The logical address space of the process is 8 pages and the size ofeach page is
ID: 3616499 • Letter: T
Question
The logical address space of the process is 8 pages and the size ofeach page is 1 Kb. There are 64 frames in the main memory. The sizeof a frame is equivalent to the size of a page. With the giveninformation, answer to the following questions:
1 Calculate thesize of Logical Address? 2 Find the size ofthe Logical Memory? 3 Calculate the size of Physical Address? 4 Determine the sizeof the Physical Memory?
The logical address space of the process is 8 pages and the size ofeach page is 1 Kb. There are 64 frames in the main memory. The sizeof a frame is equivalent to the size of a page. With the giveninformation, answer to the following questions:
1 Calculate thesize of Logical Address? 2 Find the size ofthe Logical Memory? 3 Calculate the size of Physical Address? 4 Determine the sizeof the Physical Memory? 4 Determine the sizeof the Physical Memory?
Explanation / Answer
Logical address space has 8bits..therefore it requires 3 bits to denote the page [since 23 =8] Size of each page is 1Kb i.e. 1024 words..therefore it requires 10bits to denote the words . [since 1024 = 210] 1) Therefore size of logical address = 3 bits (for page) + 10bits ( for words) = 13 bits 2) size of logical memory = 8 pages each of size 1kb = 8Kb 3) 64 frames in the main memory. Therefore it requires 6 bits to denote 64 frames[since 26 =64] it requires 10 bits to denote words in eachframe [ since size of frame = size of page = 1Kb] Therefore size of physical address = 6 (denotesframes) + 10 (words in page) = 16 bits 4) size of physical memory = 64 frames of size 1Kb each =64Kb
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