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The portion of trap.h and syscall.h files for xv6 are shown…

The portion of trap.h and syscall.h files for xv6 are shown in the table below.

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For a segment that grows negatively, if a VA’s offset is out…

For a segment that grows negatively, if a VA’s offset is out-of-bounds then a page fault will occur.

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The valid bit in the TLB means the same as the valid bit in…

The valid bit in the TLB means the same as the valid bit in the entries of the page table.

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Now consider a function where we print the array contents in…

Now consider a function where we print the array contents in column order   void print_col_array(int *arr, int cols, int rows){    for(size_t i=0; i

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The PCB holds the most recent instruction pointer of a proce…

The PCB holds the most recent instruction pointer of a process while it is actively running on the CPU.

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On the same 32-bit machine, if the address space for a proce…

On the same 32-bit machine, if the address space for a process is divided into pages of size 4KB and each page table entry is 8 bytes, how large would a linear page table be for each process?

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When prog is executed, which of the following outputs is imp…

When prog is executed, which of the following outputs is impossible?

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In kernel mode the MMU checks the VA’s offset against the ba…

In kernel mode the MMU checks the VA’s offset against the base and bounds for that segment.

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If the scheduler is running the SJF algorithm, what is the a…

If the scheduler is running the SJF algorithm, what is the average response time of this workload?

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CS 537: Introduction to Operating Systems Fall 2024: Midterm…

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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