Exercise #1: 

a. 20
	logical page number = 20 / 8192 = 0
	offset in page = 20 - 0 * 8192 = 20
	physical frame = 13   (from page table)
	physical address: 13 * 8192 + 20 = 106516

b. 9100
	logical page number = 9100 / 8192 = 1
	offset in page = 9100 - 8192 * 1 = 908
	physical frame = 8   (from page table)
	physical address: 8 * 8192 + 908 = 66444

c. 50321
	logical page number = 50321 / 8192 = 6
        page fault!
	

d. 81589
	logical page number = 81589 / 8192 = 9
	offset in page = 81589 - 8192 * 9 = 7861
	physical frame = 2   (from page table)
	physical address: 2 * 8192 + 7861 = 24245


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Exercise #2:

a. There are 32 pages, so the logical page number is 5-bit
   There are 2^{11} bytes in a page, so the offset is 11-bit
   Therefore, logical addresses are 16-bit

b. There are 8 frames, so there are 2^{3} * 2^{11} = 2^{14} bytes in RAM. 
   Therefore, physical addresses are 14-bit


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Exercise #3:

a. A process can have an address space up to 2^{32} bytes, and each page
   is 2^{13} bytes. Therefore, an address space can have up to 2^{32 - 13}
   = 2^{19} pages. Therefore, there would be 2^{19} entries in a
   conventional 1-level page table.


b. There are 2^{31} bytes in RAM, so there are 2^{31 - 13} = 2^{18} frames in RAM. 
   Therefore, an inverted page table would have 2^{18} entries.

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Exercise #4:

a. The logical address space contains 2^{38} bytes, and a page is 2^{15} bytes. Therefore
   there are 2^{38 - 15} = 2^{23} pages in the logical address space.

b. - The offset is 15-bit (because a page is 2^{15}-byte)
   - A frame is 2^{15} bytes and a page table entry is 8 bytes. Therefore, a frame can contain
     2^{15 - 3} = 2^{12} page table entries. This is the "inner" page number
   - The "outer" page number is then whatever remains: 38 - 15 - 12 = 11. 

                        ----------------------------
    The split is thus:  | 11-bit | 12-bit | 15-bit |
                        ----------------------------

c. The process address space itself contains 2^{33}/2^{15} = 2^{18} pages
   Each inner page table points to 2^{12} pages, so we need 2^{18} / 2^{12} = 2^{6} of them
   And then we need one outer page table page.

    Total:   2^{18} + 2^{6} + 1