Consider the following statements regarding page table and inverted page table.

  1. A Computer System with 48-bit logical address space and 1Kbytes page size. The system has 1 Gbytes of physical memory. This implies that there are 512 K entries in an inverted page table.
  2. A Computer System with 32-bit logical address space and 4Kbytes page size. The system has 1 Gbytes of physical memory and page table entries size is 4 bytes. This implies that the size of page table is 2MB.

Which of the following option is correct?

(A) only statement 1 is correct.
(B) both statement 1 and 2 are correct
(C) only statement 2 is correct
(D) neither statement 1 nor 2 is correct


Answer: (D)

Explanation:

  1. Number of entries in Inverted page table = physical address space / page size = 230/10 = 20 = 1024 K entries.
  2. Since we have a virtual address space of 2^32 and each page size is 2^12, we can store (2^32/2^12) = 2^20 pages. Since each entry into this page table has an address of size 4 bytes, then we have 2^20*4 = 4MB. So the page table takes up 4MB in memory.

So, neither statement 1 nor 2 is correct.

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  • Last Updated : 29 Oct, 2018

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