Operating System

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Operating System
OPERATING SYSTEM
By Student's Name
Course Code and Name
Professor’s Name
University Name
City, State
Date of Submission

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Question 1
a). First-fit algorithm:
212K is put in 500K partition
417K is put in 600K partition
112K is put in 288K partition (new partition 288K = 500K - 212K)
426K must wait
Worst fit algorithm:
212K is put in 600K partition
417K is put in 500K partition
112K is put in 388K partition
426K must wait
Best-fit algorithm:
212K is put in 300K partition
417K is put in 500K partition
112K is put in 200K partition
426K is put in 600K partition
Best-fit makes the most efficient use of memory.

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b).
i) We utilize the following scheme
P1 P2 d
10 10 12
The 12 least significant digits in this address, allow access for 2 ^12 bytes -4Kb. These are
pointed to by any of the 2 ^10 entries of p2. In total, a second level page table can point to 2 ^
22 bytes -4Mb. Each such page table is pointed to by a first level table entry. In this scenario,
we need four page tables’ pages, a single first level page table (also known as the directory),
that points to level page tables (Han et al, 2015)

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ii).
iii) i) The physical memory frame is 20 nanoseconds.
There is no TLB
Two accesses of memory: page look up followed by the exact access = 2 x 20 ns= 40
nanoseconds (Levis et al, 2015, pp 9).
ii) There exists a TLB with an access speed of 0.05 nanoseconds
2x 0.05 =0.1