- Email: [email protected]
On optimal maintenance of a series system
Microelectron.Reliab.,Vol.23,No. 5.pp. 827-831.1983.
0026-2714/8353.00+ .00 © 1983PergamonPressLtd.
Printedin GreatBritain.
ON OPTIMAL MAINTENANCE OF A SERIES SYSTEM ASHOK KUMAR and V. B. KAPOOR Defence Science Centre, Metcalfe House, Delhi-110054, India and M. C. GUPTA 10/2 Shakti Nagar, Delhi-110007, India
(Received for publication 2 February 1983) Abstract--Maintenance of an n-unit series system with general distributions for failure-time and repair times has been discussed. The system has been identified by up and down states. It is assumed that there are a finite number of alternatives to a decision maker in each state. The system earns at a fixed rate (possibly negative) in each state and fixed transition costs are incurred at the time of transitions. The problem is to choose a vector of alternatives (one alternate in each state) so that the overall expected profit is maximized. A procedure to determine optimal maintenance policy is developed based on Howard's Policy Iteration scheme. Numerical examples are included to illustrate the results.
INTRODUCTION
Define the following system states to identify the system at any instant.
The problems of the maintenance of equipments/ systems is quite important to modern business and industrial houses. The system designer/analyst is often confronted with the problem of running the system in the most efficient manner. Profit appears to be a more appropriate parameter in evaluation studies of systems as compared to other parameters, such as reliability, availability, M T S F , etc. Many authors have concentrated their attention on analyzing their models under different conditions with a view to obtaining these parameters. A detailed list of references is contained in K u m a r and Agarwal [1]. Mine and Kawai I-2] discussed the maintenance policies for a 2-unit parallel redundant system which maximizes expected profit for an infinite time span. K u m a r [3] obtained an analytic expression for expected profit in a standby system under different operating environments. There may be several alternatives available to the systems analyst for maintaining the system in up and down states. Different types of failure demand different types of repair. A maintenance policy is a set of alternatives (one alternative from each state). The optimal maintenance policy ( O M P ) is one which maximises expected profit of the system. The present paper attempts to obtain optimal maintenance policy for an n-unit series system. Howard's Policy Iteration method [4] has been employed to achieve the results.
So all the n units (components) are operating. S~ the system is in the failed state due to the failure of the ith unit (component) and is under repair. (i = 1,2 .... ,n). The system is up in S O and it is down in S~. Let there be several alternatives available in each state to maintain the system. F o r instance, when the system is in So, there may be ordinary maintenance (O,.), costly maintenance (Cm) or very costly maintenance (VC,~). Whichever maintenance scheme is followed, the system will fail eventually. However, the mean time to failure increases as we go from O,, to C,, and from C~" to VCm. Similarly, when the system is in S~ [failed and ith unit (component) is under repair] there may be several alternatives for repair: a costly repair may ensure a faster repair. Symbolically, let there be N alternatives in each state (N varies from state to state). Let us define a maintenance policy as the set of maintenance alternatives--one from each state. An optimal maintenance policy is the one which maximizes the expected profit of the system. Let us assume the following cost structure for the system: yk earning rate of the system in S O when the kth maintenance alternative is adopted. y~ loosing rate of the system in S~ when the kth repair alternative is adopted (i = 1, 2 . . . . . n) yik fixed transition reward (cost) for making a transition from S~ to Sj and alternative k is chosen in S~.
SYSTEM MODEL AND THE COST STRUCTURE Let there be an n-unit (component) series system with units (components) labelled 1, 2 . . . . . n. The system fails if any one of the units (components) fails. Let the failure time p.d.f, of the ith unit (component) be f~(t). The system can be repaired and let the repair time p.d.f, of the ith unit (component) be 9~(t).
OPTIMIZATION PROBLEM
Under the above cost structure, it is proposed 827
828
ASHOK KUMAR,
V. B. KAPOORand M. C. GUPTA
to determine the optimal maintenance policy for the n-unit series system which maximizes the steady state profit of the system. We develop, below, the solution procedure based on Howard's Policy Iteration method. We deal with two cases.
Case 1: when payments to be received in future are not discounted. Case 2: when a discounting factor is also taken into account.
Following Howard [4], we know that for large t
vi(t ) = vi+gt where vi(t) is the total expected earning of the system in time t, if the system starts in S~ at time t = 0. v~is the transient part of the profit and g is total expected earning of the system in steady state. Also
i = 1,2,...,n
where qiPi "~ 2 Pijrij + Yi~i • J
Here #~ is the mean unconditional waiting time in S~ and pij is the one-step transition probability for making a transition from S i to Sj. Following Howard's [4] Policy Iteration method, which is given below, we determine the optimal policy numerically after assigning appropriate values to the parameters involved.
Policy Evaluation For the present policy solve ViWgPi = qipi-k-2pijvj J
i = 1,2,...,n
with v, = 0, for g and relative values V l , / ) 2 , . . . , 1)n- 1
Policy Improvement Find the alternative k in each state i that maximizes k 1[k 7 qi + 7 X / E P i f J - vii ~i Lj
Step 1: define the system in terms of its states and transitions between them. Specify the cost structure and alternatives in each state. Step 2: determine the transition probabilities and mean unconditional waiting times under each alternative. Step 3: Select one alternative in each state for which 1 ~
k k
k
qgi = ~ 2.,PijrU + Yi i.t~ j
Case 1: undiscounted cost structure
Viq-gfli -~ q i l z i + 2 p i j v j J
In fact, the above procedure involves following steps to determine the optimal maintenance policy.
d
using the relative values vi of the previous policy. Make this alternative the new decision in state i. Repeat for all states to find the new policy.
is maximum. Name it the initial policy. Step 4: Solve the following set of n equations for the policy obtained at step 3. i = 1,2 . . . . . n
Vi+gPi = qi#i+~pqvj J
for v~, vz,..., v,_ a by setting v, = 0. Step 5: By using values of v x, v2 . . . . . v,_ t and g obtained at step 4, compute the test quantity k
1I-
k
7
for each alternative in each state. Select the alternative in each state for which the test quality is maximum. Step 6: Examine whether the new policy is different from the initial policy. If yes go to step 4, otherwise the optimal policy is obtained. In what follows we shall illustrate the results by a numerical example. Let us consider a system with two types of failure, Type 1 and Type 2. Thus, we have the following system states, S O system is operative $1 system is failed and is under repair due to Type 1 failure. S z system is failed due to Type 2 failure and is under repair. Let the failure time distribution be exponential with rate 2 and the probability that the failure is of Type 1 or 2 be pl or P2. Let the repair-time p.d.f, of Type 1 and 2 be 91(0 and 92(0, respectively. In fact, the system may be thought of as composed of 2-units operating in series with failure time distributions exponential with rate 21 and 22, such that 2 = 21 +22 and px = 21/2 and P2 = 22/2. Let repair-time p.d.f. gi(t)(i = 1, 2) be given by 91(t) = r/2te -~' g2(t) = ½r/at2e -~t
withmean with mean
ml = 2/r/ m2 =
3p/.
Let there be two alternatives in So: (i) 0,, ordinary maintenance; (ii) C,, costly maintenance. Similarly let there be two alternatives in S~: (i) 0 r ordinary repair; (ii) Mr minor repair. State $2 has three alternatives: (i) Or ordinary repair; (ii) M, minor repair; (iii) C, costly repair.
Optimal maintenance k
~2
¢tk
--
- l0
- 15
10.000
248.800
0.6
--
- 10
-15
20.000
199.350
--
--
- 20
--
--
5.000
- 24.000
1
--
--
-15
--
--
3.333
54.500
-50)
1
-
-
-20
-
-
150.000
-50.133
-75)
1
' --
--
-15
--
--
60.000
-75.250
0.10, - 150)
1
--
--
- 10
--
--
30.000
-150.333
States
Alternatives
So
10~,
(0.10,
-,
250)
--
0.6
0.4
2 Cm
(0.05,
-,
200)
--
0.4
1 O,
(-,
0.4,
- 20)
1
2 M,
(-,
0.6,
-50)
1 Or
(-,
0.02,
2 M,
(-,
0.05,
3 C,
(-,
S1
S2
Various" state alternative combinations and their resultant values have been computed and presented above. T h u s , we find t h a t t h e first a l t e r n a t i v e in e a c h s t a t e h a s t h e m a x i m u m value, s o t h e initial p o l i c y for t h e p o l i c y i t e r a t i o n p r o c e s s is (1 1 1). N o w w i t h this p o l i c y we s o l v e t h e set o f e q u a t i o n s 1)i .+ 9 p k = q- k~. i k "4- ~'~ _k 1)j /,pij
829
with
a g m n we f o r m v a r i o u s and g and compute
v t = 5160 9 = 30.8 with I)2 = 0 .
N o w w i t h t h e s e v a l u e s of vi's we a g a i n c a l c u l a t e t h e test q u a n t i t i e s , w h i c h a r e g i v e n below.
which implies that
State
Alternatives
So
1 2
15.0 30.85
S1
1 2
30.8 27.7
$2
1 2 3
- 13.87 15.32 31.0
0.6Vl - vo - 109 = - 2488.00 v~ - v o + 59 = - 120.0 - v o + 1509 = - 7519.95. S o l v i n g t h e s e e q u a t i o n s we g e t (with v 2 = 0), v o = 6303.45 v 1 = 6242.00 9 = -8.11. W i t h t h e s e v a l u e s o f v~'s we c a l c u l a t e t h e test q u a l i t y k
1V
e q u a t i o n s i n v o l v i n g Vo, Vl, v2
v o = 5434
vz = 0
J
qk
k
-1
Test quantity
F r o m t h e a b o v e , we find t h a t t h e p o l i c y c o m e s o u t to be (2 1 3), w h i c h is t h e s a m e as t h e earlier o n e a n d so t h i s is t h e o p t i m a l policy a n d t h e g a i n for t h e s y s t e m is 30.8 u n i t s of m o n e y .
Case 2: discounted cost structure for all t h e a l t e r n a t i v e s in e a c h state. T h e v a l u e s o f t h e test q u a n t i t i e s a r e g i v e n below. State SO
S1
S2
Alternatives
F o l l o w i n g H o w a r d [4], we k n o w t h a t for l a r g e t a n d d i s c o u n t i n g f a c t o r ct > 0 1
vi = ~ pijrijh*(a) + Yi- (1 - h* (a) + ~ pijh*(a)vj j
Test quantity
1
- 8.11
2
8.66
1 2
- 8.11 -30.66
1 2 3
-8.11 29.81 59.98
F r o m t h i s we find t h a t t h e 2 n d a l t e r n a t i v e in S o, t h e 1st a l t e r n a t i v e in S 1 a n d t h e 3rd a l t e r n a t i v e in $2 h a v e t h e m a x i m u m value, t h e r e f o r e t h e i m p r o v e d p o l i c y is w i t h t h e s e a l t e r n a t i v e s (2 1 3). W i t h t h i s p o l i c y
Ct
j
w h e r e v~ is t h e e x p e c t e d p r e s e n t v a l u e o f infinite t i m e o p e r a t i o n in s t a t e i, hij(t) is t h e h o l d i n g t i m e p.d.f, o f t h e s y s t e m in S i before m a k i n g a t r a n s i t i o n to S j,
hi(t) = ~ pijhij J and
Re (s) > O.
f*(s) = f o e-~'f(t)dt Let
1
Pi(°O = ~ Pijh~(ct)rij+ Yi- (i - h~(a)) j
O~
ASHOKKUMAR,V. B. KAPOORand M. C. GUPTA
830
Step 4: Solve the following set of n equations for the policy obtained at step 3.
then
vi = Pi(e) + Y', Pijh*(cOv~• J
vi = Pi(cQ+~pijh*(e)vj
Step 5: By using values of v~'s obtained at step 4, c o m p u t e the test quantity. Pi(~) + ~, Pijh*(o;)vj J
Policy Evaluation
for each alternative in each state. Select the alternative in each state for which the test quantity is maximum. Step 6: Examine whether the new policy is different from the initial policy. If yes go to step 3, otherwise the optimal policy is obtained.
F o r the present policy solve
v i = Pi(~) + ~,pljh~(oOvj J
i = 1,2 . . . . . n
/
Following H o w a r d ' s [4] Policy Iteration method, which is given below, we determine the optimal policy numerically after assigning appropriate values to the parameters involved.
i= 1,2,...,n
for
In what follows we shall illustrate the results by a numerical example. Let us consider a series system. The system failure can be thought of as failure due to unit i or 2, as in the previous example. The life time p.d.f, of the system has been assumed to be Erlangian. The repair time p.d.f, of Type 1 follows an Erlangian distribution and that of Type 2 follows exponential distribution. Let the discounting factor a = 0.4. Thus we have
V l ~ V 2 ~ • . . ~ Vn
Policy Improvement Find the alternative in each state that maximizes
P~(~) + y' p~jh*k(o~)vj J
h~j(~)
using the present values v~ of the previous policy. M a k e this alternative the new decision in state i. Repeat for all states to find the new policy.
Poj 22, + a /
h*j(o:):l( rl" ~ P~j \ ~ + ~ /
r=5
and The above procedure involves following steps to determine the optimal maintenance policy. In this illustration also, there are the same set of alternatives in each state as in the previous example. Various state alternative combinations and their resultant values have been c o m p u t e d and are presented below.
Step 1: Define the system in terms of its states and transitions between them. Specify the cost structures, discounting factor and alternatives in each state. Step2: D e t e r m i n e the transition probabilities, holding time p.d.f.'s and their Laplace transforms evaluated at ~. Step3: Select one alternative in each state for which
Thus we find that the first alternative in each state has the m a x i m u m value, therefore the initial policy for the policy iteration process is (1 1 1). N o w with this policy we form the equation
1
Pi(~) = 2 Pijrijh*(ct) +Yi- (1 - h*(c~)) j
O~
v i = Pi(cO+ ~, pijh*(°Ovi J
is maximum. N a m e it initial policy.
State
Alternatives
p~ko
P~I
p~2 (~,
So
1 0 m
--
0.6
0.4 (O.lO, - ,
250)
2 C.,
--
0.4
0.6 (0.05, - ,
1 O,
1
--
--
(-,
2 Mr
1
1 Or
1
2 M,
1
3 Cr
1
S1
S2
n,
y~)
k r,o
r~,
#2
hZ~(~) h~?(~) h~k(~) h?k(~)
--
-I0
-15
200) --
-10
-15
0.4,
-20) -20
--
--
0.4016
--
--
0.04016 - 37.9520
- 5 0 ) - 15
--
--
0.5345
-
-
0.5345 -67.2050
--
P~(~)
0 . 1 8 5 0.2775 0.2220
483.4750
0.100 0.0667 0.0800
451.0000
-
-
--
(-,
0.6,
-
-
--
(-,
0.02, -50)
-20
--
--
0.4760
--
--
0.4760 -120.0200
--
--
(-,
0.05, - 7 5 ) - 1 5
--
--
0.1111
--
--
0.1111 -168.3352
--
--
(-,
0.10,-150) - 1 0
-
-
0.2000
--
--
0.2000 -302.0000
Optimal maintenance which implies vo = 483.475+0.111v a +0.111vz
831
Again, with these values of vi's we calculate test quantities which are given below.
/)1 = - 37.952 + 0.4016v o
State
Alternatives
v2 = - 120.02 + 0.476v o.
SO
1 2
494.6537 463.0283
S1
1 2 1 2 3
160.7009 197.1874 -96.4745 -113.3799 - 203.0706
Test quantity
Solving the above set of equations we get /)o = 485.2532 S2
v 1 = 156.9256 v2 = -96.9220. N o w with these values of vi we calculate the test quantities pijhij (o~)l)j J for each alternative in every state. The values of test quantities are given below. State
Alternatives
SO
1 2
Test quantity 490.1354 461.4002
$1
1 2
156.9256 192.1628
$2
1 2 3
-96.9220 - 114.4236 -200.9494
F r o m the above we see that the first alternative, in S O and $2 and the 2nd alternative in $1 form the new policy, i.e. our new policy is (1 2 1). With this policy again we form the various equations involving vi's and calculate vo = 494.6472 vl = 197.1839 v z = -96.4748.
M.R 23/-% E
F r o m this we find that the resultant policy comes out to be (1 2 1), which coincides with the earlier policy, which evidently is the optimal operating policy in the present case. Following this policy we ensure m a x i m u m gain for the system. Acknowledgements--The authors are grateful to Dr. A. K. Sreedhar, Director, Defence Science Centre, Delhi, for granting permission to present/publish the paper at a Seminar "Reliability Engineering and Management Applications" at the Military College, EME, Secundrabad. Thanks are also due to Dr. D. Ray, Scientist "E", for the help in preparing the paper.
REFERENCES
1. A. Kumar and M. Aggarwal, A review of standby redundant systems, IEEE Trans. Reliab. R-29, 290 294 (1980). 2. H. Mine and H. Kawai, An optimal maintenance policy for a 2-unit paralleled system with degraded states, IEEE Trans. Reliab. R-23, 81 85 (1974). 3. A. Kumar, Profit evaluation in some repairable redundant systems, Z. angew. Math. Mech. 57, 485-489 (1977). 4. R. A. Howard, Dynamic Probabilistic Systems, Vol. II. John Wiley, New York (1971).
0026-2714/8353.00+ .00 © 1983PergamonPressLtd.
Printedin GreatBritain.
ON OPTIMAL MAINTENANCE OF A SERIES SYSTEM ASHOK KUMAR and V. B. KAPOOR Defence Science Centre, Metcalfe House, Delhi-110054, India and M. C. GUPTA 10/2 Shakti Nagar, Delhi-110007, India
(Received for publication 2 February 1983) Abstract--Maintenance of an n-unit series system with general distributions for failure-time and repair times has been discussed. The system has been identified by up and down states. It is assumed that there are a finite number of alternatives to a decision maker in each state. The system earns at a fixed rate (possibly negative) in each state and fixed transition costs are incurred at the time of transitions. The problem is to choose a vector of alternatives (one alternate in each state) so that the overall expected profit is maximized. A procedure to determine optimal maintenance policy is developed based on Howard's Policy Iteration scheme. Numerical examples are included to illustrate the results.
INTRODUCTION
Define the following system states to identify the system at any instant.
The problems of the maintenance of equipments/ systems is quite important to modern business and industrial houses. The system designer/analyst is often confronted with the problem of running the system in the most efficient manner. Profit appears to be a more appropriate parameter in evaluation studies of systems as compared to other parameters, such as reliability, availability, M T S F , etc. Many authors have concentrated their attention on analyzing their models under different conditions with a view to obtaining these parameters. A detailed list of references is contained in K u m a r and Agarwal [1]. Mine and Kawai I-2] discussed the maintenance policies for a 2-unit parallel redundant system which maximizes expected profit for an infinite time span. K u m a r [3] obtained an analytic expression for expected profit in a standby system under different operating environments. There may be several alternatives available to the systems analyst for maintaining the system in up and down states. Different types of failure demand different types of repair. A maintenance policy is a set of alternatives (one alternative from each state). The optimal maintenance policy ( O M P ) is one which maximises expected profit of the system. The present paper attempts to obtain optimal maintenance policy for an n-unit series system. Howard's Policy Iteration method [4] has been employed to achieve the results.
So all the n units (components) are operating. S~ the system is in the failed state due to the failure of the ith unit (component) and is under repair. (i = 1,2 .... ,n). The system is up in S O and it is down in S~. Let there be several alternatives available in each state to maintain the system. F o r instance, when the system is in So, there may be ordinary maintenance (O,.), costly maintenance (Cm) or very costly maintenance (VC,~). Whichever maintenance scheme is followed, the system will fail eventually. However, the mean time to failure increases as we go from O,, to C,, and from C~" to VCm. Similarly, when the system is in S~ [failed and ith unit (component) is under repair] there may be several alternatives for repair: a costly repair may ensure a faster repair. Symbolically, let there be N alternatives in each state (N varies from state to state). Let us define a maintenance policy as the set of maintenance alternatives--one from each state. An optimal maintenance policy is the one which maximizes the expected profit of the system. Let us assume the following cost structure for the system: yk earning rate of the system in S O when the kth maintenance alternative is adopted. y~ loosing rate of the system in S~ when the kth repair alternative is adopted (i = 1, 2 . . . . . n) yik fixed transition reward (cost) for making a transition from S~ to Sj and alternative k is chosen in S~.
SYSTEM MODEL AND THE COST STRUCTURE Let there be an n-unit (component) series system with units (components) labelled 1, 2 . . . . . n. The system fails if any one of the units (components) fails. Let the failure time p.d.f, of the ith unit (component) be f~(t). The system can be repaired and let the repair time p.d.f, of the ith unit (component) be 9~(t).
OPTIMIZATION PROBLEM
Under the above cost structure, it is proposed 827
828
ASHOK KUMAR,
V. B. KAPOORand M. C. GUPTA
to determine the optimal maintenance policy for the n-unit series system which maximizes the steady state profit of the system. We develop, below, the solution procedure based on Howard's Policy Iteration method. We deal with two cases.
Case 1: when payments to be received in future are not discounted. Case 2: when a discounting factor is also taken into account.
Following Howard [4], we know that for large t
vi(t ) = vi+gt where vi(t) is the total expected earning of the system in time t, if the system starts in S~ at time t = 0. v~is the transient part of the profit and g is total expected earning of the system in steady state. Also
i = 1,2,...,n
where qiPi "~ 2 Pijrij + Yi~i • J
Here #~ is the mean unconditional waiting time in S~ and pij is the one-step transition probability for making a transition from S i to Sj. Following Howard's [4] Policy Iteration method, which is given below, we determine the optimal policy numerically after assigning appropriate values to the parameters involved.
Policy Evaluation For the present policy solve ViWgPi = qipi-k-2pijvj J
i = 1,2,...,n
with v, = 0, for g and relative values V l , / ) 2 , . . . , 1)n- 1
Policy Improvement Find the alternative k in each state i that maximizes k 1[k 7 qi + 7 X / E P i f J - vii ~i Lj
Step 1: define the system in terms of its states and transitions between them. Specify the cost structure and alternatives in each state. Step 2: determine the transition probabilities and mean unconditional waiting times under each alternative. Step 3: Select one alternative in each state for which 1 ~
k k
k
qgi = ~ 2.,PijrU + Yi i.t~ j
Case 1: undiscounted cost structure
Viq-gfli -~ q i l z i + 2 p i j v j J
In fact, the above procedure involves following steps to determine the optimal maintenance policy.
d
using the relative values vi of the previous policy. Make this alternative the new decision in state i. Repeat for all states to find the new policy.
is maximum. Name it the initial policy. Step 4: Solve the following set of n equations for the policy obtained at step 3. i = 1,2 . . . . . n
Vi+gPi = qi#i+~pqvj J
for v~, vz,..., v,_ a by setting v, = 0. Step 5: By using values of v x, v2 . . . . . v,_ t and g obtained at step 4, compute the test quantity k
1I-
k
7
for each alternative in each state. Select the alternative in each state for which the test quality is maximum. Step 6: Examine whether the new policy is different from the initial policy. If yes go to step 4, otherwise the optimal policy is obtained. In what follows we shall illustrate the results by a numerical example. Let us consider a system with two types of failure, Type 1 and Type 2. Thus, we have the following system states, S O system is operative $1 system is failed and is under repair due to Type 1 failure. S z system is failed due to Type 2 failure and is under repair. Let the failure time distribution be exponential with rate 2 and the probability that the failure is of Type 1 or 2 be pl or P2. Let the repair-time p.d.f, of Type 1 and 2 be 91(0 and 92(0, respectively. In fact, the system may be thought of as composed of 2-units operating in series with failure time distributions exponential with rate 21 and 22, such that 2 = 21 +22 and px = 21/2 and P2 = 22/2. Let repair-time p.d.f. gi(t)(i = 1, 2) be given by 91(t) = r/2te -~' g2(t) = ½r/at2e -~t
withmean with mean
ml = 2/r/ m2 =
3p/.
Let there be two alternatives in So: (i) 0,, ordinary maintenance; (ii) C,, costly maintenance. Similarly let there be two alternatives in S~: (i) 0 r ordinary repair; (ii) Mr minor repair. State $2 has three alternatives: (i) Or ordinary repair; (ii) M, minor repair; (iii) C, costly repair.
Optimal maintenance k
~2
¢tk
--
- l0
- 15
10.000
248.800
0.6
--
- 10
-15
20.000
199.350
--
--
- 20
--
--
5.000
- 24.000
1
--
--
-15
--
--
3.333
54.500
-50)
1
-
-
-20
-
-
150.000
-50.133
-75)
1
' --
--
-15
--
--
60.000
-75.250
0.10, - 150)
1
--
--
- 10
--
--
30.000
-150.333
States
Alternatives
So
10~,
(0.10,
-,
250)
--
0.6
0.4
2 Cm
(0.05,
-,
200)
--
0.4
1 O,
(-,
0.4,
- 20)
1
2 M,
(-,
0.6,
-50)
1 Or
(-,
0.02,
2 M,
(-,
0.05,
3 C,
(-,
S1
S2
Various" state alternative combinations and their resultant values have been computed and presented above. T h u s , we find t h a t t h e first a l t e r n a t i v e in e a c h s t a t e h a s t h e m a x i m u m value, s o t h e initial p o l i c y for t h e p o l i c y i t e r a t i o n p r o c e s s is (1 1 1). N o w w i t h this p o l i c y we s o l v e t h e set o f e q u a t i o n s 1)i .+ 9 p k = q- k~. i k "4- ~'~ _k 1)j /,pij
829
with
a g m n we f o r m v a r i o u s and g and compute
v t = 5160 9 = 30.8 with I)2 = 0 .
N o w w i t h t h e s e v a l u e s of vi's we a g a i n c a l c u l a t e t h e test q u a n t i t i e s , w h i c h a r e g i v e n below.
which implies that
State
Alternatives
So
1 2
15.0 30.85
S1
1 2
30.8 27.7
$2
1 2 3
- 13.87 15.32 31.0
0.6Vl - vo - 109 = - 2488.00 v~ - v o + 59 = - 120.0 - v o + 1509 = - 7519.95. S o l v i n g t h e s e e q u a t i o n s we g e t (with v 2 = 0), v o = 6303.45 v 1 = 6242.00 9 = -8.11. W i t h t h e s e v a l u e s o f v~'s we c a l c u l a t e t h e test q u a l i t y k
1V
e q u a t i o n s i n v o l v i n g Vo, Vl, v2
v o = 5434
vz = 0
J
qk
k
-1
Test quantity
F r o m t h e a b o v e , we find t h a t t h e p o l i c y c o m e s o u t to be (2 1 3), w h i c h is t h e s a m e as t h e earlier o n e a n d so t h i s is t h e o p t i m a l policy a n d t h e g a i n for t h e s y s t e m is 30.8 u n i t s of m o n e y .
Case 2: discounted cost structure for all t h e a l t e r n a t i v e s in e a c h state. T h e v a l u e s o f t h e test q u a n t i t i e s a r e g i v e n below. State SO
S1
S2
Alternatives
F o l l o w i n g H o w a r d [4], we k n o w t h a t for l a r g e t a n d d i s c o u n t i n g f a c t o r ct > 0 1
vi = ~ pijrijh*(a) + Yi- (1 - h* (a) + ~ pijh*(a)vj j
Test quantity
1
- 8.11
2
8.66
1 2
- 8.11 -30.66
1 2 3
-8.11 29.81 59.98
F r o m t h i s we find t h a t t h e 2 n d a l t e r n a t i v e in S o, t h e 1st a l t e r n a t i v e in S 1 a n d t h e 3rd a l t e r n a t i v e in $2 h a v e t h e m a x i m u m value, t h e r e f o r e t h e i m p r o v e d p o l i c y is w i t h t h e s e a l t e r n a t i v e s (2 1 3). W i t h t h i s p o l i c y
Ct
j
w h e r e v~ is t h e e x p e c t e d p r e s e n t v a l u e o f infinite t i m e o p e r a t i o n in s t a t e i, hij(t) is t h e h o l d i n g t i m e p.d.f, o f t h e s y s t e m in S i before m a k i n g a t r a n s i t i o n to S j,
hi(t) = ~ pijhij J and
Re (s) > O.
f*(s) = f o e-~'f(t)dt Let
1
Pi(°O = ~ Pijh~(ct)rij+ Yi- (i - h~(a)) j
O~
ASHOKKUMAR,V. B. KAPOORand M. C. GUPTA
830
Step 4: Solve the following set of n equations for the policy obtained at step 3.
then
vi = Pi(e) + Y', Pijh*(cOv~• J
vi = Pi(cQ+~pijh*(e)vj
Step 5: By using values of v~'s obtained at step 4, c o m p u t e the test quantity. Pi(~) + ~, Pijh*(o;)vj J
Policy Evaluation
for each alternative in each state. Select the alternative in each state for which the test quantity is maximum. Step 6: Examine whether the new policy is different from the initial policy. If yes go to step 3, otherwise the optimal policy is obtained.
F o r the present policy solve
v i = Pi(~) + ~,pljh~(oOvj J
i = 1,2 . . . . . n
/
Following H o w a r d ' s [4] Policy Iteration method, which is given below, we determine the optimal policy numerically after assigning appropriate values to the parameters involved.
i= 1,2,...,n
for
In what follows we shall illustrate the results by a numerical example. Let us consider a series system. The system failure can be thought of as failure due to unit i or 2, as in the previous example. The life time p.d.f, of the system has been assumed to be Erlangian. The repair time p.d.f, of Type 1 follows an Erlangian distribution and that of Type 2 follows exponential distribution. Let the discounting factor a = 0.4. Thus we have
V l ~ V 2 ~ • . . ~ Vn
Policy Improvement Find the alternative in each state that maximizes
P~(~) + y' p~jh*k(o~)vj J
h~j(~)
using the present values v~ of the previous policy. M a k e this alternative the new decision in state i. Repeat for all states to find the new policy.
Poj 22, + a /
h*j(o:):l( rl" ~ P~j \ ~ + ~ /
r=5
and The above procedure involves following steps to determine the optimal maintenance policy. In this illustration also, there are the same set of alternatives in each state as in the previous example. Various state alternative combinations and their resultant values have been c o m p u t e d and are presented below.
Step 1: Define the system in terms of its states and transitions between them. Specify the cost structures, discounting factor and alternatives in each state. Step2: D e t e r m i n e the transition probabilities, holding time p.d.f.'s and their Laplace transforms evaluated at ~. Step3: Select one alternative in each state for which
Thus we find that the first alternative in each state has the m a x i m u m value, therefore the initial policy for the policy iteration process is (1 1 1). N o w with this policy we form the equation
1
Pi(~) = 2 Pijrijh*(ct) +Yi- (1 - h*(c~)) j
O~
v i = Pi(cO+ ~, pijh*(°Ovi J
is maximum. N a m e it initial policy.
State
Alternatives
p~ko
P~I
p~2 (~,
So
1 0 m
--
0.6
0.4 (O.lO, - ,
250)
2 C.,
--
0.4
0.6 (0.05, - ,
1 O,
1
--
--
(-,
2 Mr
1
1 Or
1
2 M,
1
3 Cr
1
S1
S2
n,
y~)
k r,o
r~,
#2
hZ~(~) h~?(~) h~k(~) h?k(~)
--
-I0
-15
200) --
-10
-15
0.4,
-20) -20
--
--
0.4016
--
--
0.04016 - 37.9520
- 5 0 ) - 15
--
--
0.5345
-
-
0.5345 -67.2050
--
P~(~)
0 . 1 8 5 0.2775 0.2220
483.4750
0.100 0.0667 0.0800
451.0000
-
-
--
(-,
0.6,
-
-
--
(-,
0.02, -50)
-20
--
--
0.4760
--
--
0.4760 -120.0200
--
--
(-,
0.05, - 7 5 ) - 1 5
--
--
0.1111
--
--
0.1111 -168.3352
--
--
(-,
0.10,-150) - 1 0
-
-
0.2000
--
--
0.2000 -302.0000
Optimal maintenance which implies vo = 483.475+0.111v a +0.111vz
831
Again, with these values of vi's we calculate test quantities which are given below.
/)1 = - 37.952 + 0.4016v o
State
Alternatives
v2 = - 120.02 + 0.476v o.
SO
1 2
494.6537 463.0283
S1
1 2 1 2 3
160.7009 197.1874 -96.4745 -113.3799 - 203.0706
Test quantity
Solving the above set of equations we get /)o = 485.2532 S2
v 1 = 156.9256 v2 = -96.9220. N o w with these values of vi we calculate the test quantities pijhij (o~)l)j J for each alternative in every state. The values of test quantities are given below. State
Alternatives
SO
1 2
Test quantity 490.1354 461.4002
$1
1 2
156.9256 192.1628
$2
1 2 3
-96.9220 - 114.4236 -200.9494
F r o m the above we see that the first alternative, in S O and $2 and the 2nd alternative in $1 form the new policy, i.e. our new policy is (1 2 1). With this policy again we form the various equations involving vi's and calculate vo = 494.6472 vl = 197.1839 v z = -96.4748.
M.R 23/-% E
F r o m this we find that the resultant policy comes out to be (1 2 1), which coincides with the earlier policy, which evidently is the optimal operating policy in the present case. Following this policy we ensure m a x i m u m gain for the system. Acknowledgements--The authors are grateful to Dr. A. K. Sreedhar, Director, Defence Science Centre, Delhi, for granting permission to present/publish the paper at a Seminar "Reliability Engineering and Management Applications" at the Military College, EME, Secundrabad. Thanks are also due to Dr. D. Ray, Scientist "E", for the help in preparing the paper.
REFERENCES
1. A. Kumar and M. Aggarwal, A review of standby redundant systems, IEEE Trans. Reliab. R-29, 290 294 (1980). 2. H. Mine and H. Kawai, An optimal maintenance policy for a 2-unit paralleled system with degraded states, IEEE Trans. Reliab. R-23, 81 85 (1974). 3. A. Kumar, Profit evaluation in some repairable redundant systems, Z. angew. Math. Mech. 57, 485-489 (1977). 4. R. A. Howard, Dynamic Probabilistic Systems, Vol. II. John Wiley, New York (1971).