1. Introduction

* 한라대학교 경상학부

If a machine fails during operation, it is repaired or replaced by new one.  A typical maintenance policy for a repairable machine is the minimal repair policy.  By the minimal repair policy, a machine is minimally repaired upon failures, and replaced by new one at regular intervals[2, 4, 6] or the number of failures reaches to a certain amount[8, 11].  Since a minimal repair involves only that amount of work which is necessary to restore the failed machine to the average operating condition of its age, it is more appropriate to conduct at minor failures than critical failures.

When a group of identical machines are under operation, which is common in many practical situations, a popular maintenance policy is a group maintenance policy.  The idea of group maintenance policy is based on the set-up cost savings earned by performing maintenance all machines at a maintenance set-up.  Under the classical group maintenance policy, if a machine fails during operation, it is left idle until the next group maintenance time.  At the group maintenance time, failed machines are replaced by new ones and the preventive maintenance (PM) is serviced for the survivals to restore it to an as-good-as-new condition.  The group maintenance is performed when a prescribed time interval elapses[7], or when the number of failed machines reaches to a certain amount[5, 1].  In the (m,T) group maintenance model of Ritchken and Wilson[9], a group maintenance is performed either when an operation time T elapses or the number of failed machines reaches to m out of N machines, whichever comes first.

In the previous group maintenance policies, the failed machines are left idle until the next group maintenance set-up.  Even when a minor failure occurs to a machine early in the operation, it is forced to wait to the next group maintenance time.  If minimal repairs are performed immediately to minor failures during operation, the non-serious failures will not need a long wait.

In this paper, the (m,T) group maintenance model of Ritchken and Wilson[9] is extended by classifying failures into two modes, critical and minor failures.  The proposed policy is as follows.  There are N machines of the same type under operation.  If a minor failure occurs, minimal repair is performed immediately to continue its operation.  When a critical failure occurs to a machine, it is left idle until the next group maintenance incurring downtime cost.  Group maintenance is set up either when the operation time T elapses or the number of critical failures reaches to m, [eqn: 1 <= m <= N], whichever comes first.  At the group maintenance, the critically failed machines are replaced by new ones, and PM is serviced for the survivals to restore it to an as-good-as-new condition.  Then a new cycle begins.

The proposed policy is different from Ritchken and Wilson[9] in that, unlike their model in which failed machines are left idle regardless of the failure mode, minimal repairs are performed at minor failures during operation.  If all failures are critical, no minimal repair is performed, and failed machines are left idle.  In this case, the proposed policy reduces to Ritchken and Wilson[9].  If all failures are minor, no downtime cost is incurred, and the proposed policy becomes Barlow and Hunter[2].

Notations

[eqn: F(t), ~f(t),~r(t)]  cdf, pdf, failure rate

function of time-to-failure of a

machine

[eqn: { c}_{s }]  set-up cost of a group maintenance

[eqn: { c}_{m }]  minimal repair cost for a minor

failure

[eqn: { c}_{r }(i)]  group replacement cost for i

critically-failed machines

[eqn: { c}_{p }(i)]  group PM service cost for i

survived machines

[eqn: { c}_{d }(i)]  downtime cost per unit time for

i critically-failed machines

2. Average Cost Rate under the Extended (m,T) Policy

If a machine fails during operation, it is inspected whether it is critical or minor failure.  We assume that we have a clear criterion on judging whether a failure is critical or minor.  Let

[eqn: p(t)] = the conditional probability that a

failure at time t is critical,

[eqn: q(t)] = the conditional probability that a

failure at time t is minor,

[eqn: p(t)+q(t)=1].

Then, by the decomposition property of a non-homogeneous Poisson process, the critical and minor failure rates are [eqn: p(t)r(t)] and [eqn: q(t)r(t)] respectively[3].

Let Y denotes the time to the critical failure of a machine.  Then the survival function of Y is

[eqn: P LEFT { Y >= y RIGHT } == 1-G(y)]

[eqn: =exp LEFT [ - INT _{ 0}^{y }p(x)r(x)dx RIGHT ]].    (1)

Under the proposed policy in which the critically failed machines are left idle, define two random variables as follows:

[eqn: { T}_{i }]  = time to the ith critical failure,

[eqn: 1 <= i <= N],

[eqn: N(t)] = the number of critical failures

during [0,t).

Since the event [eqn: LEFT {{ T}_{m }>t RIGHT }] is equivalent to the event[eqn: LEFT { N(t)<m RIGHT }], the survival function of the time to the mth critical failure,[eqn: { T}_{m }], is represented as

[eqn: P LEFT { { T}_{m }>t RIGHT } =P LEFT { N(t)<m RIGHT }]                   [eqn: = SUM from { {i }=0} to m-1 P LEFT { N(t)=i RIGHT }]

[eqn: = SUM from { {i }=0} to m-1 PMATRIX { { N}# {i }} { LEFT [ G(t) RIGHT ] }^{i } { LEFT [ 1-G(t) RIGHT ] }^{N-i }]

[eqn: = SUM from { { i}=0} to m-1 A(i,t)],                    (2)

where,

[eqn: A(i,t)==SUM from { {i }=0} to m-1 PMATRIX { { N}# {i }} { LEFT [ G(t) RIGHT ] }^{i }]

[eqn: TIMES { LEFT [ 1-G(t) RIGHT ] }^{N-i }].

According to the proposed policy, all machines are renewed at each group maintenance time.  Since a group maintenance is performed either when the operation time T elapses or the mth critical failure occurs, the length of a renewal cycle is [eqn: min LEFT { T,{ T}_{m } RIGHT }].  Therefore the expected length of a cycle is

[eqn: E LEFT [ cycle RIGHT ]=E LEFT [ min LEFT { T, { T}_{m } RIGHT } RIGHT ]]

[eqn: = INT _{ 0}^{T }P LEFT { { T}_{m }>t RIGHT } dt]

[eqn: = SUM from { {i }=0} to m-1 INT _{ 0}^{T }A(i,t)dt].         (3)

The expected cost during a cycle consists of five parts: The group maintenance set-up cost, the cost of downtime loss, the group replacement cost of critically failed machines, the PM service cost for the survivals, and the minimal repair cost for the minor failures.

2.1 The expected downtime cost

Proposition 1.  The downtime during a cycle is

[eqn: SUM from { { i}=1} to m-1 i LEFT [ min LEFT ( T,{ T}_{i+1 } RIGHT )]

[eqn: -min LEFT ( T, { T}_{i } RIGHT ) RIGHT ]].   (4)

Proof.  Derivation of (4) is attributed to Ritchken and Wilson[9], and is explained as follows.  Referring to Figure 1, the downtime during a cycle is

0       if[eqn: T~<= { T}_{1 }]

[eqn: LEFT ( T- { T}_{1 } RIGHT )]     if[eqn: { T}_{1 } < T<= { T}_{2 }]

[eqn: LEFT ( { T}_{2 } - { T}_{1 } RIGHT ) + 2LEFT ( T- { T}_{2 } RIGHT )]

if[eqn: { T}_{2 } < T<= { T}_{3 }]

..... [eqn: LEFT ( { T}_{2 } - { T}_{1 } RIGHT ) +2 LEFT ( { T}_{3 }- {T }_{2 } RIGHT )+]

[eqn: ...+(m-1) LEFT ( T- { T}_{m-1 } RIGHT )]

if [eqn: { T}_{m-1 } < T<= { T}_{m }]

[eqn: LEFT ( { T}_{2 } - { T}_{1 } RIGHT ) +2 LEFT ( { T}_{3 }- {T }_{2 } RIGHT )+]

[eqn: ...+(m-1) LEFT ( { T}_{m } - { T}_{m-1 } RIGHT )]

if   [eqn: T~> { T}_{m }]

It can be verified that (4) represents the downtime during a cycle for possible ranges of T.         ■

Taking expectation of (4), the expected downtime cost during a cycle[eqn: { K}_{d }]  is

Figure 1. Graphical representation of

time to critical failure

[eqn: { K}_{d }`= SUM from { { i}=1} to m-1 {{c }_{d } } (i) E LEFT [ min LEFT ( T, { T}_{i+1 } RIGHT )]

[eqn: -min LEFT ( T, { T}_{i } RIGHT ) RIGHT ]]

[eqn: = SUM from { { i}=1} to m-1 {{c }_{d } } (i) LEFT [ SUM from { { j}=0} to i INT _{ 0}^{T } A(j,t)dt]

[eqn: - SUM from { { j}=0} to i-1 INT _{ 0}^{T } A(j,t)dt RIGHT ]]

[eqn: = SUM from { { i}=1} to m-1 {{c }_{d } } (i) INT _{ 0}^{T } A(i,t)dt].      (5)

2.2 The expected group replacement cost of

critical failures

The group replacement cost during a cycle depend on the length of[eqn: { T}_{m }].  If[eqn: { T}_{m } <=T], exactly m machines are replaced.  If[eqn: { T}_{m } >T], then fewer than m machines are replaced.  Therefore the expected group replacement cost[eqn: { K}_{r }] is

[eqn: { K}_{r }` = { c}_{r }(m) SUM from { { i}=m} to N P LEFT { N(t)=i RIGHT }]

[eqn: + SUM from { { i}=1} to m-1 { c}_{r }(i) P LEFT { N(t)=i RIGHT }]

[eqn: `= { c}_{r }(m) SUM from { { i}=m} to N A(i,T)]

[eqn: + SUM from { { i}=1} to m-1 { c}_{r }(i) A(i,T)]. (6)

2.3 The expected group PM service cost

If[eqn: { T}_{m } <=T], exactly N-m machines are serviced to restore it to an as-good-as-new condition.  If[eqn: { T}_{m } >T], then more than N-m machines are replaced.  Thus the expected group PM service cost[eqn: { K}_{p }] is

[eqn: { K}_{p }` = { c}_{p }(N-m) SUM from { { i}=m} to N P LEFT { N(t)=i RIGHT }]

[eqn: + SUM from { { i}=0} to m-1 { c}_{p }(N-i)P LEFT { N(T)=i RIGHT }]

[eqn: ` = { c}_{p }(N-m) SUM from { { i}=m} to N A(i,T)]

[eqn: + SUM from { { i}=0} to m-1 { c}_{p }(N-i) A(i,T)].  (7)

2.4 The expected minimal repair cost for

minor failures

The expected cost of minimal repairs during a cycle is the expected number of minimal repairs during a cycle times the cost of a minimal repair.  Let

[eqn: H(t)= INT _{ 0}^{t } q(x)r(x)dx]           (8)

and [eqn: h(t)=dH(t)/dt=q(t)r(t)].  Then (8) represents the expected number of minimal repairs (minor failures) during [0,t), and [eqn: h(t)] represents the minor failure rate function[3].

Proposition 2. The conditional expected number of minimal repairs during a cycle is

[eqn: (N-m+1)min LEFT { H(T), H( { T}_{m }) RIGHT }]

[eqn: + SUM from { {i }=1} to m-1 min LEFT { H(T), H( { T}_{i }) RIGHT }] (9)

Proof.  Referring again to Figure 1, the conditional expected number of minimal repairs during a cycle depends on the cycle termination point as follows:

[eqn: NH(T)]    if [eqn: T~<= { T}_{1 }]

[eqn: (N-1)H(T)+H LEFT ( { T}_{1 } RIGHT )][eqn: `]if[eqn: { T}_{1 } < T<= { T}_{2 }]

.....

[eqn: (N-m+1)H(T)+H( { T}_{1 })+]

[eqn: ...+H( { T}_{m-1 })][eqn: `]

if [eqn: { T}_{m-1 } < T<= { T}_{m }]

[eqn: (N-m+1)H( { T}_{m } )+H( { T}_{1 })+]

[eqn: ...+H( { T}_{m-1 })]    if [eqn: T~> { T}_{m }]

It can be easily verified that (9) represents the conditional expected number of minimal repairs during a cycle for possible ranges of T.        ■

To evaluate the expectation of (9), we first observe that, using differentiation by parts,

[eqn: E LEFT [ min LEFT { H(T), H( { T}_{m }) RIGHT } RIGHT ]]

[eqn: =H(T)P LEFT { { T}_{ m}>T` RIGHT }]

[eqn: + INT _{ 0}^{T }H(t)dP LEFT { { T}_{m }<=t` RIGHT }]

[eqn: =H(T)P LEFT { { T}_{ m}>T` RIGHT }]

[eqn: +H(T)P LEFT { { T}_{m }<=t` RIGHT }]

[eqn: - INT _{ 0}^{T }h(t) P LEFT { { T}_{m }<=t` RIGHT } dt]   [eqn: = INT _{ 0}^{T }h(t)P LEFT { { T}_{m }>t` RIGHT } dt]

[eqn: = SUM from { { i}=0} to m-1 INT _{ 0}^{T }h(t)A(i,t)dt].             (10)

Substituting i in place of m in (10) and summing over i from 1 to m-1,

[eqn: SUM from { { i}=1} to m-1 E LEFT [ min LEFT { H(T), H( { T}_{i }) RIGHT } RIGHT ]]

[eqn: = SUM from { { i}=1} to m-1 SUM from { { j}=0} to i-1 INT _{ 0}^{T }h(t)A(j,t)dt][eqn: = SUM from { { i}=0} to m-2 (m-1-i) INT _{ 0}^{T }h(t)A(i,t)dt].  (11)

Using (10) and (11) with some algebraic manipulation, the expectation of (9) is

[eqn: (N-m+1)E LEFT [ min LEFT { H(T), H( { T}_{m }) RIGHT }RIGHT ]]

[eqn: + SUM from { {i }=1} to m-1 E [min LEFT { H(T), H( { T}_{i }) RIGHT }]]

[eqn: = (N-m+1)SUM from { { i}=0} to m-1 INT _{ 0}^{T }h(t)A(i,t)dt]

[eqn: + SUM from { { i}=0} to m-2 (m-1-i) INT _{ 0}^{T }h(t)A(i,t)dt]

[eqn: = SUM from { { i}=0} to m-1 (N-i) INT _{ 0}^{T }h(t)A(i,t)dt].

Therefore, the expected cost of minimal repairs during a cycle is

[eqn: { K}_{m } = { c}_{m } SUM from { { i}=0} to m-1 (N-i)]

[eqn: TIMES INT _{ 0}^{T }h(t)A(i,t)dt].      (12)

Summing the cost components (5), (6), (7), (12),  and dividing it by the expected length of a cycle (3), the expected cost rate under the extended (m,T) policy is, from the renewal reward theorem[10],

[eqn: C(m,T)= LEFT [ { c}_{s }+ { K}_{ r}+ { K}_{p }+ { K}_{d }]

[eqn: + { K}_{m } RIGHT ] /E[cycle]][eqn: over {E[cycle] }]      [eqn: over {E[cycle] }](13)

3. Special Cases of the

Extended (m,T) Policy

3.1 [eqn: p(t)=1]

In this case, all failures are critical and no minimal repair is performed.  Thus the model reduces to Ritchken and Wilson[9].   Inserting [eqn: p(t)=1] into (1), the distribution function of time to the critical failure becomes

[eqn: G(t)=1-exp LEFT [ - INT _{ 0}^{t }r(x)dx RIGHT ] =F(t)]

which is the distribution function of time to failure of a machine.  Replacing [eqn: G(t)] in (2) by[eqn: F(t)], the cost components (3), (5), (6), and (7) correspond to the cost components equations (3), (4), (6), and (5) of Ritchken and Wilson[9] respectively.  The minimal repair cost is zero from (8) since [eqn: q(t)=0].  Therefore the cost rate function (13) reduces to the cost rate function (7) of Ritchken and Wilson[9].

3.2 [eqn: p(t)=0]

This case implies that all failures are minor and minimal repairs are performed for the failures.  Thus no machine is left idle by the critical failure.  Inserting [eqn: p(t)=0] into (1), we get [eqn: G(t)=0].  Therefore (3) becomes

[eqn: E[cycle]= SUM from { { i}=0} to m-1 INT _{ 0}^{T } A(i,t)dt]

[eqn: = INT _{ 0}^{T }A(0,t)dt= INT _{ 0}^{T }dt=T].     (14)

Further, it is easily verified that the cost components become

[eqn: { K}_{d } =0,~ { K}_{r } =0,~ { K}_{p }= { c}_{p }(N),~]

[eqn: { K}_{m }= { c}_{m }NH(T)]          (15)

where, [eqn: H(T)= INT _{ 0}^{T }r(t)dt] from (8) since[eqn: q(t)=1] in this case.  Inserting (14) and (15) into (13),

[eqn: C(inf,T)= LEFT [ { c}_{s }+ { c}_{ p}(N)]

[eqn: + { c}_{m }N INT _{ 0}^{T }r(t)dt RIGHT ] /T]

which is the cost rate expression of Barlow and Hunter[2]'s minimal repair policy.

4. Numerical Examples

Since the decision variables m and T in the cost function (13) are expressed in a complicated manner, it is difficult to obtain the closed form expressions of the optimum values of m and T which minimize (13).  Instead, the numerical experience for Weibull  time to failure of a machine is presented in this section.

Assuming a constant critical failure probability [eqn: p(t)=p], and inserting the Weibull failure rate [eqn: r(t)= lambda beta {lambda t }^{beta -1 }] into (1), we get [eqn: G(t)=1-exp LEFT { -p {( lambda t)}^{ beta } RIGHT }].  The minor failure rate function and the number of minimal repairs during a cycle (8) becomes respectively

[eqn: h(t)=(1-p) beta lambda { ( lambda t)}^{ beta -1 }]

[eqn: H(t)=(1-p) { ( lambda t)}^{ beta }].

Suppose that N=5 machines are under operation and the proportion of critical failures is p=0.3.  The Weibull parameters and the cost parameters are:

[eqn: lambda =0.3/year],[eqn: beta]=2,[eqn: { c}_{ s}=10],[eqn: { c}_{d }(i)=10i],[eqn: { c}_{p } (i)=20i],[eqn: { c}_{r }(i)=30i],[eqn: { c}_{m }=1].

Figure 2. The behaviour of cost rate

function(m=4).

A computer program written in Quick BASIC was run on an IBM PC to compute the expected cost rate (13).  The result shows that (13) is unimodal in T for a given value of m(Figure 2).

Repeating the computation with the various values of m from 1 to 5, the optimum values of policy variables are obtained as[eqn: { T}^{* }=4.30] year and[eqn: { m}^{* }=2nd] critical failure with the corresponding cost rate[eqn: { C}^{* } =31.68].

In Table 1, the optimum values of m and T with the corresponding cost rate for various values of p are presented.  The result reveals that as the proportion of critical failures p increases,[eqn: { m}^{* }] and[eqn: { C}^{* }] increase accordingly, but[eqn: { T}^{* }] decreases.

5. Conclusions

In this paper, a group maintenance model with two modes of failures was  presented and analyzed.  Classifying failures into critical or minor failures,  more practical maintenance decisions could be made.  It was shown that the presented model includes earlier models as special cases.

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