Question

Consider a logical address space of 8 pages; each page is 2048 byte long, mapped onto a physical memory of 64 frames.

(i) How many bits are there in the logical address and how many bits are there in the physical address?

(ii) A 6284 bytes program is to be loaded in some of the available frames ={10,8,40,25,3, 15,56,18,12,35} . Show the contents of the program's page table.

(iii) What is the size of the internal fragmentation?

(iv) Convert the following logical addresses 2249 and 5245 to physical addresses.

Answer 1

Question gives, logical address space = 8 pages, each page size = 2048 bytes, physical memory = 64 frames. (i) Addressing with in a 2048-word page requires 11 bits (Since, 2¹¹ = 2048). Since, the logical address space consists of 8 pages, 2³ = 8 pages. The logical address must be 11 + 3 = 14. Similarly, since there are 64 = 2⁶ physical pages, physical address = 11 + 6 = 17. (ii) Given program(length) size = 6284 bytes. Given, available frames = {10, 8, 40, 25, 3, 15, 56, 18, 12, 35}. The given program can be divided into 6284/2048 = 4 pages.