Question

Chapter 8, Consider a computer system with a 32-bit logical addressand 4-KB page size. The system supports up to 512 MB of physicalmemory. How many entries arethere in each of the following?

a.A conventional single-level page table.

b.An inverted page table.

Answer 1

32 -bit ==>2^32 Bytes --logical memory space

given page size is 4KB = 2^12Bytes

given physical memory size =512MB

                                                    =2^29bytes

so number of pages =2^32/2^12

                                    =2^20 pages

number of frames =2^29/2^12

                                 =2^17

a.number of entries in A conventional single-level page table are=2^20 (number of pages)

b.number of entries in a inverted page table =2^17 (number of frames)

Answer 2

its simple!
a) 2^32/2^12=2^20 (page size=2^12)
b)2^29/2^12=2^17 (512MB=2^29)

Answer 3

Logical address =32-bitPage Size = 4KBPhysical Memory = 512MBNo. Of entries in the page table inconventional single-level page table

-With 32 bit virtual addresses and 4KB pages, 512MB physicalmemory,

512MB˜102WORDS

Therearepage table entries
No. Of entries in inverted pagetableOneentry for each frame of physical memoryPhysicalMemory= 512MBNo. of entries=512MB/8 ==67108864Thus,no.of entries = 67108864I hope this will behelpful.