Read the following and answer the questions : Consider a machine with 16 GB main memory and 32‐bits virtual address space, with page size as 4KB. Frame size and page size is same for the given machine.
If a process of size 34KB is to be executed on this machine, then what will be the size of internal fragmentation for this process?
2KB
The process needs \(\lceil 34/4\rceil=9\) pages, allocating \(9\times4=36\) KB. Internal fragmentation is \(36-34=2\) KB.
Important Questions from Frame Allocation - Teaching
Find number of pages required for the given virtual address space
- The number of bits reserved for the frame offset is ______
For the machine described in the passage, assume byte addressing and a single-level page table containing one 4-byte entry for every virtual page in the complete 32-bit virtual address space. Each entry includes the physical frame number and control bits. What is the page-table storage size, excluding any additional overhead?
Use binary units: 1 KB = \(2^{10}\) bytes, 1 MB = \(2^{20}\) bytes, and 1 GB = \(2^{30}\) bytes.
