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When we are doing inventory optimization, say implementing a base-stock policy, one of the KPIs we monitor is the average on-hand inventory, which we use to calculate the average inventory holding cost.

For a base-stock policy with base-stock level $S$, if executed perfectly, the average on-hand inventory is the sum of the safety stock and the cycle stock, where cycle stock is defined as half of the average demand during the review period.

Now my question is, is there a similar way to calculate the average on-hand inventory for a periodic review inventory policy such as the $(s, S)$-policy? Recall a $(s,S)$-policy with review period $R$ operates as follows: every $R$-period, we review the current inventory position, which includes both the on-hand and in-transit inventory, if the inventory position is less than $s$ (the reorder-point), then we order inventories to bring the inventory position to level $S$ (the order-up-to level), otherwise we do not order.

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I do not believe there is a simple way to calculate this. The inventory level in an $(s,S)$ policy has a probability distribution that is not easy to characterize, and therefore neither is the expected inventory level.

See Zheng and Federgruen (1991) or Veinott (1966) for some theory on $(s,S)$ policies. I don't think either of these papers characterizes the inventory level r.v. directly. But if it were easy to characterize that r.v., then it wouldn't have been such hard work for these papers to evaluate the expected cost.

(Interestingly, for $(r,Q)$ policies, the inventory level distribution is very easy: The IL is uniformly distributed on $[r,Q]$ (for continuous demands) or on $[r+1,Q]$ (for discrete demands).)

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  • $\begingroup$ Thank you for your reply Professor Snyder! I was thinking about using the power estimation (or similar estimation methods that exploit the similarities between a $(r,Q)$ and $(s,S)$ policy) to approximate the parameters $s,S$. Can we then use $s+(S-s)/2$ as the average inventory level? $\endgroup$
    – TTY
    Jan 10 at 1:13
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    $\begingroup$ $s + (S-s)/2$ seems like a reasonable approximation, regardless of what method you use to set $s$ and $S$. $\endgroup$ Jan 10 at 14:43
  • $\begingroup$ Great, I will use that then, thank you again for your reply! $\endgroup$
    – TTY
    Jan 11 at 2:55

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