A Comparative Study of Profit Analysis of Two Reliability Models on a 2-unit PLC System

Transcription

1 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril A Comarative Study of Profit Analysis of Two Reliability Models on a 2-unit PLC System Dr. Bhuender Parashar, Dr. Nitin Bhardwaj Abstract - A two-unit PLC system is analysed with two different situations resulting in to two models. In first model two identical units are used in hot standby and no riority regarding oeration/ reair is set for any of the units. While in the other model two units are used in hot standby on master-slave basis. The slave unit can also fail but generally its failure rate is lower than that of the master unit. Priority is given to the reair of minor fault over the major fault. In other cases of failure, riority for reair is given to the master unit. Also, riority for oeration is given to the master unit. For various measures of system effectiveness the exressions are obtained and a comarative study of both the models is done for the rofit with resect to various arameters followed by the conclusion. Index Terms - Relaibility, Semi-markov, Regenerative Point, PLC, Hot Standby, Comarative Study, Profit 1 INTRODUCTION In the field of reliability engineering, a number of researchers like [1] to [5] have analysed various models under different assumtions and collecting real data. System based on a articular model cannot be considered as best. A model may be better in some situations and may be worse in some other situations when comared with some other models. Keeing this in view, the resent study deals with the comarison between two models at a time to see which is better than the other under the stated situations. The comarison is done grahically considering the articular case that the time to reair/relacement are exonential as taken in the concerned models. Grahs are lotted taking the values of rates, costs and robabilities estimated based uon the data collected from an industry. Values of some other rates/costs have been assumed ver used. The system is analysed by making use of semi-markov rocesses and regenerative oint technique and exressions for different measures of the system effectiveness are obtained. 2. MODEL - I 2.1 Assumtions 1. Initially one unit is oerative and the other is hot standby. 2. Failure times are assumed to have exonential distribution as the other times have general distributions. 3. There are two tyes of failure - minor failures (reairable) and major failures (irrearable). 4. After each reair, the system works as good as new one. In this model, it is considered that both the oerative as well as the standby unit are identical and no articular unit is given any riority for oeration/ reair. All failures are reaired by an exert reairman. 2.2 Notations O - oerative unit hs - hot standby unit - constant failure rate of the unit - constant failure rate of the hot standby unit - robability of minor failure q1 - robability of minor failure (reairable) q2 - robability of major failure (irrearable) Fre - unit is under reair in case of minor failure FRe - reair by the reairman is continuing from the revious state Fre - unit is under relacement in case of major failure FRe - relacement is continuing from the revious state wre - failed unit waiting for reair from the reairman wre - failed unit waiting for relacement from the reairman g1(t), G1(t) -.d.f. and c.d.f. of reair time of unit having minor failure h(t), H(t) -.d.f. and c.d.f. of relacement time of unit having major failure w(t), W(t) -.d.f. and c.d.f. of waiting time. - Stieltjes transform 2.3 Transition Probabilities and Mean Sojourn Times A transition diagram showing the various states of transition of system is shown as in Fig. 1. The eochs of entry into states 0, 1 and 2 are regenerative oints and thus these states are regenerative states. The transition robabilities are given below: 01 = + q1, 02 = q2, 13 = ( + q1)[1 g1()], 14 = q2[1 g1()], 10 = g1(), 20 = h(), 25 = ( + q1)[1 h()], htt://

2 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril = q2[1 h()], (3) 11 ( + q1)[1 g1()], (4) 12 q2[1 g1()], (5) 21 ( + q1)[1 h()], (6) 22 q2[1 h()] By these transition robabilities, it can be verified that = 1 (3) (4) = 1 = (5) (6) = 1 = The mean sojourn time (i) in the regenerative state i are given by 1 1 g 1 (λ ) 1 h (λ ) 0 = λ α, 1 = λ, 2 = λ The unconditional mean time taken by the system to transit for any regenerative state j when it (time) is counted from the eoch of entrance into state i is mathematically stated as ' td Qij (t) - qij (0) mij = 0 Thus, m01 + m02 = 0 m10 + m13 + m14 = 1 m20 + m25 + m26 = 2 G m10+m11(3)+m12(4) = 0 1(t) dt = 1 (say) H m20+m21(5)+m22(6) = 0 (t) dt = 2 (say) (o, h s ) 0 (+q 1 ) (+) g 1 (t) (F re, o) (+q 1 ) 1 3 g 1 (t) (F Re, w re ) h(t) q 2 (+) 4 q 2 (F Re, w re ) 5 h(t) g 1 (t) (+q 1 ) (F Re, w re ) 2 q 2 6 (F re, o) h(t) (F Re, w re ) Fig. 1 U State Failed State Regeneration Point htt://

3 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril Mean Time to System Failure Regarding the failed state as absorbing states and emloying the arguments used for regenerative rocesses, we have the following recursive relation for i(t). 0(t) = Q01(t) 1(t) + Q02(t) 2(t) 1(t) = Q10(t) 0(t) + Q13(t) + Q14(t) 2(t) = Q20(t) 0(t) + Q25(t) + Q26(t) Now, taking L.S.T. of the above equations and solving them for 0(s), the mean time to system failure (MTSF) when the system starts from the state 0, is 1 - φ 0 (s) N lim MTSF = s0 s D Where N = and D = Availability Analysis A0(t) = M0(t) + q01(t) A1(t) + q02(t) A2(t) A1(t) = M1(t) + q11(3)(t) A1(t) + q12(4)(t) A2(t) + q10(t) A0(t) A2(t) = M2(t) + q21(5)(t) A1(t) + q22(6)(t) A2(t) + q20(t) A0(t) M0(t) = e - (+) t, M1(t) = e - t G 1(t), M2(t) = e - t H (t) Taking L.T. of the above equations and solving them for A0(s), and then in steady-state, availability of the system is given by: N1 lim s A 0 (s) A0 = s 0 = D 1 N1 = 0 [(1 11(3)) (1 22(6)) 12(4) 21(5)] + 1 [01 (1 22(6)) 02 21(5) ] + 2 [01 12(4) + 02 (1 11(3))] D1 = 0 [10 21(5) (4)] + 1 [21(5) ] + 2 [12(4) ] 2.6 Busy Period Analysis of Exert Reairman (Reair Time Only) Using the robabilistic arguments, we have the following recursive relations for BRi(t) : BR0(t) = q01(t) BR1(t) + q02(t) BR2(t) BR1(t) = W1(t)+q11(3)(t) BR1(t)+q12(4)(t) BR2(t)+q10(t) BR0(t) BR2(t) = q21(5)(t) BR1(t) + q22(6)(t) BR2(t) + q20(t) BR0(t) W1(t) = G 1 (t) Taking L.T. of the above equations and solving them for BR0(s), and then in steady-state, the total fraction of time for which the system is under reair by exert reairman, is given by : lim s BR 0 (s) BR0 = s 0 =, N2 = 1 [ (5)] N2 D 1 and D1 is already secified. 2.7 Busy Period Analysis of Exert Reairman (Relacement Time Only) Using the robabilistic arguments, we have the following recursive relations for BRPi(t) : BRP0(t) = q01(t) BRP1(t) + q02(t) BRP2(t) BRP1(t) = q11(3)(t) BRP1(t) + q12(4)(t) BRP2(t)+q10(t) BRP0(t) BRP2(t) = W2(t) + q21(5)(t) BRP1(t) + q22(6)(t) BRP2(t) + q20(t) BRP0(t) W2(t) = H (t) Taking L.T. of the above equations and solving them for BRP0(s), and then in steady-state, the total fraction of time for which the system is under relacement by exert reairman, is given by : N3 lim s BRP 0 (s) BRP0 = s 0 = D 1 Where N3 = 2 [12(4) ] and D1 is already secified. 2.8 Exected Number of Visits by Exert Reairman Using the robabilistic arguments, we have the following recursive relations for Vi(t) : V0(t) = Q01(t) [1+V1(t)]+Q02(t) [1+V2(t)] V1(t) = Q10(t) V0(t)+Q11(3)(t) V1(t)+Q12(4)(t) V2(t) V2(t) = Q20(t) V0(t)+Q21(5)(t) V1(t)+Q22(6)(t) V2(t) Taking L.S.T. of the above equations and solving them for V0(s), and then in steady-state, the number of visits er unit time is given by : N4 lim s V 0 (s) V0 = s 0 = D 1 N4 = (4) (5) secified. and D1 is already 2.9 Exected Number of Relacements Using the robabilistic arguments, we have the following recursive relations for RPi(t) : RP0(t) = Q01(t) RP1(t) + Q02(t) [1+RP2(t)] RP1(t) = Q10(t) RP0(t) + Q11(3)(t) RP1(t) + Q12(4)(t) [1+RP2(t)] htt://

4 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril RP2(t) = Q20(t) RP0(t) + Q21(5)(t) RP1(t) + Q22(6)(t) [1+RP2(t)] Taking L.S.T. of the above equations and solving them for RP0(s), and then in steady-state, the number of relacements er unit time is given by : N5 lim s RP 0 (s) RP0 = s 0 = D 1 N5 = 12(4) and D1 is already secified Profit Analysis Profit (P1) = C0 (A0) C2 (BR0) C3 (BRP0) C4 (V0) C5 (RP0) C0 = Revenue er unit utime. C2 = Cost er unit utime for which the exert reairman is busy for reair. C3 = Cost er unit utime for which the exert reairman is busy for relacement. C4 = Cost er visit of reairman. C5 = Cost er unit relacement 3. MODEL - II 3.1 Assumtions In this model, another situation is analysed PLCs are used as hot standby on the basis of master-slave concet. The slave unit can also fail but generally its failure rate is lower than that of the master unit. Priority is given to the reair of minor fault over the major fault. In other cases of failure, riority for reair is given to the master unit. Also, riority for oeration is given to the master unit. Rest of the assumtions are as same as in model-i. 3.2 Notations Mo - master unit is oerative So - slave unit is oerative Shs - slave unit is hot standby - constant failure rate of the master unit - constant failure rate of the slave unit Mre - master unit is under reair of reairman in case of minor failure MRe - reair of master unit by the reairman is continuing from the revious state Mre - master unit is under relacement in case of major failure MRe - relacement of master unit is continuing from the revious state Mwre - failed master unit waiting for reair from the reairman Mwre - failed master unit waiting for relacement from the reairman Sre - slave unit is under reair of reairman in case of minor failure SRe - reair of slave unit by the reairman is continuing from the revious state Sre - slave unit is under relacement in case of major failure SRe - relacement of slave unit is continuing from the revious state Swre - failed slave unit waiting for reair from the reairman Swre - failed slave unit waiting for relacement from the reairman Rest other notations are same as used in model-i. 3.3 Transition Probabilities and Mean Sojourn Times A transition diagram showing the various states of transition of system is shown as in Fig. 2. The eochs of entry into states 0, 1, 2, 3 and 4 are regenerative oints and thus these states are regenerative states. The transition robabilities are given below : 01 = q λ α, 02 = ( q 1) α λ α 2 α q λ α, ( q 1) λ λ α 03 =, 04 =, 15 = ( + q1) [1 h()], 16 = q2 [1 h()], 10 = h(), 27 = q2 [1 h()], 28 = ( + q1) [1 h()], 20 = h(), 39 = q2 [1 g1()], 3,10 = ( + q1) [1 g1()], 30 = g1(), 4,11 = ( + q1) [1 g1()], 4,12 = q2 [1 g1()], 40 = g1(), (6) 12 q2 [1 h()], (5) 14 ( + q1) [1 h()], (7) 21 q2 [1 h()], (8) 23 ( + q1) [1 h()], 2 λ (9) 32 q2 [1 g1()], (10) 34 ( + q1) [1 g1()], (11) 43 ( + q1) [1 g1()], (12) 41 q2 [1 g1()] By these transition robabilities, it can be verified that = 1 htt://

5 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril = 1 = = 1 = ,10 = 1 = 30 + (6) (5) (7) (8) (9) (10) (11) (12) ,11 + 4,12 = 1 = The mean sojourn time (i) in the regenerative state i are given by 0 = 2 = 1 λ α, 1 = 1 (α ) h α g 1 1 (α ) α, 3 = 1 (λ ) h λ g 1 1 (λ ) λ 4 = The unconditional mean time taken by the system to transit for any regenerative state j when it (time) is counted from the eoch of entrance into state i is mathematically stated as td Q (t) - q (0) ij ' ij mij = 0 Thus, m01 + m02 + m03 + m04 = 0 m10 + m15 + m16 = 1 m20 + m27 + m28 = 2 m30 + m39 + m3,10 = 3 m40 + m4,11 + m4,12 = 4 H m10 + m14(5) + m12(6)= 0 (t) dt = 1 (say) H m20 + m21(7) + m23(8)= 0 (t) dt = 1 G m30+ m32(9) + m34(10)= 0 1(t)dt =2 (say) G m40 + m43(11) + m41(12) = 0 1(t) dt = Mean Time to System Failure Regarding the failed state as absorbing states and emloying the arguments used for regenerative rocesses, we have the following recursive relation for i(t). 0(t) = Q01(t) 1(t) + Q02(t) 2(t) + Q03(t) 3(t) + Q04(t) 1(t) = Q10(t) 4(t),, 0(t) + Q15(t) + Q16(t) htt:// 2(t) = Q20(t) 0(t) + Q27(t) + Q28(t) 3(t) = Q30(t) 0(t) + Q39(t) + Q3,10(t) 4(t) = Q40(t) 0(t) + Q4,11(t) + Q4,12(t) Now, taking L.S.T. of the above equations and solving them for 0(s), we obtain N 0(s) 0(s) 0(s) = D 0 q01 q15 q16 q02 q 27 q 28 N (s) (s) [ (s) (s)] (s) [ (s) (s)] [ (s) (s)] q (s) q q3,10 q (s)[ q (s) q (s)] 04 4,11 4,12 D (s) 1 - (s) (s) (s) (s) 0 q01 q10 q02 q20 - q (s) q (s) - q (s) q (s) Now the mean time to system failure (MTSF) when the system starts from the state 0, is 1 - φ0 (s) N lim s0 MTSF = s D N = D = Availability Analysis A0(t) = M0(t) + q01(t) A1(t) + q02(t) A2(t) + q03(t) A3(t) + q04(t) A4(t) A1(t) = M1(t) + q10(t) A0(t) + q12(6)(t) A2(t) + q14(5)(t) A4(t) A2(t) = M2(t) + q20(t) A0(t) + q21(7)(t) A1(t) + q23(8)(t) A3(t) A3(t) = M3(t) + q30(t) A0(t) + q32(9)(t) A2(t) + q34(10)(t) A4(t) A4(t) = M4(t) + q40(t) A0(t) + q41(12)(t) A1(t) + q43(11)(t) A3(t) M0(t) = e -(+)t M1(t) = e - t H (t) M2(t) = e - t H (t) M3(t) = e - t G 1(t) M4(t) = e - t G 1(t) Taking L.T. of the above equations and solving them for A0(s), we get : N 1(s) D (s) A0(s) = 1 In steady-state, availability of the system is given by : A0 = s (s) lim A 0 s 0 = N1 D 1

6 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril Fig. 2 N1 = 0[{1 12(6) 21(7)} {1 34(10) 43(11)} - 14(5) 41(12) {1-23(8) 32(9)} - 23(8){32(9) + 12(6) 34(10) 41(12)} + 43(11){ 34(10) 02-14(5) 32(9) 21(7)}] + 1 [{ (12)} {1-23(8) 32(9)} + 34(10) 41(12) { (8)} + 21(7) (10) 43(11) 01(02 1) + 32(9) 21(7) { (11)}] + 2 [{ (9)} {1 14(5) 41(12)} (6) {1 34(10) 43(11)} + 41(12) 12(6) { (10)} + 43(11) 32(9){ (5)}] + 3 [{ (8)} {1 14(5) 41(12)} 14(5) 43(11) { (7)} 43(11) 04 {1 12(6) 21(7)} + 12(6) 23(8) { (12)} - 12(6) 21(7) ] + 4 [{ (10)} {1 12(6) 21(7)} + 34(10) 23(8) { (6)} - 23(8) 32(9) (5) 01{1-23(8) 32(9)} + 14(5) 21(7) { (9)}] D1 = 0 [10{1-34(10)43(11) 32(9) 23(8)} (5){1-32(9) 23(8)} + 12(6) 20 {1 34(10) 43(11)} + 14(5) 43(11){ (9)} + 12(6) 23(8){ (10)}] + 1 [01{1-34(10)43(11) 32(9) 23(8)} (12) {1-32(9) 23(8)} (7){1-34(10) 43(11)} + 21(7) 32(9){ (11)} + 41(12) 34(10) { (8)} + 02{1-34(10)43(11) 41(12) 14(5)} (9) {1 41(12) 14(5)} (6){1-34(10)43(11)} + htt:// 43(11) 32(9){ (5)} + 41(12) 12(6) { (10)}] + 2 [03{1 12(6) 21(7) 03 41(12)} (10) {1 12(6) 21(7)} (8) {1 14(5) 41(12)} + 12(6) 23(8){ (12)} + 43(11) 14(5) { (7)} + 04{1 21(7)12(6) 32(9) 23(8)} (10) {1-21(7) 12(6)} (5){1 23(8)32(9)} + 21(7) 14(5){ (9)} (8) {12(6 ) 34(10) 32(9) 14(5)}] 3.6 Busy Period Analysis of Exert Reairman (Reair Time Only) Using the robabilistic arguments, we have the following recursive relations for BRi(t) : BR0(t) = q01(t) BR1(t) + q02(t) BR2(t) + q03(t) BR3(t) + q04(t) BR4(t) BR1(t) = q10(t) BR0(t) + q12(6)(t) BR2(t) + q14(5)(t) BR4(t) BR2(t) = q20(t) BR0(t) + q21(7)(t) BR1(t) + q23(8)(t) BR3(t) BR3(t) = W3(t) + q30(t) BR0(t) + q32(9)(t) BR2(t) + q34(10)(t) BR4(t) BR4(t) = W4(t) + q40(t) BR0(t) + q41(12)(t) BR1(t) + q43(11)(t) BR3(t)

7 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril W3(t) = G 1 (t) = W4(t) Taking L.T. of the above equations and solving them for BR0(s),we get : N2(s) D (s) BR0(s) = 1 In steady-state, the total fraction of time for which the system is under reair by exert reairman, is given by : lim s BR 0 (s) N D BR0 = s 0 = 1, N2 = 2 [(1+ 34(10))(01 12(6) 23(8) (8) (6) 21(7)) + (1 + 43(11)) (01 14(5) (5) 21(7) + 04) - 14(5) 41(12)( (8)) + 14(5) 32(9) (03 21(7) (8)) - 04 (12(6) 41(12) + 23(8) 32(9) + 21(7) 12(6)] and D1 is already secified. 3.7 Busy Period Analysis of Exert Reairman (Relacement Time Only) Using the robabilistic arguments, we have the following recursive relations for BRPi(t) : BRP0(t) = q01(t) BRP1(t) + q02(t) BRP2(t) + q03(t) BRP3(t) + q04(t) BRP4(t) BRP1(t) = W1(t) + q10(t) BRP0(t) + q12(6)(t) BRP2(t) + q14(5)(t) BRP4(t) BRP2(t) = W2(t) + q20(t) BRP0(t) + q21(7)(t) BRP1(t) + q23(8)(t) BRP3(t) BRP3(t) = q30(t) BRP0(t) + q32(9)(t) BRP2(t) + q34(10)(t) BRP4(t) BRP4(t) = q40(t) BRP0(t) + q41(12)(t) BRP1(t) + q43(11)(t) BRP3(t) W1(t) = H (t) = W2(t) Taking L.T. of the above equations and solving them for BRP0(s),we get : N 3(s) D (s) BRP0(s) = 1 In steady-state, the total fraction of time for which the system is under relacement by exert reairman, is given by : lim s BRP 0 (s) BRP0 = s 0 = 1, N3 = 1 [(1-34(10) 43(11)) ( (6) (7)) + 32(9) (04 43(11) + 03) (1 + 21(7)) (9) (14(5) 43(11) - 23(8)) + 41(12){(1 + 12(6)) ( (10)) (8) 34(10) (8) 32(9) (5) (5) 32(9)}] and D1 is already secified. 2 N D 3.8 Exected Number of Visits By Exert 3 htt:// Reairman Using the robabilistic arguments, we have the following recursive relations for Vi(t) : V0(t) = Q01(t) [1+V1(t)] + Q02(t) [1+V2(t)] + Q03(t) [1+V3(t)] + Q04(t) [1+V4(t)] V1(t) = Q10(t) V0(t)+ Q12(6)(t) V2(t) + Q14(5)(t) V4(t) V2(t) = Q20(t) V0(t)+ Q21(7)(t) V1(t) + Q23(8)(t) V3(t) V3(t) = Q30(t) V0(t)+ Q32(9)(t) V2(t) + Q34(10)(t) V4(t) V4(t)= Q40(t) V0(t)+Q41(12)(t) V1(t)+Q43(11)(t) V3(t) Taking L.S.T. of the above equations and solving them for V0(s), we get : N4(s) D (s) V0(s) = 1 In steady-state, the number of visits er unit time is given by : lim s V 0 (s) N D V0 = s 0 = 1 N4 = 1-34(10) 43(11) - 23(8) 32(9) 12(6) 21(7) + 12(6) 21(7) 34(10) 43(11) 21(7) 32(9) 14(5) 43(11) 41(12) 12(6) 23(8) 34(10) - 14(5) 41(12) + 41(12) 14(5) 32(9) 23(8) and D1 is already secified. 3.9 Exected Number of Relacements Using the robabilistic arguments, we have the following recursive relations for RPi(t) : RP0(t)=Q01(t) [1+RP1(t)]+Q02(t) [1+RP2(t)]+Q03(t) RP3(t) + Q04(t) RP4(t) RP1(t)=Q10(t) RP0(t)+Q12(6)(t) [1+RP2(t)]+ Q14(5)(t) RP4(t) RP2(t)=Q20(t) RP0(t)+Q21(7)(t) [1+RP1(t)]+ Q23(8)(t) RP3(t) RP3(t)=Q30(t) RP0(t)+Q32(9)(t) [1+RP2(t)]+ Q34(10)(t) RP4(t) 4

8 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril RP4(t)=Q40(t) RP0(t)+Q41(12)(t) [1+RP1(t)]+Q43(11)(t ) RP3(t) Taking L.S.T. of the above equations and solving them for RP0(s), we get : RP0(s) = N 5(s) D 1(s) In steady-state, the number of relacements er unit time is given by : lim s RP 0 (s) N D RP0 = s 0 = 1 N5 = ( ) [(1-34(10)43(11))(1 12(6)21(7)) - 14(5) 41(12) (1-32(9)23(8)) + 23(8)(- 32(9) - 41(12)34(10)12(6)) + 43(11)(02 34(10) - 14(5)32(9)21(7)] + 12(6)[(1-32(9)23(8)) (0441(12) + 01) (7) + 41(12) 34(10)(02 23(8) + 03) - 34(10) 43(11) 01(1-02) + 32(9)21(7) (04 43(11) + 03)] + 21(7)[(1-14(5) 41(12) ) (03 32(9) + 02) (6)(1-34(10)43(11)) + 41(12) 12(6)(34(10) ) + 32(9)43(11) (14(5) )] + 32(9) [(1-14(5)41(12) )(0223(8) + 03) - 14(5)43(11) ( (7)) 0443(11)(1 12(6) 21(7)) - 12(6)21(7) + 12(6)23(8) (04 41(12) + 01)] + 41(12) [(1-12(6)21(7) )(03 34(10) + 04 ) + 34(10)23(8) ( (6)) (5) (1 23(8)32(9)) + 14(5)21(7) ( (9) ) (8) 32(9)] and D1 is already secified. 5 C5 = Cost er unit relacement. 4. Comarative Analysis Let Pi be the rofit of the model discussed in the ith model (i = 1,2) resectively. 4.1 Comarison Between Model-I and Model II Fig. 3 shows the behaviour of the difference of rofits (P1 P2) with resect to failure rate (α) for different values of the revenue er unit u time (C 0). It can be interreted from the grah that the difference of rofits (P1 P2) decreases with increasing failure rate (α) and has higher values for higher values of the revenue er unit u time (C 0). Further, more conclusions can be drawn as follows: (i) For C 0 = 55, the difference of rofits (P1 P2) > or = or < 0 if α < or = or > So, the model-i is better or worse than that of model-ii according as α < or > In case of α = , both the models are equally good. (ii) For C 0 = 60, the difference of rofits (P1 P2) > or = or < 0 if α < or = or > Therefore, the model of model-i is better or worse than that of model-ii according as α < or > Both the models are equally good for α = (iii) For C 0 = 65, the difference of rofits (P1 P2) > or = or < 0 if α < or = or > So, the model of model-i is better or worse than that of model -II according as α < or > For α = , both the models are equally good. Fig. 4 shows the behaviour of the difference of rofits (P1 P2) with resect to cost er visit for different values of the revenue er unit u time (C 0). It can be observed from the grah that the difference of rofits (P1 P2) decreases with increase in the cost er visit and has higher values for higher values of the revenue er unit u time (C 0). More conclusions can be drawn as follows: 3.10 Profit Analysis Profit (P2) = C0 (A0) C2 (BR0) C3 (BRP0) C4 (V0) C5 (RP0) C0 = Revenue er unit utime. C2 = Cost er unit utime for which the exert reairman is busy for reair. C3 = Cost er unit utime for which the exert reairman is busy for relacement. C4 = Cost er visit of reairman. DIFFERENCE OF PROFIT (P 1 -P 2 ) vs. FAILURE RATE (α) FOR DIFFERENT VALUES OF THE REVENUE PER UNIT UP TIME (C 0 ) 6 4 C 0 = 5 5 C 0 = 6 0 C 0 = 6 5 htt:// 2 (P 1 -P 2 )

9 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril =0.193, q 1 =0.706, q 2 =0.101, = , 1 =0.3206, 2 =0.3417, 3 =0.1336, C 2 =1000, C 3 =1000, C 4 =5000, C 5 = Failure Rate (α) Fig DIFFERENCE OF PROFIT (P 1 - P 2 ) vs. COST PER VISIT FOR DIFFERENT VALUES OF THE REVENUE PER UNIT UP TIME (C 0 ) =0.193, q 1 =0.706, q 2 =0.101, = , = , 1 =0.3206, 2 =0.3417, 3 =0.1336, C 2 =1000, C 3 =1000, C 5 = C0 = 8 C0 = 9 C0 = 10 ) P 2 - (P COST PER VISIT Fig. 4 (i) For C 0 = 8, the difference of rofits (P1 P2) > or = or < 0 if cost er visit < or = or > INR So, the model of model-i is better or worse than that of model-ii according as cost er visit < or > INR In case cost er visit = INR 28510, both the models are equally good. (ii) For C 0 = 9, the difference of rofits (P1 P2) > or = or < 0 if cost er visit < or = or > INR So, the model of model-i is better or worse than that of model-ii according as cost er visit < or > INR Both the models are equally good if cost er visit = INR (iii) For C 0 = 10, the difference of rofits (P1 P2) > or = or < 0 if cost er visit < or = or > INR So, the model of model-i is better or worse than that of model-ii according as cost er visit < or > INR If cost er visit = INR 60014, both the models are equally good. htt://

10 International Journal of Scientific & Engineering Research, Volume 4, Issue 4, Aril Conclusion After comarison between the two models, we conclude that there are different situations in a articular model can be referred over the other model. Deending uon resources available and situations, the organization can adot the model which is more rofitable to it. References [1] Attahiru Sule Alfa, W.L. and Zhao, Y.Q., Stochastic analysis of a reairable system with three units and reair facilities. Microelectron. Reliab., 38(4), (1998). [2] Tuteja. R.K., Taneja, G. and Uasana Vashistha, Costbenefit analysis of a system oeration and sometimes reair of main unit deends on sub-unit, Pure and Alied Mathematika Sciences, LIII(1-2), 41-61(2001). [3] Guta, K.C., Sindhu, B.S. and Taneja, G. L., Profit analysis of a system with atience time and two tyes of visiting reairman. Journal of Decision and Mathematical Sciences, 7(1-3), (2002). [4] Tuteja. R.K., Taneja, G. and Parashar B., Probabilistic Analysis of a PLC Hot Standby System with Three Tyes of Failure and Two Tyes of Reair Facility, Pure and Alied Mathematika Sciences,Vol. LXIV, No.1-2, Pg (2006). [5] B. Parashar, Statistical Comarative Study of Profit Analysis of Two Different Models for Two units PLC Systems. IFRSA s International Journal of Comuting (IIJC), Vol. 1, Issue 2, Pg (2011). htt://