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Recent questions tagged demand-paging

1 vote
2 answers
1
The address sequence generated by tracing a particular program executing in a pure demand paging system with $100$ records per page, with $1$ free main memory frame is recorded as follows. What is the number of Page Faults? $0100,0200,0430,0510,0530,0560,0120,0220,0240,0260,0320,0370.$ $15,4$ $6,4$ $7,2$ $4,6$
asked Apr 1 in Operating System Lakshman Patel RJIT 72 views
0 votes
1 answer
2
The address sequence generated by tracing a particular program executing in a pure demand paging system with $100$ records per page, with $1$ free main memory frame is recorded as follows. What is the number of Page Faults? $0100,0200,0430,0510,0530,0560,0120,0220,0240,0260,0320,0370.$ $15,4$ $6,4$ $7,2$ $4,6$
asked Apr 1 in Operating System Lakshman Patel RJIT 31 views
0 votes
0 answers
3
In a swapping system with variable partitions, segments have probability -s/10 distribution ( e / 10) where s is the segment size in kilobytes. Holes -h/5 have probability distribution ( e / 5) where h is the hole size in kilobytes. What is the average ... q2soln.html i'm stuck at how average segment segment size and average hole size is calculated? please comment if you have got any insight.
asked Dec 25, 2018 in Operating System aambazinga 67 views
2 votes
1 answer
4
Consider the following extract from a program, written in a C-like language, that computes the transpose of a matrix. for (i=0; i<N; i++) for (j=0; j<N; j++) B[i,j]=A[i,j]; $A$ and $B$ are $N \times N$ matrices with floating point entries that are ... bytes Each of $A$ and $B$ is stored starting from the beginning of a page None of the pages allocated to $A$ and $B$ are initially in memory.
asked Sep 18, 2018 in Operating System jothee 128 views
1 vote
2 answers
5
Consider a demand paged memory system, page table is held in registers. It takes 800 nsec to service a page fault if empty page is available or replaced page is not modified and 950 nsec if the replaced page is modified, main memory access time is 120 nsec. If page to be replaced is modified 85% of time and page faultrate is 20% then average memory access time is ________. (Upto 1 decimal place)
asked Aug 14, 2018 in Operating System jhaanuj2108 597 views
0 votes
1 answer
6
In a demand-paged system, it takes 100 nanoseconds to access memory. The page table has 8 entries and is held in registers. It takes 10 milliseconds to service a page fault if an empty frame is available or if the victim frame is not dirty. If the ... victim frame is dirty 80% of the time, find the maximum page fault rate for which the effective memory access time remains within 200 nanoseconds
asked Dec 20, 2017 in Operating System kauray 296 views
6 votes
2 answers
7
A demand paging system has page fault service time as 125 time units if page is not dirty and 400 times units of page fault service time if it is a dirty page. Memory access time is 10 time units. The probability of a page fault is 0.3. In case of page fault, the ... dirty is P. It is observed that average access time is 50 time units. Then, the value of P is ______? [upto four decimal places]
asked Oct 27, 2017 in Operating System shivangi5 499 views
3 votes
2 answers
8
The address sequence generated by tracing a particular program executing in a pure demand paging system with 200 records per page with 1 free main memory frame is recorded as follows: 0100, 0139, 0209, 0430, 0237, 0578, 0500, 0730, 0799, 0600 The number of page faults are ________.
asked Oct 9, 2017 in Operating System Tuhin Dutta 270 views
0 votes
2 answers
9
In Dynamic Loading:- We load the particular module into the main memory when it is needed. In Demand Paging:- We load the particular page into the main memory when it is needed. Both don't seems to be same??
asked Sep 12, 2017 in Operating System Shubhanshu 794 views
2 votes
1 answer
10
In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. ... is modified 70% of the time ? What is the maximum acceptable page-fault rate for an effective access time of no more than 200 nanoseconds?
asked Sep 1, 2017 in Operating System set2018 818 views
1 vote
0 answers
11
Can anyone pls explain what is the concept applied here http://www.cs.jhu.edu/~yairamir/cs418/os6/sld010.htm
asked Jun 20, 2017 in Operating System Aashish S 311 views
1 vote
3 answers
12
Consider the following extract from a program, written in a C-like language, that computes the transpose of a matrix. for (i = 0; i < N; i++) for (j = 0; j < N; j++) B[i,j] = A[j,i]; A and B are N N matrices with floating point entries that are ... entry: 8 bytes Each of A and B is stored starting from the beginning of a page None of the pages allocated to A or B are initially in memory
asked Feb 24, 2017 in Operating System Devasish Ghosh 362 views
9 votes
4 answers
13
Suppose: TLB lookup time = 20 ns TLB hit ratio = 80% memory access time = 75 ns swap page time = 500,000 ns 75% of pages are dirty OS uses a 3 level page table What is the effective access time (EAT) if we assume the page fault rate is 15% ?
asked Jan 25, 2017 in Operating System biranchi 2k views
0 votes
1 answer
14
TLB lookup time (th) = 20 ns TLB hit ratio (ht) = 99% Memory access time (ts) = 100 ns Page fault rate (hp) =0.05% Swap page time (in or out) (tp) = 5000,000 ns What is the effective access time (EAT) if we assume that all pages currently in main memory are dirty? 5118.91 ns 51.21 ns 5120.9 ns None of the above
asked Jan 24, 2017 in Operating System vnc 301 views
4 votes
1 answer
15
Consider a demand-paging system with a paging disk that has average access and transfer time of 20 milliseconds. Addresses are translated through a page table in main memory, with an access time of 1 microsecond per memory access. Thus, each memory reference through ... remaining, 10 percent (or 2 percent of the total) cause page faults. What is the effective memory access time in milliseconds?
asked Jan 23, 2017 in Operating System biranchi 2.3k views
1 vote
1 answer
16
Which of the following programming techniques and structures are good for demand-paged environment a)Stack b)Hashed symbol table c)Sequential search d)Pure code Only a Both a, b Both a, c please explain with the reason
asked Dec 4, 2016 in Operating System Akriti sood 694 views
2 votes
1 answer
18
In a demand paging memory system, page table is held in registers. The time taken to service a page fault is $8$ m.sec. if an empty frame is available or if the replaced page is not modified, and it takes $20$ m.secs., if the replaced page is modified. What is the average access time to service ... to be replaced is modified $70$% of the time ? $11.6$ m.sec. $16.4$ m.sec. $28$ m.sec. $14$ m.sec.
asked Jul 23, 2016 in Operating System makhdoom ghaya 2.2k views
1 vote
2 answers
19
A demand paging system,with page table held in integer,takes $5ms$ to service a page fault if an empty page is available,or if the page to be replaced is not dirty.It takes $15ms$ if the replaced page is dirty. Memory access time is $1\mu s.$ Assume ... . The approximate maximum acceptable page fault rate to meet this time requirement will be______$\%?$ $\text{(Correct to two decimal places).}$
asked Jan 13, 2016 in Operating System khushtak 253 views
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