in question b

(1)As a result, the column query's output in the figure is 0110 0111 0111.With segmented offsets, the quadratic addressing approach is also used. As a result, the cubic value is 0110 0111 0111 + 0000 0100 0010 = 0110 1011 1001. (it is in the binary form)

(2)Base address from Segmentation Table = 0110 0111 0111

in question d

(1)In the final sector, the results of the sixth site chair are pooled with both the 7th-page frame and the 8th-page frame. As a result, the address is 0111. This local address is sequenced with the book offset, which can be comparable to 0111 1011 1001

(2)Physical Address = 0111 0111 0111

I do not know why I got two different answer, please help and explain.

Consider a byte-addressable computer system with a 16-bit virtual address, total physical memory size 4KB, page size is 256Bytes, the maximum size of segment is 1KB. Given the following segment table, page table and a 16 bits hexadeciaml logical address "OC42", complete the address translation diagram below. (a) segment table index (b) linear address (i.e. in binary form) (c) page table index (d) physical address (i.e. in binary form) OC42 Segmentation table (segment base adde.) 0000,0110,0100 0010,0100,0000 0011,1010,0011 0110,0111,0111 0001,0001,0100 0100,0111,0110 0101,0110,1111 Index number? page table (frame number 0 4 1 11 6 3 7 (m) Index number? linear address? (iv) physical address? 10 0110,00 10,0000 00100110,0000