# Recent questions tagged demand-paging 1 vote
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? $0100,0200,0430,0510,0530,0560,0120,0220,0240,0260,0320,0370.$ $15,4$ $6,4$ $7,2$ $4,6$
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? $0100,0200,0430,0510,0530,0560,0120,0220,0240,0260,0320,0370.$ $15,4$ $6,4$ $7,2$ $4,6$
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.
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.
1 vote
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)
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
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]
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 ________.
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??
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?
1 vote
11
Can anyone pls explain what is the concept applied here http://www.cs.jhu.edu/~yairamir/cs418/os6/sld010.htm
1 vote
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
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% ?
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
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?
1 vote
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
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.
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______$\%?$ $\text{(Correct to two decimal places).}$