5. consider a logical address space of 32 pages with 1024 bytes per page; mapped onto a physical memory of 16 frames. a. how many bits are required in the logical address? b. how many bits are required in the physical address?

The logical address space is 32 pages, and we need 5 bits plus 25 to represent 32 pages. The offset or "d" parameter can be represented by 11 bits 211=2048 because each page holds 2048 words. Therefore, 5 plus 11 equals a total of 16 bits to indicate a logical address. The length of physical addresses is 44 bits, and each page has 4 protection bits.

What length in bits does a physical page number have?

Using the slide from before: Virtual page number has a size of 20 bits. Physical page numbers are rounded up to 32 bits from 18 bits plus the valid bit. The offset refers to the number of bits needed to address a byte within a page. Therefore, 57 bits for the page number and 7 bits for the offset. Put the bottom nibble 0H in segment address and add offset address to get the whole physical address. The construction of a 20-bit physical address is depicted in the figure. Assume that the Base Address of the Data Segment is 1000h and that the Offset (0020h) of the Data Segment contains the data you require.

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