Performance Analysis of Network System Configured as Series-Parallel Subject to Different Repair Policies: Copula Approach to Joint Probability Distribution

Ismail Muhammad Musa1*, Ibrahim Yusuf2

department of Mathematical Sciences, Al-Qalam University katsina, Nigeria ^Department of Mathematical Sciences, Bayero University Kano, Nigeria

beanec28@gmail.com iyusuf.mth@buk.edu.ng

Abstract

A network system is designed in this paper. It has two labs connected in parallel with four units each; working under 3-out-of-4: G; policy, two servers connected in parallel working under 1-out-of-2: G; policy and a router. The labs, servers and the router are connected in series all together. By making use of a supplementary variable and Laplace transforms to varies measures of system reliability, the network system has been studied and evaluation of Sensitivity, Availability, Reliability, (Mean time to failure) MTTF, and cost analysis for particular values of the failure and repair rates is made. Conclusion has been done with computed results demonstrated by tables, figures and graphs.

Keywords: Network, System, Availability; Reliability; Gumbel Hougard, seriesparallel.

1. Introduction

It is very vital that a network system or any industrial system should be defendable and consistent; such system is more desirable than any other. The availability and reliability of that system plays a big role in its production output, performance, industrial growth and expected profit. In trying to enhance the performance of such desirable network/industrial systems many researchers evaluate different systems. Kabiru and Singh [1] made a reliability assessment of complex system consisting two subsystems connected in series configuration using Gumbel-Hougaard family copula distribution. Yusuf and Ismail [6] focused on reliability analysis of communication network system with redundant Relay Station under Partial and Complete Failure.

Singh and Poonia [14] study the performance analysis of a complex repairable system with two subsystems in series configuration with an imperfect switch. Using transition diagrams and systems of first order differential difference equations Yusuf and Sani in [2] developed and solved a system recursively to obtain the steady-state availability, busy period of repair men and profit function. Kakkar and Chitkara [15] consider finding the reliability of two unit parallel industrial system. Lado and Singh [4] develop a model of a complex repairable system having two subsystems A and B which is connected in a series configuration. Kakkar and Chitkara [9] present reliability of two dissimilar parallel units in the present of preventive maintenance. Yusuf and Mahmud [3] use Markov model of a system derived through the system state transition table and differential

Ismail Muhammad Musa, Ibrahim Yusuf

PERFORMANCE ANALYSIS OF NETWORK SYSTEM CONFIGURED AS .. c c ' u°

Volume 15, September 2020

SERIES-PARALLEL SUBJECT TO DIFFERENT REPAIR POLICIES...___

difference equations which are further used to evaluate the system availability and mean time to system failure and profit.

Using Copula Singh and Gulati [13] studied the performance analysis of complex system in series configuration under different failure and repair discipline. Garg [18] use fuzzy Kolmogorov's differential equations to make an approach for analyzing the reliability of industrial system. Yusuf and Bala [21] made an analysis of reliability characteristics of a parallel system with external supporting devices for operation.

2. State Description, Notation, and Assumptions 2.1 State description

So: This state represents a fully functional system with three working units and one standby unit in

both labs, one operational server and one standby server, and working router.

S1: This represents operational state with one unit down in lab 1.

S2: This represents failure state due to failure of two units in lab 1.

S3: This represents operational state with one unit down in lab 2.

S4: This represents failure state due to failure of two units in lab 2.

S5: This represents operational state with one server down.

S6: This represents failure state due to failure of both servers.

S7: This represents failure state due to failure of the main router.

S8: This represents operational state with one unit in lab 1 down and one server under repair. S9: This represents operational state with one unit in lab 2 down and one server under repair. S10: This represents operational with one server down and a unit in lab 2 under repair. S11: This represents operational with one server down and a unit in lab 1 under repair.

2.2 Notations

• t Stands for Time variable on a time scale.

• s Stands for Laplace transform variable for all expressions.

• p1 Stands for Failure rate of any unit in lab 1.

• p2 Stands for Failure rate of any unit in lab 2.

• p3 Stands for Failure rate of any server.

• p4 Stands for Failure rate of the router.

• <p(a) Stands for all Repair rates.

• Pi (t) Stands for the probability that the system is in Si state at instants for i = 0 to 9.

• P(t) Stands for Laplace transformation of the state transition probability P(t).

• Pi (a, t) Stands for the probability that a system is in state Si for i = 1,2, ...,11, the system is running under repair and elapse repair time is (a, t) with repair variable a and time variable t.

• Ep(t) Stands for Expected profit during the time interval [0, t).

• K1, K2 Stands for Revenue and service cost per unit time respectively.

2.3 Assumptions

It's assumed that at the beginning all the units in labs, servers and the router are working perfectly. At least three units from the two labs, one server and a router are needed for the system to operate. Failure rates are regarded as the same and may follow exponential distribution. Likewise repairs follow general distribution.

Figure 1: Transition Diagram

3. Mathematical Model Formulation

Bearing in mind the probability and continuity arguments, train set of difference differential equations and the present mathematical model are lump together as:

(yt + 4ßi + 4ß2 + 2ß3 + ß4) p0(t) = f0°° 0(a)Pl(a, t)da + f0°°^(a)p2(a, t)da + f0°° 0(a)p3(a, t)da + ÇKa)P4(a, t)da + f°° 0(a)p5(a, t)da + C[i(a)p6(a, t)da + f°°n(a)p7(a, t)da

(1)

(Î + Î+3Pi + 2ß3 +ß4 + 0(a)) Pi(a, t) = 0 (2)

(Tt + i + »(a))P2M = 0 (3)

(jt + Ta + 3ß2+ 2ßs +ß4 + 0(a)) Ps (a, t) = 0 (4)

ii + i + »(a))P4(a,t) = ° (5)

(i + i + 4ßi + 4ß2 +ßs+ß4 + 0(a)) p5(a, t) = 0 (6)

fe+i+^W^0 (7)

(Tt + Î + »(a))P7(a,t) = 0 (8)

(yt + ^ + ßs+ß4 + 0(a))p8(a,t) = 0 (9)

(jt + ^-a+ßs+ß4 + 0(a))p9(a,t) = 0 (10)

(Tt + Ta + 3ß2+ß4 + 0(a))pio(a,t) = 0 (11)

(yt + ^+3ß1+ß4 + 0(a))p11(a,t) = O (12)

Boundary condition

Pi(a,t) = 4ßiP0(t) (13)

p2(a,t) = 3ßiPi(a,t) (14)

p3(a,t) = 4ß2p0(t) (15)

p4(a,t) = 3ß2p3(a,t) (16)

ps(a,t) = 2ß3p0(t) (17)

p6(a,t) = ß3ps(a,t) (18) P7 (a, t) = ß4 (p0 (t) + pi (a, t) + p3 (a, t) + p5 (a, t) + p8 (a, t) + p9 (a, t) + pw (a, t)+pn (a, t))

(19)

Pg(a,t) = 2ß3Pi(a,t) (20)

p9(a,t) = 2ß3P3(a,t) (21)

Pio(a,t) = 4ß2P5(a,t) (22)

Pii(a.t) = 4ßips(a,t) (23)

Solution of the Model:

Using initial condition, Po (0) =1and Laplace transformation of equations (1) to (23) we have the following:

(s + + 4p2 + 2p3 + p4)p^(s) = r 0(a)p±(a, s)da + Cp(a)^(a, s)da + C 0(a)p3<(a, s)da +

Ç Ka)p4(a,s)da + f™ 0(a)p5(a,s)da + Ç p(a)p6(a,s)da + Ç p(a)p7 (a, s)da

(24)

{Tt + Î+3Pi + 2ß3 +ß4 + 0(a)) Pi(a,s) = 0 (25)

(s+-^ + p(a))p2(a,s) = 0 (26)

(s+ya + 3ß2+ 2ß3 +ß4 + 0(a)) p3(a, s) = 0 (27)

(s+-^ + p(a))p4(a,s) = 0 (28)

(s+ya + 4ßi + 4ß2 +ß3+ß4 + 0(a)) P5(a, s) = 0 (29)

(s+ß(a))p6(a,s) = 0 (30)

(s+p(a))p7(a,s) = 0 (31)

(s+-^ + ß3+ß4 + 0(a))p8(a,s) = 0 (32)

(s+ya + ß3+ß4 + 0(a)) p9(a, s) = 0 (33)

(s+ya + 3ß2+ß4 + 0(a)) PÏ7(a, s) = 0 (34)

(s+j-a + 3ßi+ß4 + 0(a)) pi7(a, s) = 0 (35)

Laplace transform of boundary conditions

pi(0,s) = 4ßiTo(s) (36)

p2(0,s) = 3ßipi(0,s) (37)

p7(0,s) = 4ß2P7(s) (38)

p4(0,s) = 3ß2p3(0,s) (39)

P7(0,s) = 2ß3p7(s) (40)

P7(0,s) = ß3p7(0,s) (41)

P7(0, s) = ß4 (p0(s) + Pi(0, s) + P3(0, s) + P5(0, s) + P8(0, s) + P9(0, s) + Pi777(0, s) + Pi7(0, s))

(42)

p8(0,s) = 2ß3pi(0,s) (43)

Ismail Muhammad Musa, Ibrahim Yusuf

PERFORMANCE ANALYSIS OF NETWORK SYSTEM CONFIGURED AS .. c c t , ,o i,8

Volume 15, September 2020

SERIES-PARALLEL SUBJECT TO DIFFERENT REPAIR POLICIES...___

P<}(0, s) = 2ßsp3(0,s) (44)

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PÏô(0,s) = 4ß2pJ(0,s~) (45)

Üi;(0,s) = 4ßip5(0,s) (46)

The solution of (24) to (35) with (36) to (46) results as;

(47)

P(S)=M1 (^(s+sß^ßM} (48)

iW D(s) V s+3ß1 + 2ß3+ß4 J V '

(49)

^-^t:™'} (5»)

nc^m (51)

( \ = 2ÈL [i-S^(s+4ßi+4ß2 + ß3 + ß4)^

( ) = D(s) ( s+4ß1+4ß2 + ß3 + ßi } ( )

^-fj^} (53)

JT^ - ßl(i+4ßi+4ß2 + 2ß3 + i6ßiß3 + i6ß2ß3) p-V^j

-pr(s) = SJiß3Ci-sixJ^3^} S (55)

8 D(s) l s + ß3+ß4 J V '

= (56)

KM-**. (57)

~p—r\ - 8ßlß3 (i-S$(s+3ß1 + ßi)}

P,

(s) = 8_Ml(i-s*(s+s^} (58)

iiW D(s) l s+ß3+ß4 J V '

Where D(s) = s + 4pi + 4p2 + 2p3 + p4- {^[^(s + + 2p3 + p4)} + 12p12[S^(s)} + 4p2[S^(s + 3p2 + 2p3 + p4)} + 12p22[S^(s)} + 2fi3[Stp(s + 4pi + 4p2 + p3 + p4)} + 2p32[S^s)} + p4 (1 + 4pi + 4p2 + 2p3 + I6P1P3 + 16p2p3)&(5)}} (59)

The followings are Laplace transformations of the state transition probabilities when the system is in initial, partial failure and failed condition at any time:

TÇ(s) = p-(s) + Pi (s) + P^(s) + P~5(s) + p-(s) + P9 (s) + T\0(s) + P^(s) (60)

Jçw=w (1+4,1 r^ri+4*2 v:^::?4}+

{ S + 4H1 + 4P2 + P3 + P4 } + 8p1p3 { S + P3 + P4 } + 8p2p3 { S + P3 + P4 } + 8l2f. {i-^!!^ (61)

4. Analytical Study of the Model for Particular Case

Availability Analysis

WithSrf, (s) = S,. (s) = —,1 = 1 ^^ = — and considering the values of different

^ s+<p ^ s+p s s+<p s s+p. °

parameters as = 0.03, p2 = 0.02, p3 = 0.05, p4 = 0.06, $ = ^ = 1 in (61), the expression for availability by taking the inverse Laplace transform is obtained as:

Pup(t) = 0.03892291366e-2 83830i077t - 0.06425625912e-i48977640St 0.00241062060e-i28332776it + 0.00023451Q3193e-i 22973i692t + 1.036198Q63e-00i7i6306460t 0.00204112976e-i i50000000t - 0.006655302231e-i ii0000000t

Table 1 and Figure 2 below shows different values of Pup(t) using (63) and time variable t= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,

Table 1: Variation of Availability with respect to time

Time(t) Availability

0 1

2 0.9970

4 0.9672

6 0.9348

8 0.9033

10 0.8728

12 0.8433

14 0.8149

16 0.7874

18 0.7608

1,2 1 0,8 1 0,997 09672

> 0,6

S CD 0,4 $ Availability

rs ä 0,2 0

0 2 4 6 8 10 12 14 16 18 Time (t)

Figure 2: Availability as function of time

Reliability Analysis

All repair rates, 0, set to zero in equation (61), the values of failure rates as ft = 0.03, ft = 0.02, ft = 0.05, ft = 0.06 and then computing inverse Laplace transform, express the reliability for the system as;

fl(t) = -2.889523810e-03600000000t + 0.03809523810e-0 1500000000t + 0.08000000000e-0 1100000°00t + 0.5714285714e-02200000000t + 1.2000000000e-02500000000t +

2 g-0.3100000000t (64)

Table 2 and Figure 3 below shows different values of Pup(t) using (63) and time variable t= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Table 2: Computation of reliability for different values of time

Time(t) Reliability

0 1

2 0.8577

4 0.6451

6 0.4553

8 0.3106

10 0.2081

12 0.1382

14 0.0917

16 0.0610

18 0.0409

Figure 3: Reliability as function of time Mean Time to Failure (MTTF) Analysis

All repairs are set to zero in equation (49), and then taking limit, as s tends to zero express MTTF as:

MTTF = lim P~~fs) = 1 ^ + 4li + + zfe + 8life + +

sm J 401 + 4ft,+ 203 +04 ( 301 + 203 + 04 302 + 203 + 04 401 + 402 + 03 + (S4 £3+ (S4 £3+(S4

-«+JW (65)

301+04 3^1+^4/ V >

Table 3 and figure 4 below express the variation of MTTF with respect to failure rates, by fixingft = 0.02,ft = 0.05,ft = 0.06 and varying ft as 0.01,0.02,0.03,0.04,005,0.06,0.07,0.08,0.09, fixing ft = 0.03, ft = 0.05, ft = 0.06 and varying ft as 0.01,0.02,0.03,0.04,005,0.06,0.07,0.08,0.09, fixingft = 0.03, ft = 0.02, ft = 0.06 and varying ft as 0.01,0.02,0.03,0.04,005,0.06,0.07,0.08,0.09, fixing ft = 0.03, ft = 0.02, ft = 0.05, and varying ft as0.01,0.02,0.03,0.04,005,0.06,0.07,0.08,0.09, in (61).

Table 3: Computation of MTTF corresponding to the various values of failure rates

Failure rate MTTF ft MTTF ft MTTF ft MTTF ft

0.01 8.0406 7.2282 8.6971 9.6404

0.02 7.4263 6.8927 8.0926 8.9180

0.03 6.8927 6.5597 7.6128 8.3040

0.04 6.4307 6.2478 7.2208 7.7725

0.05 6.0297 5.9624 6.8927 7.3061

0.06 5.6798 5.7036 6.6126 6.8927

0.07 5.3726 5.4699 6.3694 6.5231

0.08 5.1012 5.2586 6.1555 6.1903

0.09 4.8600 5.0673 5.9650 5.8889

40 35 30 25 20 15 10 5 0

MTTF ß4

MTTF ß3

MTTF ß2

MTTF ß1

0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,09

Figure 4: MTTF corresponding to the various values of failure rates Sensitivity Analysis corresponding to (MTTF)

Table 4 and figure 5 below, express the sensitivity in MTTF of the system, using the partial differentiation of MTTF with respect to the failure rates of the system and setting parameters as ftx = 0.03,ft = 0.02, ft = 0.05, ft = 0.06.

Table 4: MTTF sensitivity as function of time

Failure rate ö(MTTF)/ ft ö(MTTF)/ ft ö(MTTF)/ ft ö(MTTF)/ ft

0.01 -65.5677 -32.4004 -68.4440 -78.8498

0.02 -57.2902 -33.9081 -53.4467 -66.2544

0.03 -49.6020 -32.4031 -43.1042 -56.9357

0.04 -42.9726 -29.8976 -35.6883 -49.6544

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0.05 -37.3926 -27.1880 -30.1944 -43.8144

0.06 -32.7291 -24.5889 -26.0096 -39.0221

0.07 -28.8269 -22.2133 -22.7442 -35.0194

0.08 -25.5471 -20.0894 -20.7423 -31.6289

0.09 -22.7743 -18.2092 -28.7236 -28.7236

Figure 5: MTTF sensitivity as function of time

Cost Analysis

£p(0 = Ki/>up(t)dt-tf2t (65)

Equation (65) can be used to attain expected profit of the system in the interval [0,t), if service facilities is always available. Using (61) and (65), we obtain:

£p(t) = tf1(-0.01371345485e-2 838301077t + 0.04313147859e-1489776405t + 0.001878413822e-128332776" - 0.0001907069004e-1229731692t - 60.37376699e-0017163064600t + 0.001768793892 e-1 150000000t + 0.005995767776e-1 11000000M + 60.335) -

(66)

Table 5 and Figure 6 express the expected profit by setting = land = 0.6,0.5,0.4,0.3,0.2 and 0.1 respectively and varying time t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 in (66).

Table 5: Expected profit as function of time

Time(t) Ep(t);K2=0.6 Ep(t);K2=0.5 Ep(t);K2=0.4 Ep(t);K2=0.3 Ep(t);K2=0.2 Ep(t);K2=0.1

0 0 0 0 0 0 0

2 0.8016 0.0016 1.2016 1.4016 1.6016 1.8016

4 1.5672 1.9672 2.3672 2.7672 3.1672 3.5672

6 2.2690 2.8690 3.4690 4.0690 4.6690 5.2690

8 2.9069 3.7069 4.5069 5.3069 6.1069 6.9069

10 3.4828 4.4828 5.4828 6.4828 7.4828 8.4828

12 3.9987 5.1987 6.3987 7.5987 8.7987 9.9987

14 4.4567 5.8567 7.2567 8.6567 10.0567 11.4567

16 4.8588 6.4588 8.0588 9.6588 11.2588 12.8588

18 5.2069 7.0069 8.8069 10.6069 12.4068 14.2069

70

60

Ep(t);K2=0.1

Ep(t);K2=0.2

Ep(t);K2=0.3

Ep(t);K2=0.4

Ep(t);K2=0.5

Ep(t);K2=0.6

0 2 4 6 8 10 12 14 16 18

Time (t)

Figure 6: Expected profit as function of time

5. Conclusions

To make a conclusion on the performance of the system in this study, a study of the reliability measures on repair and failure rates with different values has been made. A provision of information on the availability of the system changing with time at fixed failure rates with different values has been made in table 1 and figure 2. It has been evidently found that gently and slowly the availability of system decreases and the probability of the failure increases at fixed failure rates p1 = 0.03, p2 = 0.02, p3 = 0.05, p4 = 0.06, and at a long run it will become steady to zero value. Therefore, at any chosen time and set of parametric values, one can easily say the future behavior of the system.

Setting repairs to zero, an analysis of the system reliability has been provided in table 2 and figure 3. By studying the availability and reliability of the system, one can conclude that providing repair to the system is better than replacement.

With other parameters considered as constant, mean-time-to-failure (MTTF) in variation of P1,P2,P3 and P4 respectively has been provided in figure 4. It also signifies that the variation of P1,P2,P3 and p4 are responsible for the better performance of the system.

Figure 5 shows the sensitivity analysis of system.

Figure 6 and table 5 shows the calculations of the profit at fixed revenue cost K1 = 1 per unit time and service costK2 = 0.6,0.5,0.4,0.3,0.2 and 0.1. The result shows that when the service cost K2 fixed at minimum value 0.1, the expected profit increases with respect to the time. In the end, one can observe that profit decrease whenever service cost increase.

References

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Received: July 23, 2020 Approved: September 20, 2020