58
472.4
6.54
514.6
114
E[N(X)] = I r P{N(X) = r}
r=0
Using (5.22) in the last expression, the following is obtained
E[N (X) ] = l rP
r=0
CO
= l
r=l
rP
r
= Pi +
y rP
Lx t
r=2
Substituting (5.25) in (5.27) yields
co r1
E[N(X) ] = a, + 7 r[a Y g P ]
1 Lr, r rm m
r=2 m=l
= a
oo oo XT1
+ l ra l l rg P
1 r L0 S rm m
r=2 r=2 m=l
oo ri
= I ra J Y (r m + m)g P
^ r* Zj l. N r m
r=l r=2 m=l
rm m
co co r1
= y ra y y (r m)g P
r Ln L. rm :
r=l r=2 m=l
m
co ri
 y y mg p
l l rm m
r=2 m=l
oo rl
Expanding Y Y (r m)g P it is observed that
Lr,L^ rm m
r=2 m1
CO
l
r=2
rl
l
m=l
(r m)g
P
rm m
(5.26)
(5.27)
(5.28)
41
Simplifying
X+dx ^ 
g(u)R(u)du
X
X
v(u)R(u)du 
1 0
X
g(u)R(u)du
X+d.x
v(u)R(u)du
X
X
X+dx
> [KR(X) KR(X + dx)]
v(u)R(u)du KR(X)
X+dx
X
Rearranging (2.18) and using
yields
X+dx
X
v(u)R(u)du
(2.18)
d[R(u)] = R(X + dx) R(X)
X
g(u)R(u)du
> K
v(u)R(u)du
X+dx
d[R(u)]
X
X
X
v(u)R(u)du
X+dx
+ [KR(X) +
g(u)R(u)du]
0
X
Dividing both sides of the last expression by
in the following inequality
X+dx X+dx
g(u)R(u)du > K  d[R(u)]
X
v(u)R(u)du
v(u)R(u)du results
X
KR (X) +
g(u)R(u)du
X
v(u)R(u)du
X+dx
v(u)R(u)du
LX
(2.19)
Using (2.5) and (2,14) in (2.19) yields
X+dx
X+dx
g(u)R(u)du K
X
X
[i + A(u)]R(u)du
r X+dx
, g(X) KA(X) Ki
v(X)
v(u)R(u)du
X
LIST OF TABLES (Continued)
Table Page
4.5 Table of Optimum Values for E,Xpolicy for a
Concave Curve p(E) as in Figure 4.5.......... 100
B.l Table of S(Y,B,A) for B = .10 128
B.2 Table of S(Y,B,A) for B .20 130
B.3 Table of S(Y,B,A) for B = .30... 132
B.4 Table of S(Y,B,A) for B = .40 134
B.5 Table of S(Y,B,A) for B .50 136
B.6 Table of S(YsB,A) for B = .60 138
B.7 Table of S(Y,B,A) for B = .70 140
B.8 Table of S(Y,B,A) for B = .80 142
B.9 Table of S(Y,B,A) for B = .90 144
viii
98
c X + AEX
K P (a) c + AE + c P (a)
fl(0) = S (X, a) 1 + PQ(a)P;L(a)
(4.31)
As in the case of the polynomial (4.21), for the polynomial
(4.27), necessary or sufficient conditions for an optimal E,Xpolicy
to exist cannot be obtained. Quantitative results are determined in
a similar fashion to that described in Section 4.3b.
The results obtained for the polynomial (4.27) are similar
to that obtained for the polynomial (4.21). One notable difference
does occur in the results obtained for these polynomials. For a
particular curve p(E) and a set of critical variables, the amount of
partial repair corresponding to the polynomial (4.27) is less than or
equal to the amount of partial repair corresponding to the polynomial
(4.21). In addition, there is a decrease in the replacement age X,
and an increase in f^(0).
Table 4.4 illustrates the sensitivity of the parameters for
the E,Xpolicy for a convex curve p(E), as in Figure 4.6. For a concave
curve p(E) as in Figure 4.5, Table 4.2 illustrates the change from
major repair to minimal repair as the variables are varied one at a
time. Comparing the results of Tables 4.2 and 4.5, it is observed
that the switch from major to minimal repair occurs at a lower scaling
factor for the polynomial (4.27) than for the polynomial (4.21).
Polynomials (4.21) and (4.27) have the same operating cost
g(x) = cjX + cQ and differ in the expenditure repair function C(E,u)
(i.e., C(E,u) = E and C(E,u) = Eu), which in general is
C(E,u) = Eum m = 0,1,2,... (4.32)
Thus, the amount of partial repair is a nonincreasing function of the
integer m.
69
ment age X to exist for the linear hysteresis problem where a(x) = ax,
 2
/. (x) = A = constant, K(X) = K = constant, and g(x) = c^x + c^x + c^
are that c. > 0 for i=l,2 and
i
(2gX + y)[S(X,a) 1] + (K gX2 YX)Sx(X,a) = 0 (3.43)
where X is the unique solution to (3.43).
Proof:
If the initial equipment age x is such that x < X, then it
is clear from substituting (3.42) into (3.38) that if an optimal
replacement age X exists, it minimizes both f^(0) and f^(x) for all x.
Differentiating the expression for f^(0) in (3.42) with respect to X
and equating the result to zero, a sufficient condition for the
existence of an optimal equipment replacement age X, say X is
obtained and is given by the equation in X,s
(2gX* + y)[S(X*,a) 1] + (K gX*2 yX*)Sv(X*,a) = 0 (3.44)
Note that (3.44) is identical to (3.43). Dividing (3.44) by
yS (X*,a) and rearranging yields
X
Â£ x*2 + X* ^ X* + 1
Y Y
_
Letting 6 = g/y in the last expression, the following is obtained
<5X*2 + X (26X* + 1)
Taking the second derivative of f^(0) with respect to X
S(X*,a) 1
Sx(X*,a)
K
Y
(3.45)
S(X*,a) 1
Sx(X*,a)
K
Y
yields
6
new equipment. This type of repair and replacement problem shall be
referred to as Problem II or the major repair problem.
An extension to Problems I and II is a repair and replacement
problem which lies between these problems. Repairing a failed
equipment beyond a minimal repair but not up to a major repair is
often feasible. Minimal repair often entails the partial disassembly
of the equipment. Instead of simply repairing or replacing the failed
part, it may be economically feasible to repair or replace parts which
have not failed, since the initial labor cost for disassembling the
equipment has already been paid. After such a repair, the service
age of the equipment is less than its service age prior to commencing
the repair. The partial reduction of service age can be a function
of repair costs, length of downtime, and service age at the instant
of failure. In addition, replacements are made with identical new
equipment. This type of repair and replacement problem shall be
referred to as Problem III or the hysteresis problem. It can even be
rationalized that a minimal repair results in some recovery of service
age.
In order to determine when to repair or replace an equipment,
it is necessary to utilize an inspection policy. This policy is used
to determine the state, either operating or failed, and the service
age of the equipment. Inspection policies can be continuous or discrete.
Continuous inspection policies involve constant monitoring of the
equipment. Discrete inspection policies are customarily periodic
although some random policies have been examined in the literature.
Equipment aging can be either discrete or continuous. Discrete
APPENDIX A
Consider the equations
f(x) A(x)f1(x) + B(x)f1(a(x)) = 0 (A.l)
f^(x) A(x)f^(x) + B(x)f^(a(x)) = C(x) (A.2)
The properties of f^(x) are continuity, monotone increasing in x, and
existence of the first derivative. The functions A(x) and B(x) are
continuous in x. C(x) is a known function of x and a(x) is a bounded
and continuous function of x. The following theorem is now proved.
Theorem A.l
If f^(x) = yx(x) is any solution (called the complementary
solution) to the homogeneous equation (A.l) and f^(x) = y2(x) is any
solution (called the particular solution) to the nonhomogeneous
equation (A. 2), then f^(x) = y^(x) + y^ (x) is a solution to (A. 2).
Proof:
Since y^(x) is a solution to (A.l), the following holds
yj_(x) A(x)Yl(x) + B(x)y;L(a(x)) = 0 (A.3)
Further since y2(x) is a solution to (A.2)
y^(x) A(x)y2(x) + B(x)y2(a(x)) = C(x) (A.4)
Adding (A.3) and (A.4) and recalling that the derivative of a sum is
the sum of the derivatives, the following is obtained
122
Table 3.7. Table of Optimum X and Corresponding f^(0) for
g(x) = 1.75x^ + 7.5x When a = 0.25
K
0
XhT(i)
0.20
0.10
0.40
0.10
0.20
0.30
0.60
0.10
0.20
0.30
0.40
0.50
0.80
0.10
0.20
0.30
0.40
0.50
0.60
0.70
50
X
O j
rH
Ml ,
2.86
276.4
3.12
73.6
3.06
114.1
3.00
235.7
3.41
34.9
3.35
44.9
3.30
61.8
3.24
95.7
3.19
197.7
3.73
19.6
3.68
23.5
3.63
29.0
3.58
37.4
3.52
51.3
3.47
79.4
3.42
163.8
150
X
fl(0)
4.58
464.1
5.24
106.1
5.12
167.1
5.00
351.2
6.00
43.2
5.89
56.3
5.78
78.5
5.67
123.1
5.56
257.8
6.80
21.7
6.71
26.2
6.62
32.5
6.53
42.2
6.43
58.4
6.33
91.0
6.23
189.5
250
X
f1(0)
5.71
581.9
6.72
120.4
6.55
191.2
6.39
405.7
7.86
45.3
7.73
59.3
7.59
82.9
7.45
130.6
7.30
275.1
9.04
21.9
8.93
26.5
8.82
32.9
8.71
42.8
8.59
59.3
8.47
92.6
8.34
193.0
350
X
fl(0)
6.60
670.3
7.93
128.5
7.74
205.1
7.53
437.9
9.42
46.0
9.27
60.3
9.11
84.5
8.95
133.5
8.77
281.7
10.90
21.9
10.78
26.5
10.65
33.0
10.53
42.9
10.39
59.5
10.26
92.9
10.11
193.8
O''
37
f1(0)
KR(X) +
X
g(u)R(u)du
0
X
v(u)R(u)du
.
0
(2.11)
A central theorem for age replacement with major repair and downtime
is now presented.
Theorem 2.1
g(u) KX (u) Ki an ^ Â£..*.
If < (u > 0) is strictly increasing to infinity,
v(u) 
then a unique finite optimal equipment replacement age X exists which
is the solution to the equation
X
v(u)
0
g(X) KX(X) Ki
v(X)
g(u) KX(u) Ki
v(u)
R(u)du = K
and if the initial age, x, satisfies x <_ X then an optimal Xpolicy
exists for the major repair problem with arbitrary repair time distribution.
Proof:
If the initial equipment age x is such that x < X, then it is
clear from substituting (2.11) into (2.6) that if an optimal replace
ment age X exists, it minimizes both f (0) and f^(x) for all x.
Differentiating (2.11) with respect to X and equating the result to
zero yields
[KR(X) + g(X)R(X)]
v(u)R(u)du
X
 v(X)R(X)[KR(X) +
g(u)R(u)du] = 0
(2.13)
18
age at which the equipment is replaced, the following functional
equation is obtained.
K(x) + h(0)
f(x) = min <
h(x)
x = X (replace)
0 < x < X (do not replace)
Several conclusions concerning decision regions and continuity
of f(x) at x = X can be drawn from this functional equation. Given an
age x, a replace policy is preferred to a do not replace policy if
and only if
K(x) 4 h(0) < h(x)
Similarly, a do not replace policy at age x is preferred to a replace
policy if and only if
K(x) + h(0) > h(x)
These two conclusions generate the decision regions of do not replace
and replace. The cost curves and the alternating decision regions
are illustrated in Figure 1,1. At x = X, the decision maker is
indifferent to either a policy of replace or do not replace if and
only if
K(x) + h(0) = h(x)
For this last conclusion to be true, it is necessary that the function
f(x) be continuous at x = X.
Continuity of the function f(x) at x = X implies that the limit
of f(x) as x approaches X from the left and the limit of f(x) as
x approaches X from the right are equal. Therefore the cost curves
50
which, providing x ^ 0, results in the difference equation
(j+l)a
(3.9)
J+l
The solution to (3.9) obtained recursively is
(i + A AhT(i)ak) j _> 1
where is an arbitrary constant whose value is to be determined.
From (3.8) then
CO
(i + A AhT(i)ak)
yl(x) = a0 1 + J
(3.10)
j=l
Define 0 = i + A and
00 3
v xJ
r tt (0 Ah (i)ak)
k=0
S (x, 0 Ah (i) ,a) = 1 + l ~
j=l J'
(3.11)
The properties of the function S(x,8,Ah^(i)a) are discussed
in detail in Appendix B. The pertinent ones for this chapter are
(i)for all nonnegative real values of x, S(x,0,Ah (i),a)
is a convergent power series;
(ii) all derivatives of S(x,0,Ah^,(i) ,a) with respect to x
exist and are positive for all nonnegative real values
of x; and
(iii) [S(x,8,Ah (i),a) 1] is positive for all positive
real values of x.
Equation (3.10) can be rewritten using (3.11) as
yx(x) = aoS(x,0,AhT(i),a)
(3.12)
73
Substituting (3.53) in (3.47) yields
d Â£1(>
dX2
X=X?
2BSx(X*,a) + (2gX* + Y)Sxx(X*,a)
SÂ¡XX^)[S(X*,a) 1)
Since 6 = B/y, then
d2fx(0) (26X* + DS^iX*^) 26Sx(X5V,a)
dX2' a = yS^(X*, a) [ S (X*, a) 1]
Using (3.52) in the last expression, it follows that
d f1(0)
dX
> 0
X=X*
This proves existence and concludes the proof of Theorem 3.3. The
condition c. > 0 for i=l,2 is one of a pair of sufficient conditions
i
for a unique finite optimal equipment replacement age X to exist.
The following is a numerical example of the linear hysteresis
2
problem with g(x) = c^x + c^x + c^.
Example 2
Consider the problem posed in Example 1, with one exception.
The combined repair and operating cost in $ per year of service for
2
the equipment when service age is u is 1.73 u + 7.62 u + 6.5.
From the problem statement, it is known that K = $350,
X = .1/year, i = .l$/($year), 0 = .2/year, a = 1/2, hT(i) = 1,
Ah (i) = .1/year, P (1/2) = .15, c^ 1.73, = 7.62, and Cq = 6.5.
Then solving numerically equation (3.43), it is determined that the
4
the relevant costs over a given length of time. In the renewal
theoretic approach, the focus is on the expected number of occurrences
(or renewals) of two or more events over a given length of time. As
a result of this approach, many interesting statistics concerning the
stochastic process of service age in the repairreplacement problem
are obtained. Since associated with each occurrence of an event is a
cost, expressions for the expected cost for a given length of time and
the expected cost per unit time are readily obtained. In contrast, the
functional equation approach accumulates future expected costs in a
cost function. Unless these future costs are discounted to the
present (concept of present value), the cost function would grow without
bound for an infinite time span. For a finite time span, it is still
important to consider discounting since future alternatives are being
examined in terms of present day values. As a result, expressions
such as total expected discounted cost for either a finite or infinite
time span arise.
When equipment is not subject to failure, the element of
uncertainty is removed from the optimization criterion. This results
in a criterion of total discounted cost for the functional equation
approach. The functional equation approach will be discussed in
greater detail in Section 1.4.
1.3 Literature Review
The literature of equipment repair and replacement has
differentiated according to the amount of effort expended on the repair
of a failed equipment. The amount of repair can take on the range of
possibilities from minimal repair to major repair or overhaul. Minimal
110
AX(la )
e
(la)(la3)
X
[e
AX
Ax.(1a)
Ae dx^]
aAX(la)
' (1a)
P{N(X) = 2}
(5.15)
Using (5.8) in (5.15), the following is obtained
P{N(X) = 3} =
AX fX Ax.(1a3)
Ae dx.
(la)(la)
0
AX(1a )
e
(1a)(1a2)
P{N(X) = 1}
aAX(la)
(1a)
P{N(X) = 2}
Or
AX
P{N(X) = 3} =
(1a)(1a2)(1a3)
[1 eXX(1a >]
AX(la )
e
(1a)(1a2)
P{N(X) = 1}
aAX(la)
(1a)
P{N(X) = 2}
(5.16)
Now
P{N(X) = 4} = P{W5 > X, W4 < X, W3 < X, W2 < X, Wx < X}
= e
AX
X ,, 4. Xax., 3n Xa x^ax^ 2.
Ax (1a ) r 1 Ax(la ) 12 Ax0(la )
Ae 1  Ae
0
0 0
Xa3x a2x2ax ,
le 4
J
0
Ae
dx., dxdxdx.
12 3 4
LIST OF TABLES
Table Page
1.1
Significant Papers for Problems I and II...
16
3.1
Table of Optimum X and Corresponding f^(0)
for
g(x) = lOx When a = 0.0....................
62
3.2
Table of Optimum X and Corresponding f^(0)
g(x) = lOx When a = 0.25.........
for
63
3.3
Table of Optimum X and Corresponding f^(0)
g(x) = lOx When a = 0.50.
for
64
3.4
Table of Optimum X and Corresponding f^(0)
g(x) = lOx When a = 0.75.
for
65
3.5
Table of Optimum X and Corresponding f^(0)
g(x) = lOx When a = 1.0
for
66
3.6
Table of Optimum X and Corresponding f^(0)
g(x) = 1.75x^ + 7.5x When a = 0.0..........
for
75
3.7
Table of Optimum X and Corresponding f^(0)
g(x) = 1.75x^ + 7.5x When a = 0.25..
for
76
3.8
Table of Optimum X and Corresponding f^(0)
g(x) = 1.75x^ + 7.5x When a = 0.50
for
77
3.9
Table of Optimum X and Corresponding f^(0)
g(x) = 1.75x^ + 7.5x When a = 0.75
for
78
3.10
Table of Optimum X and Corresponding f^(0)
g(x) = 1.75x^ + 7.5x When a = 1.0..........
for
79
4.1
Table of Optimum Values for E,Xpolicy for
Convex Curve p(E) as in Figure 4.6.........
a
94
4.2
Table of Optimum Values for E,Xpolicy for
Concave Curve p(E) as in Figure 4.5
a
95
4.3
Table of Optimum Values for E,Xpolicy for
Linear Curve p(E) as in Figure 4.1.........
a
97
4.4
Table of Optimum Values for E,Xpolicy for
Convex Curve p(E) as in Figure 4.6
a
99
vii
Table 4.5. Table of Optimum Values for E,Xpolicy for a
Concave Curve p(E) as in Figure 4.5.
q(x) = lOx + A Ex
K
e
A
AhT(i)
Scaling Factor
E
X
a
f1(0)
100
0.4
0.1
0.1
20
20.00
5.56
0.00
89.20
100
0.4
0.1
0.1
25
0.00
5.74
1.00
91.24
100
0.4
0.3
0.3
25
25.00
4.35
0.00
360.59
100
0.4
0.3
0.3
30
0.00
4.83
1.00
383.18
100
0.6
0.3
0.3
20
20.00
5.35
0.00
85.30
100
0.6
0.3
0.3
25
0.00
5.74
1.00
91.24
200
0.6
0.3
0.3
25
25.00
8.51
0.00
96.63
200
0.6
0.3
0.3
30
0.00
9.12
1.00
103.90
200
0.6
0.45
0.3
15
15.00
8.82
0.00
92.59
200
0.6
0.45
0.3
20
0.00
9.12
1.00
103.90
60
Using (3.36) in (3.35) and if > 0, the following is obtained
dfj(0)
dX2
cisxx^x ,a')
P1(a)Sv(X',a)[sr,a) 1]
> 0
x=x
X
which proves existence. Then the condition c^ 0 is one of a pair
of sufficient conditions for a unique finite optimal equipment replace
ment age X to exist.
The following is a numerical example of the linear hysteresis
problem with g(x) = c^x 4 c^.
Example 1
Consider a continuously operating equipment subject to a
constant failure rate and with an expected number of failures of
10 per year. Following each failure, the equipment is instantly
repaired. Repair of an equipment results in a recovery of usable
service life, where the amount of recovery is proportional to the age
of the equipment. Assume that a repair recovers half of the service
life used by the equipment. The combined repair and operating cost
in $ per year of service for the equipment when service age is u is
10.04 u + 7.2. It is anticipated that when the existing equipment
is in need of being replaced, the new equipment will have identical
characteristics and a fixed cost of $350 will be incurred. Assuming
a fixed interest rate of 10% per year, it is required to determine
the service life at which it is more economical to replace the
existing equipment rather than to continue operating and repairing it.
From the problem statement, it is known that K = $350,
36
Let
X
1 R(X) 
hT(i)
X(u)R(u)du
X
d[R(u)] hT(i)
X(u)R(u)du
X
X
[i + X(u)]R(u)du hT(i)
X(u)R(u)du
X
{i + X(u)[1 h (i)]}R(u)du
; 1
0
v(u) = i + X(u)[l hT(i)] u _> 0
(2.9)
This is the expression that Sivazlian [21] defined for j(u) in the
minimal repair problem (see 1.3a). Then
1 R(X) hT(i)
X
X(u)R(u)du
0
X
v(u)R(u)du
0
(2.10)
Substituting (2.10) in (2.8) yields
X
K(X)R(X) +
g(u)R(u)du
fi(0) sr
v(u)R(u)du
For the remainder of this chapter, the analysis is restricted
to the case where the fixed cost of replacement K(X) is independent
of the replacement age X (i.e., no salvage feature incorporated in
the replacement cost). Since K(X) = K = constant, then
Table 3.2. Table of Optimum X and Corresponding f^(0) for
g(x) = lOx When a = 0.25
K
50 150 250 350
0
XhT(i)
X
f1(0)
X
f1(o)
X
fx(o)
X
f1(0)
0.20
0.10
3.43
262.9
6.35
392.9
8.63
456.0
10.68
493.8
0.40
0.10
3.88
68.9
8.05
85.0
11.87
88.1
15.63
88.7
0.20
3.77
107.2
7.69
135.1
11.25
141.0
14.75
142.4
0.30
3.65
222.6
7.33
286.9
10.64
302.0
13.89
306.1
0.60
0.10
4.44
32.2
10.29
34.7
16.04
34.8
21.79
34.8
0.20
4.32
41.7
9.92
45.3
15.42
45.5
20.92
45.5
0.30
4.20
57.6
9.54
63.2
14.79
63.5
20.04
63.5
0.40
4.09
89.5
9.17
99.5
14.17
100.0
19.17
100.0
0.50
3.97
185.5
8.80
209.1
13.54
210.4
18.29
210.5
0.80
0.10
5.11
18.1
12.88
18.4
20.63
18.4
28.38
18.4
0.20
4.98
21.8
12.50
22.2
20.00
22.2
27.50
22.2
0.30
4.86
26.9
12.13
27.6
19.38
27.6
26.63
27.6
0.40
4.74
34.8
11.75
35.7
18.75
35.7
25.75
35.7
0.50
4.62
47.9
11.38
49.4
18.13
49.4
24.88
49.4
0.60
4.50
74.2
11.00
76.9
17.50
76.9
24.00
76.9
0.70
4.38
153.5
10.63
160.0
16.88
160.0
23.13
160.0
46
Barlow and Hunter's [1] result for the major repair problem with zero
downtime for repair is recovered if c(u) = 0, k(u) = k = constant, and
E(T) = 0 (zero downtime). In addition, from Corollary 2.4, the
existence and uniqueness of an optimal solution is assured provided that
k > K and A(u) is strictly increasing to infinity. Scheaffer's [20]
result is recovered by setting c(u) cu k(u) = k = constant, and
E(T) = 0.
124
as the method of integration in series has resulted in a complementary
solution being obtained for the case a(x) = ax, A(x) = X. The
complementary solution is assumed to be of the form
00
y (x) = l a xj
j=0 J
Upon appropriately substituting the assumed form of the solution in
the homogeneous equation, the values of the constants cu are determined
in terms of a^, an arbitrary constant.
115
51P1
+ BP + 232P1
+ 3^3 + 232P2 + 3g3P1
+ 3xP4 + 2B2P3 + 333P2 + 434P1
= I re y P,
S r i i
r=l i=l
r1
Similarly, expanding Â£ Â£ m3 P
2

