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The safety and availability of cold standby systems are often critically dependent on the successful activation and subsequent operation of their standby components. In this study, an availability estimation model is developed for a repairable 1outof3 cold standby system and applied to a real industrial scenario involving a main electrical power network, a local electricity generator, and a domestic electricity generator. A repairable 1outof3 cold standby system comprising three components is evaluated, and Markov models for system reliability are introduced. The analysis reveals a relatively high availability, indicating that the main electrical power network (MEPN) is essential. Furthermore, the effects of component capacity partitioning on system availability are investigated through numerical analysis. The results demonstrate that a 1outof3 system exhibits improved performance and stability due to its higher capacity level.
1outof3 cold standby system, markov model, steadystate availability
System availability is of paramount importance for operations under challenging environments. In the digital economy, critical facilities such as data centers, communication centers, and financial services necessitate continuous, highquality electrical supplies. To ensure exceptional availability, the power supply must be extremely reliable [1]. The main electrical power network (MEPN) in Iraq plays a vital role in daily life, powering various electrical tools that make life more convenient. However, due to frequent blackouts in the MEPN, people have turned to private generators, either local ones that supply power to each block or domestic ones used in individual households. Although these generators present their own challenges, they serve as essential alternatives during extensive blackouts in the MEPN. Consequently, these three power sources operate in tandem, with repair processes required when their functionality falters. This leads to parttime operation availability for the system. Achieving the desired level of availability involves providing sufficient redundancy, reducing the probability of failure, and minimizing repair time [2]. Availability is contingent upon the types of breakdowns included in the analysis [3].
Redundancy is a valuable tool for enhancing a system's availability by adding plugins. The plugins' status determines the type of redundancy, which can be classified into active redundancy, standby redundancy, and active/standby redundancy. Standby redundancy systems are configurations where a system is considered to have failed when all units have failed. In such systems, only one unit (or a defined number) operates at a time, while the other units await activation upon the operating unit's failure [4], as illustrated in Figure 1. This system comprises three units: A represents the initially operating unit, B and C are standby units, and S is the changeover device (switch). Standby systems can be further classified into cold standby, hot standby, and warm standby [5].
Figure 1. 1outof3 standby system
This study focuses on cold standby systems. In such systems, the standby units do not carry any load during the waiting period before activation. They remain in a dormant mode with a zero failure rate, while active units experience a higher failure rate of () [2, 4]. A comprehensive literature review reveals various studies examining the advantages and characteristics of cold standby systems.
Chung [6] clarified that active and cold standby units exhibit different constant failure rates, with failed system repair times having arbitrary distributions. Asker [4] investigated a twounit repairable standby system with a changeover device (switch) functioning either perfectly or imperfectly, in conjunction with a cold or partly loaded standby unit, resulting in eight distinct models for the system. Wang and Loman [1] proposed a generalized repairable system reliability model, KoutofN, with M cold standby units, which was subsequently applied by the General Electric Company to design an onsite power plant with N active generators (operating in derating mode) and backup generators in standby. This design aimed to provide an extremely reliable power source. PérezOcón and MontoroCazorla [7] presented a system comprising n units, with only the working unit being susceptible to failure, while the other units are in repair, cold standby, or waiting for repair.
Zhang and Wang [8] examined a twocomponent cold standby repairable system with one repairman and the use of priority, assuming that component M after repair is not "as good as new". Manglik and Ram [9] analyzed the reliability of a fourcomponent system arranged in series, where subsystems A, B, and C consist of single units, and subsystem D comprises three units: one active and two in cold standby arranged in parallel. Bao and Cui [10] proposed a new repairable system based on a Markov repairable twoitem cold standby system, in which the effects of system failure could be neglected. Zhai et al. [11] explored a 1outofn cold standby system, scheduling backups to ensure a standby component can effectively take over the task when the online component fails.
Grida et al. [2] addressed the effects of plugin economy of scale on achieving high availability levels. Peng et al. [12] investigated a cold standby system with two different components, utilizing the highly applicable phasetype (PH) distribution to describe the life and maintenance time of system components in a unified manner and constructing a systems' reliability model for broader applicability. Batra et al. [13] optimized the number of standby units for a system with one operative unit, using SemiMarkov processes and the regenerative point technique. Krishnan [14] provided a survey of reliability studies on koutofn: G systems, exploring various cases and applications. Grida et al. [15] employed the Markov modeling technique to compute the reliability and mean time to failure for nonrepairable systems with varying failure rates.
Savita et al. [16] conducted a study on a system with two distinct units, one of good quality and another of substandard quality, analyzing the system to determine reliability measures using the SemiMarkov process and the Regenerative technique. RuizCastro [17] examined matrix analysis methods for modeling complex discrete cold standby systems subject to multiple events, facilitating algorithmic and computational development of multistate complex systems. The presented method enabled the analysis of optimization problems in multistate complex systems, providing results that demonstrate the profitability of preventive maintenance and revealing the optimal number of operational units. Patawa et al. [18] analyzed the behavior of two dissimilar units in a cold standby repairable system with a waiting time facility, establishing that the Bayesian method under suitable prior is a practical and straightforward approach for analyzing redundant repairable systems with a waiting time facility. Danjuma et al. [19] investigated the reliability of a system with three components (A, B, and C) coupled in series and parallel, evaluating the system using the Markov birthdeath process and deriving expressions for availability and mean time to system breakdown. The results indicated that system effectiveness indicators, such as availability and mean time to system failure, increase with repair rates and decrease with failure rates.
In light of these studies and frequent outages in the Main Electrical Power Network (MEPN) in Iraq, which has led to the reliance on local and domestic electricity generators, this paper evaluates a cold standby system model of 1outof3. The system comprises three components representing the main electrical power network, a local generator, and a domestic generator. The 1outof3 cold standby system prioritizes the use and maintenance of components, with the backup itself being a 1outof3 system. A set of assumptions, including cold standby, perfect switching, and a 1outof3 system for the backup, are considered. The entire process is demonstrated and explained using a Markov Transition Chart, as shown in Figure 2.
The remainder of the paper is organized as follows: Section 2 provides a detailed description of the considered system. Section 3 describes the system state using the transition matrix. Section 4 presents the availability of the two cold standby systems with repairable components and calculates the stationary availability of the system. Section 5 examines the effect of component capacity partitioning on system availability for a 1outof3 cold standby system. Finally, Section 6 discusses the findings and offers conclusions.
Figure 2. Markov transition diagram for 1outof 3 cold standby system
In the current study, we have evaluated a repairable 1outof3 cold standby system, which comprises of three components: A is the initial operational component, B, C are the two cold standby components. As shown, component A is the (MEPN), component B is the local power generator, and component C is the household electricity generator. The assumptions are detailed as follows:
1. The switching of the system to be perfect. Hence, failure of the active component A is detected immediately, and B standby component is activated with probability 1 (It does not fail during its operation and does not fail in switching from normal operating component to the standby component).
2. At the beginning of the operation of the system, component A starts operation first, if it is in a good state while the components B and C in standby state.
3. Component A has priority in use and maintenance, and components B and C are repaired sequentially. In other words when component A is satisfactorily repaired when component B is in service state, component B is replaced by it. So component A enters the service state, and component B enters the standby state or fails.
4. When a component fails, it is instantaneously replaced by one of the standby components.
5. Each component has a constant operating failure rate (λ), and a constant repair rate (μ).
6. The time of repair is exponentially distributed with repair rate (μ).
7. When repair action is completed, component A is placed in operating state and the components B and C are placed in standby state.
8. System failure occurs when the operating component A fails before repairing the other components B or C.
9. The failed state of the system is state (4) as shown in Figure 2, when all components have failed, one of them is repaired instantaneously, which is (A) and the system is thus transitioned to state (5).
The transition of a system from one state to another is best described by the transition matrix, as in Table 1.
Using the time derivative of state probabilities and a Markov transition diagram to examine the system states depicted in Figure 2 and Table 1, we can construct the following formulas as the state probabilities of the system.
Table 1. Transition probability matrix

P_{0} 
P_{1} 
P_{2} 
P_{3} 
P_{4} 
P_{5} 
P_{6} 
P_{7} 
P_{0} 
 λ_{A} 
λ_{A} 
0 
0 
0 
0 
0 
0 
P_{1} 
µ_{A} 
 (λ_{B+} µ_{A}) 
λ_{B} 
0 
0 
0 
0 
0 
P_{2} 
0 
0 
 (λ_{C+} µ_{A}) 
µ_{A} 
λ_{C} 
0 
0 
0 
P_{3} 
µ_{B} 
0 
λ_{A} 
 (λ_{A+} µ_{B}) 
0 
0 
0 
0 
P_{4} 
0 
0 
0 
0 
 µ_{A} 
µ_{A} 
0 
0 
P_{5} 
0 
0 
0 
0 
λ_{A} 
 (λ_{A}+ µ_{B}) 
µ_{B} 
0 
P_{6} 
µ_{C} 
0 
0 
0 
0 
0 
 (λ_{A}+ µ_{C}) 
λ_{A} 
P_{7} 
0 
0 
0 
0 
λ_{B} 
0 
µ_{A} 
 (λ_{B}+ µ_{A}) 
$\frac{d p_1(t)}{d t}=\lambda p_0(\lambda+\mu) p_1$, (2)
$\frac{d p_2(t)}{d t}=\lambda p_1(\lambda+\mu) p_2+\lambda p_3$, (3)
$\frac{d p_3(t)}{d t}=\mu p_2(\lambda+\mu) p_3$, (4)
$\frac{d p_4(t)}{d t}=\lambda p_2\mu p_4+\lambda p_5+\lambda p_7$, (5)
$\frac{d p_5(t)}{d t}=\lambda p_4(\lambda+\mu) p_5$, (6)
$\frac{d p_6(t)}{d t}=\mu p_5(\lambda+\mu) p_6+\mu p_7$, (7)
$\frac{d p_7(t)}{d t}=\lambda p_6(\lambda+\mu) p_7$. (8)
Under steady state, the time derivatives of state probability are:
$p_0=\left[\frac{\lambda+\mu}{\lambda}\right] p_1$, (9)
$p_2=\left[\frac{\lambda(\lambda+\mu)}{\lambda^2+\lambda \mu+\mu^2}\right] p_1$, (10)
$p_3=\left[\frac{\lambda \mu}{\lambda^2+\lambda \mu+\mu^2}\right] p_1$, (11)
$\begin{aligned} p_4=\left[2 \frac{\lambda^2}{\mu^2}+\frac{\lambda^3}{\mu^3}\right. & \frac{\lambda^3}{\mu^3+\lambda \mu^2}\frac{2 \lambda^2\lambda \mu}{\lambda^2+\lambda \mu+\mu^2}\frac{\lambda^3}{\lambda^2 \mu+\lambda \mu^2+\lambda^3}+\frac{\lambda}{\mu}\frac{\lambda^2}{\mu^2+\lambda \mu} \left.+\frac{\lambda^3+\lambda^2 \mu}{\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3}\right] p_1\end{aligned}$, (12)
$p_5=\left[\frac{\lambda^2}{\mu^2}\frac{\lambda^2\lambda \mu}{\lambda^2+\lambda \mu+\mu^2}+\frac{\lambda}{\mu}\frac{\lambda^2}{\lambda \mu+\mu^2}+\frac{\lambda^2 \mu}{\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3}\,\,\right] p_1$, (13)
$p_6=\left[\frac{\lambda}{\mu}\frac{\lambda \mu}{\lambda^2+\lambda \mu+\mu^2}\right] p_1$, (14)
$p_7=\left[\frac{\lambda^2}{\lambda \mu+\mu^2}\frac{\lambda^2 \mu}{\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3}\,\,\right] p_1$, (15)
Combining Eqns. (9)(15) and condition
$\sum_{i=0}^7 p_i(t)=1$ (16)
$p_1=\frac{\lambda \mu^3}{\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4}$ , (17)
So
$p_0=\frac{\mu^3(\lambda+\mu)}{\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4}$ , (18)
$p_2=\frac{\lambda^2 \mu^3(\lambda+\mu)}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}$ , (19)
$p_3=\frac{\lambda^2 \mu^3}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}$ , (20)
$\begin{gathered}p_4=\frac{\lambda^4+2 \lambda^3 \mu+\lambda^2 \mu^2}{\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ \frac{\lambda^4 \mu^3}{\left(\lambda^2 \mu+\lambda \mu^2+\mu^3\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\frac{\lambda^3 \mu^3}{\left(\mu^2+\lambda \mu\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ \frac{\lambda^4\mu^3}{\left(\mu^3+\lambda \mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ +\frac{\lambda^3 \mu^4+\lambda^4 \mu^3}{\left(\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ \frac{\lambda^2 \mu^32 \lambda^3 \mu^3}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\end{gathered}$, (21)
$\begin{gathered}p_5=\frac{\lambda^3 \mu+\lambda^2 \mu^2}{\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\frac{\lambda^3 \mu^3}{\left(\mu^2+\lambda \mu\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ +\frac{\lambda^3 \mu^4}{\left(\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ \frac{\lambda^2 \mu^32 \lambda^3 \mu^3}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\end{gathered}$, (22)
$p_6=\frac{\lambda^2 \mu^2}{\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\frac{\lambda^2 \mu^4}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}$ , (23)
$p_7=\frac{\lambda^3 \mu^3}{\left(\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\frac{\lambda^3 \mu^4}{\left(\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}$ (24)
Availability is the probability that the system is operating at a specified time (t), which is always associated with the concept of maintainability. Availability depends on both failures and relies on repair rates [9].
4.1 Calculate the stationary availability of system
The system is in operation when it is in either the state $p_0$, $p_1$, $p_2$, $p_3$, $p_4$, $p_5$, $p_6$ and $p_7$. Therefor, the general form to calculate the stationary availability of system is:
$\begin{gathered}\mathrm{V}_{\mathrm{AA}}=\left[p_0(\infty)+p_1(\infty)+p_2(\infty)+p_3(\infty)\right. \left.+p_5(\infty)+p_6(\infty)+p_7(\infty)\right] .\end{gathered}$ (25)
Using Eqns. (17)(20) and Eqns. (22)(25) can be rewritten as:
$\begin{aligned} & \mathbf{V}_{\mathbf{A A}}=\left[\frac{\mu^3(\lambda+\mu)}{\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4}+\frac{\lambda \mu^3}{\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4}\right. \\ & +\frac{\lambda^2 \mu^3(\lambda+\mu)}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}+\frac{\lambda^2 \mu^4}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ & +\frac{\lambda^3 \mu+\lambda^2 \mu^2}{\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\frac{\lambda^3 \mu^3}{\left(\mu^2+\lambda \mu\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ & +\frac{\lambda^3 \mu^4}{\left(\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ & +\frac{\lambda^2 \mu^2}{\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ & \frac{\lambda^2 \mu^4}{\left(\lambda^2+\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ & +\frac{\lambda^3 \mu^3}{\left(\lambda \mu+\mu^2\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)} \\ & \left.\frac{\lambda^3 \mu^4}{\left(\lambda^3+2 \lambda^2 \mu+2 \lambda \mu^2+\mu^3\right)\left(\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4\right)}\right] \\ & \end{aligned}$.
So,
$\mathrm{V}_{\mathrm{AA}}=\frac{\mu^4+2 \lambda \mu^3+\lambda^3 \mu+2 \lambda^2 \mu^2}{\lambda^4+2 \lambda^3 \mu+2 \lambda^2 \mu^2+2 \lambda \mu^3+\mu^4}$ (26)
Term (Γ) is defined as the ratio of repair rate to component failure rate and is used to analyze the impact of component capacity partitioning on system availability: $\Gamma=\mu / \lambda$.
Hence, we will calculate and analyze the effect of component capacity partitioning on system availability 1outof3 cold standby system from Eq. (26).
$V_{A A}=\frac{\Gamma^4+2 \Gamma^3+\Gamma+2 \Gamma^2}{1+2 \Gamma+2 \Gamma^2+2 \Gamma^3+\Gamma^4}$ (27)
Taking the individual values of different (Γ) the ratio of repair rate to component failure rate and substitute into an Eq. (27). Table 2 shows the availability values.
Table 2. Calculates the availability
Γ values 
V_{AA} values 

Γ_{ 1} 
1 
V_{AA1} 
0.7500000 
Γ_{ 2} 
1.5 
V_{AA2} 
0.87692307 
Γ_{ 3} 
11 
V_{AA3} 
0.9993169 
Γ_{ 4} 
21 
V_{AA4} 
0.9998971 
Γ_{ 5} 
31 
V_{AA5} 
0.9999675 
Γ_{ 6} 
41 
V_{AA6} 
0.9999858 
Γ_{ 7} 
51 
V_{AA7} 
0.9999926 
Γ_{ 8} 
61 
V_{AA8} 
0.9999956 
Figure 3. The steady state availability with respect to components setting up and Γ
Taking the marital values of different (Γ) the ratio of repair rate to component failure rate and substitute into an Eq. (27). And we get Table 3 to see the availability values.
The following chart shows the values of the second table graphically,
Figure 4. The steady state availability with respect to components setting up and Γ
Table 3. Calculates the availability
Γ values 
V_{AA} values 

Γ_{ 1} 
20 
V_{AA1} 
0.9998812 
Γ_{ 2} 
30 
V_{AA2} 
0.9999641 
Γ_{ 3} 
40 
V_{AA3} 
0.9999847 
Γ_{ 4} 
50 
V_{AA4} 
0.9999921 
Γ_{ 5} 
60 
V_{AA5} 
0.9999954 
In this paper, we have a model for a 1outof3 cold standby system developed and studied on the real industrial application of the (MEPN), local electricity generator and domestic electricity generator. Various reliability indices are evaluated such as availability for the considered system by employing Markov Process. From the results and analysis of the designed system, one can conclude the following:
(i) Figure 3 and Figure 4 show the expected stable availability of a 1outof3 cold standby system. In terms of repair failure rate, a 1outof3 system performed better due to its higher level of redundancy. To have a higher repair failure rate, the performance of 1outof3 is better.
(ii) The analysis of the models shown have relatively high availability according to the model and that the (MEPN) cannot be dispensed with.
I would like to acknowledge and give my warmest thanks to my supervisor (Dr. Hussein K. Asker) who made this work possible. His guidance and advice carried me through all the stages of writing my paper.
Matlab (R2014a) was used in the current study to examine the results of all the formulas mentioned above. Moreover, the (R2014a) software was used to draw the diagrams (3A and 3B).:
Code Matlab
P00=solve("X∗P0−(X+Y)∗P1=0","P0")
P22=solve(′Y∗P2−(X+Y)∗P3=0"," P2")
P33=X∗P1−(X+Y)∗P22+X∗P3
P333=solve(P33," P3")
P222=subs(P22, P3, P333)
P66=solve("−X∗P0+Y∗P1+Y∗P3+Y∗P6=0","P6")
P666=subs(P66, P3, P333)
P6666=expand(subs(P666, P0, P00))
P77=solve(′X∗P6−(X+Y)∗P7=0′," P7")
P777=expand(subs(P77, P6, P6666))
P55=solve("Y∗P5−(X+Y)∗P6+Y∗P7=0","P5")
P555=subs(P55, P6, P6666)
P5555=expand(subs(P555, P7, P777))
P44=solve("Y∗P4−(X+Y)∗P5"," P4")
P444=expand(subs(P44, P5, P5555))
P11=P00+P1+P222+P333+P444+P5555+P6666+P777
P111=solve(P11=1,P1)
PP0=subs(P00,P1,P111)
PP2=subs(P222,P1,P111)
PP3=subs(P333,P1,P111)
PP4=subs(P444,P1,P111)
PP5=subs(P5555,P1,P111)
PP6=subs(P6666,P1,P111)
PP7=subs(P777,P1,P111)
KKK=PP0+P111+PP2+PP3+PP5+PP6+PP7
KKK2=expand(PP0+P111+PP2+PP3+PP5+PP6+PP7)
KKK3=expand(KKK2+PP4)
P44444=1PP4
clear all;clc;
X=[20 30 40 50 60]
Y=(X.4+2.∗(X.3)+X+2.∗(X.2))./(1+2.∗X+2.∗(X.2)+2.∗(X.3)+(X.4))
plot(X, Y)
X=[1 1.5 11 21 31 41 51 61]
Y=(X.4+2.∗(X.3)+X+2.∗(X.2))./(1+2.∗X+2.∗(X.2)+2.∗(X.3)+(X.4))
plot(X, Y)
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