I don't know this board but maybe I can answer some questions:
- If it filled the gap starting from 8000h to 0B000h, only 192 KB are needed. And that number fits inside one bank of 256 KB.
- There is only one dip "up", number 5. 5, 6, 7 and 8 could make "1000" = 8. Could these four bits be the starting address? Just a guess...