Question

How do modern operating systems solve this? [3 marks] i) This is a snapshot of a page table and a translation look aside buffer (TLB) of an operating system (Assume that these are the only populated entries). Toble 1: Page Table Entry Virtual Page Page Frame Time Loaded Time R bit M bit number Referenced 2 0 60 161 0 1 1 1 130 0 160 1 2 26 162 0 1 30 3 20 163 1 Table 2: TLB entry Virtual page number Page Frame 1 1 30 3 (a) What is the purpose of TLB? (b) [2 marks] Describe the process happening in belween a generation of page request and accessing the page reference when following pages are requested by the operating system? i. Page 9 il. page 0 il. page 30 6 marks] (c) Answer the question with reference to the Table 1: Page Table Entry i. What is R bit? [1 mark] 1 mark] i) Company X is designing an operating system for a critical computer system where the data consist of multiple components such as data segment which allows (read/write) and a security critical code segment which is (read-only). A major requirement of the system is to secure the code i. What is M bit? segment from malware and unauthorized access What should be the virtual address translating scheme, segmentation or paging? Give 1 reason for your answer? (2 marks iv) In a bite- addressable machine with 16 bit addresses and a page size of 4K (a) How many pages can there be in the virtual address space? (b) What are the page number and offset for the following addresses? (1 mark] (4 marks L 25035 H 804

answer for all questions....

Answer 1

a)  Translation Lookaside Buffer (TLB) is a special cache used to keep track of recently used transactions from the page table. TLB contains page table entries that have been most recently used. So basically it is a cache of the page table.

b) When any page request happens in the system then it first checks into the TLB whether a page is present or not? then check into the page table then if it is present in the memory then find corresponding frame number and find location in the memory but if the page number is not present in the page table then it can conclude that the page is not present in the memory so miss of the memory is happening and bring the page in the memory and then that page is used for further computation.

i) page 9

Page 9 is not present in the TLB so check the PAGE TABLE and it is also not present in the Page table so not present in the memory. so bring the page 9 into the memory and make the entry in the page table but before that one page from the page table is swapped to secondary memory and make a room for page 9.

ii) page 0

Page 0 is not present in the TLB so it is not a very frequent page so check in the page table and page 0 is present in the page table and corresponding page frame of the memory is 2. and a modified bit is 0 so from frame 2 we can directly access the page 0.

iii) page 30

page 30 is present in the TLB and Modified (M) bit is 1 so that is modified than an earlier copy of the frame so consider modified memory of page 30.

c) M and R bit

1) R(reference bit)

The referenced bit will say whether this page has been referred to in the last clock cycle or not. It is set to 1 by hardware when the page is accessed.

2) M(modified bit)

The modified bit indicates whether the page has been modified or not. Modified means sometimes when we try to change or write something on to the page. If a page is modified, we should replace that page with some other page, then the modified information should be kept on the hard disk or it has to be written back or it has to be saved back. It is set to 1 by hardware on write-access to page which is used to avoid writing when swapped out. A dirty bit is known as a modified bit.

iii) company should use the segmentation type of memory management scheme.

because here we do not want to change the company side code view so in the segmentation type the user view function is considered for the paging. segmentation is more appropriate in this case.

iv) In a bit addressable scheme the machine has a 16-bit size of the address while page size is 4k which is 12 bit so total bit for the page size is 16-12=4 so there are 2^4 pages possible, 16 pages are there so page table contains the 16 possible entries.

a) 16 pages are there in the virtual address space.

b) find the page number and offset

i) 25035

for 25035 decimal number corresponding, a binary number is 0110000111001011

first 4 bit for page number so that is 0110 and last 12 bit for offset that is 000111001011

so page number is 6 and offset 459.

ii) 804

0000001100100100 is a binary number corresponding to the 804 decimal number.

where first 4 is age number i.e. 0000 that is 0 is page number and last 12 bit is an offset 001100100100 i.e 804

so address 804 the page number is 0 and offset is 804.