If the optimal base stock level of process 1 is 0, then the average cost is identical when ( t_{1} ) is changed from 0 to ( d ).
From discussion of "Optimal base stock level", if ( t_{1} ge d ) then optimal base stock level of process 1 is ( 0 ), and if ( t_{1} < d ) then there are two possible cases on the optimal base stock levels.
In particular, when parameters satisfy ( rho le h_{1}/ ({h_{1} + b} ) ), optimal base stock level of process 1 is zero and it is optimal to process an item at processes 1 and 2 continuously.
Note that when the optimal base stock level of process 1 is zero, there is an optimal positive interval for release lead time of process 2. When demand lead time can be long enough, then it is optimal to have no base stocks and wait for several time and process an item at processes 1 and 2 continuously.
In case 1, if demand lead time is no more than 3.442 optimal base stock of process 1 is 1, and otherwise it becomes 0. In case 2, when demand lead time is from 0.000 to 0.198 or from 0.821 to 0.893, the optimal base stock level of process 1 is 0, and in case 3 it is 0 for any demand lead time.
The optimal base stock level of process 1 is less than 2. It is because that by (1), if the inventory level of process 1 is positive, then the starting epoch for processing an item in process 2 ordered by demand information is ( t_{1} ) periods later from the start time for process 1 ordered by the same order.
If the optimal base stock level of process 1 is one, then optimal ( t_{1} ) is 0, which means the optimal release lead time of process 1 is the same as process 2. Then by (1), processes 1 and 2 start process items at the same time.
In "Optimal release lead time of process 2", optimal release lead time of process 2 is developed, and in "Optimal release lead time of process 1", optimal release lead time of process 1 is discussed, and thus pairs of optimal base stock levels and optimal release lead time are derived for given demand lead time.
