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The new stationary policy R is obtained by choosing R i = ai for all i I with the convention that R i is chosen equal to the old action Ri when this action minimizes the policy-improvement quantity Step 3 (convergence test) If the new policy R = R, then the algorithm is stopped with policy R Otherwise, go to step 1 with R replaced by R The policy-iteration algorithm converges after a nite number of iterations to an average cost optimal policy We defer the proof to Section 67 The policy-iteration algorithm is empirically found to be a remarkably robust algorithm that converges very fast in speci c problems The number of iterations is practically independent of the number of states and varies typically between 3 and 15, say Also, it can be roughly stated that the average costs of the policies generated by policy iteration converge at least exponentially fast to the minimum average cost, with the greatest improvements in the rst few iterations Remark 641 The average cost optimality equation Since the policy-iteration algorithm converges after nitely many iterations, there exist numbers g and i , i I , such that i = min ci (a) g + pij (a) j , i I (643) a A(i)

It is the modest man from Wadena, Minnesota, who s now charged with leading the company into its second century, a century that promises to be every bit as unpredictable as the company s first Nonetheless, Walgreens can already identify many of the forces that will shape its strategy in the coming decades, including intensified competition Having outlasted once venerable chains like Shapiro, Liggett s, and Thrifty s, among others, and surpassed Eckerd, Rite Aid, and CVS, it s fair to say Walgreens has held its own and has to be considered the greatest drugstore chain over the past century (When longtime rival Rite Aid was posting record profits at the end of the millennium, Walgreens executives couldn t quite figure out how they

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This functional equation is called the average cost optimality equation Using Theorem 621, we can directly verify that any stationary policy R for which the action Ri minimizes the right-hand side of (643) for all i I is average cost optimal To see this, note that i = ci (Ri ) g + and i ci (a) g +

pij (Ri ) j ,

Split leg[Leg=2; CS=2] Split leg result Connect to resource[CS=1] Play announcement[CS=1;]

(644)

pij (a) j ,

(645)

The equality (644) and Theorem 621 imply that g(R ) = g Let R be any stationary policy Taking a = R i in (645) for all i I and applying Theorem 621, we nd g(R) g In other words, g(R ) g(R) for any stationary policy R This shows not only that policy R is average cost optimal but also shows that the constant g in (643) is uniquely determined as the minimal average cost per time

Specialized resource report Disconnect forward connection with argument[CS=1] Move leg[Leg=2] Move leg result Continue with argument[CS=1] Called party is reconnected to calling party

unit It is stated without proof that the function i , i I , in (643) is uniquely determined up to an additive constant Next the policy-iteration algorithm is applied to compute an average cost optimal policy for the control problem in Example 611 Example 611 (continued) A maintenance problem It is assumed that the number of possible working conditions equals N = 5 The repair costs are given by Cf = 10, Cp2 = 7, Cp3 = 7 and Cp4 = 5 The deterioration probabilities qij are given in Table 641 The policy-iteration algorithm is initialized with the policy R (1) = (0, 0, 0, 0, 2, 2), which prescribes repair only in the states 5 and 6 In the calculations below, the policy-improvement quantity is abbreviated as Ti (a, R) = ci (a) g(R) +

.

pij (a)vj (R)

when the current policy is R Note that always Ti (a, R) = vi (R) for a = Ri Iteration 1 Step 1 (value determination) The average cost and the relative values of policy R (1) = (0, 0, 0, 0, 2, 2) are computed by solving the linear equations v1 = 0 g + 09v1 + 01v2 v2 = 0 g + 08v2 + 01v3 + 005v4 + 005v5 v3 = 0 g + 07v3 + 01v4 + 02v5 v4 = 0 9 + 05v4 + 05v5 v5 = 10 g + v6 v6 = 0 g + v1 v6 = 0, where state s = 6 is chosen for the normalizing equation vs = 0 The solution of these linear equations is given by g(R (1) ) = 05128, v1 (R (1) ) = 05128, v2 (R (1) ) = 56410, v3 (R (1) ) = 74359, v4 (R (1) ) = 84615, v5 (R (1) ) = 94872, v6 (R (1) ) = 0

Table 641 i\j 1 2 3 4 1 090 0 0 0 The deteriorating probabilities qij 2 010 080 0 0 3 0 010 070 0 4 0 005 010 050 5 0 005 020 050

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