Your Answer : (Not Answered)

Correct Answer : B

This could lead to Deadlock.

Process P[i] locks m[i] and then m[(i+1) mod 4]. For P[0] to P[3] this gives a circular chain: P[0] holds m[0] and wants m[1], P[1] holds m[1] and wants m[2], P[2] holds m[2] and wants m[3], and P[3] holds m[3] and wants m[0]. If all four acquire their first mutex before any gets its second, each waits for the next forever: hold-and-wait with circular wait, a deadlock.

Thrashing is excessive paging and has nothing to do with mutexes, and since a deadlock is possible, 'starvation but not deadlock' and 'none of the above' are wrong.

Your Answer : (Not Answered)

Correct Answer : D

The totals are 236 and 236 cylinders.

SSTF always serves the nearest pending request. From 53 the order is 65, 67, 37, 14, 98, 122, 124, 183:

12 + 2 + 30 + 23 + 84 + 24 + 2 + 59 = 236 cylinders.

SCAN moves towards cylinder 0 first, serving 37 and 14 on the way, reaches 0, then reverses and sweeps up to the last request, 183:

(53 - 0) + (183 - 0) = 53 + 183 = 236 cylinders.

640 is the first-come-first-served total for this queue, and 252 is not produced by either algorithm here, so the other options mix in the wrong figures.

Your Answer : (Not Answered)

Correct Answer : C

The program would take 45 sec.

Each page fault costs 2001 microseconds instead of 1, so it adds 2000 microseconds over a normal instruction.

\[ \begin{aligned} \text{fault time} &= 15000 \times 2000\ \mu s = 30\ s \\ \text{normal time} &= 60 - 30 = 30\ s \end{aligned} \]

With twice the memory, the interval between faults doubles, so the number of faults halves to 7,500, while the normal execution time is unchanged:

\[ \begin{aligned} \text{new time} &= 30 + 7500 \times 2000\ \mu s \\ &= 30 + 15 = 45\ s \end{aligned} \]

30 seconds would be halving the whole run time, wrongly treating the normal instructions as reduced too, and 60 seconds ignores the saving.

Your Answer : (Not Answered)

Correct Answer : B

Starvation can occur with (iii), (iv) and (v).

Under priority scheduling, a low-priority process can wait indefinitely if higher-priority processes keep arriving. Shortest process next and shortest remaining time first always favour short jobs, so a long process can be postponed forever if short jobs keep coming.

First-come-first-served and round robin cannot starve a process. FCFS serves every process in arrival order, and round robin gives each ready process a time slice in turn, so every process eventually runs. Options including (i) or (ii) are therefore wrong.

Your Answer : (Not Answered)

Correct Answer : D

The availability is approximately 99.97%.

Availability is the fraction of time the system is up:

\[ \begin{aligned} A &= \frac{MTBF}{MTBF + MTTR} \\ &= \frac{43200}{43200 + 10} \\ &= 0.99977 \approx 99.97\% \end{aligned} \]

Here 30 days = 30 × 24 × 60 = 43,200 minutes, so both times are in minutes. The other options are far too low: 96.97% would mean about 22 hours of downtime a month, not 10 minutes.

Your Answer : (Not Answered)

Correct Answer : A

The diagram represents a Centralized system.

All the processing is done by one computer system, C, and the dumb terminals, T, have no processing power of their own. They only send keystrokes to the central computer and display what it returns. A single computer serving many terminals is the classic centralized (time-sharing mainframe) arrangement.

A distributed system would need several computers, each with its own processing, cooperating over a network. Local-access and remote-access describe how a user reaches a computer, not the architecture shown, in which everything depends on the one central system.

Your Answer : (Not Answered)

Correct Answer : A

The true statements are (1) and (3).

link(sfname, tfname) creates a new directory entry, tfname, pointing to the same inode as the existing file sfname, and increments the inode's link count. So after the call both names refer to the same inode (1), and sfname must already exist before the call (3).

Statement (2) is false: the call creates a second, separate directory entry, so the two names have different directory entries that share one inode. Statement (4) is false too: if tfname already exists the call fails, so any option containing (2) or (4) is wrong.

Your Answer : (Not Answered)

Correct Answer : A

The call that does not return to the calling point is exec.

A successful exec replaces the calling process's program with a new one: the old code, data and stack are discarded, so there is nothing to return to and execution begins at the new program's entry point. exec returns only if it fails.

fork returns twice, once in the parent (with the child's process ID) and once in the child (with 0), and ioctl returns a status to its caller. longjmp is a library function, not a system call, and it jumps back to a point saved earlier by setjmp.

Your Answer : (Not Answered)

Correct Answer : B

The MS-DOS counterpart of tar is backup.

Unix tar (tape archive) bundles files and directories into a single archive, traditionally for backup to tape. MS-DOS provided the BACKUP command (with RESTORE) to copy files and directory trees from a hard disk to backup media in a form that could later be restored.

DIR only lists directory contents, COPY copies individual files without creating an archive, and EDIT is a text editor.

Your Answer : (Not Answered)

Correct Answer : A

The option is -i.

grep -i pattern file ignores the difference between upper and lower case, so 'unix', 'Unix' and 'UNIX' all match.

-l lists only the names of the files containing a match, and -c prints the count of matching lines instead of the lines. -p is not a standard grep option.

Your Answer : (Not Answered)

Correct Answer : A

The size of the swap space is decided at the time of operating system installation.

Swap space is disk space set aside for pages or processes moved out of main memory. On most systems it is a separate disk partition or a preallocated swap file, created and sized when the disk is partitioned during installation of the operating system.

It is not fixed afresh at every start-up or when the swapper is loaded; those simply use the space already configured. The hardware does not decide it either, although the amount of physical memory guides the administrator's choice.

Your Answer : (Not Answered)

Correct Answer : B

The page fault rate is \((X - M) / (D - M)\).

If \(p\) is the page fault rate, a fraction \(p\) of accesses take \(D\) units and the rest take \(M\) units, so the measured average is

\[ \begin{aligned} X &= (1 - p)M + pD \\ X - M &= p(D - M) \\ p &= \frac{X - M}{D - M} \end{aligned} \]

Check: if no access faults, \(X = M\) and \(p = 0\); if every access faults, \(X = D\) and \(p = 1\). The option \((D - M)/(X - M)\) is the reciprocal and exceeds 1. \((D - X)/(D - M)\) is the hit rate \(1 - p\), and \((X - M)/(D - X)\) is the ratio of faults to hits.

Your Answer : (Not Answered)

Correct Answer : A

The maximum number of page table entries is \(2^{21}\).

A single-level page table needs one entry for each page of the logical address space. The page size is \(2048 = 2^{11}\) bytes, so 11 of the 32 logical address bits are the offset within a page:

\[ \text{entries} = \frac{2^{32}}{2^{11}} = 2^{21} \]

The 20-bit physical address decides the number of frames, \(2^{20}/2^{11} = 2^{9}\), and the width of each entry, but not how many entries there are. \(2^{20}\) wrongly takes the physical address width, and \(2^{22}\) and \(2^{24}\) assume smaller pages.

Your Answer : (Not Answered)

Correct Answer : C

The minimum values are V1 = 12 and V2 = 7.

I-1. A binary semaphore initialised to 1 lets only one thread into the critical section at a time, so every update is applied in full and the result is the same in every run:

\[ V_{1} = 10 + 5 - 3 = 12 \]

I-2. A counting semaphore initialised to 2 lets two threads update X together, so one update can overwrite another (X = X+1 is a read, an add and a write). To make X as small as possible: a decr thread reads X = 10 and pauses inside. Meanwhile the five incr threads pass one by one through the second slot, raising X to 15. The paused decr then writes 9, wiping out all five increments. The other two decr threads run normally: 8, then 7.

No run can end below 7, because the three decrements can lower X by at most 3 from 10. So V2 = 7. 15 and 7 as V1 are impossible under mutual exclusion, and 8 is reachable but is not the minimum.

Your Answer : (Not Answered)

Correct Answer : C

The initialization is \(S_{1} = 1;\ S_{2} = 0;\ S_{3} = 0\).

The sequence must start with B, so only T2, which waits on \(S_{1}\), may run first: \(S_{1} = 1\) and the others 0. Then T2 prints B and signals \(S_{3}\), which lets only T1 print C; T1 signals \(S_{2}\), which lets only T3 print A; T3 signals \(S_{1}\) and the cycle repeats. At every moment exactly one semaphore is 1, so the order BCA is forced.

In the other options more than one semaphore starts at 1. With \(S_{2} = 1\), T3 could print A first; with \(S_{3} = 1\), T1 could print C first; and option D, with \(S_{1} = 0\), cannot print B first at all.

Your Answer : (Not Answered)

Correct Answer : C

The solution violates the progress requirement.

This is strict alternation. Mutual exclusion holds: turn is either 0 or 1, and each process can enter only when turn has its own value, so both can never be inside together. Bounded waiting also holds, because after a process leaves it hands the turn to the other, so neither can enter twice while the other waits.

Progress fails. Suppose Process 0 runs its critical section and sets turn = 1, and Process 1 has no wish to enter. If Process 0 now wants to enter again, it must wait for Process 1 to take its turn, which may never happen. A process outside its critical section is blocking another from entering, which is exactly what the progress requirement forbids.

Your Answer : (Not Answered)

Correct Answer : A

The possible results are (a = 4, b = 4); (a = 3, b = 3); (a = 4, b = 3).

Each thread's two statements run in order, giving six interleavings. What matters is whether each update to a comes before or after the other thread's update to a, and the same for b:

• a: a + 1 then 2a gives 4; 2a then a + 1 gives 3.
• b: b + 1 then 2b gives 4; 2b then b + 1 gives 3.

T1 updates a before b and T2 updates b before a. If T1's a-update comes after T2's a-update, T2 has already finished, so T1's b-update is also last: (3, 3). Otherwise a = 4 and b can be either. So (4, 4), (4, 3) and (3, 3) occur, but (3, 4) cannot, which rules out options B and C. Values of 2 would require an update to be lost, impossible with atomic statements.

Your Answer : (Not Answered)

Correct Answer : D

The burst times that are NOT possible are P = 3, Q = 7, R = 7, S = 3.

With a quantum of 4 and the order P, Q, R, S, these bursts run as P (3, done), Q (4, 3 left), R (4, 3 left), S (3, done), Q (3, done), R (3, done). The switches are P to Q, Q to R, R to S, S to Q and Q to R, so there is no switch from R to Q at all, against the 'exactly one' required.

Each of the other sets meets every condition. For P = 4, Q = 10, R = 6, S = 2 the run is P, Q, R, S, Q, R, Q, giving one S to Q, one R to Q, two Q to R and no S to P. P = 2, Q = 9, R = 5, S = 1 runs in the same order (P, Q, R, S, Q, R, Q), and so does P = 4, Q = 12, R = 5, S = 4, where R finishes its last unit before Q's final quantum. In each, Q still needs CPU time after R's second turn, which is what produces the single R to Q switch.

Your Answer : (Not Answered)

Correct Answer : D

'Data Compilation' is not a system call category. The usual categories are process control, file manipulation, device manipulation, information maintenance and communication. Process control, device manipulation and file manipulation are named in the options, so the fourth option is the odd one out. The system calls of an operating system let programs request services such as creating a process, opening a file or reading from a device.

Your Answer : (Not Answered)

Correct Answer : D

The exclamation mark (!) is not an input/output redirection symbol. In the Linux shell the symbol < redirects standard input from a file, > redirects standard output to a file overwriting it, and >> appends the output to a file. The '!' is used for history expansion and for negation, not for redirecting input or output. For example, command > file.txt sends the output to a file, command < file.txt takes input from it, and command >> file.txt adds to it.