CODEWITHSUNDEEP
memory-managementoperating-systempaging
icynoodle
icynoodle

Reputation: 13

frames in system/ Paging

can someone explain to me how to solve this problem?

A main memory has 128Mbytes and the size of a page is 2Kbytes. Consider that the memory is byte addressable. A process P has 6 logical pages (first logical page has the address 0).

a. What is the number of frames in this system?

b. How many bits are allocated to the page address?

Upvotes: 0

Views: 225

Answers (2)

Yomna
Yomna

Reputation: 1

Since physical memory = 128 MB ➔ # frames = 128 MB / 2 KB = 2^15 frames ➔ number of bits for frame number = 15 bits

Upvotes: 0

Robin Caron
Robin Caron

Reputation: 627

a. Amount of main memory is 128MB which is 128 * 1024KB = 131,072KB. Since each page is 2KB that means you have 131,072KB / 2KB/page = 65,536 pages (or frames)

b. Page address is 2KB which 2^11 bytes. Assuming a system where the pages are in bytes that would imply that the number of bits for the page address is 11.

Upvotes: 0

Related Questions