Answer page completely and show how you get results (show all calculations). Tha
ID: 3935866 • Letter: A
Question
Answer page completely and show how you get results (show all calculations).
Thank you.
The total number of pages that make up the executable program is shown (no fragmentation) The total size of a frame is equal to 512 bytes. The CPU needs to access data that is at the logical address 0 0 1 1 0 1 0 1 0 1 0 1 (12 bits) which is in frame 15 The size of a word in the system is 32 bits. What is the logical page number? What is the logical page offset? What is the total size in bytes of the program? What is the actual physical address in binary (12 bits) What is the total size of a page in bytes in this system?Explanation / Answer
Answer:
A) 001 :1
B) 341
C) 8*512 B : 4KB
E) 512 B
101 010 101
Page# page offset (341)
Program has 8 pages (3 bits)
And one page size : 512B
Total :8*512B :4KB
D) For P. A should be
1111 | 101 010 101 , but it is becoming 13 bits
E) 512B