CC BY
22
Parallel System Analysis with Priority and Inspection Using Semi-Markov Approach
Neetu Dabas
Department of Statistics, AIAS, Amity University, Noida, Uttar Pradesh-201 313 Email: neetudabas2@gmail. com
Reetu Rathee
Department of Statistics, AIAS, Amity University, Noida, Uttar Pradesh-201 313 Email: rrathee@amity.edu
Abstract
In this paper a parallel system has been discussed with the idea of priority to preventive maintenance over replacement. The system has two identical units and facility of inspection is given to the failed unit before repair/replacement. There is a single server who play four-in-one role of inspection, replacement, repair and preventive maintenance and comes immediately when required. Units are failed with constant rate whereas failure time is random. The distribution of time for repair activities is arbitrary and there rates follow exponential distribution. The random variable associate with different rates are stochastically independent. Mathematical expression for several reliability terms like MTSF, availability, busy period analysis for server , expected number of visits by the server and cost benefit are obtained by using semi-markov process and regenerative point technique. Graphs are drawn to find the effect of various parameters on MTSF, Availability and profit.
Keywords: parallel system, priority, preventive maintenance, replacement, inspection
I. Introduction
The world is moving day by day towards the smart technology. Advance development of technology has significantly increased cost and complexity of industrial systems. Thus it has become an essential to operate industrial systems with minimum down time in order to achieve optimized production, increase profit and to avoid the losses. Hence, the need for reliability modeling and analysis of complex industrial systems is inevitable. Reliability analysis of parallel systems has been broadly studied by many researchers because parallel configuration is more reliable then series. Dhillon and Viswanath [6] analyzed a parallel system with the common-cause failure. Sridharan and Kalyani [7] gives common-cause failure analysis of a two non-identical unit parallel system using GERT technique.
A parallel (2n-2) system was investigated by E.Papageorgious and G. Kokolakis [9] where two units start operation simultaneously and any one of them was replaced by one of (n-2) warm standby units upon failure. Reliability of parallel systems was studied based on multiple competing dependent failure process by Sanling and David [10]. This problem was formulated for
the conditioning on shock sets because not all shocks cause degradation. N. Sharma and J.P.Singh Joorel [8] investigate a two unit parallel system with inspection and preparation time for replacement. To improve the reliability PM and Inspection are widely used in modern engineering systems. PM can help to extend the life time of an equipment, increase the productivity and hence decrease unexpected maintenance spending. Inspection aims to find the defect of the system and type of the defect. Goel and Gupta [13] considered a two identical unit parallel system with the concept of PM, inspection and two types of repair. They assumed that the time to failure, commencement to PM and inspection are constant while repair and maintenance times are arbitrarily distributed. M.K.Kakkar and J.Bhatti [12] purposed a two dissimilar parallel unit framework under the presumption that the unit may also fail during the preventive maintenance (PM). Wang and Lin [5] found a methodology to optimize the non-periodic maintenance for a seriesparallel system. Shruti [11] analyzed a stochastic model with two units subject to routine inspection, maintenance and replacement. In this research routine inspection is conducting over operative unit and after inspection either the unit is maintained or it failed. Repair and replacement of the unit is based upon guarantee period of the equipment. In many research priority concept is also used to make the system better in performance and hence more profitable. P. Kumar, A. Bharti, and A. Gupta [3] investigated and analyze a two unit parallel system in which priority was given to one unit over other. In this system priority unit was repairable and non-priority unit was non-repairable and preference to repair of priority unit was given over replacement of non-priority unit. R.Rathee and D.Pawar [4] introduced a reliability model of a parallel system in which priority to repair over replacement was given using maximum operation and repair times.A.Kumar and S.C.Malik [1] considered a computer system with two identical units and in each unit H/W and S/W components works together. Priority was given to PM of the unit over S/W replacement under certain assumptions. C.Aggarwal and N.Ahlawat [2] done the profit analysis of a standby system with priority to PM over repair by considering rest of the server between repairs.
In the present study a parallel system is investigate under some assumptions. Priority is given to PM over replacement with inspection. Inspection is done to find the type of failure. Semi-Markovian approach and regenerative-point-technique are used to obtain numerical expressions for various reliability terms such as MTSF, Availability, busy period analysis for server, expected number of visits by the server and cost benefit. Graphical interpretation is done to visualize the effect of several parameters (related to repair activities) on obtained reliability measurable terms.
For practical implication of system one of the example is a parallel compressor rack system. Jim Coats [14] gives the commercial and industrial applications of parallel compressor racks. There are several advantages of this system like low installation cost, capacity control, redundancy and maximum efficiency etc.
Figure 1: Parallel compressor Rack
II. Notations for System Model
5j : States o/ tfte sytem (i = 1,2,____,15)
À : Constant failure rate
a/ô/«0 : repair/replacement/PM rate of the system
upm/UPM: Unit is under PM/continuosly under PM
wpm/WPM: Unit is waiting for PM/continue waiting for PM
FUi/FWi: Failed unit under inspection/waiting for inspection
FUI/FWI: Failed unit continuously under inspection/waiting for inspection
FUr/FUrp : Failed unit under repair/replacement
FUR/FURP: Failed unit continuously under repair/replacement
mean soyonrn time in tfte state 5; m^: Contribution to mean sojourn time in state Si when the system
transits directly to state Sj h(t)/f(t)/r(t)/g(t): pdf of the inspection/repair/replacement/PM time
III. System Description and Assumptions
A parallel system with two identical units is studied under some practical assumptions which are given below:
• Initially both the units are in operative condition
• Failure rate is constant
• Only one server operator is taken to do all repair activities
• Repair is perfect or units restore in initial condition after repair
• Post failure inspection is done to find unit is repairable or replaced by new
• Time taken for repair activities is arbitrary and there rates follow exponential distribution
• The random variable associate with different rates are stochastically independent
Table 1: Description of the states
States Description
50 Both the units are in normal mode
51 One unit is operative and other is failed under inspection
52 Resume for PM
S3 One is working and other is failed under replacement
S4 One unit is continuously under inspection from previous state and other is waiting
for inspection
S5 One is working and other is failed under repair
S6 One unit is continuously under inspection from previous state and other is waiting
for PM
S7 One unit is working and other under PM
S8 One unit is continuously under replacement from previous state and other is waiting
for inspection
S9 One unit is under repair and other is continuously waiting for inspection from
previous state
S10 One unit is under replacement and other is continuously waiting for inspection from
previous state
Sil One unit is continuously under repair from previous state and other is waiting for
inspection
Sl2 One unit is continuously under repair from previous state and other is waiting for PM
S13 One unit is under repair and other is continuously waiting for PM from previous state
S14 One unit is under PM and other is continuously waiting for replacement from
previous state
S15 One unit is continuously under PM from previous state and other is waiting for
inspection
Figure 2: Transition State Diagram
Neetu Babas, Reetu Rathee
PARALLEL SYSTEM ANALYSIS WITH PRIORITY RT&A, No 2 (68) AND INSPECTION USING SEMI-MARKOV APPROACH_Volume 17, 2022
IV. Formulation and Stochastic Analysis of the Model I. Transition Probabilities & Mean Sojourn Times
Steady- state transition probabilities from regenerative state i to state j are given by the formula PO =^(œW<T%(t)dt (1)
Here Çi_,(t) / are the c.d.f/p.d.f from state i to state j in (0,t) time.
_ 2A _ a0
P01= 217^0 ' Po2= 217^0
Pl3 = M*(À +a<), Pi4 = ^ (1-ft*(À +a<)), Pis = aft*(À +a<), pw = ^ (1- h*(X +a<))
P30 = r*(À+«o), P3a=P3i.8^^^(1- r*(X+«o)), P3,i4 = (1 - r* (^+0)
P49 = P6,i3 = a , P4,i0 = P6,i4 = b
P50=/*(^+«0) , P5,ii = P5i.ii =1^(1 - f*ft+«0)) , P5,i2 = P57.i2 ^ (1 - f*(*+«o))
P70 = g*W , P7,i5 = P7i.i5 = 1 - ,
Pii 49 ^17^(1 -h*(À+Ko)) ,Pii.4,i0 = I^^(1-h*(À+«o)) ,
Pi,i46=£0^(1-h*(À+Ko)) , Pi7.6,i3=£^(1- h*(À+«o)) ,
P27 = p8i = P9i = Pi0,i = pii,i = Pi2,7 = Pi3,7 = Pi4,3 = Pi5,i = 1
It can verified that
P01 + Po2 = Pi3 + Pl5 + Pl1.49 + Pii.4,10 + Pi7.6,13 + Pi,14.6=
P50 + P51.11 + P57.12 = P30 + P31.8 + P3,14 = P49 + P4,10 = P6,13 + P6,14 = P70 + P71.15 = P27 = P81 = P91 = Pi0,1 = Pi1,1 = Pi2,7 = Pi3,7 = Pi4,3 = Pi5,1 = 1
And are given by the formula
Pi = £ (0 = J"" P (T > 0 dt = Z (2)
and
¿[Qi/W] , _
= js |5 = 0 (3)
11 1
2A + a0 A + a0 A + a0
^ = + «>), ^7 = 1(1- g*W)
1 Afc 1 a
K = h-+ ^-- + - + — ](1-ft*(A + a0))
P1 LA + a0 a(A + a0) y a v
^3 =-^L(1-ra + a0)) , ^5 = =1
II. Reliability & Mean Time to System Failure (MTSF)
Let <Pt (t) be the cdf of first passing time from the state St to the state in which failure occur and we take absorbing state as the failed state. So, the expressions for ^ (t) from which MTSF of discussed system is obtained are given as
(0 = Qij (0 © (t) + Zi,k Qik (t) (4)
Where j is the operating regenerative state to which the given regenerative state i can transit and k is the failed state to which state i can directly transit. If we take LST of above relation (4) and solved them for 0** (s), we have
fl.^1-^ (5)
Neetu Babas, Reetu Rathee
PARALLEL SYSTEM ANALYSIS WITH PRIORITY RT&A, No 2 (68) AND INSPECTION USING SEMI-MARKOV APPROACH_Volume 17, 2022
The system reliability is obtained by taking Inverse Laplace transform of (5) and MTSF is given by the formula
MTSF =limi-^ = N (6)
S D w
Where,
N = Mo + M1P01 + M3P01P13 + M5P01P15 and D = 1 - P01P13P30 - P01P15P50 (7)
III. Analysis of Availability
Let ^¿(t) be the probability of system working at time 't' w.r.t the condition that system goes to regenerative state 5j at t = 0. We have the relations for ^¿(t) as
4(0 = ^(t) + ^^(t)©^) (8)
Where j is any successive regenerative state to which the regenerative state i can transit through n transitions.
M; (t) is the probability that the system in up state Si up to the time t without visiting to any other regenerative state.
M0(t) = e"(2A+ ^ (9)
M1(t) = e"(A+K° )ùÏÏ(t) (10)
M3(t) = e"(A+K° (11)
M5(t) = e"(A+K° )iF(t) (12)
M7(t) = e-(Ai)G(t) (13)
Now, if we use LT of (8) and solved it for ^0(s).We get the result for steady state availability as A,(œ) = lims^0(s)=^ (14)
Dl
Where
W1=^o^+(^1+M5P15)ß+M3C+^7Ö (15)
£>1 = (Mo + M2P02M + G"1 + ("sP15)ß + + ^ + ^14^ (16)
IV. Busy Period Analysis for Server
Let ß/ (t),ßf (t), ßfp(t), ßf (t)be the probability of busy period of server during inspection, repair, replacement and PM at instant't' with the given condition that the system go to regenerative state Si at t=0. The recursive relations for ß/ (t), ßf (t), ßfp (t), ßf (t) are as follows:
= WW) + ^^«©^(o (17)
sf(t) = WW) + ^^«©B/W (18)
ßfp(t) = WKt) + ^qf (t)©Bj,p(t) (19)
= ^(t) + ly?^ (t)©Bj'(t) (20)
Where j is any successive regenerative state to which the regenerative state i can transit through n transitions.
Wi(t) is the probability of server busyness at state Si due to repair activities at time t without making any transition to any other regenerative state or returning to the same via one or more non regenerative state. Here,
w1(t) = e-^+^FCt) + (Ae-(A+K0)t©1)7T(^ + (a0e-(A+K0)t©1)7T(^ (21)
Ws(t) = e^+^F^ + (^^^©l)^ + (a0 e-^^©^/^) (22)
W3(t) = e-^+^ÂCt) + (Ae-(A+Ko)t©1)ÎT(^ (23)
W2(t) = G(t) = W14(t), W7(t) = e-(A)tG(t) + (Ae-(A)t©1)G(F)
Take LT of (17) to (20) and solving for 50*(5),#(T(s),£oP (s), £o*(s) -The busy time in inspection, repair, replacement and preventive maintenance for server is given by
B0(œ) = limsB0*(5)=^2 (24)
Bofl(œ) = lims<"(s)=? (25)
C(œ) = limsC*(s)=| (26)
BoP(œ)=lims<"(s)=^ (27) Here,
w2 = W1*(0)B,^3 = W5*(0)p155,N4 = Wj*(0)C
= W2*(0)p024 + W7*(0)D + W|*4(0)£ and Di is mentioned above. (28)
V. Expected Number of Visits by The Server
Consider /0(t),fi0(t),fip0(t) ,Pm0(t) as the expected number of visits make by the server for inspection, repair, replacement and PM in (0, t] .We have the following recursive relations for UO,*o(0,*Po(0 ,^o(0 are
/i(0 = ZyGS°(0© (C + Ij (t)) (29)
Ai(t) = ZyCto© (C + Rj (t)) (30)
RVi(0 = çf (0© (C + Rpj (t)) (31)
^(0 = çf (t)© (C + Pmj (t)) (32)
Where j is any successive regenerative state to which the regenerative state i can transit through n transitions and and C = 1 if j is the state where the server does the job afresh, otherwise C = 0.
Take LST of above equations solving for /0*(s),fiÔ*(5),^Po*(s), ^mo*(s) .The expected number of inspections, repairs, replacements, and preventive maintenance by the server is given by (per unit time)
/o(œ) = lims/0*(s)=^ (33)
floM=limsfl,r(s)=^ (34)
Ol
flpoK) = limsflp;*(s)=^ (35)
s^o Ol
Pm0 (m) = lim sPrnJ* (s) = — (36)
s^0 Dl
Where,
W6 = ß, N7 = (Pll.49 + P15 + Pl7.6,13)ß, ^8 = (Pl1.4,10 + Pl,14.6 + Pl3^,
N7 = p02^ + D + E and D1 is already mentioned. (37) Here A, B, C, D & E are
A = (1 - P3,14)(Pl5P50 + P70 (Pl5P57.12 + Pl7.6,is)) + (Pl3 + Pl,14.6)P30 (38)
B = (1-P3,14)(1-P02P70) (39)
C = (Pi3 +Pi,i4.6)(1 -P02P70) (40)
D = (1 - P3,14)(Pl5 + Pl7.6,13 - P01P15P50 - PisPsi.lO + P02P30 (Pl3 + Pl,14.6) (4^
E = (Pi3P3,i4 + Pi,i4.6)(1 - P02P70) (42)
VI. Profit Analysis
In steady state the profit function of the system model can be obtained as
i3BoP - k4Bom - k5!o
P = koAo - k1B0 - k2B0R - k3B0Rp - k4B0Pm - ksIo - k6Ro - k7Rpo - k8Pmo (43)
Here,
P = Profit function of system model
= Revenue per unit up — time of the system , fc2 , fc3 , fc4 = Cost per unit time of the server when it is busy in
inspection, repair, replacement, preventive maintenance , fc6 , fc7 , = Cost per unit time for inspection, repair, replacement, preventive maintenance
V. Analytical Study of the Model
In the present study concept of priority to PM over replacement with inspection is used. To see the applicability of this situation particular cases are taken for the included parameters like F(t) =« e-Kt,r(t) = pe-|3t,h(t) = ye-Yt ,g(t) = 0e-et. Results are obtained in form of tables and graphs by taking random values for the given parameters. On the basis of these cases numerical and graphical results are obtained which shows the effect of these parameters on MTSF, Availability and Profit function of the system. From the obtained results we conclude that PM does not effect the MTSF and as the failure rate increases the availability and the profit of the system is decreases. When the repair activities rate is increases then availability and profit is also increases. From the results we obtained that the system is highly profitable if we increase the PM rate at the very first stage when failure rate is very low but as the failure rate increases the system get more profit by enhancing the inspection rate. Tabular and graphical results are given below:
Table 2: Values of MTSF w.r.t various parameters
Failure rate a=2.1,ß=2,a=0.6,b=0.4, Y=1.3,a0=3 a= 4.1 ß=5 Y= 3 a0= 3.1 a= 0.4, b= 0.6
0.1 0.33274 0.33275 0.33275 0.33279 0.32204 0.33274
0.2 0.33113 0.33119 0.33119 0.33130 0.32056 0.33113
0.3 0.32875 0.32887 0.32886 0.32908 0.31838 0.32875
0.4 0.32577 0.32596 0.32594 0.32627 0.31563 0.32577
0.5 0.32235 0.32260 0.32258 0.32302 0.31247 0.32234
0.6 0.31859 0.31890 0.31887 0.31943 0.30899 0.31858
0.7 0.31458 0.31495 0.31492 0.31558 0.30527 0.31457
0.8 0.31040 0.31082 0.31079 0.31153 0.30138 0.31039
0.9 0.30610 0.30657 0.30653 0.30736 0.29737 0.30609
MTSF Vs Failure Rate
0,34000
0,33000
^,32000
in
H
£»,31000 0,30000 0,29000
a=2.1,ß=2,a=0.6,b=0.4,y=1.3,a0=3 a= 4.1 ß=5 -Y= 3 a0= 3.1
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 Failure Rate
Figure 3: MTSF VS Failure Rate
Table 3: Values of Availability w.r.t various parameters
Failure Rate a=2.1,ß=2,a=0. 6,b=0.4,y=1.3, a0=3,6=1.4 a=4.1 ß=5 a=0.4,b=0.6 Y=3 a0=3.1 6=2
0.1 0.53503 0.53911 0.54102 0.52635 0.55017 0.53213 0.56972
0.2 0.45652 0.47175 0.46481 0.45459 0.48943 0.45348 0.48158
0.3 0.42387 0.43952 0.43365 0.41815 0.46238 0.42072 0.45061
0.4 0.39668 0.41256 0.40718 0.38870 0.43924 0.39350 0.42408
0.5 0.37357 0.38955 0.38436 0.36427 0.41912 0.37041 0.40101
0.6 0.35358 0.36960 0.36443 0.34357 0.40138 0.35048 0.38067
0.7 0.33606 0.35206 0.34683 0.32572 0.38556 0.33303 0.36257
0.8 0.32052 0.33647 0.33114 0.31012 0.37131 0.31758 0.34631
0.9 0.30661 0.32248 0.31705 0.29630 0.35836 0.30376 0.33159
0,60000
0,55000
0,50000
^ 0,45000
3 0,40000 jz
a 0,35000 ss
¡> 0,30000 <
0,25000
Availability Vs Failure Rate
a=2.1,ß=2,a=0.6,b=0.4,y=1.3,a0=3,6=1.4
--a=4.1
-_ß=5 x a=0.4,b=0.6
-a0=3.1
x 8=2
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 Failure Rate
Figure 4: Availability VS Failure Rate Table 4: Values of Profit w.r.t various parameters
Failure Rate a=2.1,ß=2,a=0. 6,b=0.4,y=1.3, a0=3,6=1.4 a=4.1 ß=5 a=0.4,b=0.6 Y=3 a0=3.1 6=2
0.1 7355.86 7416.93 7463.24 7208.33 7569.95 7312.12 7759.37
0.2 6014.32 6239.56 6166.20 5957.81 6474.39 5970.56 6258.77
0.3 5391.47 5618.42 5571.51 5271.16 5911.14 5347.74 5648.91
0.4 4873.10 5098.06 5067.35 4713.77 5427.64 4830.27 5130.70
0.5 4433.01 4653.80 4633.45 4249.77 5006.14 4391.57 4683.55
0.6 4053.32 4268.64 4255.27 3855.69 4634.00 4013.52 4292.79
0.7 3721.35 3930.44 3922.08 3515.48 4301.91 3683.31 3947.67
0.8 3427.88 3630.28 3625.81 3217.79 4002.92 3391.63 3640.12
0.9 3166.00 3361.48 3360.25 2954.37 3731.65 3131.53 3363.95
Profit Vs Failure Rate
a=2.1,ß=2,a=0.6,b=0.4,y=1.3,a0=3,6=1.4 --a=4.1
7800
ß=5
£-5800
o
6800
a=0.4,b=0.6
V=3
a0=3.1
8=2
P4800
3800
2800
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 Failure Rate
Figure 5: Profit VS Failure Rate
VI. Discussion
The performance of the system model for different values of failure and repair rates is shown by the table 2, table 3, and table 4 of MTSF, availability and profit analysis w.r.t. various parameters. Also, the behavior of the MTSF, availability and profit analysis depicted by figure 3, figure 4 and figure 5 respectively.
From the table 2 and figure 3 we conclude that the MTSF is decrease with the increase of the failure rate. If we fix all the parameters and change the value of repair rate a=4.1, replacement rate p=5 and inspection rate y=3 one by one the MTSF is increase. And the MTSF is decrease if the preventive maintenance is increase by ao = 3.1 and the other parameters are fixed.
From the table 3 and figure 4 we analyze the availability of the system. By fixing all the parameters and changing in other parameter one by one we found that availability is increase in we change in a=4.1, replacement rate p=5, inspection rate y=3 and preventive maintenance rate 9=2. The availability is decline with the increase of preventive maintenance by ao = 3.1 and inter change the repair and replacement rate (a=0.4 and b=0.6).
From the table 4 and figure 5 we conclude that profit of the function is decrease by increasing the preventive maintenance by ao = 3.1 and inter change the repair and replacement rate (a=0.4 and b=0.6) one by one with the fixed of all other parameters. If we change in a=4.1, replacement rate p=5, inspection rate y=3 and preventive maintenance rate 9=2 found that profit of the system model is enhance as compared to the fixed values of all parameters.
References
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