Assume 4KB pages, 20b virtual addresses and 24b physical addresses. For the foll

Question

Assume 4KB pages, 20b virtual addresses and 24b physical addresses. For the following page table, answer the following questions. Only a portion of the (large) page table is shown. The portion shown is sufficient for this problem. Answer each question based on the page table given. Do not attempt to update the page table as you go.

Virtual Page # Valid Physical Page #

0x00 1 0xC01

0x11 0 0x005

0x23 1 0x6A3

0x24 1 0x700

0x25 1 0x702

0x75 0 0x55D

0xA0 1 0x701

0XA1 1 0xA98

0XA2 1 0x724

0xFF 0 0x702

a. How many bits is the virtual page number?

b. How many bits is the physical page number?

c. For each of the following virtual addresses, indicate if they would be a page fault or not.

Virtual Address Page Fault (Y/N)

0x00000

0x75138

0x24FFC

0xFFFFF

d. For each of the following virtual addresses, indicate what the physical address is

Virtual Address Physical Address

0xA1308

0xA2800

0x23456

Explanation / Answer

a. bits is the virtual page number ----- 2 *4 = 8 bits (1 hexadecimal number = 4 bits)

b. bits is the physical page number ---- 3*4 = 12 bits

c. page fault or not.

Virtual Address Page Fault (Y/N)

0x00000 ---- N (valid bit is 1)

0x75138 ---- Y(valid bit is 0)

0x24FFC ---- N(valid bit is 1)

0xFFFFF----- Y(valid bit is 0)

d. physical address mapping is done using corresponding addresses in page table

Virtual Address Physical Address

0xA1308 --- 0xA98308(0XA1 ----> 0xA98 + offset(308))

0xA2800 ---- 0x724800(0XA2 -----> 0x724 +offset(800))

0x23456 ----- 0x6A3456(0x23 ----> 0x6A3+ offset(456))

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