CC BY
21
The Seasonal Effect of Working Conditions of an
Ice-cream Plant
Upasana Sharma1 and Drishti 2*
1,2Department Statistics, Punjabi University, Patiala- 147002, India usharma@pbi.ac.in, drish2796@gmail.com
Abstract
An ice-cream plant's workings are analyzed in the summer and winter seasons of the paper. The ice-cream unit along with the other three units i.e., flavoring, freezing and combined flavouring and freezing units are always operational in summers, due to the high demand, while in winters the combined flavoring and freezing unit is kept in cold standby as a backup in case there is a demand for ice-cream. In this work, the semi-Markov process and the regenerative point technique have been used to analyze the system. Numerical analysis has been conducted using MATLAB. A variety of measures have been developed to evaluate the effectiveness of a system. The Code Blocks have been used in interpreting the graph in the specific case presented. All evaluation is based on the milk production data collected by the plant. Improvements to the system performance will lead to increased profits. Similar techniques can be applied to other systems.
Keywords: Seasonal functioning, semi-Markov process, Regenerative point technique, profit.
1. Introduction
The primary goal of any industry is to upgrade production through technological interventions and so, is the goal of the dairy industry i.e to improve their production operations to attain competitiveness. Physical models for predicting ice cream's thermal properties were developed by [1]. Analysis of reliability modeling for 2-out-of-3 redundant system was done by [2]. Profit analysis of a two unit standby oil delivering system where priority is given to partially failed unit over the completely failed unit for repair was analyzed by [3]. Reliability models for the fertilzer industry were pioneered by [4], [5]. Availability optimization of ice cream making unit of milk plant was discussed by [6]. Optimized scheduling, production planning and RAM study of an ice-cream plant was given by [7], [12] respectively. Reliability and profit where stand by units functions to accommodate the required demand was analyzed by [8] for system evaluation. Availability analysis of a skim milk powder and profit analysis of a butter-oil production in dairy industry was discussed by [9] respectively. Profit analysis of a system where operation is affected by temperature was discussed by [10]. Description of the four subsystems of the butter-oil production process- the melting vats, the boilers, the clarifier and the settling tanks was given by [11]. The probability of a three-unit induced draft fan system with one standby unit in a working condition was given by [13]. Modeling of two-Unit cold standby system was discussed by [14]. On the basis of progressively censored first-failure data, a problem of estimating parameters for an exponential distribution class and hazard rate functions is studied by [15]. But, none of them have discussed the working of an ice-cream plant. So, in this paper functioning of an ice-cream plant is discussed. The production of functional ice-cream plant consists of one main unit and three units grouped in parallel but in series with the unit 1. The main unit (unit 1) consists of heating, emulsifying, pasteurization, homogenization and ageing. Unit 2 is flavoring unit, unit 3 is freezing unit and unit 4 is combined flavoring and freezing. In summers, due to
high demand the whole system is operative whereas in winters, the system goes to cold standby state and undergoes maintenance. It only operates when the demand occurs. In that case the units 2, 3 along with the unit 1 operates and the unit 4 is in cold standby state and operates on the failure of either of the units 2, 3 or on the failure of both. The system goes to a failed state a) on the failure of the unit 1 or b) on the failure of unit 2 with unit 4 or c) on the failure of unit 3 with unit 4.
Semi-Markov process and regenerative point technique is used to obtain measures of system effectiveness in steady state that include MTSF, availability of the system, busy period of repairman for repair and maintenance, expected number of repair and maintenances, profit of the system.
2. Methods
The stages of methodlogy carried out is given below:
1. In the beginning, industry data on failure rates and maintenance was collected over a five-year period.
2. A comprehensive understanding of how the unit operates is the second step. Through that, reliability models are generated.
3. MATLAB is used to obtain reliability measures using semi-Markov processes and regenerative point techniques that include:
• Transition probabilities and mean sojourn time in steady state
• MTSF of the system.
• Long term availability for the system.
• Bus period analysis of the repairman.
• Expected number of repairs.
• Additionally, the system's profit potential is analyzed graphically.
4. In the following step, graphical analysis is performed by using excel and code blocks on a particular example of exponential distribution.
5. Furthermore, reliability can be improved by identifying key machines and faults, making better decisions, and formulating better strategies.
3. Annotations and Symbols
Table 1
Notations of the model
Notations Descriptions
a Failure rate of unit 1.
ai Failure rate of unit 2.
a2 Failure rate of unit 3.
a3 Failure rate of unit 4.
a4 Maintenance rate of the unit.
a, ß Rate of going to winters and summers respectively.
6 Repair rate of unit 1.
01 Repair rate of unit 2.
62 Repair rate of unit 3.
63 Repair rate of unit 4.
Notations of the model
Notations Descriptions
04 Repair rate after maintenance.
G(t), g(t) c.d.f and p.d.f of repair time of unit 1.
Gi (t), gi (t) c.d.f and p.d.f of the repair time of unit 2.
G2 (t), g2 (t) c.d.f and p.d.f of the repair time of unit 3.
G3 (t), g3 (t) c.d.f and p.d.f of the repair time of unit .
G4 (t), g4 (t) c.d.f and p.d.f of the maintenance time.
★ Symbol for convolution.
© Laplace Stieltjes Treansform.
Ml (t) Probability that the system is in up state at time t.
W (t) Probability that the system is busy for repair at time t.
Al (t) Availability of the system at time t.
Bl (t) Busy period for repair/maintenance the system at time t.
Vl (t) Expected number of repairs/maintenances.
Descrpition
States Description and symbols of the model
S These are the states when the system is operative, j=0,15,17,18.
S4, S5, S6 These are the states when the system works in a reduced capacity.
S2 This is the cold stanby state.
Si These are the failed states, i=3,7,8,9,10,11,12,13,14,16,19,20,21,22.
S, W Symbols for summer and winter respectively.
o(u1,2,3,4) The units are in operating state.
o(u1,4) Units 1 and 4 are in operating state.
o(u1,2,3) Units 1,2,3 are in operating state.
csu4 Unit 4 is in cold standby state.
fric Ice-cream unit is under repair due to failure in unit 1.
umic Ice-cream unit is under repair.
fRu4 Unit 4 is under continuous repair.
fru2,fru3,fru4 Units 2,3,4 are under repair respectively.
fwru2,fwru3 Units 2,3 are waiting for repair.
4. Model Descriptions and Assumptions
Reliability analysis is done of the working of an ice-crream plant w.r.t. seasons. At an initial stage the system is operative, in summers since the demand is high all the units are operative whereas in winters the system is in cold standby state and only operates when there is some demand.The system consists of four units, unit 1 is the main unit of the system where the process starts after that it moves to unit 2 i.e., the flavouring unit and after that unit 3 i.e., the freezing unit. Unit 4 is combined freezing and flavouring unit. In summers, the system operates at reduced capacity when any of the units 2,3,4 fails. The system goes to failed state on the failure of unit 1; unit 2, 4 and unit 3, 4.
Following are the assumptions of the system:
• The system is operating initially.
• A distribution of exponential failure times is assumed for all failure times.
• Unit 1 and unit 4 receive the most priority for repair.
States always restores the system to its original functionality after every repair.
Figure 1: State Transition Diagram
5. System Effectiveness Measures
In this model the states So, Si, S15, S17, si8 are the operating states. States S4, S5, S6 are the states operating in a reduced capacity. S2 is a cold standby state, rest are the failed states.
5.1. Mean time to system failure (MTSF)
System effectiveness measures have been achieved using semi-Markov processes and regenerative point techniques. A mean time to failure (MTSF) is determined for the system when considering the failed state as an absorbent state. In terms of probabilistic arguments, we can get the following recursive relation for fa (t):
fa (t) = Ln Qln (t)®fan (t) + Le Qle (t) where Sn indicates an un-failed regenerative state into which the given regenerative state Si can transit and Se indicates a failed state into which the state Si can transit directly. By applying the Laplace-Stieltjes Transform (L.S.T.) to the relationships given by the above equation and solving them for fa***(t), we are able to calculate:
c (t) = Ng
The mean time to system failure (MTSF), when the system started at the beginning of state So is: MTSF=/0~ R(t)dt = lims^0 R* (s) Using L' Hospital rule and putting the value of fa*** (s) we get
MTSF = T0 = lims^0 = i-fi£l = DD where
N=^0 (pi3 + pi4p47 + pi4p48 + pi5p59 + pi5p5,10 + pi6p6,ii + pi6p6,12 + pi6p6,13 - pi3pi5,2p2,15 -
pi3pi5,17pi7,15 - pi3pi5,18pi8,15 - pi4p47pi5,2p2,15 - pi4p48pi5,2p2,15 - pi5p59pi5,2p2,15 -
pi5pi5,2p2,15p5,10 - pi6pi5,2p2,15p6,ii - pi6pi5,2p2,15p6,12 - pi6pi5,2p2,15p6,13 - pi4p47pi5,17pi7,15 -
P14P48P15,17P17,15 - P15P59P15,17P17,15 - P14P47P15,18P18,15 - P14P48P15,18P18,15 -
P15P59P15,18P18,15 - P15P5,10P15,17P17,15 - P15P5,10P15,18P18,15 - P16P6,11 P15,17P17,15 -
P16P6,12P15,17P 17,15 - P16P6,13P 15,17P17,15 - P16P6,11 P15,18P18,15 - P16P6,12P15,18P18,15 -
P16P6,13P15,18P 18,15) + m (P01 - P01 P15,2P2,15 - P01P 15,17P17,15 - P01 P15,18P18,15) + H2 (P02 -
P02P14P41 - P02P15P51 - P02P16P61 - P02P15,17P17,15 - P02P 15,18P18,15 + P02P14P41 P15,17P17,15 +
P02 P15 P51P 15,17 P 17,15 + P02 P16 P61P 15,17 P17,15 + P02 P14 P41 P15,18 P18,15 + P02 P15 P51 P15,18 P18,15 +
P02 P16 P61P 15,18P 18,15) + U4 ( P01P14 - P01P14 P15,2 P2,15 - P01P14 P15,17P17,15 - P01P14 P15,18P18,15 ) +
U5 (P01P15 - P01P15P15,2P2,15 - P01P15P15,17P17,15 - P01P15P15,18P18,15 ) + H (P01P16 -
P01P16P15,2P2,15 - P01P16P 15,17P17,15 - P01P16P15,18P18,15 ) + ^15 (P02P2,15 - P02P14P41 P2,15 -
P02P15P51 P2,15 - P02P16P61 P2,15 ) + U17 (P02P2,15P15,17 - P02P14P41 P2,15P15,17 -
P02P15P51 P2,15P15,17 - P02P16P61 P2,15P15,17) + №(P02P2,15P15,18 - P02P14P41 P2,15P15,18 -
P02P15P51 P2,15P15,18 - P02P16P61 P2,15P15,18 )
D=P14P41 P15,2P2,15 - P16P51 - P16P61 - P15,2P2,15 - P15,17P17,15 - P15,18P18,15 - P14P41 +
P16 P51 P15,2 P2,15 + P16 P61 P15,2 P2,15 + P14 P41 P15,17 P17,15 + P16 P51 P15,17 P17,15 + P16 P61 P15,17 P17,15 +
P14 P41 P15,18 P18,15 + P16 P51 P15,18 P18,15 + P16 P61 P15,18 P18,15 + 1
6. Cost Measures
6.1. Long Term Availability of the System in Summers at Full Capacity
Using the theory of regeneration process, using A/ (t) where l = 0,1 as the probability that the system will be in upstate at instant t given that it is in state i at t=0, we can find that this value will satisfy the following recursive relations:
A/(t) = M/(t)+ E„q/„(t) * A„ (t) In this case, Sn can represent any state to which S/ can transit. M/ (t) is the probability that the system will be accessible at time t before visiting any other state. M0 = e-(x+^)t, M1 = e-(A+A1+A2)t
Taking Laplace transform of equations and solving for A0 we obtain:
(s) = N1(s) Ao(s) =
Steady state availability is given by:
A0 = lims^0 sAO (s) = NN1
where N1 =
U0 + m P01 - U0P13 - U0P47 - U0P48 - U0P59 - H0P5,10 - H0P6,11 - P01P47 - P01P48 - H0P14P41 + F0P13P47 + F0P13P48 - F1P01P59 - h0P15P51 + h0P13P59 - H0P16P61 + H0P47P59 + h0P48P59 -
P01 P5,10 + h0P13P5,10 + U0P47P5,10 + h0P48P5,10 - P01 P6,11 + h0P13P6,11 + h0P47P6,11 + h0 P48 P6,11 + H0 P59 P6,11 + h0 P5,10 P6,11 + P01P47 P59 + P01P48 P59 + h0 P15 P47 P51 + h0 P14 P41P59 + h0P15P48P51 - h0P13P47P59 - h0P13P48P59 - h0P16P41P64 + h0P16P47P61 + h0P16P48P61 -h0P16P51P65 + h0P16P59P61 + P01P47P5,10 + P01P48P5,10 + h0P14P41 P5,10 - h0P13P47P5,10 -h0P13P48P5,10 + h0P16P61 P5,10 + P01P47P6,11 + P01P48P6,11 + h0P14P41 P6,11 - H0P13P47P6,11 -h0P13P48P6,11 + P01P59P6,11 + h0P15P51 P6,11 - h0P13P59P6,11 - h0P47P59P6,11 - h0P48P59P6,11 +
P01 P5,10P6,11 - h0P13P5,10P6,11 - h0P47P5,10P6,11 - H0P48P5,10P6,11 + h0P16P47P51P65 + h0P16P41P59P64 + U0P16P48P51P65 - h0P16P47P59P61 - h0P16P48P59P61 + h0P16P41P64P5,10 -h0P16P47P61 P5,10 - U0P16P48P61 P5,10 - P01P47P59P6,11 - P01P48P59P6,11 - h0P15P47P51 P6,11 -h0P14P41P59P6,11 - U0P15P48P51 P6,11 + h0P13P47P59P6,11 + H0P13P48P59P6,11 - P01P47P5,10P6,11 -
P01P48P5,10P6,11 - h0P14P41 P5,10P6,11 + U0P13P47P5,10P6,11 + h0P13P48P5,10P6,11 D = (p-3P13 + ^1)(P51 - P47P51 - P48P51 - P51 P6,11 + P47P51 P6,11 + P48P51 P6,11) + (^4 + H7P47 + №P48)(P14 + P16P64 - P14P59 - P14P5,10 - P14P6,11 - P16P59P64 - P16P64P5,10 + P14P59P6,11 + P14P5,10P6,11) + (H5 + H9P59 + h10P5,10)(P15 + P16P65 - P15P47 - P15P48 - P15P6,11 - P16P47P65 -P16P48P65 + P15P47P6,11 + P15P48P6,11) + (k + h11 P6,11)(P16 - P16P47 - P16P48 - P16P59 -P16 P5,10 + P16 P47P59 + P16 P48 P59 + P16 P47P5,10 + P16 P48 P5,10 )......(1)
6.2. Long Term Availability of the System in Summers at Half Capacity
Using the theory of regeneration process, using Al (t) where l = 4,5,6 as the probability that the system will be in upstate at instant t given that it is in state i at t=0, we can find that this value will satisfy the following recursive relations:
Al (t) = Ml (t)+ Lnqln (t) * An (t) In this case, Sn can represent any state to which Si can transit. Ml (t) is the probability that the system will be accessible at time t before visiting any other state. M4 = e-(A+A3)fGi~(t), M5 = e-(A+A3)fG2~(t), M6 = e-(A+Ai+A2)fG3"(t) Taking Laplace transform of equations and solving for A0 we obtain:
A * (s) = N2(s)
A0 (s) = Di(s)
Steady state availability is given by:
A0 = lims^0 sA* (s) = N
where
N2 = -p0i(V5pi5p47 - F5pi5 - kpi6 - f4pi4 + F5pi5p48 + kpi6p47 + kpi6p48 + f4pi4p59 + kpi6p59 - f4pi6p64 - f5pi6p65 + f4pi4p5,10 + kpi6p5,10 + f4pi4p6,ii + f5pi5p6,ii - kpi6p47p59 -kpi6p48p59 + f5pi6p47p65 + f5pi6p48p65 + f4pi6p59p64 - kpi6p47p5,10 - kpi6p48p5,10 + f4pi6p64p5,10 - f5pi5p47p6,ii - f5pi5p48p6,ii - f4pi4p59p6,ii - f4pi4p5,10p6,ii ) Di is already defined in equation (i).
6.3. Busy Period Analysis for Repair in Summers
Using the theory of regeneration process, using Bl(t) where l = 4,5,6,7,8,9,10,11 as the probability that the system is under repair at an instant t given that it is in state i at t=0, we can find that this value will satisfy the following recursive relations:
Bl (t) = Wl (t) + Lnqln (t) * Bn (t) In this case, Sn can represent any state to which Si can transit. Wl (t) is the probability that the system will be busy for repair at time t before visiting any other state. W3 = W7 = W9 = Wii = G{t), W4 = e-(A+A3) Gi(t), W5 = e-(A+A3) G{(t), W6 = e-(A+Ai+A2)G3(t) + (Aie-(A+Ai+A2) * l)G3~(t) + (a2e-(A+Ai+A2) * l)G3~(t), W8 = Wi0 = Wi2 = W13 = G3~(t)
Taking Laplace transform of equations and solving for B0 we obtain:
B* (s) = N3(s) B0 (s) = D1(s)
Steady state availability is given by:
B0 = lims^0 sB0 (s) = N
where N3 =
-p0i(pi3p47f3 - pi4f4 - pi5f5 - pi6k - pi4p47f7 - pi4p48f8 - pi5p59^9 - pi6p64^4 - pi6p65f5 -
pi5p5,10fl0 - pi6p6,iifll - pi6p47p64f7 - pi6p48p64^8 - pi3№ + pi5p47^5 + pi6p47k + pi3p48^3 -
pi6p59p65f9 + pi5p48f5 + pi6p48k + pi3p59^3 + pi4p59^4 + pi6p59k - pi6p65p5,l0fl0 + pi3p5,l0f3 +
pi4 p5,l0 f4 + pi6 p5,l0 k + pi3 p6,llf3 + pi4 p6,ll^4 + pi5 p6,ll^5 + pi5 p47 p59^9 + pi6 p47 p65^5 +
pi5 p48 p59 f9 + pi6 p48 p65 f5 + pi4 p47 p59 f7 + pi4 p48 p59 f8 + pi6 p59 p64 f4 + pi5 p47 p5,l0 fl0 +
pi5 p48 p5,l0 fl0 + pi4 p47 p5,l0 f7 + pi4 p48 p5,l0 f8 + pi6 p64 p5,l0 f4 + pi6 p47 p6,ll fll + pi6 p48 p6,ll fll +
pi6 p59 p6,ll fll + pi4 p47 p6,ll f7 + pi4 p48 p6,ll F8 + pi5 p59 p6,ll^9 + pi6 p5,l0 p6,ll fll +
pi5p5,l0p6,llfl0 + pi6p47p59p65f9 + pi6p48p59p65^9 + pi6p47p59p64^7 + pi6p48p59p64f8 -
pi3p47p59f3 - pi6p47p59k - pi3p48p59f3 - pi6p48p59k + pi6p47p65p5,l0fl0 + pi6p48p65p5,l0fl0 +
pi6p47p64p5,l0f7 + pi6p48p64p5,l0f8 - pi3p47p5,l0^3 - pi6p47p5,l0k - pi3p48p5,l0^3 -
pi6p48p5,l0k - pi3p47p6,llf3 - pi5p47p6,llf5 - pi3p48p6,llf3 - pi5p48p6,llf5 - pi3p59p6,llf3 -
pi4p59p6,ll f4 - pi3p5,l0p6,ll № - pi4p5,l0p6,llf4 - pi6p47p59p6,ll Fll - pi6p48p59p6,ll fll -
pi5p47p59p6,llf9 - pi5p48p59p6,llf9 - pi4p47p59p6,llf7 - pi4p48p59p6,llf8 - pi6p47p5,l0p6,llfll -
pi5p47p5,l0p6,llfl0 - pi6p48p5,l0p6,llFll - pi5p48p5,l0p6,llFl0 - pi4p47p5,l0p6,ll^7 -
pi4 p48 p5,l0 p6,ll f8 + pi3 p47 p59 p6,llf3 + pi3 p48 p59 p6,llf3 + pi3 p47 p5,l0 p6,llf3 + pi3 p48 p5,l0 p6,ll№)
Di is already defined in equation (l).
6.4. Expected Number of Repairs in Summers
Leting V/(t) be the expected number of repairs in 0 < l < t such that it is given the system entered the state S/ at t=0, we get
V/(t) = En Q/n(t)[h/ + V/(t)]; l=4,5,6,7,8,9,10,11
h i 1,when state S/ is the regenerative state / 0, otherwise Taking LST of equations, we get:
voo (s) = N
The equation describing the number of repairs per unit time in steady state
V0 = lims^0 svOO (s) = D
where
N4 = - P01( P13P47 - P14P41 - P15 P51 - P16 P61 - P14 P47 - P14P48 - P15 P59 - P16 P64P41 -
P16P65P51 - P15P5,10 - P16P6,11 - P16P47P64 - P16P48P64 - P13 + P15P47P51 + P16P47P61 + P13P48 -
P16P59P65 + P15P48P51 + P16P48P61 + P13P59 + P14P59P41 + P16P59P61 - P16P65P5,10 + P13P5,10 +
P14 P5,10 P41 + P16 P5,10 P61 + P13 P6,11 + P14 P6,11 P41 + P15 P6,11 P51 + P15 P47 P59 + P16 P47 P65 P51 +
P15 P48 P59 + P16 P48 P65 P51 + P14 P47 P59 + P14 P48 P59 + P16 P59 P64 P41 + P15 P47 P5,10 + P15 P48 P5,10 +
P14 P47 P5,10 + P14 P48 P5,10 + P16 P64 P5,10 P41 + P16 P47 P6,11 + P16 P48 P6,11 + P16 P59 P6,11 + P14 P47 P6,11 +
P14 P48 P6,11 + P15 P59 P6,11 + P16 P5,10 P6,11 + P15 P5,10 P6,11 + P16 P47 P59 P65 + P16 P48 P59 P65 +
P16P47P59P64 + P16P48P59P64 - P13P47P59 - P16P47P59P61 - P13P48P59 - P16P48P59P61 +
P16P47P65P5,10 + P16P48P65P5,10 + P16P47P64P5,10 + P16P48P64P5,10 - P13P47P5,10 - P16P47P5,10P61 -
P13P48P5,10 - P16P48P5,10P61 - P13P47P6,11 - P15P47P6,11 P51 - P13P48P6,11 - P15P48P6,11 P51 -
P13P59P6,11 - P14P59P6,11 P41 - P13P5,10P6,11 - P14P5,10P6,11 P41 - P16P47P59P6,11 - P16P48P59P6,11 -
P15P47P59P6,11 - P15P48P59P6,11 - P14P47P59P6,11 - P14P48P59P6,11 - P16P47P5,10P6,11 -
P15P47P5,10P6,11 - P16P48P5,10P6,11 - P15P48P5,10P6,11 - P14P47P5,10P6,11 - P14P48P5,10P6,11 +
P13 P47 P59 P6,11 + P13 P48 P59 P6,11 + P13 P47 P5,10 P6,11 + P13 P48 P5,10 P6,11)
D1 is already defined in equation (1).
6.5. Availability of the System in Winters
The availability A^ of the system in winters is:
AW = lims^0(sAOw) = D2
where
M0 = e-(x+^), M15 = e-(7+A+A1+A2) t, m17 = e-(A+A3)fG2"(t), M18 = e-(A+A3 )tG3~(t)
N5 = H0 + H15P02P2,15 - H0P2,14 - H0P15,2P2,15 - H0P15,16 - H0P15,17P17,15 - H0P15,18P18,15 -H0P17,19 - H0P 17,20 - H0P 18,21 - H0P18,22 + H17P02P2,15P15,17 + H18P02P2,15P15,18 + H0P2,14P15,16 + H0P2,14P15,17P17,15 + H0P2,14P15,18P18,15 - H15P02P2,15P17,19 - H15P02P2,15P17,20 + H0P2,14P17,19 + H0P15,2P2,15P17,19 + H0P2,14P17,20 + H0P15,2P2,15P17,20 - H15P02P2,15P18,21 - H15P02P2,15P18,22 + H0 P2,14 P18,21 + H0 P15,2 P2,15 P18,21 + H0 P2,14 P18,22 + H0 P15,2 P2,15 P18,22 + H0 P15,16 P17,19 + H0 P15,16 P 17,20 + H0 P 15,18 P17,19 P18,15 + H0 P15,16 P18,21 + H0 P15,18 P17,20 P18,15 + H0 P15,16 P18,22 + H0 P15,17 P 17,15 P 18,21 + H0 P15,17 P17,15 P18,22 + H0 P17,19 P18,21 + H0 P17,19 P18,22 + H0 P17,20 P18,21 + H0P17,20P 18,22 - H18P02P2,15P15,18P17,19 - H18P02P2,15P15,18P17,20 - H17P02P2,15P15,17P18,21 -H17P02P2,15P15,17P18,22 - H0P2,14P15,16P17,19 - H0P2,14P15,16P17,20 - H0P2,14P15,18P17,19P18,15 -H0P2,14P15,16P18,21 - H0P2,14P15,18P17,20P18,15 - H0P2,14P15,16P18,22 - H0P2,14P15,17P17,15P18,21 -H0 P2,14 P15,17 P17,15 P18,22 + H15 P02 P2,15 P17,19 P18,21 + H15 P02 P2,15 P17,19 P18,22 +
H15P02P2,15P17,20P18,21 - H0P2,14P17,19P18,21 - H0P15,2P2,15P17,19P18,21 + H15P02P2,15P17,20P18,22 -H0P2,14P17,19P18,22 - H0P2,14P17,20P18,21 - H0P15,2P2,15P17,19P18,22 - H0P15,2P2,15P17,20P18,21 -H0P2,14P17,20P18,22 - H0P15,2P2,15P17,20P18,22 - H0P15,16P17,19P18,21 - H0P15,16P17,19P18,22 -H0P15,16P 17,20P 18,21 - H0P15,16P17,20P18,22 + H0P2,14P15,16P17,19P18,21 + H0P2,14P15,16P17,19P18,22 + H0 P2,14 P15,16 P17,20 P18,21 + H0 P2,14 P15,16 P17,20 P18,22
D2 = (H14P2,14 + H2)(P15,2 - P15,2P17,19 - P15,2P17,20 - P15,2P18,21 - P15,2P18,22 + P15,2P17,19P18,21 + P15,2P 17,19P18,22 + P15,2P17,20P18,21 + P15,2P17,20P18,22) + (^15 + H16P15,16)(P17,15 - P2,14P17,15 -
P17,15P 18,21 - P17,15P 18,22 + P2,14Pl7,15Pl8,21 + P2,14Pl7,15Pl8,22) + (^17 + ^19P 17,19 + №P17,20 )(P15,17 - P2,14P15,17 - P15,17P18,21 - P15,17P18,22 + P2,14P15,17P18,21 + P2,14P15,17P18,22 ) + (№ + U21 P18,21 + U22P 18,22)(P15,18 - P2,14P15,18 - P15,18P17,19 - P15,18P17,20 + P2,14P15,18P17,19 + P2,14 P 15,18 P17,20).......(2).
6.6. Busy Period for Repair in Winters
The busy period for repair of the system in winters is:
BWR = lims^o (sB^R ) = g
where
W17 = e-(x+x3) G2~(t), W18 = e-(x+x3) G3~(t), W19 = W21 = G(t), W20 = W22 = G3(t)
N6 = -P02(P2,15P15,16P17,19U16 - P2,15P15,17^17 - P2,15P15,18^18 - P2,15P15,17P17,19^19 -
P2,15P 15,17P17,20^20 - P2,15P15,18P18,21 U21 - P2,15P15,18P18,22^22 - P2,15P15,16^16 +
P2,15 P 15,18 P17,19№ + P2,15 P15,16 P17,20 ^16 + P2,15 P15,18 P17,20 ^18 + P2,15 P15,16 P18,21^16 +
P2,15 P 15,17 P18,21^17 + P2,15 P15,16 P18,22 ^16 + P2,15 P15,17P18,22 ^17 + P2,15 P15,18 P17,19 P18,21 ^21 +
P2,15 P 15,18 P17,19 P18,22 U22 + P2,15 P15,18 P17,20 P18,21 ^21 + P2,15 P15,18 P17,20 P18,22^22 +
P2,15 P 15,17 P17,19 P18,21 U19 + P2,15 P15,17 P17,20 P18,21 ^20 + P2,15 P15,17P17,19 P18,22^19 +
P2,15P 15,17P17,20P18,22^20 - P2,15P15,16P17,19P18,21 ^16 - P2,15P15,16P17,19P18,22^16 -
P2,15P 15,16 P17,20 P18,21 U16 - P2,15 P15,16 P17,20P18,22 ^16 )
D2 is already defined in equation (2).
6.7. Busy Period for Maintenance in Winters
The busy period for maintenance BWM of the system in winters is:
dWM _ 1:,,, (cjz*wM\ _ N7
B0 = lims^0(sBo ) = D
where W14 = G4~(t)
N7 = P02P2,14^14(P15,16P 17,19 - P15,17P17,15 - P15,18P18,15 - P17,19 - P17,20 - P18,21 - P18,22 -P15,16 + P15,16 P17,20 + P15,18 P 17,19 P18,15 + P15,16 P18,21 + P15,18 P17,20 P18,15 + P15,16 P18,22 + P15,17 P 17,15 P 18,21 + P 15,17 P17,15 P18,22 + P17,19 P18,21 + P17,19 P18,22 + P17,20 P18,21 + P17,20 P18,22 -P15,16 P 17,19 P 18,21 - P 15,16 P 17,19 P18,22 - P15,16 P17,20 P18,21 - P15,16 P17,20 P18,22 + 1) D2 is already defined in equation (2).
6.7.1. Expected Number of Repairs in Winters
The expected number of repair VWR of the system in winters is:
V0WR = lims^0(sVTR ) = D2
where
N8 = -P02(P2,15P15,16P17,19 - P2,15P15,17P17,15 - P2,15P15,18P18,15 - P2,15P15,17P17,19 -P2,15P 15,17P17,20 - P2,15P15,18P18,21 - P2,15P15,18P18,22 - P2,15P15,16 + P2,15P15,18P17,19P18,15 + P2,15 P 15,16 P17,20 + P2,15 P15,18 P17,20 P18,15 + P2,15 P15,16 P18,21 + P2,15 P15,17 P18,21 P17,15 + P2,15 P 15,16 P18,22 + P2,15 P15,17 P18,22 P17,15 + P2,15 P15,18 P17,19 P18,21 + P2,15 P15,18 P17,19 P18,22 + P2,15 P 15,18 P17,20 P18,21 + P2,15 P15,18 P17,20 P18,22 + P2,15 P15,17 P17,19 P18,21 + P2,15 P15,17 P17,20 P18,21 + P2,15P 15,17P17,19P18,22 + P2,15P15,17P17,20P18,22 - P2,15P15,16P17,19P18,21 - P2,15P15,16P17,19P18,22 -P2,15P 15,16 P17,20 P18,21 - P2,15 P15,16P17,20P 18,22) D2 is already defined in equation (2).
6.7.2. Expected Number of Maintenances in Winters
The expected number of maintenances V0"M of the system in winters is:
V0WM = lims^0(sVTM ) = N2
where
N9 = P02P2,14(P 15,16P 17,19 - P 15,17Pl7,15 - Pl5,18Pl8,15 - Pl7,19 - Pl7,20 - Pl8,21 - Pl8,22 - Pl5,16 + Pl5,16 P 17,20 + Pl5,18 P 17,19 Pl8,15 + Pl5,16 Pl8,21 + Pl5,18 Pl7,20 Pl8,15 + Pl5,16 Pl8,22 + Pl5,17 Pl7,15 Pl8,21 + Pl5,17P 17,15P 18,22 + Pl7,19Pl8,21 + Pl7,19Pl8,22 + Pl7,20Pl8,21 + Pl7,20Pl8,22 - Pl5,16Pl7,19Pl8,21 -Pl5,16P 17,19P 18,22 - P 15,16Pl7,20Pl8,21 - Pl5,16Pl7,20Pl8,22 + 1) D2 is already defined in equation (2).
7. Transition Probabilities and Mean Sojourn Time
P01 = 0+6, P02 = 0+6, Pl3 = A+A1TA2+A3, P14=_J_, Pl5 = a+a1 +A2+A3 , Pl6 =
A+Aj+A2+A3
A+A1+A2+A3, P2,14 = a4A+Y, P2,15 = A4+Y, P41 = gl (a + A3), P47 = J+J) (1 - gl (a + A3)), P48 = A+J (1 - gl (a + a3 )), P51 = g2 (a + a3), P59 = J-J (1 - g2 (a + a3)), P5,10 = A+J) (! - g2 (a + a3)), P61 = g3 (a + al + a2), P6,ll = jtJ+j- (1 - g3 (a + al + a2)), P6,12 = p642) = j++jjtxi(1 - g3(a + A1 + A2)),P6,13 = p653) = STJ+Ai(j - g3(a + Al + J2)) Pl5,2 =
5+a+a +A2, P 15,17 = 5+a+i1+a2, p15,18 = ¿tata^a^ p17,15 = g2 (a + A3), p17,19 =
J-J (1 - g2 (a + a))), Pl7,20 = J+J (1 - g2 (a + a))), P 18,15 = gl (a + a)), Pl8,21 =
J+A) (1 - gl (a + a))), Pl8,22 = j+J (1 - gi (a + a))) mean sojourn times are as follows:
¥0 = a+6, ¥ = A+A1+A2+A3 , ¥2 = 3+tAj, V4 = J+J) (1 - g1(a + A3)), ^5 =
A+A3 (1 - g2 (a + A3)), ¥6 = J+J+J2 (1 - g) (a + J1 + A2)), ¥15 = 3+J+J^, ¥17 = J^ (1 - g2 (a + a))), ¥18 = jt^ (1 - gl (a + a)))
If time is measured from the epoch of entry into state Sn, the unconditional mean transit time of the system from any state S/ is:
№/„=/°° fdQ/„ (t) = -qf„'(0)
where,
№01 + №02 = ¥0, ^13 + №14 + №15 = ¥1, №2,14 + №2,15 = ¥2, №41 + №47 + №48 =
¥4, №51 + №59 + №5,10 = ¥5, №61 + №6,11 + №6,12 + №6,13 = ¥6, №61 + №6,11 + №642) + №653 = K, №15,2 + №15,16 + №15,17 + №15,18 = ¥15, №17,15 + №17,19 + №17,20 = ¥17, №18,15 + №18,21 + №18,22 = ¥18
K = (JTjW + A+JJj /0° tg)(t)dt + ATA^ /0° e-(A+Al+A2)g3(t)dt -(atA-S^ /0° e-(ATAl+A2)g)(t)dt
From the above transition probabilites it is verified that:
P01 + P02 = 1, P13 + P14 + P15 = 1, P2,14 + P2,15 = 1, P41 + P47 + P48 = 1, P51 + P59 + P5,10 =
1, P61 + P6,11 + P6,12 + P6,13 = 1, P61 + P6,11 + P642) + P653) = 1, Pl5,2 + Pl5,16 + Pl5,17 + Pl5,18 = 1, Pl7,15 + Pl7,19 + Pl7,20 = 1, Pl8,15 + Pl8,21 + Pl8,22 = 1
8. Profit analysis
The profit incurred to the system is:
P = A0C0 + AlCl + AWC2 - (B0C) + BWRC4 + B0WMC5 + V0C6 + vwrc7 + V0WMC8) Where
C0, C1 are the revenues generated in summers when the system operates at full and half capacity respectively. C2 is the revenue generated in winters.
C3, C4 is the cost per unit time when the repairman in busy for reapir in summers and winters respectively.C5 is the cost per unit time when the repairman is busy for maintenance. C6, C7 cost per repair in summers and winters respectively.C8 cost per maintenance.
Rates of the system Associated Values (per hr)
Failure of unit 1 0.000778752
Failure rate of unit 2 0.0007787652
Failure rate of unit 3 0.00077528
Failure rate of unit 4 0.00053279
Maintenance rate of the unit 0.000594484
Repair rate of unit 1 0.065642055
Repair rate of unit 2 0.0432694
Repair rate of unit 3 0.014205127
Repair rate of unit 4 0.02134
Rate of maintenance 0.00279431
Rate of going to summers 0.00019841
Rate of going to winters 0.0002777
Rate of going to operating state .0033
Rate of going to standby state .00027
Figure 2: Values computed from the data collected
Costs of the system Associated Values (Rs.)
Revenue per up time in summers when system 256000
works at full capacity
Revenue per up time in summers when system 125000
works at halfcapacity
Revenue per up time in winters 21000
Cost per unittime when repairman is busy for 10000
repair in summers
Cost per unittime when repairman is busy for 5000
repair in winters
Cost per unittime when repairman is busy for 12500
maintenance ¡n winters
Cost per repair in summers 18000
Cost per repair in winters 9000
Cost per maintenance 22000
Figure 3: Values computed from the data collected
9. Graphical Analysis and Conclusion using Particular Case
Let us assume an exponential distribution for all the repair rates such that
g(t) = ee-e(t), gi (t) = 91e-ei(t), g2 (t) = 02e-ez(f), g3(t) = d3 e-e3(t),g4(t) = e-e4(t)
Voi = j++p,' Vo2 = j+p' pi3 = A+Aj+A2+A3' V14=_h_' Vi5 = A+A1 +A2+a3 ' Vi6 =
A+Ai +A2+A3
A+A1+3A2+A3' V2'14 = aa+y' V2'15 = Ai+7' V4i = e1+A1+A3' V47 = e1+A+A3' p48 = e1+A+A3' V5i =
e2 V _ a V _ A3 V __e3_ V __a_ V - V(12) _
e2+A+A3'V59 = e2+A+A3' V5'10 = e2+a+a3' V6i = e3+a+a1+a2' V6'11 = e3+A+A1+a2' V6'12 = V64 =
_A1_ V _ V(13) _ _A2_ V _ _d_ V _ A2 V _
e3+A+A1+A2' V6'13 = V65 = e3+A+A1+A2' v15'2 = d+A+Aj+A2' №17 = d+A+Aj+A2' №18 =
d+A+A11 +A2' Pi7'15 = e2+A+A3 ' Pi7'19 = e2+A+A3' №20 = e2+A+A3' Pi8'15 = e1+A+A3' Pi8'21 = e2+A+A3 ' Pi8'22 = 62+A+A3' ^0 = a+p 'Ui = A+Aj +1A2+A3' ^2 = 3+A4' U4 = a+a3 (1 - gl (A +
13))' U5 = A+A3+e2' U6 = A+A!+A2+e3 Ui5 = d+A+Aj +A2' Ui7 = A+A3+e2' Ui8 = A+A3+ei
MTSF v/s failure rate A varying failure rate At
I 110m
FailLreralcA ^^^^
Figure 4: MTSF v/s Failure rate
Profit v/s failure rate A varying failure rate Ai
Failure rate Ai
Figure 5: Profit v/s Failure rate
Profit v/s Cost C| varying cost CE
Cost Ci
Figure 6: Profit v/s Cost
Figures 4, 5 are the MTSF and profit graphs showing a similar trend against the failure rate Ai varying failure rate A.
It shows that as the failure rate A or Ai increases the MTSF and profit decreases. cut off points for figure 6 are as follows:
Table 2: Cut-off Points
Cost C3RS. Revenue per up timeRs.
50000 70000 19736.6955 33564.2752
In table 2 the cut-off points have been shown and from figure 6 it is also clear that with increase in the cost Ci the profit of the system increases. A number of research papers have been reviewed in this review that have contributed greatly to the field of reliability engineering over the years. The authors have conducted a substantial literature review with an aim of providing reliability engineers and industry leaders with recommendations on improving system reliability. As a final point, we see a wide range of potential uses for the new methods, techniques, and models. The system analysis was performed using a semi-Markov process and regenerative point technique. Code Blocks and Excel are used for graphical analysis, while MATLAB is used for calculation.In conclusion, researchers should consider cost factors as well as reliability factors in order to attain maximum reliability at a minimal cost in the future.
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