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Cost analysis of a 2-unit standby redundant electronic system with critical human errors
Microelectron. Reliab., Vol. 26, No. 5, pp. 841-846, 1986.
0026-2714/8653.00 + .00 Pergamon Journals Ltd.
Printed in Great Britain.
COST ANALYSIS OF A 2 - U N I T S T A N D B Y R E D U N D A N T E L E C T R O N I C SYSTEM WITH CRITICAL H U M A N ERRORS P. P. GUPTA,R. K. GUPTAand R. K. SHARMA Reliability Group, Mathematics Department, M.M. (P.G.) College, Modinagar (U.P.), India
(Receivedfor publication 17 December 1985) Abstract--In this paper, investigations have been carried out for the evaluation of availability and expected profit during the operable stage of a standby redundant, electronic system, incorporating the concept of human failure. The system can be in any of the three states: good, degraded and failed. One repair facility is available for the repair of a unit in failed or degraded state. The system cannot be repaired when it fails due to critical human errors. The repair of the system in any state follows general distribution. To make the system more applicable to practical lifeproblems, time dependent probabilities have been evaluated so as to forecast the expected profit and the operational availability of the system at any time.
INTRODUCTION
Many authors [1, 2] have evaluated the cost analysis of various complex systems, but so far, very few researchers have attempted the problem of evaluating the availability and the cost analysis of a 2-unit standby redundant system under critical human errors. According to Meister [3], 20-30% of system failures are due to human errors which may be due to (a) misinterpretation of the system, (b) wrong action and (c) maintenance errors. Recently Dhillon I-4] considered the reliability of a system with critical human errors, under the assumption that repair rate follows exponential distribution. However, in practical life models repair need not always follow exponential distribution. Keeping the above facts in mind, the authors in this paper have, therefore, considered a 2-unit standby redundant electronic system which may be in any of three states: good, degraded and failed. The failure of the system due to human errors is also considered in two good states of the system.
(ii) the system is in failed state when both units fail completely. (5) At an instant, only one change can take place in the state of the system. (6) A unit can fail in the degraded state too, with different failure rate than that in the good state of the system. (7) The system has one repair facility which cannot be availed in the degraded state D2. (8) The system cannot be repaired when it has failed due to critical human errors. (9) Repairs are to be like new and never damages anything. (10) The failure and repair times for the system follow exponential and general distributions, respectively.
NOTATION
(1)
ASSUMPTIONS
(1) The system has two identical units connected in standby redundancy. (2) (2) Initially the system is in good state. (3) A unit of the system can fail partially or completely. The system is said to be in degraded state if it works with a partially failed unit. (4) The system has three states: good, degraded and (3) failed. (a) The system is in good state when it operates with a good unit, the other standby unit may be good or failed. (4) (b) The system is in degraded state when it operates with a degraded unit, the other standby unit may be good or failed. (c) (i) The system can fail due to critical human errors in any of the two good states of the system or 841
G1, G 2 good states of the system
G1 system operating with one unit which is good, the standby unit is also good G2 system operating with the standby unit which is good, the other unit has failed and is under repair D I , D 2 degraded states of the system D1 system operating with a unit which is degraded, the other unit is good D2 system operating with a standby unit which is degraded, the other unit is failed and is under repair F, F1 failedstates of the system F failed state of the system due to critical human error in good state G1 or G2 F 1 failedstate of the system when both units (main and standby) fail 2, 2D, 2'D constant failure rates of a unit 2 failure rate of a good unit to failed state 2D failure rate of a good unit to degraded state 2'D failure rate of a degraded unit to failed state
P. P. GUPTA,et el.
842
F kh,~L____ PF (t) h
~-~
t
hh
)
,4
( "G,"II \
o,
[
/
,.,
J
4~o]y )
_
t
'
J
Fig. 1. Possible states and transitions.
(5)
(6)
2hx,2h 2 constant failure rates of the system due to Si(r), ~)~r) i=G2,r=x i = D2, r = y i = Fl,r = z
(7)
Pj(t)
critical human errors in good state G1 and G2 p.d.f, and hazard rate for repair time of the system repair of the system in good state G2, repair is completed in elapsed repair time x repair of the system in degraded state D2, repair is completed in elapsed repair time y repair of the system in failed state Ft, repair is completed in elapsed repair time z Pr{system in statej at time t}
j = G1, G2, D~,Dz, F1,F
(8)
{fo
Si(r) = ~bi(r)-exp -
$i(r)dr
2'D + dpD2(y) Po~(Y, t) = [ ~ t + &-~+ ~)r,(z)]Pr,(z,t,=O, 8
P r(t) = PG,(t)2hl + PG~(t)2h2.
(4) (5) (6)
BOUNDARY CONDITIONS
PG~(O, t)
= 2PG,(t) +
2'DPD~(t),
(7)
Pp.(O, t) = 2DPG~(t),
(8)
Pr,(O, t) = 2Pad(t) + 2'DPD~(t)
(9)
INITIAL CONDITIONS
i = G 2, D 2, F 1.
Pi(O) =
i # GI.
(1o)
DIFFERENCE DIFFERENTIAL EQUATIONS
SOLUTION OF THE PROBLEM
By elementary probability and continuity arguments, the difference-differential equations for the Stochastic process, which is continuous in time and discrete in space, are given by
Taking Laplace transform of the equations (1)-(9) and using initial condition (10), we get
[_s+ 2 + 2D+ 2hl]P~(s) = 1+
[~+2+2D+2hl]PG,(t)
+
+
o
PF,(Z, t)dPF~(z)dz,
(1)
PG2(X,s)d?G2(X)dx
o
Pv,(z, s)4oF,(z)dz,
(11)
S + ~ x + 2 + 2D + 2h2 + d)G2(X) l~a2(X,S ) = 0,(12) Is + 2'D] PD, (s) = 2D. PG, (S) +
+ 2 ' D Po,(t) = 2D'Po,(t)+
do
Po2(y,s)ckD2(y)dy,
(13)
PD~(y,t)d#o~(y)dy,
(3)
[S+~y+2'D+ckD~(y)]PD~(y,s)=O,
(14)
Cost analysis
I
s+
+
= o,
/15)
843
where
Al(s)
SPF(S) = 2ha" PGI(s) + 2h2' P~,(s),
(16)
PG,(O,s) = 2PG~(s)+A'D'PD~(S),
(17)
Pp,(O, s) = 2D" PG,(s),
(18)
PF~(O,s) = 2" PG2(s)+ 2'D" Po,(s).
~(19)
Integrating (12), (14) and (15) and using (17), (18) and (19) respectively, we get PG~(x, s) = [ 2 P ~ (s) + 2'OPo~ (s)] x exp [ - (s + 2 + 2D + 2h2)x ]
x exp(- f: ~,~,(x)dx),
2D[1 - S-6,(s + 2 + 2D + ).h2)] [1 - go,(s + 2'D) (28) (s + 2 + 2D + ).h2)(s + J.'D) 2 Therefore
Po~(s) = A2(s)" PG~(S),
(29)
where Az(S) -~
2D + 2"(s + 2'D)" AI(s)
(30)
(s + 2'D) [ 1 - ).'DAI(S)]'
From equations (25) and (29), we get
PG2(s) = A3(s)'PG,(s), (20)
(31)
where A3(S)
Po, (Y, s) = 2DP62 (s)' exp [ - (s + 2'D)y]
[2 + 2'D' A2(s)] • [1 - g6~(s + 2 + 2D + ,2.h2)]
(s + J. + AD + ).h2) From equations (24) and (30), we get
Pv~(z, s) = [2" P~,(s) + 2'DPo~(s)]
x exp(-sz).exp(-f:4,~(z)dz). (22) Substituting from (20) and (22) in equation (11), on simplification we get (s + 2 + 2D + 2h 1)/~o1(s) = 1 + [2" POl(S) + 2'D" enl (S)]
Po~(s) = A4(s)PG~(s), A4(s) = 2D" A3(s)" [1-So2(s+2'D)]/(s+A'O).
fo
A(s) = (s + A + AD + 2h1) - [ 2 + 2'D" Az(S)] • ~'~(s + 2 + 2 D +
(s + 2 + 2D + 2h2)
PFI(S) = As(s)/[s" A(s)], (25)
PF~(S) = Ptq(z, s)dz
(39)
EVALUATION OF LAPLACE TRANSFORM OF UP AND DOWN STATE PROBABILITIES
(26)
The Laplace transform of the probability of operational availability and non-availability at time t are given as under
Using equation (21) in equation (13), we get (s + 2'O)Pn,(s) = 20- PG~(s)+ 2DPG,(s) x [1-~o~(s+A'D)/(s+2'D).
(38)
where As(s ) = [-2. A3(s ) + 2'D. a4(s)] [1 - gFI(S)].
and
= [2P~,(s)+2'D'Po,(s)]['I-~v~(s) 1.
(37)
Also, using equations (31) and (33) in equation (26), we get
= [2.P~,(s)+2'O.Po,(s)] [ 1 - SG~(s + 2 + 2D + 2h2) ]
(36)
Using equations (31) and (35) in equation (16), we get
PF(S) = [2hi + 2h2" A3(s)]/[s" A(s)].
PG~(x, s) dx
x
J-h2)
- [2" A3(s) + 2'D" Aa(s)] ' gF,(s). (24)
Using equation (20), on simplification we get
Po~(s) =
(35)
where
Po~(y, s)dy = 2D. PG~(s) x [1 - gn=(s + 2'O)]/(s + 2'0).
(34)
Using equations (29), (31) and (33) in equation (23), we get
Pc~(s)= 1/A(s),
(23)
Using equation (21), on simplification we get Po~(s) =
(33)
where
x S6~(s + 2 + 2D + ,~h2) + [2Jffo2(s)
+2'O'-Po,(s)'gv,(s).
(32)
Pup(S) = P61(s) + P6~(s) + Po~(s) + Po~(s)
(27)
Substituting for P6,(s) from equation (25) in (27), on simplification we get
Po,(s) = 2D" P6~(s)/(s + 2'D) + [~.P~,(s) + 2'0" PD~(S)] "Z l(S),
= [ 1 + A2(s) + A3(s) + A4(s)]/A(s)
(40)
and Pdown(S) = -PFI(S)+ Pr(s) = I-2hl + 2h2" A3(s) + A5(s)'I/[s" A(s)']. (41)
P.P. GUPTA,et el.
844 ERGODIC BEHAVIOUR OF THE SYSTEM
Using the corollary of Abel's theorem lim s. F(s) = lira. F(t) = F (say). .~0
t~0
Pr,(s) = As(s)/[sA(s)],
(47)
PF(S) = [2hl +/ih2" A3(s)]/[s" A(s)],
(48)
Pup(S) = [1 + A2(s)+ A3(s)+ A4(s)]/A(s)
(49)
Provided the limit on the right hand side exists, the time independent probabilities are obtained as follows
and
Pup = lim- [s'Pup(S)]
A i(s) = 2o/n(s),
s~0
Pdown(S) = [2hl +/ih2" A3(s) + As(s)]/[s" A(s)],
A2(s ) = 2D" [B(s) +/i'(s + 2'O)]/[(s + 2'O)Bl(s)],
= lim "s" [1 +A2(s)+Az(s) s~0
(51)
A3(S) = (s + ~I'D + (ao2)B2(s)/Bl(S),
+ a4(s)]/A(s) = 0.
(42) A4(s)
Because A(s) -7s 0 and Pdow, = 1, Pup = 1.
2D" B2(s) - - , B1(s)
(52)
A 5(s) = s[2" (s + 2'D + ~D2) + 2'D"/ID]
PARTICULAR CASE
× B:(s)/[(s +
When repair follows exponential time distribution
4~F,)'BI(S)], (53)
A(s) = (s +/i + liD +/ihl) - [ / i + / I ' D ' A2(s)]
Setting
Si(s)=
(50)
where
(~i
-
s +dpi
for
i=G2,D2
and
× ckcj(s+2+/iO+/ih2+cka2)
FI,
-- I/i" A3(s) + / I ' D " A4(S)] • ff)F,/(S
we get
-~- ~)rt),
(54)
B(s) = (s +/i +/ID + 2h2 + dpa~) PG,(S) = l/a(s),
(43)
P~(s) = A3(s)/A(s),
(44)
Pp,(s) = Az(s)/a(s),
(45)
PD2(S) = A4(s)/A(s),
(46)
x(s+/i'D+OD)(s+/i'D),
(55)
n ( s ) - 2'0" 20
(56)
Bz(S) = 2. (s + 2'D) + 2'D. 2D.
(57)
Bl(s) =
and
EVALUATION OF INVERSE LAPLACE TRANSFORM OF Pap(S) AND Pdon(S)
Setting ~bo2 = ~ba= = 0.8, ~br, = 0.6, 2D = 2'D = 0.05, 2 = 0.04, 2hl = 2h: = 0.2 in equations (43)-(57) and simplifying, we get (s + 0.6)(s + 1.09)(s 4 + 2.13s 3 + 1.2650s 2 + 0.154275s + 0.0046294) Pup(S) =
(s + 0.026)(s + 0.064)(s + 0.29)(s + 0.446)(s + 1.309) x
1
[ s - ( - O . 9 4 2 5 + O.2881679i)] " [ s - ( - 0 . 9 4 2 5 - 0 . 2 8 8 1 6 7 9 i )
(58)
and
PDI(s) =
(0.05)(s + 0.6)(s + 1.09)(s 3 + 1.99s 2 + 1.0635s + 0.048325) " {(s+O.O26)(s+O.O64)(s+O.29)(s+O.446)(s+ 1.309)
(59)
x [s-(-O.9425+O.2881679i)].[s-(-0.9425-0.2881679i)]} Taking the inverse Laplace transform of equations (58) and (59), we get Pup(t) = 0.1728679 exp ( - 0.260 + 0.0959487 exp ( - 0.0640 + 1.0979981 e x p ( - 0.290-0.5301422 exp ( - 0 . 4 4 6 0 + 0.0669847 exp ( - 1.309t) + 2 exp ( - 0.9425t) × [0.0481714 cos (0.28816790 + 0.1168938 sin (0.28816790]
(60)
and
Pol(t) = O.1316093 exp ( - 0.0260 + 0.0934836 exp ( - 0.0640 - 0.3003391 exp ( - 0.29t) + 0.0827971 exp ( - 0.4460 - 0.0044971 exp ( - 1 . 3 0 9 0 - 2 exp ( - 0.94250 × [0.0026614 cos (0.28816790 + 0.00309 sin (0.2881679t)], where, of course, Pup(t)+ Pdown(t) = 1.
(61)
Cost analysis
845
COST FUNCTION ANALYSIS
Table 2
OF THE SYSTEM
Expected profit G(t) The s-expected up time of the system during (0, t] is
S1. no.
E(t) = ~o P.p(t)dt and the s-expected busy period of the service facility in (0, t] is #8(0 =
Po,(t).
--
Hence, the expected net gain function is given by
G(t) =
expected total revenue function in (0, t] -
Time t
1
0
2 3 4 5 6 7 8 9 10 ll
1 2 3 4 5 6 7 8 9 I0
C 2 = 0.5
C 2 = 0.10
C 2 - 0.05
0.0000000 0.4352562 0.7053963 0.8460866 0.9470489 0.9605711 0.9193612 0.8327057 0.7081851 0.5522155 0.3697642
0.0000000 0.8273402 1.4747398 1.9989349 2.4418053 2.8087030 3.1171644 3.3778506 3.5994313 3.7890003 3.9522488
0.0000000 0.8763507 1.5709077 2.3240476 2.7286498 3.0397195 3.3918898 3.6959937 3.9608371 4.1935984 4.4000594
expected cost due to repair in (0, t] repair cost per unit time (spent in the service facility), respectively.
= C 1 [ - 6.6487653 exp (--0.0260 -
NUMERICAL COMPUTATION
1.4991984 exp ( -- 0.064t)
Availability analysis
- 3.7862003 exp (-- 0.290 + 1.1886596exp(-0.4460
Setting t = 0, 1,2 .... ,10 in equation (60), we get Table 1.
- 0.0511724 exp ( - 1.309t)
Cost analysis
+ 10.9595146 - exp ( - 0.94250
Setting C~ = 1, t = 0, 1,2,..., 10, in equation (62) the net gain for C2 = 0.05, 0.1, 0.5 are shown in Table 2.
x {0.1628388 cos (0.2881679t) + O.1982627 sin (0.28816790}]
-- C2[t-
5.6621355 + 5.0618961
INTERPRETATION
x exp ( - 0.026t) + 1.4606812 exp ( - 0.0640
OF THE RESULTS
Table 1 computes the availability of the system at an instant t and Fig. 2 shows the availability vs. time. The availability of the system decreases very slowly and over a long time showing that the system remains available for a long period of time. Table 2 computes the expected profit during the
- 1.035652 exp ( - 0.290 + 0.1856437 exp ( - 0.446t) - 0.0034355 exp ( - 1.3090 - exp (-- 0.94250 x {0.006998 cos (0.28816790 + 0.0044172 sin (0.28816790}],
1.00~
(62)
0.900
where Cl and C2 are the revenue per unit up time and
A
Table I.
I~
°o=\
X
S1. no.
Time t
1
0
2 3 4 5 6 7 8 9 10 ll
1 2 3 4 5 6 7 8 9 l0
Pup(t) 1.0000000 0.7967283 0.6658796 0.5672142 0.4879072 0.4230221 0.3698577 0.3263895 0.2909030 0.2619238 0.2381771
Pdow.(t) 0.0000000 0.2032717 0.3341204 0.4327858 0.5120928 0.5769779 0.6301423 0.6736105 0.7090970 0.7380762 0.7618229
0.600 ,.Q
<
\x
0.500
\x
0.400 0.300 0.2001 I 0 1
\x
I
I
I
J
I
I
h
I
I
2
3
4
5
6
7
8
9
10
Time Fig. 2. Availability vs. time.
846
P.P. GUPTA,et al.
5.000
C2 = 0.05 / x ~
interval (0, t] for the fixed value of the revenue per unit time. A critical examination of the graph (Fig. 3), expected profit vs. time, reveals that the expected profit increases for a long time when C 2 ~< 0.1.
4.000
Acknowledgements--R. K. Gupta and R. K. Sharma are highly grateful to C.S.I.R., New Delhi, for granting and providing financial assistance to the Research Project, under which the work was initiated and carried out.
A v
3.000 "0
REFERENCES
2.000 X uJ
1.000
0.000 0
I
I
I
i
I
I
I
I
J
1
2
3
4
5
6
7
8
9
Time
Fig. 3. Expected profit vs. time.
10
1. P. R. Parthasarathy, Cost analysis for 2-unit system, IEEE Trans. Reliab. R-28, 268-269 (1979). 2. P. P. Gupta and S. C. Agarwal, Cost-function analysis of a 3-state repairable system, Microelectron. Reliab. 24, 51-53 (1984). 3. D. Meister, The problem of human-initiated failure, Eighth National Symposium on Reliability and Quantity Control (1962). 4. B. S. Dhillon and R. B. Mishra, Reliability evaluation of system with critical human error, Microelectron. Reliab. 24, 743-756 (1984).
0026-2714/8653.00 + .00 Pergamon Journals Ltd.
Printed in Great Britain.
COST ANALYSIS OF A 2 - U N I T S T A N D B Y R E D U N D A N T E L E C T R O N I C SYSTEM WITH CRITICAL H U M A N ERRORS P. P. GUPTA,R. K. GUPTAand R. K. SHARMA Reliability Group, Mathematics Department, M.M. (P.G.) College, Modinagar (U.P.), India
(Receivedfor publication 17 December 1985) Abstract--In this paper, investigations have been carried out for the evaluation of availability and expected profit during the operable stage of a standby redundant, electronic system, incorporating the concept of human failure. The system can be in any of the three states: good, degraded and failed. One repair facility is available for the repair of a unit in failed or degraded state. The system cannot be repaired when it fails due to critical human errors. The repair of the system in any state follows general distribution. To make the system more applicable to practical lifeproblems, time dependent probabilities have been evaluated so as to forecast the expected profit and the operational availability of the system at any time.
INTRODUCTION
Many authors [1, 2] have evaluated the cost analysis of various complex systems, but so far, very few researchers have attempted the problem of evaluating the availability and the cost analysis of a 2-unit standby redundant system under critical human errors. According to Meister [3], 20-30% of system failures are due to human errors which may be due to (a) misinterpretation of the system, (b) wrong action and (c) maintenance errors. Recently Dhillon I-4] considered the reliability of a system with critical human errors, under the assumption that repair rate follows exponential distribution. However, in practical life models repair need not always follow exponential distribution. Keeping the above facts in mind, the authors in this paper have, therefore, considered a 2-unit standby redundant electronic system which may be in any of three states: good, degraded and failed. The failure of the system due to human errors is also considered in two good states of the system.
(ii) the system is in failed state when both units fail completely. (5) At an instant, only one change can take place in the state of the system. (6) A unit can fail in the degraded state too, with different failure rate than that in the good state of the system. (7) The system has one repair facility which cannot be availed in the degraded state D2. (8) The system cannot be repaired when it has failed due to critical human errors. (9) Repairs are to be like new and never damages anything. (10) The failure and repair times for the system follow exponential and general distributions, respectively.
NOTATION
(1)
ASSUMPTIONS
(1) The system has two identical units connected in standby redundancy. (2) (2) Initially the system is in good state. (3) A unit of the system can fail partially or completely. The system is said to be in degraded state if it works with a partially failed unit. (4) The system has three states: good, degraded and (3) failed. (a) The system is in good state when it operates with a good unit, the other standby unit may be good or failed. (4) (b) The system is in degraded state when it operates with a degraded unit, the other standby unit may be good or failed. (c) (i) The system can fail due to critical human errors in any of the two good states of the system or 841
G1, G 2 good states of the system
G1 system operating with one unit which is good, the standby unit is also good G2 system operating with the standby unit which is good, the other unit has failed and is under repair D I , D 2 degraded states of the system D1 system operating with a unit which is degraded, the other unit is good D2 system operating with a standby unit which is degraded, the other unit is failed and is under repair F, F1 failedstates of the system F failed state of the system due to critical human error in good state G1 or G2 F 1 failedstate of the system when both units (main and standby) fail 2, 2D, 2'D constant failure rates of a unit 2 failure rate of a good unit to failed state 2D failure rate of a good unit to degraded state 2'D failure rate of a degraded unit to failed state
P. P. GUPTA,et el.
842
F kh,~L____ PF (t) h
~-~
t
hh
)
,4
( "G,"II \
o,
[
/
,.,
J
4~o]y )
_
t
'
J
Fig. 1. Possible states and transitions.
(5)
(6)
2hx,2h 2 constant failure rates of the system due to Si(r), ~)~r) i=G2,r=x i = D2, r = y i = Fl,r = z
(7)
Pj(t)
critical human errors in good state G1 and G2 p.d.f, and hazard rate for repair time of the system repair of the system in good state G2, repair is completed in elapsed repair time x repair of the system in degraded state D2, repair is completed in elapsed repair time y repair of the system in failed state Ft, repair is completed in elapsed repair time z Pr{system in statej at time t}
j = G1, G2, D~,Dz, F1,F
(8)
{fo
Si(r) = ~bi(r)-exp -
$i(r)dr
2'D + dpD2(y) Po~(Y, t) = [ ~ t + &-~+ ~)r,(z)]Pr,(z,t,=O, 8
P r(t) = PG,(t)2hl + PG~(t)2h2.
(4) (5) (6)
BOUNDARY CONDITIONS
PG~(O, t)
= 2PG,(t) +
2'DPD~(t),
(7)
Pp.(O, t) = 2DPG~(t),
(8)
Pr,(O, t) = 2Pad(t) + 2'DPD~(t)
(9)
INITIAL CONDITIONS
i = G 2, D 2, F 1.
Pi(O) =
i # GI.
(1o)
DIFFERENCE DIFFERENTIAL EQUATIONS
SOLUTION OF THE PROBLEM
By elementary probability and continuity arguments, the difference-differential equations for the Stochastic process, which is continuous in time and discrete in space, are given by
Taking Laplace transform of the equations (1)-(9) and using initial condition (10), we get
[_s+ 2 + 2D+ 2hl]P~(s) = 1+
[~+2+2D+2hl]PG,(t)
+
+
o
PF,(Z, t)dPF~(z)dz,
(1)
PG2(X,s)d?G2(X)dx
o
Pv,(z, s)4oF,(z)dz,
(11)
S + ~ x + 2 + 2D + 2h2 + d)G2(X) l~a2(X,S ) = 0,(12) Is + 2'D] PD, (s) = 2D. PG, (S) +
+ 2 ' D Po,(t) = 2D'Po,(t)+
do
Po2(y,s)ckD2(y)dy,
(13)
PD~(y,t)d#o~(y)dy,
(3)
[S+~y+2'D+ckD~(y)]PD~(y,s)=O,
(14)
Cost analysis
I
s+
+
= o,
/15)
843
where
Al(s)
SPF(S) = 2ha" PGI(s) + 2h2' P~,(s),
(16)
PG,(O,s) = 2PG~(s)+A'D'PD~(S),
(17)
Pp,(O, s) = 2D" PG,(s),
(18)
PF~(O,s) = 2" PG2(s)+ 2'D" Po,(s).
~(19)
Integrating (12), (14) and (15) and using (17), (18) and (19) respectively, we get PG~(x, s) = [ 2 P ~ (s) + 2'OPo~ (s)] x exp [ - (s + 2 + 2D + 2h2)x ]
x exp(- f: ~,~,(x)dx),
2D[1 - S-6,(s + 2 + 2D + ).h2)] [1 - go,(s + 2'D) (28) (s + 2 + 2D + ).h2)(s + J.'D) 2 Therefore
Po~(s) = A2(s)" PG~(S),
(29)
where Az(S) -~
2D + 2"(s + 2'D)" AI(s)
(30)
(s + 2'D) [ 1 - ).'DAI(S)]'
From equations (25) and (29), we get
PG2(s) = A3(s)'PG,(s), (20)
(31)
where A3(S)
Po, (Y, s) = 2DP62 (s)' exp [ - (s + 2'D)y]
[2 + 2'D' A2(s)] • [1 - g6~(s + 2 + 2D + ,2.h2)]
(s + J. + AD + ).h2) From equations (24) and (30), we get
Pv~(z, s) = [2" P~,(s) + 2'DPo~(s)]
x exp(-sz).exp(-f:4,~(z)dz). (22) Substituting from (20) and (22) in equation (11), on simplification we get (s + 2 + 2D + 2h 1)/~o1(s) = 1 + [2" POl(S) + 2'D" enl (S)]
Po~(s) = A4(s)PG~(s), A4(s) = 2D" A3(s)" [1-So2(s+2'D)]/(s+A'O).
fo
A(s) = (s + A + AD + 2h1) - [ 2 + 2'D" Az(S)] • ~'~(s + 2 + 2 D +
(s + 2 + 2D + 2h2)
PFI(S) = As(s)/[s" A(s)], (25)
PF~(S) = Ptq(z, s)dz
(39)
EVALUATION OF LAPLACE TRANSFORM OF UP AND DOWN STATE PROBABILITIES
(26)
The Laplace transform of the probability of operational availability and non-availability at time t are given as under
Using equation (21) in equation (13), we get (s + 2'O)Pn,(s) = 20- PG~(s)+ 2DPG,(s) x [1-~o~(s+A'D)/(s+2'D).
(38)
where As(s ) = [-2. A3(s ) + 2'D. a4(s)] [1 - gFI(S)].
and
= [2P~,(s)+2'D'Po,(s)]['I-~v~(s) 1.
(37)
Also, using equations (31) and (33) in equation (26), we get
= [2.P~,(s)+2'O.Po,(s)] [ 1 - SG~(s + 2 + 2D + 2h2) ]
(36)
Using equations (31) and (35) in equation (16), we get
PF(S) = [2hi + 2h2" A3(s)]/[s" A(s)].
PG~(x, s) dx
x
J-h2)
- [2" A3(s) + 2'D" Aa(s)] ' gF,(s). (24)
Using equation (20), on simplification we get
Po~(s) =
(35)
where
Po~(y, s)dy = 2D. PG~(s) x [1 - gn=(s + 2'O)]/(s + 2'0).
(34)
Using equations (29), (31) and (33) in equation (23), we get
Pc~(s)= 1/A(s),
(23)
Using equation (21), on simplification we get Po~(s) =
(33)
where
x S6~(s + 2 + 2D + ,~h2) + [2Jffo2(s)
+2'O'-Po,(s)'gv,(s).
(32)
Pup(S) = P61(s) + P6~(s) + Po~(s) + Po~(s)
(27)
Substituting for P6,(s) from equation (25) in (27), on simplification we get
Po,(s) = 2D" P6~(s)/(s + 2'D) + [~.P~,(s) + 2'0" PD~(S)] "Z l(S),
= [ 1 + A2(s) + A3(s) + A4(s)]/A(s)
(40)
and Pdown(S) = -PFI(S)+ Pr(s) = I-2hl + 2h2" A3(s) + A5(s)'I/[s" A(s)']. (41)
P.P. GUPTA,et el.
844 ERGODIC BEHAVIOUR OF THE SYSTEM
Using the corollary of Abel's theorem lim s. F(s) = lira. F(t) = F (say). .~0
t~0
Pr,(s) = As(s)/[sA(s)],
(47)
PF(S) = [2hl +/ih2" A3(s)]/[s" A(s)],
(48)
Pup(S) = [1 + A2(s)+ A3(s)+ A4(s)]/A(s)
(49)
Provided the limit on the right hand side exists, the time independent probabilities are obtained as follows
and
Pup = lim- [s'Pup(S)]
A i(s) = 2o/n(s),
s~0
Pdown(S) = [2hl +/ih2" A3(s) + As(s)]/[s" A(s)],
A2(s ) = 2D" [B(s) +/i'(s + 2'O)]/[(s + 2'O)Bl(s)],
= lim "s" [1 +A2(s)+Az(s) s~0
(51)
A3(S) = (s + ~I'D + (ao2)B2(s)/Bl(S),
+ a4(s)]/A(s) = 0.
(42) A4(s)
Because A(s) -7s 0 and Pdow, = 1, Pup = 1.
2D" B2(s) - - , B1(s)
(52)
A 5(s) = s[2" (s + 2'D + ~D2) + 2'D"/ID]
PARTICULAR CASE
× B:(s)/[(s +
When repair follows exponential time distribution
4~F,)'BI(S)], (53)
A(s) = (s +/i + liD +/ihl) - [ / i + / I ' D ' A2(s)]
Setting
Si(s)=
(50)
where
(~i
-
s +dpi
for
i=G2,D2
and
× ckcj(s+2+/iO+/ih2+cka2)
FI,
-- I/i" A3(s) + / I ' D " A4(S)] • ff)F,/(S
we get
-~- ~)rt),
(54)
B(s) = (s +/i +/ID + 2h2 + dpa~) PG,(S) = l/a(s),
(43)
P~(s) = A3(s)/A(s),
(44)
Pp,(s) = Az(s)/a(s),
(45)
PD2(S) = A4(s)/A(s),
(46)
x(s+/i'D+OD)(s+/i'D),
(55)
n ( s ) - 2'0" 20
(56)
Bz(S) = 2. (s + 2'D) + 2'D. 2D.
(57)
Bl(s) =
and
EVALUATION OF INVERSE LAPLACE TRANSFORM OF Pap(S) AND Pdon(S)
Setting ~bo2 = ~ba= = 0.8, ~br, = 0.6, 2D = 2'D = 0.05, 2 = 0.04, 2hl = 2h: = 0.2 in equations (43)-(57) and simplifying, we get (s + 0.6)(s + 1.09)(s 4 + 2.13s 3 + 1.2650s 2 + 0.154275s + 0.0046294) Pup(S) =
(s + 0.026)(s + 0.064)(s + 0.29)(s + 0.446)(s + 1.309) x
1
[ s - ( - O . 9 4 2 5 + O.2881679i)] " [ s - ( - 0 . 9 4 2 5 - 0 . 2 8 8 1 6 7 9 i )
(58)
and
PDI(s) =
(0.05)(s + 0.6)(s + 1.09)(s 3 + 1.99s 2 + 1.0635s + 0.048325) " {(s+O.O26)(s+O.O64)(s+O.29)(s+O.446)(s+ 1.309)
(59)
x [s-(-O.9425+O.2881679i)].[s-(-0.9425-0.2881679i)]} Taking the inverse Laplace transform of equations (58) and (59), we get Pup(t) = 0.1728679 exp ( - 0.260 + 0.0959487 exp ( - 0.0640 + 1.0979981 e x p ( - 0.290-0.5301422 exp ( - 0 . 4 4 6 0 + 0.0669847 exp ( - 1.309t) + 2 exp ( - 0.9425t) × [0.0481714 cos (0.28816790 + 0.1168938 sin (0.28816790]
(60)
and
Pol(t) = O.1316093 exp ( - 0.0260 + 0.0934836 exp ( - 0.0640 - 0.3003391 exp ( - 0.29t) + 0.0827971 exp ( - 0.4460 - 0.0044971 exp ( - 1 . 3 0 9 0 - 2 exp ( - 0.94250 × [0.0026614 cos (0.28816790 + 0.00309 sin (0.2881679t)], where, of course, Pup(t)+ Pdown(t) = 1.
(61)
Cost analysis
845
COST FUNCTION ANALYSIS
Table 2
OF THE SYSTEM
Expected profit G(t) The s-expected up time of the system during (0, t] is
S1. no.
E(t) = ~o P.p(t)dt and the s-expected busy period of the service facility in (0, t] is #8(0 =
Po,(t).
--
Hence, the expected net gain function is given by
G(t) =
expected total revenue function in (0, t] -
Time t
1
0
2 3 4 5 6 7 8 9 10 ll
1 2 3 4 5 6 7 8 9 I0
C 2 = 0.5
C 2 = 0.10
C 2 - 0.05
0.0000000 0.4352562 0.7053963 0.8460866 0.9470489 0.9605711 0.9193612 0.8327057 0.7081851 0.5522155 0.3697642
0.0000000 0.8273402 1.4747398 1.9989349 2.4418053 2.8087030 3.1171644 3.3778506 3.5994313 3.7890003 3.9522488
0.0000000 0.8763507 1.5709077 2.3240476 2.7286498 3.0397195 3.3918898 3.6959937 3.9608371 4.1935984 4.4000594
expected cost due to repair in (0, t] repair cost per unit time (spent in the service facility), respectively.
= C 1 [ - 6.6487653 exp (--0.0260 -
NUMERICAL COMPUTATION
1.4991984 exp ( -- 0.064t)
Availability analysis
- 3.7862003 exp (-- 0.290 + 1.1886596exp(-0.4460
Setting t = 0, 1,2 .... ,10 in equation (60), we get Table 1.
- 0.0511724 exp ( - 1.309t)
Cost analysis
+ 10.9595146 - exp ( - 0.94250
Setting C~ = 1, t = 0, 1,2,..., 10, in equation (62) the net gain for C2 = 0.05, 0.1, 0.5 are shown in Table 2.
x {0.1628388 cos (0.2881679t) + O.1982627 sin (0.28816790}]
-- C2[t-
5.6621355 + 5.0618961
INTERPRETATION
x exp ( - 0.026t) + 1.4606812 exp ( - 0.0640
OF THE RESULTS
Table 1 computes the availability of the system at an instant t and Fig. 2 shows the availability vs. time. The availability of the system decreases very slowly and over a long time showing that the system remains available for a long period of time. Table 2 computes the expected profit during the
- 1.035652 exp ( - 0.290 + 0.1856437 exp ( - 0.446t) - 0.0034355 exp ( - 1.3090 - exp (-- 0.94250 x {0.006998 cos (0.28816790 + 0.0044172 sin (0.28816790}],
1.00~
(62)
0.900
where Cl and C2 are the revenue per unit up time and
A
Table I.
I~
°o=\
X
S1. no.
Time t
1
0
2 3 4 5 6 7 8 9 10 ll
1 2 3 4 5 6 7 8 9 l0
Pup(t) 1.0000000 0.7967283 0.6658796 0.5672142 0.4879072 0.4230221 0.3698577 0.3263895 0.2909030 0.2619238 0.2381771
Pdow.(t) 0.0000000 0.2032717 0.3341204 0.4327858 0.5120928 0.5769779 0.6301423 0.6736105 0.7090970 0.7380762 0.7618229
0.600 ,.Q
<
\x
0.500
\x
0.400 0.300 0.2001 I 0 1
\x
I
I
I
J
I
I
h
I
I
2
3
4
5
6
7
8
9
10
Time Fig. 2. Availability vs. time.
846
P.P. GUPTA,et al.
5.000
C2 = 0.05 / x ~
interval (0, t] for the fixed value of the revenue per unit time. A critical examination of the graph (Fig. 3), expected profit vs. time, reveals that the expected profit increases for a long time when C 2 ~< 0.1.
4.000
Acknowledgements--R. K. Gupta and R. K. Sharma are highly grateful to C.S.I.R., New Delhi, for granting and providing financial assistance to the Research Project, under which the work was initiated and carried out.
A v
3.000 "0
REFERENCES
2.000 X uJ
1.000
0.000 0
I
I
I
i
I
I
I
I
J
1
2
3
4
5
6
7
8
9
Time
Fig. 3. Expected profit vs. time.
10
1. P. R. Parthasarathy, Cost analysis for 2-unit system, IEEE Trans. Reliab. R-28, 268-269 (1979). 2. P. P. Gupta and S. C. Agarwal, Cost-function analysis of a 3-state repairable system, Microelectron. Reliab. 24, 51-53 (1984). 3. D. Meister, The problem of human-initiated failure, Eighth National Symposium on Reliability and Quantity Control (1962). 4. B. S. Dhillon and R. B. Mishra, Reliability evaluation of system with critical human error, Microelectron. Reliab. 24, 743-756 (1984).