## Developer capacity, D-23, and dilution

Among the accusations ever leveled at me, exhibiting a command of math and chemistry are not to be found. I'm running a series of tests with D-23, from N+2 to extra-compensating, in both tray and rotary tank (Jobo on a Uniroller). I'd like to be sure I'm not introducing an unrealized variable in my dilutions. My weekly available time for photography, at present, is constrained.
D-23
D-23 capacity, straight, is listed at the equivalent of 16 4x5s per liter (4 8x10s, 320 sq. in.). My question is the minimum amount of this stock solution required when it's diluted; I've been unable to find a clear answer. Specifically, I would think that a liter a working solution of 1:1 would then process 8 (one-shot), one of 1:2 would process 4 and one of 1:3 would process 2. Therefore, my logic runs, a could process only 1 sheet in a half-liter of 1:3; that would be 4 ounces of stock in that working solution to process a single 4x5 sheet. Is that right?
What I have seen often in posts on various sites, is suggestions like, "Well, I would double the total working solution at that dilution, to avoid early exhaustion..."
Since my calculation methods are not infrequently flawed, I will grateful for any scientific correction.