VALUE-ITERATION ALGORITHM

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where Ru and Rv are the actions which are prescribed by policy R(n) in the states u and v Step 4(b) For each i I , change Vn (i) according to Vn (i) := Vn 1 (i) + {Vn (i) Vn 1 (i)} Step 4(c) n := n + 1 and go to step 1 In the case of a tie when selecting in step 4(a) the state u for which the minimum in mn is obtained, it is conventional to choose the minimizing state of the previous iteration when that state is one of the candidates to choose; otherwise, choose the rst state achieving the minimum in mn The same convention is used for the maximizing action v in Mn The choice of the dynamic relaxation factor is motivated as follows We change the estimate Vn (i) as V n (i) = Vn 1 (i) + {Vn (i) Vn 1 (i)} for all i in order to accomplish at the (n + 1)th iteration that cu (Ru ) +

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in the implicit hope that the difference between the new upper and lower bounds Mn+1 and mn+1 will decrease more quickly Using the relation mn = Vn (u) Vn 1 (u) = cu (Ru )+ j puj (Ru )Vn 1 (j ) Vn 1 (u) and the similar relation for Mn , it is a matter of simple algebra to verify from the above condition the expression for We omit the easy proof that > 0 Numerical experiments indicate that using a dynamic relaxation factor in value iteration often greatly enhances the speed of convergence of the algorithm The modi ed value-iteration algorithm is theoretically not guaranteed to converge, but in practice the algorithm will usually work very well It is important to note that the relaxation factor is kept outside the recursion equation in step 1 so that the bounds mn and Mn in step 2 are not destroyed Although the bounds apply, it is no longer true that the sequences {mn } and {Mn } are monotonic To conclude this section, we apply value iteration to two examples The rst example concerns the maintenance problem from Example 611 and the second example illustrates the usefulness of value iteration for the computation of performance measures for a single Markov chain Example 611 (continued) A maintenance problem For the maintenance problem the recursion equation (661) becomes

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Vn (N ) = Cf + Vn 1 (N + 1), Vn (N + 1) = 0 + Vn 1 (1) We have applied the standard value-iteration algorithm to the numerical data from Table 641 For each stationary policy the associated Markov chain {Xn } is aperiodic Taking V0 (i) = 0 for all i and the accuracy number = 10 3 , the algorithm is stopped after n = 28 iterations with the stationary policy R(n) = (0, 0, 0, 1, 2, 2) together with the lower and upper lower bounds mn = 04336 and Mn = 04340 The average cost of policy R(n) is estimated by 1 (mn + Mn ) = 2 04338 and this cost cannot deviate more than 01% from the theoretically minimal average cost In fact policy R(n) is optimal as we know from previous results obtained by policy iteration To get a feeling of how strongly the required number of iterations depends on , we applied standard value-iteration for = 10 2 and = 10 4 as well For these choices of the accuracy number , standard valueiteration required 21 and 35 iterations respectively Example 661 A nite-capacity queue with deterministic arrivals Consider a single-server queueing system having a nite waiting room for K customers (including any customer in service) The arrival process of customers is deterministic Every D time units a customer arrives A customer nding a full waiting room upon arrival is lost The service times of the customers are independent random variables having an Erlang (r, ) distribution What is the long-run fraction of customers who are lost Taking the constant interarrival time as time unit, the fraction of lost customers can be seen as an average cost per time unit when a cost of 1 is incurred each time an arriving customer nds the waiting room full The queueing process embedded at the arrival epochs can be described by a Markov process by noting that the Erlang (r, ) distributed service time can be seen as the sum of r independent phases each having an exponential distribution with mean 1/ A customer is served by serving its phases one at a time The queueing problem can now be converted into a Markov decision model with a single action in each state The state of the system is observed at the arrival epochs and the set of possible states of the system is given by I = {0, 1, , Kr} State i corresponds to the situation that i uncompleted service phases are present just prior to the arrival of a new customer In each state i there is a single action to be denoted by a = 0 The action a = 0 in state i corresponds to the acceptance of the newly arriving customer when i Kr r and corresponds to the rejection of the customer otherwise The one-step costs ci (a) are given by ci (a) = 0 1 if i Kr r, if i > Kr r

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