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Using the Lagrange method this problem can be numerically solved Rare events In many applied probability problems one has to study rare events For example, a rare event could be a system failure in reliability applications or buffer over ow in nite-buffer telecommunication problems Under general conditions it holds that the time until the rst occurrence of a rare event is approximately exponentially distributed Loosely formulated, the following result holds Let {X(t)} be a regenerative process having a set B of (bad) states such that the probability q that the process visits the set B during a given cycle is very small Denote by the random variable U the time until the process visits the set B for the rst time Assuming that the cycle length has a nite and positive mean E(T ), it holds that P {U > t} e tq/E(T ) for t 0; see Keilson (1979) or Solovyez (1971) for

fter opening his first store in 1901, it took Charles Walgreen Sr a full 8 years of hard work to secure his second small store, but only 20 more years to create the foundation for an empire The 1910s and 1920s were probably the most demanding decades in Walgreens history All four family members (Charles Sr, Myrtle, and offspring Chuck and Ruth) worked ungodly hours to feed the chain s burgeoning growth But those might have been the company s most exciting years, too, filled with boundless optimism, energy, and inno43

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a proof The result that the time until the rst occurrence of a rare event in a regenerative process is approximately exponentially distributed is very useful It gives not only quantitative insight, but it also implies that the computation of the mean of the rst-passage time suf ces to get the whole distribution In the next example we obtain the above result by elementary arguments Example 224 A reliability problem with periodic inspections High reliability of an electronic system is often achieved by employing redundant components and having periodic inspections Let us consider a reliability system with two identical units, where one unit is in full operation and the other unit is in warm standby The operating unit has a constant failure rate of 0 and the unit in standby has a constant failure rate of 1 , where 0 1 < 0 Upon failure of the operating unit, the standby unit is put into full operation provided the standby is not in the failure state Failed units are replaced only at the scheduled times T , 2T , when the system is inspected The time to replace any failed unit is negligible A system failure occurs if both units are down It is assumed that ( 0 + 1 )T is suf ciently small so that a system failure is a rare event In designing highly reliable systems a key measure of system performance is the probability distribution of the time until the rst system failure To nd the distribution of the time until the rst system failure, we rst compute the probability q de ned by q = P {system failure occurs between two inspections} To do so, observe that a constant failure rate for the lifetime of a unit implies that the lifetime has an exponential distribution with mean 1/ Using the fact that the minimum of two independent exponentials with respective means 1/ 0 and 1/ 1 is exponentially distributed with mean 1/( 0 + 1 ), we nd by conditioning on the epoch of the rst failure of a unit that q=.

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(n) (n) (n) (n) 1 ( D/c)[1 + zk (zk ) (zk )] ln(zk ) + 2 ik/c (n) (n) 1 ( D/c)zk (zk )

(0) where (z) is the derivative of (z) The starting value zk for the Newton Raphson iteration has to be chosen properly To make an appropriate choice for (0) zk , we have a closer look at the equation (G3) Let us rewrite this equation as ln(z) = ( D/c){ (z) 1} + 2 ik/c and analyse it for the case of light traf c with 0 Then the solution of the equation tends to e 2 ik/c Inserting z = e 2 ik/c on the right-hand side of the equation for ln(z) yields (0) zk = exp ( D/c){ (e 2 ik/c ) 1} + 2 ik/c

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