Operating System Test 5

Concept:

The working set model: The memory that is being accessed frequently by an application.

Same thing is done by principle of locality.

Explanation:

There are two types of principle of locality

1. Spatial locality: whenever we are looking for an element the chances are that, the particular element will be present in a close proximity around the one which we have previously referred.
2. Temporal locality: Least recently used element is going to be used again.

Principle of locality is the correct answer, hence option 3 is the correct answer.

Data:

number of pages = 128

size of page/frame = 1024 word

number of frames = 64

Calculation:

size of logical address = number of pages × size of page

size of logical address = 128 × 1024 words = 2¹⁷  words

number of bits = log₂217  = 17 bits

size of physical address = number of frames × size of a frame

size of physical address = 64 × 1024 words = 2¹⁶ word

number of bits = log2216  = 16 bits

Data:

PAS = 32 bits

Page size = 4 KB

Page table size = 72 MB

Formula:

number of frames = \(\frac{Physical \;address}{page\;size}\)

Calculation:

number of frames = \(\frac{2^{32}}{2^{12}} = 2^{20}\)

translation = 20 bits

entry of page table = 2 + 1 + 1 + 20 = 24 bits = 3 byte

number of pages = \(\frac{page\;table\;size}{page \;table\; entry \;size} = \frac{72 \;MB}{3} =24 \;M\)

Virtual address = number of pages × page size = 24 × 2²⁰ × 4 KB 

Virtual address = 96 × 230 B

length of virtual address space = log₂ 96 × 230 = 37 bits

Data:

main memory = 100 ns

TLB  = 10 ns

EAT = 150

Formula:

EAT = h_t × (t_b + t_m) + (1 – h_t) (t_b + (n + 1) t_m)

Calculation:

150 ns = h_t × (10 ns + 100 ns) + (1 – h_t) (10 ns + 3 × 100 ns)

150  = 110 h_t + 310  – 310 h_t

160 ns = 200 ht

ht = 0.8  

ht = 80%  

Important Points:

EAT → Average effective memory access time

ht → Hit ratio of TLB

tb → TLB access time

t_m → Main memory access time

Concept:

Best fit allocation:

The best fit allocation strategy chooses the smallest available memory partition that can satisfy the memory requirement. It creates the smallest hole.

First fit allocation:

The first fit chooses the first available memory partition that can satisfy the requirement.

Worst fit allocation:

The worst fit allocation strategy chooses the largest available memory partition that can satisfy the memory requirement. It creates the largest hole.

Next fit allocation:

It works same as First Fit, the only difference it maintain a pointer to all last allocated memory space to the process and begins it search from there if new request is arrived, unlike first fit which start will initial memory space.

Explanation:

Option 1 and Option 4 : FALSE

The hole created by first fit may or may not be larger than the hole created by next fit

Option 2: FALSE

The hole created by worst fit is always larger than or equal to the hole created by first fit

Option 3: TRUE

The hole created by best fit could never be larger than the hole created by first fit.

Data:

Size of one page = 32 KB = 215 B

Virtual address space = 48 bits

number of entries in TLB = 1024

2 -way set associative

Formula:

set = \(\frac{entries}{k-way}\)
Virtual address = tag + set + page offset (in bits)

Calculation:

set = \(\frac{1024}{2} = 2^{9}\)

48 = tag + 9 + 15

∴ tag = 24

Data

1 word = 4 byte

Logical address = 2³² word

page size = 2¹⁰ KB = 2⁸ word

page table entry (PTE) = 8 bytes = 2 word

Formula:

Page table size (PTS) = number of pages × PTE

Calculation:

number of pages = \(\frac{{{2^{32}}}}{{{2^{8}}}}\; = {2^{24}}\)

PTS = 2²⁴ × 2 = 32 MW

Important points

1 MW = 2²⁰ W