CS 537: Introduction to Operating Systems Fall 2024: Midterm Exam #1 This exam is closed book, but you may use 1 sheet of notes. No calculators may be used. You have 1 hour and 30 minutes to complete this exam. Unless stated (or implied) otherwise, you should make the following assumptions: The OS manages a single uniprocessor (single core) All memory is byte addressable Page table and page directory entries require 4 bytes Data is allocated with optimal alignment, starting at the beginning of a page Assume leading zeros can be removed from numbers (e.g., 0x06 == 0x6). Hex numbers are represented with a proceeding “0x” The following might help you with some calculations: 0x100 = 28 = 256 210 = 1024 212 = 4096 210 bytes = 1KB 220 bytes = 1MB This exam has 60 questions. Each question has the same number of points. Good luck!
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On the same system, how many pages are there in the virtual…
On the same system, how many pages are there in the virtual address space of a process?
Processes and Scheduling
Processes and Scheduling
The purpose of the ASID in a TLB entry is to reduce the size…
The purpose of the ASID in a TLB entry is to reduce the size of a process’ page table.
The purpose of the TLB is to speed address translation.
The purpose of the TLB is to speed address translation.
Sparsely used address spaces are poor candidates for multi-l…
Sparsely used address spaces are poor candidates for multi-level page tables.
What will be the first TLB hit with the same TLB size and re…
What will be the first TLB hit with the same TLB size and replacement policy as before?
For the same address space size, having smaller pages means…
For the same address space size, having smaller pages means you will have smaller page tables.
Convert the VA 0x7011 to its PA.
Convert the VA 0x7011 to its PA.
A page table entry holds the virtual page number for a given…
A page table entry holds the virtual page number for a given physical page.