Question 1
In the context of operating systems, which of the following statements is/are correct with respect to paging?
Page size has no impact on internal fragmentation.
Multilevel paging is necessary to support pages of different sizes.
Paging incurs memory overheads.
Paging helps solve the issue of external fragmentation.
Question 1 Explanation: 
  1. False, Large page size may lead to higher internal fragmentation.
  2. False, To support pages of different sizes, the Instruction set architecture should support it. Multi-level paging is not necessary.
  3. True, The page table has to be stored in main memory, which is an overhead. 
  4. True, Paging avoids the external fragmentation. 
Question 2
Three processes arrive at time zero with CPU bursts of 16, 20 and 10 milliseconds. If the scheduler has prior knowledge about the length of the CPU bursts, the minimum achievable average waiting time for these three processes in a non-preemptive scheduler (rounded to nearest integer) is _________ milliseconds.
Question 2 Explanation: 
Minimum average waiting time amongst non-preemptive scheduling is given by SJF. So, avg waiting time = (0+10+26)/3 = 12 ms.
Question 3
Which of the following standard C library functions will always invoke a system call when executed from a single-threaded process in a UNIX/linux operating system? 
Question 3 Explanation: 
  • A sleep system call takes a time value as a parameter, specifying the minimum amount of time that the process is to sleep before resuming execution. The parameter typically specifies seconds, although some operating systems provide finer resolution, such as milliseconds or microseconds.
  • strlen() is not system call.
  • The function call malloc() is a library function call that further uses the brk() or sbrk() system call for memory allocation
  • Exit also system call
Question 4
Consider the following pseudocode, where S is a semaphore initialized to 5 in line#2 and counter is a shared variable initialized to 0 in line#1. Assume that the increment operation in line#7 is not atomic.
  1. int counter = 0
  2. Semaphore S = init(5);
  3. void parop(void)
  4. {
  5.       wait(S);
  6.       wait(S);
  7.       counter++;
  8.       signal(S);
  9.       signal(S);
  10. }  
If five threads execute the function parop concurrently, which of the following program behaviour(s) is/are possible?
The value of counter is 1 after all the threads successfully complete the execution of parop.
The value of counter is 5 after all the threads successfully complete the execution of parop.
The value of counter is 0 after all the threads successfully complete the execution of parop.
There is a deadlock involving all the threads.
Question 4 Explanation: 

count =0











(1) Deadlock is possible.

For a deadlock free operation

No. of resources >= No. of process (Req-1) + 1

Here, no. of resources = 5 (semaphore value)

Each thread requires = 2 resources (wait call)

No. of threads = 5

      5 ≥ 5* (2-1)+1

         ≱ 6. So deadlock is possible.

This occurs when all 5 threads get blocked on first wait().

(2) count =5 is possible

When all threads enter CS and execute count++ sequentially.

(3) Count=1 is possible.

Assembly level:

     read R0, count

     inc R0, 1

     write count, R0

Thread - 1 reads Count=0 in R0, preempted

Thread-2 reads count=0, is r1, completes, count=1

Thread-3, 4 & 5 completes.

Thread-1 is given CPU

MC R0, 1, so R0=1

Write R0 to count.

So, Count=1. 

Question 5
Consider a linear list based directory implementation in a file system. Each directory is a list of nodes, where each node contains the file name along with the file metadata, such as the list of pointers to the data blocks.Consider a given directory foo. Which of the following operations will necessarily require a full scan of foo for successful completion?
Deletion of an existing file from foo
Opening of an existing file in foo
Creation of a new file in foo
Renaming of an existing file in foo
Question 5 Explanation: 
Renaming and creating a new file requires us to search the complete inode for duplicate names (to check if the same name already exists). There is no such problem in deleting and opening a file, as duplicate names are not possible while they are created.
Question 6
There are 6 jobs with distinct difficulty levels, and 3 computers with distinct processing speeds. Each job is assigned to a computer such that:
  • The fastest computer gets the toughest job and the slowest computer gets the easiest job.
  • Every computer gets at least one job.
The number of ways in which this can be done is ______
Question 6 Explanation: 

Let the levels be  1,2,3,4,5,6. (1 is the least difficult, 6 is the most difficult level)
Let the computers be F,M,S( fast, medium, slow).

As per the given constraint,  1 must be given to F and 6 must be given to S.
Now we are left with 2,3,4,5 and  F being assigned 1, S being assigned 6 and M being assigned none.
Another constraint is that, every computer must be assigned atleast one.
So compute with assigning one job to M, two jobs to M, three jobs to M and four jobs to M.
Assigning one job to M: we can assign 1 out of 4 jobs in (4C1 ways) and remaining 3 jobs to F,S in  2*2*2 = 8  ways. (each job has two options, either F or S),

Assigning two jobs to M: we can select two jobs from 4 in 4C2 ways and remaining 2 can be distributed to  F and S in  2*2 ways ( each job has two options either F or S)

Assigning three jobs to M: we can select 3 out of 4 in 4C3 ways remaining can be distributed to F,M in 2 ways. 

Assigning 4 jobs to M: it can be done in only one way.

Total : 4c1*8  +  4C2* 4  + 4C3*2 + 1

   = 32+24+8+1


Question 7

Which of the following statements is true?

ROM is a Read/Write memory
PC points to the last instruction that was executed
Stack works on the principle of LIFO
All instructions affect the flags
Question 7 Explanation: 
We know that stack works on the principle of LIFO.
Question 8

In a paged segmented scheme of memory management, the segment table itself must have a page table because

the segment table is often too large to fit in one page
each segment is spread over a number of pages
segment tables point to page table and not to the physical locations of the segment
the processor’s description base register points to a page table
Both A and B
Question 8 Explanation: 
The segment table is often too large to fit in one page. This is true and the segment table can be divided into pages. Thus page table for each segment table, pages are created.
Segment paging is different from paged segmentation.
Question 9

Which of the following page replacement algorithms suffers from Belady’s anamoly?

Optimal replacement
Both (A) and (C)
Question 9 Explanation: 
FIFO is suffers from Belady's Anamoly.
Question 10

Which scheduling policy is most suitable for a time shared operating system?

Shortest Job First
Round Robin
First Come First Serve
Question 10 Explanation: 
In Round robin, we use the time quantum based on this execution can be done. Then it is said to be time shared operating system.
Question 11

The sequence …………… is an optimal non-preemptive scheduling sequence for the following jobs which leaves the CPU idle for …………… unit(s) of time.

Question 11 Explanation: 
Here, in option (B) and (C) they have given CPU idle time is 0 which is not possible as per schedule (B) and (C).
So, (B) and (C) will be eliminated.
Now in (A) and (D):
For r(A),

So, idle time is between 0 to 1 which in case of option (A).
For option(D),

We can see there is no idle time at all, but in option given idle time is 5, which is not matching with our chart. So option (D) is eliminated.
Therefore, the correct sequence is option (A).
Question 12

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, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370 
Question 12 Explanation: 
Page are fitted in frames, so first we need to determine the pages but given request are just the record request in decimal. We can assume that first page to address from 0000 to 0099 and page 2 contains records from 0100 to 0199 and so on (it is given in question that each page contains 100 records) and so on. So page request string is
01, 02, 04, 04, 05, 05, 05, 01, 02, 02, 02, 03, 03.
Clearly 7 page faults.
Question 13

In a virtual memory system the address space specified by the address lines of the CUP must be __________ than the physical memory size and _______ than the secondary storage size.

smaller, smaller
smaller, larger
larger, smaller
larger, larger
Question 13 Explanation: 
Primary memory < Virtual memory < Secondary memory.
Question 14

A computer installation has 1000k of main memory. The jobs arrive and finish in the following sequences.

 Job 1 requiring 200k arrives
 Job 2 requiring 350k arrives
 Job 3 requiring 300k arrives
 Job 1 finishes
 Job 4 requiring 120k arrives 
 Job 5 requiring 150k arrives
 Job 6 requiring 80k arrives 

(a) Draw the memory allocation table using Best Fit and First fit algorithms.
(b) Which algorithm performs better for this sequence?

Theory Explanation.
Question 15

Consider the following program segment for concurrent processing using semaphore operators P and V for synchronization. Draw the precedence graph for the statements S1 to S9.

 a, b, c, d, e, f, g, h, i, j, k : semaphore;
    begin S1; V(a); V(b) end;
    begin P(a); S2; V(c); V(d) end;
    begin P(c); S4; V(c) end; 
    begin P(d); S5; V(f) end;
    begin P(e); P(f); S7; V(k) end;
    begin P(b); S3;V(g);V(h) end;
    begin P(g); S6; V(i) end;
    begin P(h); P(i); S8; V(j) end;
    begin P(j); P(j); P(k); S9 end;
Theory Explanation.
Question 16

The head of a moving head disk with 100 tracks numbered 0 to 99 is currently serving a request at tract 55. If the queue of requests kept in FIFO order is

  10, 70, 75, 23, 65 

Which of the two disk scheduling algorithms FCFS (First Come First Served) and SSTF (Shortest Seek Time First) will require less head movement? Find the head movement for each of the algorithms.

Theory Explanation.
Question 17

Consider allocation of memory to a new process. Assume that none of the existing holes in the memory will exactly fit the process’s memory requirement. Hence, a new hole of smaller size will be created if allocation is made in any of the existing holes. Which one of the following statements is TRUE?

The hole created by worst fit is always larger than the hole created by first fit.
The hole created by best fit is never larger than the hole created by first fit.
The hole created by first fit is always larger than the hole created by next fit.
The hole created by next fit is never larger than the hole created by best fit.
Question 17 Explanation: 
The size of hole created using best fit is never greater than size created by first fit. The best fit chooses the smallest available partition to fit the size of the process. Whereas, first fit and next fit doesn’t consider the size of the holes available.
Question 18

Consider the following statements about process state transitions for a system using preemptive scheduling.

    I. A running process can move to ready state.
    II. A ready process can move to running state.
    III. A blocked process can move to running state.
    IV. A blocked process can move to ready state.

Which of the above statements are TRUE?

II and III only
I, II and III only
I, II, III and IV
I, II and IV only
Question 18 Explanation: 
Question 19

Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P1,P2,P3,P4.

If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is _____.

Question 19 Explanation: 

Turn Around Time = (21 – 0) + (13 – 0) + (2 – 0) + (6 – 0), Average = 42/4 = 10.50

Turn Around Time (TAT) = (18 – 0) + (21 – 0) + (10 – 0) + (14 – 0), Average = 63/4 = 15.75
Absolute difference = |10.50-15.75| = 5.25
Question 20

Each of a set of n processes executes the following code using two semaphores a and b initialized to 1 and 0, respectively. Assume that count is a shared variable initialized to 0 and not used in CODE SECTION P.

wait (a); count = count+1;
if (count==n) signal (b);
signal (a); wait (b); signal (b);

What does the code achieve?

It ensures that all processes execute CODE SECTION P mutually exclusively.
It ensures that at most two processes are in CODE SECTION Q at any time.
It ensures that no process executes CODE SECTION Q before every process has finished CODE SECTION P.
It ensures that at most n-1 processes are in CODE SECTION P at any time.
Question 20 Explanation: 
The wait(a) ensures that the count value is correctly incremented (no race condition) if(count==n) signal (b); // This signal(b) statement is executed by the last (nth) process only. Rest of the n-1 processes are blocked on wait(b). Once the nth process makes signal(b) then the rest of the processes can proceed and enter Code section Q.
Question 21

Consider the following five disk access requests of the form (request id, cylinder number) that are present in the disk scheduler queue at a given time.

         (P, 155), (Q, 85), (R, 110), (S, 30), (T, 115) 

Assume the head is positioned at cylinder 100. The scheduler follows Shortest Seek Time First scheduling to service the requests.

Which one of the following statements is FALSE?

The head reverses its direction of movement between servicing of Q and P.
T is serviced before P.
R is serviced before P.
Q is serviced after S, but before T.
Question 21 Explanation: 
Question 22

Consider a paging system that uses a 1-level page table residing in main memory and a TLB for address translation. Each main memory access takes 100 ns and TLB lookup takes 20 ns. Each page transfer to/from the disk takes 5000 ns. Assume that the TLB hit ratio is 95%, page fault rate is 10%. Assume that for 20% of the total page faults, a dirty page has to be written back to disk before the required page is read in from disk. TLB update time is negligible. The average memory access time in ns (round off to 1 decimal places) is ______.

154.5 ns
Question 22 Explanation: 
p=0.1, 1-p=0.9
d=0.2, 1-d=0.8
EMAT = h×(T+M)+(1-h)[(1-p)×2M+p[(1-d)[D+M]+d(2D+M)]+T]
= 0.95×(20+100)+(1-0.95)[(1-0.1)×200+(0.1)[(1-0.2)[5000+100]+(0.2)(10000+100)]+20]
= 154.5 ns
Question 23

The process state transition diagram in below figure is representative of

a batch operating system
an operating system with a preemptive scheduler
an operating system with a non-preemptive scheduler
a uni-programmed operating system
Question 23 Explanation: 
Transaction from running → ready present.
So this is preemptive.
Question 24

A critical section is a program segment

which should run in a certain specified amount of time
which avoids deadlocks
where shared resources are accessed
which must be enclosed by a pair of semaphore operations, P and V
Question 24 Explanation: 
In CS, share resources are accessed.
Question 25

Which of the following is an example of spooled device?

A line printer used to print the output of a number of jobs.
A terminal used to enter input data to a running program.
A secondary storage device in a virtual memory system.
A graphic display device.
Question 25 Explanation: 
Example of spooled device is a line printer used to print the output of a number of jobs.
Question 26

The correct matching for the following pairs is

         A. Activation record	   1. Linking loader
         B. Location counter	   2. Garbage collection
         C. Reference counts	   3. Subroutine call
         D. Address relocation	   4. Assembler 
A – 3 B – 4 C – 1 D – 2
A – 4 B – 3 C – 1 D – 2
A – 4 B – 3 C – 2 D – 1
A – 3 B – 4 C – 2 D – 1
Question 26 Explanation: 
Each time a subroutine is called its activation record is created.
An assembler uses location counter value to give address to each instruction which is needed for relative addressing as well as for jump labels.
Reference count is used by garbage collector to clear the memory whose reference count be comes 0.
Linker loader is a loader which can load several compiled codes and link them together into a single executable. Thus it needs to do relocation of the object codes.
Question 27

A 1000 Kbyte memory is managed using variable partitions but to compaction. It currently has two partitions of sizes 200 Kbytes and 260 Kbytes respectively. The smallest allocation request in Kbytes that could be denied is for

Question 27 Explanation: 
200 and 260 are already hold by some other processes. Now we have to model the partition in such a way so that smallest allocation request could be denied. So, we can do the division as,

So, smallest allocation request which can be denied is 181 KB.
Question 28

A solution to the Dining Philosophers Problem which avoids deadlock is

ensure that all philosophers pick up the left fork before the right fork
ensure that all philosophers pick up the right fork before the left fork
ensure that one particular philosopher picks up the left fork before the right fork, and that all other philosophers pick up the right fork before the left fork
None of the above
Question 28 Explanation: 
In the Dining philosopher problem, each philosopher needs exactly two chopsticks to eat food but the problem is: each philosopher is going to take one chopstick at a time, which is placed at its right-hand side or at its left-hand side, but remember all should choose in the same manner like if first one chooses in a clockwise manner then each one should choose in clockwise, this type of picking cause, a circular waiting loop because each one is depending on other. This is also called as circular waiting and it leads to deadlock.
To avoid this, atleast one philosopher should choose its first chopstick in different way so that circular loop is not formed.
Question 29

Four jobs to be executed on a single processor system arrive at time 0+ in the order A, B, C, D. their burst CPU time requirements are 4, 1, 8, 1 time units respectively. The completion time of A under round robin scheduling with time slice of one time unit is

Question 29 Explanation: 
Time quantum is 1 unit.

∴ Completion time of A is 9 unit.
Question 30

A demand paged virtual memory system uses 16 bit virtual address, page size of 256 bytes, and has 1 Kbyte of main memory. LRU page replacement is implemented using a list whose current status (page number in decimal) is

For each hexadecimal address in the address sequence given below

 00FF, 010D, 10FF, 11B0

(i) the new status of the list
(ii) page faults, if any, and
(iii) page replacements, if any

Theory Explanation.
Question 31

The concurrent programming constructs fork and join are as below:

   N = 2
   M = 2
   fork L3
   fork L4
L1:join N
L2:join M
   goto L1
   goto L2
Theory Explanation.
Question 32

A computer system uses the Banker’s Algorithm to deal with deadlocks. Its current state is shown in the table below, where P0, P1, P2 are processes, and R0, R1, R2 are resources types.

(a) Show that the system can be in this state.
(b) What will the system do on a request by process P0 for one unit of resource type R1?

Theory Explanation.
Question 33

A file system with a one-level directory structure is implemented on a disk with disk block size of 4K bytes. The disk is used as follows:

 Disk-block 0: File Allocation Table, consisting of one 8-bit entry per date block, 
               representing the data block address of the next date block in the file
 Disk block 1: Directory, with one 32 bit entry per file:
 Disk block 2: Data block 1;
 Disk block 3: Data block 2; etc. 

(a) What is the maximum possible number of files?
(b) What is the maximum possible file size in blocks?

Theory Explanation.
Question 34

Locality of reference implies that the page reference being made by a process

will always be to the page used in the previous page reference
is likely to be to one of the pages used in the last few page references
will always be to one of the pages existing in memory
will always lead to a page fault
Question 34 Explanation: 
Locality of reference is also called as principle of locality. There are three different principle of locality:
(1) Temporal locality
(2) Spatial locality
(3) Sequential locality
→ In programs related data are stored in consecutive locations and loops in same locations are used repeatedly.
Question 35

The correct matching for the following pairs is

(A) Disk Scheduling        (1) Round robin
(B) Batch Processing       (2) SCAN
(C) Time sharing           (3) LIFO
(D) Interrupt processing   (4) FIFO  
A – 3 B – 4 C – 2 D – 1
A – 4 B – 3 C – 2 D – 1
A – 2 B – 4 C – 1 D – 3
A – 3 B – 4 C – 3 D – 2
Question 35 Explanation: 
Disk scheduling - SCAN
Batch processing - FIFO
Time sharing - Round Robin
Interrupt processing - LIFO
Question 36

I/O redirection

implies changing the name of a file
can be employed to use an existing file as input file for a program
implies connection 2 programs through a pipe
None of the above
Question 36 Explanation: 
Redirection is known as capturing output from a file, command, program (or) even code block within a script and sending it as input to another file, command, program (or) script.
Question 37


reduces page I/O
decreases the degree of multiprogramming
implies excessive page I/O
improve the system performance
Question 37 Explanation: 
Thrashing implies excessive page I/O.
Question 38

Dirty bit for a page in a page table

helps avoid unnecessary writes on a paging device
helps maintain LRU information
allows only read on a page
None of the above
Question 38 Explanation: 
The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device.
Question 39

An operating system contains 3 user processes each requiring 2 units of resource R. The minimum number of units of r such that no deadlocks will ever arise is

Question 39 Explanation: 
For the system to cause deadlock give each process 1 resource less than they require. Since in this case they require 2 resource each, so just give them 1 resource each. So if at max if 3 resource will be available then there can be deadlock. So by adding one more resource deadlock will never occur. So in total minimum 4 resources are required so that deadlock will never occur.
Question 40

Each Process Pi, i = 1.......9 is coded as follows

           {Critical section}

The code for P10 is identical except it uses V(mutex) in place of P(mutex). What is the largest number of processes that can be inside the critical section at any moment?

None of the above
Question 40 Explanation: 
Since the both code (i.e., P1 to P9 and P10) can be executed any number of times and code for P10 is
Now, let me say P1 is in CS then P1 is in CS
then P10 comes and executes CS (upon mutex).
Now, P2 comes (down on mutex).
Now P10 moves out of CS (again binary semaphore will be 1).
Now P3 comes (down on mutex).
Now P10 comes (upon mutex).
Now P4 comes (down mutex).

So, if we take P10 'in' and 'out' of CS recursively, all 10 processes can be in CS at same time using binary semaphore only.
Question 41

Match the pairs:

(a) Critical region     (p) Hoare's monitor
(b) Wait/Signal         (q) Mutual exclusion
(c) Working Set         (r) Principle of locality
(d) Deadlock            (s) Circular Wait
(a) - (q), (b) - (p), (c) - (r), (d) - (s)
Question 42

State whether the following statements are TRUE or FALSE with reason:

The Link-load-and-go loading scheme required less storage space than the link-and-go loading scheme.
Question 42 Explanation: 
In link and go scheme the linkage editor co-exists with program in main memory while performing the linking task.
Whereas in link, load and go scheme the linkage editor does not co-exists with program in main memory while performing linking task.
Question 43

A linker reads four modules whose lengths are 200, 800, 600 and 500 words, respectively. If they are loaded in that order, what are the relocation constants?

0, 200, 500, 600
0, 200, 1000, 1600
200, 500, 600, 800
200, 700, 1300, 2100
Question 43 Explanation: 
First module loaded starting at address 0. Size is 200. Hence it will occupy first 200 address, last address being 199.
Now 2nd will start loading at 200, since size is 800, so last address is 999.
Now 3rd module will start loading at 1000, since size is 600. So last address is 1599.
Now 4th module will start loading at 1600 and go till 2099.
Question 44

Which of the following is an example of a spooled device?

The terminal used to enter the input data for the C program being executed
An output device used to print the output of a number of jobs
The secondary memory device in a virtual storage system
The swapping area on a disk used by the swapper
Question 44 Explanation: 
Spooling is a technique in which an intermediate device such as disk is interposed between process and low speed i/o device.
For example in printer if a process attempt to print a document but printer is busy printing another document, the process, instead of waiting for printer to become available, write its output to disk. When the printer becomes available the data on disk is printed.
Spooling allows process to request operation from peripheral device without requiring that the device is ready to service the request.
Question 45

When  the  result  of  a  computation  depends  on  the  speed  of  the  processes involved there is said to be

cycle stealing
rare condition
a time lock
a deadlock
Question 45 Explanation: 
When first result depends on ordering of processes it is called race condition.
Speed of processes corresponds to ordering of processes.
Question 46

A counting semaphore was initialized to 10. Then 6P (wait) operations and 4V (signal) operations were completed on this semaphore. The resulting value of the semaphore is

Question 46 Explanation: 
Let the semaphore be S which is initially 10.
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 - 6 + 4 = 8
Question 47

A  computer  has  six  tape  drives,  with  n  processes  competing  for  them.  Each process may need two drives. What is the maximum value of n for the system to be deadlock free?

Question 47 Explanation: 
Each process needs 2 drives. So for deadlock just give each process one drive. So total 6 process can be given 1 drive each and can cause deadlock. So to break deadlock just reduce 1 process.
So maximum no. of process for the system to be deadlock free is 5.
Question 48

The overlay tree for a program is as shown below:

What will be the size of the partition (in physical memory) required to load (and run) this program?

12 KB
14 KB
10 KB
8 KB
Question 48 Explanation: 
To enable a process which is larger than the amount of memory allocated to it, we can use overlays. The idea of overlays is to heap only those instructions and data that are needed at any given time. When some instructions are needed, they are loaded into space occupied previously by instructions that are no longer needed.
For the above program, maximum memory will be required when running code portion present at leaves.
Maximum requirement
= MAX(12, 14, 10, 14)
= 14
Question 49

Consider n processes sharing the CPU in a round-robin fashion. Assuming that each process switch takes s seconds, what must be the quantum size q such that the overhead resulting from process switching is minimized but at the same time each process is guaranteed to get its turn at the CPU at least every t seconds?

q ≤ t-ns/n-1
q ≥ t-ns/n-1
q ≤ t-ns/n+1
q ≥ t-ns/n+1
Question 49 Explanation: 
Question 50

If an instruction takes i microseconds and a page fault takes an additional j microseconds, the effective instruction time if on the average a page fault occurs every k instruction is:

i + (j/k)
i + (j*k)
(i+j)/ k
Question 50 Explanation: 
Effective memory access time
= service time + page fault rate * page fault service time
= i + 1/k * j
= i + j/k

The effective instruction time, if on the average a page fault occurs every k instructions, is:
i + (j / k)
This formula accounts for the normal instruction time (i) and the additional time for a page fault (j), considering the average frequency of page faults (k instructions between page faults).
There are 50 questions to complete.

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