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58 Consider an irreducible continuous-time Markov chain with state space I and in nitesimal transition rates qij Let {pj , j I } be the equilibrium distribution of the Markov chain Assume that the Markov chain is time reversible and thus has the detailed balance property (5115) Suppose that the Markov chain is truncated to the subset A I That is, qij is changed to 0 for all i A and j A Prove that the equilibrium distribution of the truncated / Markov chain is given by pi A pi = , i A k A pk This important result is due to Kelly (1979) 59 Suppose we wish to determine the capacity of a stockyard at which containers arrive according to a Poisson process with a rate of = 1 per hour A container nding a full yard upon arrival is brought elsewhere The time that a container is stored in the yard is exponentially distributed with mean 1/ = 10 hours Determine the required capacity of the yard so that no more than 1% of the arriving containers nd the yard full How does the answer change when the time that a container is stored in the yard is uniformly distributed between 5 and 15 hours 510 Long-term parkers and short-term parkers arrive at a parking place for cars according to independent Poisson processes with respective rates 1 = 4 and 2 = 6 per hour The parking place has room for N = 10 cars Each arriving car which nds all places occupied goes elsewhere The parking time of long-term parkers is uniformly distributed between 1 and 2 hours, while the parking time of short-term parkers has a uniform distribution between 20 and 60 minutes Calculate the probability that a car nds all parking places occupied upon arrival 511 Consider the loss version of the delay model from Exercise 54 In the loss model each service request nding all c agents occupied upon arrival is lost and has no further in uence on the system Let p(i1 , i2 ) denote the long-run fraction of time that simultaneously i1 major-language customers are in service and i2 minor-language customers are in service Verify from the equilibrium equations for the state probabilities p(i1 , i2 ) that, for some constant C > 0, ( / )i1 ( 2 / 2 )i2 p(i1 , i2 ) = C 1 1 i1 ! i2 ! for all i1 , i2 with 0 i1 + i2 c Next conclude that the equilibrium distribution of the number of occupied agents is given by formula (521) with = 1 + 2 and 1/ = ( 1 / ) (1/ 1 ) + ( 2 / ) (1/ 2 ) 512 Units offered for repair arrive at a repair facility according to a Poisson process with rate There are c repairmen Each repairman can handle only one unit at a time An offered unit nding all repairmen busy is rejected and handled elsewhere The repair time of a unit consists of two phases The rst phase is exponentially distributed with mean 1/ 1 and the second one is exponentially distributed with mean 1/ 2 (a) Let p (i1 , i2 ) be the equilibrium probability of having i1 units in repair phase 1 and i2 units in repair phase 2 Verify that, for some constant C, the probability p (i1 , i2 ) = C( / 1 )i1 ( / 2 )i2 /(i1 !i2 !) for all i1 , i2 (b) What is the equilibrium distribution of the number of busy repairmen (c) What is the long-run fraction of offered units that are rejected Does this loss probability increase when the two repair phases are more variable than the exponential phases but have the same means as the exponential phases 513 Consider a continuous-review inventory system in which customers asking for a certain item arrive according to a Poisson process with rate Each customer asks for one unit of the item Customer demands occurring when the system is out of stock are lost The (S 1, S) control rule is used Under this control rule the base stock is S and a replenishment for.

Taking one year as time unit, consider the numerical example with the data = 01, = 100 and = 5 The uniformization method is applied with the choice = 100 for the uniformized leaving rate (the value 0 is taken for the leaving rate from the state crash) The calculations yield the value 0000504 for the probability 1 Q(T ) that a crash will occur in the next T = 10 years Similarly, one can calculate that with probability 03901 the sleep mode will not be reached in the next 10 years In other words, the probability of no shuttle mission in the next 10 years equals 03901 However, if one wishes to calculate the probability distribution of the number of required shuttle missions in the next 10 years, one must use the Markov reward model with lump rewards (assume a lump reward of 1 each time the process jumps from either state 2 or state 1 to the sleep mode) This Markov reward model is much more dif cult to solve and will be discussed in the next section

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A basic and practically important problem is the calculation of the transient probability distribution of the cumulative reward in a continuous-time Markov chain For example, a practical application is an oil-production platform which has to meet a contractually agreed production level over a speci ed time period The production rate is not constant but depends on the stochastically varying state of the platform In this example a continuous-time Markov chain with reward rates may be appropriate, where the reward rate r(j ) represents the production rate in state j In Section 461 we rst consider the special case of a Markov reward model with reward rates that are either 0 or 1 The cumulative reward in this model corresponds to the cumulative sojourn time of the process in a certain set of (good) states A nice and simple extension of the uniformization method can be given to compute the transient distribution of the cumulative sojourn time in some given set of states The general Markov reward model with both reward rates and lump rewards is dealt with in Section 462 A discretization algorithm will be discussed

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