# Reorder point and safety stock for very long lead times

I have previously asked this problem here-Lead time longer than cycle time, but I keep getting more confused the more I think about it. Using the Q,R inventory model, suppose my mean monthly demand $$\mu = 4000, \sigma_D = 1500$$, Order quantity $$Q = 6000$$ and lead time $$L = 3$$(months). Choosing $$\alpha = 2.33$$, I get $$SS \approx 6000$$ and $$ROP \approx 18000$$. ROP is interpreted as the inventory position here. The values seem reasonable enough.

Now suppose I take $$L = 10$$, the safety stock value shoots up to $$SS = 11,000$$ and $$ROP = 51,000$$ This definitely seems absurdly high to me. I don't get why we use the lead time as it is in the formula because at a steady-state, and assuming the demands are normally distributed, we will be receiving an order every $$Q/\mu = 6000/4000 = 1.5$$ months on an average. Why, then, does the formula say that we need to use the lead time L = 10, when we know that the inventory would be replenished much before that? In my opinion, we only need to cover-up for fluctuations in demand in that 1.5 month period. I couldn't find any text that clearly explains this dichotomy between Inventory Position and Inventory Level and any help is appreciated.

EDIT:

You seem to be conflating two ordering systems. You say you will be receiving an order every 1.5 months on average. The only way that could happen with a lead time of 10 months would be if you had more than one order outstanding at a time. This ties to your use of $$Q=6,000$$ in the second paragraph. That's the order quantity when $$L=3$$, not when $$L=10$$. So if you are going to stick to an ROP/ROQ system (order when you hit the ROP, then not again until after that order arrives), you'll order $$Q=40,000$$ units, which is 10 months of demand. If, instead, you decide to stick with $$Q=6,000$$, you're moving from an ROP system to what looks more like an EOQ system ... or at least you'll be ordering at more than one reorder point.