We are given: - Logical address space size = \( 2^{32} \) bytes - Page size = \( 4096 = 2^{12} \) bytes - Each inner page table entry = \( 8 \) bytes - Two-level hierarchical paging
Step 1: Break Down the Logical Address A logical address consists of: 1. Offset within the page: Since the page size is \( 2^{12} \), the offset requires \( 12 \) bits. 2. Inner page table index: Each inner page table entry is \( 8 \) bytes, meaning an inner page table can hold \( 2^{12} / 8 = 2^9 \) entries. This requires \( 9 \) bits. 3. Outer page table index: The remaining bits are used to index the outer page table.
Step 2: Compute the Outer Page Table Index \( b \) The total logical address is \( 32 \) bits, and it is divided as: \[ b + 9 + 12 = 32 \] Solving for \( b \): \[ b = 32 - 9 - 12 = 11 \] Thus, the answer is \( 11 \).
According to the map shown in the figure, which one of the following statements is correct?
Note: The figure shown is representative.
In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and the line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?
Note: The figure shown is representative
A disk of size 512M bytes is divided into blocks of 64K bytes. A file is stored in the disk using linked allocation. In linked allocation, each data block reserves 4 bytes to store the pointer to the next data block. The link part of the last data block contains a NULL pointer (also of 4 bytes). Suppose a file of 1M bytes needs to be stored in the disk. Assume, 1K = \(2^{10}\) and 1M = \(2^{20}\). The amount of space in bytes that will be wasted due to internal fragmentation is ___________. (Answer in integer)