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Chapt 6c

# Chapt 6c - 6.4 Sums of Independent Random Variables 1 217...

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6.4 Sums of Independent Random Variables 217 F U ( u ) = 1 - 1 summationdisplay k = 0 u ) k e - λ u k ! = 1 - e - λ u { 1 + λ u } R U ( u ) = 1 - F U ( u ) = e - λ u { 1 + λ u } (a) Since 1 = 50, we have that P [ U > 100 ] = R U ( 100 ) = e - 100 / 50 braceleftbigg 1 + 100 50 bracerightbigg = e - 2 { 1 + 2 } = 0 . 4060 (b) When we increase the mean lifetime of the component by 10%, we obtain 1 = 50 ( 1 + 0 . 1 ) = 55. Thus, with the new λ u = 100 / 55, the corresponding value of R U ( 100 ) is R U ( 100 ) = e - 100 / 50 braceleftbigg 1 + 100 50 bracerightbigg = 0 . 4574 trianglesolid That is, the probability that the system lifetime exceeds 100 hours increases by approximately 13%. Example 6.13 The time to failure of a component of a system is exponentially distributed with a mean of 100 hours. If the component fails, it is immediately replaced by an identical spare component whose time to failure is independent of that of the previous one and the system experiences no downtime in the process of component replacement. What is the smallest number of spare parts that must be used to guarantee continuous operation of the system for at least 300 hours with a probability of at least 0.95? Solution Let X be the random variable that denotes the lifetime of a compo- nent, and let the number of spare parts be n - 1. Let U be the random variable that denotes the lifetime of the system. Then U = X 1 + X 2 + ··· + X n , which is an Erlang- n random variable whose reliability function is given by R U ( u ) = e - λ u braceleftbigg 1 + λ u + u ) 2 2 ! + ··· + u ) n - 1 ( n - 1 ) ! bracerightbigg 0 . 95 Since λ = 1 / 100 we have that R U ( 300 ) = e - 3 braceleftbigg 1 + 3 + ( 3 ) 2 2 ! + ( 3 ) 3 3 ! + ( 3 ) 4 4 ! + ( 3 ) 5 5 ! + ··· + ( 3 ) n - 1 ( n - 1 ) ! bracerightbigg 0 . 95 The following Table 6.1 shows the values of R U ( 300 ) for different values of n .

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218 Chapter 6 Functions of Random Variables Table 6.1 Values of n - 1 and R U ( 300 ) n - 1 R U ( 300 ) 0 0.0498 1 0.1991 2 0.4232 3 0.6472 4 0.8153 5 0.9161 6 0.9665 Thus, we see that with n - 1 = 5 we cannot provide the required probabil- ity of operation, while with n - 1 = 6 we can. This means that we need 6 spare components to achieve the goal. trianglesolid 6.5 Minimum of Two Independent Random Variables Consider two independent continuous random variables X and Y . We are in- terested in a random variable U that is the minimum of X and Y ; that is, U = min ( X , Y ) . The random variable U can be used to represent the reliabil- ity of systems with series connections, as shown in Figure 6.12. Such systems are operational as long as all components are operational. The first component to fail causes the system to fail. Thus, if in the example shown in Figure 6.12 the times- to-failure are represented by the random variables X and Y , then S represents the time until the system fails, which is the minimum of the lifetimes of the two components. The CDF of U can be obtained as follows: F U ( u ) = P [ U u ] = P [ min ( X , Y ) u ] = P [ ( X u , X Y ) ( Y u , X > Y ) ] Since P [ A B ] = P [ A ]+ P [ B ]- P [ A B ] , we have that F U ( u ) = F X ( u ) + F Y ( u ) - F XY ( u , u ) . Also, since X and Y are independent, we obtain the CDF and PDF of U as follows: F U ( u ) = F X ( u ) + F Y ( u ) - F XY ( u , u ) = F X ( u ) + F Y ( u ) - F X ( u ) F Y ( u ) f U ( u ) = d du F U ( u ) = f X ( u ) + f Y ( u ) - f X ( u ) F Y ( u ) - F X ( u ) f Y ( u ) = f X ( u ) { 1 - F Y ( u ) } + f Y ( u ) { 1 - F X ( u ) }
6.6 Maximum of Two Independent Random Variables 219 Figure 6.12

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Chapt 6c - 6.4 Sums of Independent Random Variables 1 217...

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