Operating System Test 1

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Operating System Test 1

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Question 1
2 / -0.33

The main function of the dispatcher (the portion of the process scheduler) is:

A
swapping a process to the disk

B
assigning ready process to the CPU

C
suspending some of the processes when the CPU load is high

D
bring processes from the disk to the main memory
Solution
- Mid-term scheduler: swap the processes from the wait state to hard disk
- Short-term scheduler: selects the processes that are ready to execute and allocates the CPU to one of them
- Dispatcher: Context switches, in which the dispatcher saves the state of the previous process then the dispatcher loads the saved state of the new process selected by the short-term scheduler
- Long-term scheduler: bring the processes to main memory from the hard disk
Question 2
2 / -0.33

Consider the following set of processes {A, B, C, D}, with the arrival time and the CPU burst time given in milliseconds.
Process Id
Arrival time
Burst time
A
4
3
B
9
1
C
0
4
D
8
5

What is the average waiting time in millisecond for these processes with the shortest job first algorithm?

A
0.75

B
1

C
1.25

D
1.5

Solution

Gantt chart:
C
A
-
D
B
0 4 7 8 13 14
Process table:
Process Id
Arrival Time (AT)
Burst Time (BT)
Completion Time (CT)
Turn Around Time (TAT)
Waiting Time (WT)
A
4
3
7
3
0
B
9
1
14
5
4
C
0
4
4
4
0
D
8
5
13
5
0

Average waiting time = \(\frac{0~+4+0+0~}{4}\)
Average waiting time = 1 ms
Important Points:
TAT = CT – AT
WT = TAT – BT
Question 3
2 / -0.33

Suppose there is a paging system with a translation lookaside buffer. Assuming that the entire page table and all the pages are in the physical memory, what is the effective memory access time in ms if it takes 5 msec to search the TLB and 70 msec to access physical memory? The TLB hit ratio is 0.8.

A
89-89

B
894-891

C
897-896

D
899-892

Solution

Data:
TLB hit ratio = p = 0.8
TLB access time = 5 milliseconds
Memory access time = m = 70 milliseconds
Formula:
EMAT = p × (t + m) + (1 – p) × (t + m + m)
Calculation:
EMAT = 0.8 × (5 + 70) + (1 – 0.8) × (5 + 70 + 70)
EMAT = 89 milliseconds.
Important points:
During TLB hit
The frame number is fetched from the TLB (5 ms)
and page is fetched from physical memory (70 ms)
During TLB miss
TLB no entry matches (5 ms)
The frame number is fetched from the physical memory (70 ms)
and pages are fetched from physical memory (70 ms)
Question 4
2 / -0.33

Consider a computer system having R resources of the same type and these resources are shared by 6 processes that have maximum demands of 7, 5, 3, 8, 4, 6 respectively. Then for what minimum values of R, the system will not be in deadlock?

A
26

B
27

C
28

D
29

Solution

Concept:
Deadlock can occur If any process gets < needed (requested) resource
Max resource per process to be in deadlock = max demand – 1
For 6 process, max resource to be in deadlock = (7 – 1) + (5 – 1) + (3 – 1) + (8 - 1) + (4 - 1) + (6 - 1) =27
For 6 process, min resource not to be in a deadlock: 27 + 1 =28
Tips and Tricks:
\(\sum_i^nP_i\ -\ n +1 =(7+5+3+8+4+6)-6 + 1=28\)
where Pi is the max resource needed by the ith process and n is the total number of process
Question 5
2 / -0.33

Which of the following is TRUE about user-level threads and kernel-level threads?
S₁: Typically if a user-level thread is performing blocking system call then the entire process will be blocked
S₂: Kernel level threads are designed as dependent threads
S₃: Every thread has its own register and stack

A
S₁ and S₂ only

B
S₂ and S₃ only

C
S₁ and S₃ only

D
All are true

Solution

S₁: TRUE:
If a user-level thread block then entire process will get blocked
S₂: FALSE
Kernel level threads are independent that is why they also require more context switch than user – level threads.
S₃: TRUE
Every thread has its own register and stack
Therefore S₁ and S₃ are TRUE
Question 6
2 / -0.33

Consider a hypothetical system with byte-addressable memory, 64-bit logical addresses, 8 Megabyte page size and page table entries of 16 bytes each. What is the size of the page table in the system in terabytes?

A
32-32

B
324-321

C
327-326

D
329-322

Solution

Data
1 word (W) = 1 byte (B)
Logical address = 264 B
page size = 8 MB = 223 B
page table entry (PTE) = 16 B
Formula:
Page table size (PTS) = number of pages × PTE
Calculation:
number of pages = \(\frac{{{2^{64}}}}{{{2^{23}}}} = {2^{41}}\)
Page table size (PTS) = number of pages × y
\(PTS = {2^{41}} \times 16\;B = 2^{45} \;B\)
∴ PTS = 32 TB
Important points
1 TB = 240 B
where B stands for byte
Question 7
2 / -0.33

Suppose that jobs arrive according to the following schedule :
Process Arrival time     Run time
A 0 12
B 3 7
C 6 2
D 8 5
E 9 2
F 12 12

Now give the sequence of the processes as they gets share of CPU if the scheduler used is Shortest Remaining Time First.

A
A,B,C,B,E,D,B,A,F

B
A,B,C,B,E,B,D,B,A,F

C
A,B,C,B,E,B,D,A,F

D
A,B,C,B,B,B,E,D,A,F

Solution

Concept:
Shortest remaining time next: pre-emptive scheduling
In shortest remaining time next, when a process of less burst time(run time) arrives than in ready queue then context switch will happen
Explanation:
Time           Running                  Remaining times at the end
0 – 3 A                                 A – 9 B – 7
3 – 6             B                                 A – 9 B – 4 C – 2
6 – 8 C                                 A – 9 B – 4 D – 5
8 – 9 B                                 A – 9 B – 3 D – 5 E – 2
9 – 11 E                                 A – 9 B – 3 D – 5
11 – 12          B                                 A – 9 B – 2 D – 5 F – 12
12 – 14          B                                 A – 9 D – 5 F – 12
14 – 19          D                                 A – 9 F – 12
19 – 28          A                                 F – 12
28 – 40 F
Question 8
2 / -0.33

The process executes the code where x is an integer and x is initialized to 10
while(--x){
fork();
if(x == 2)
break;
}
The total number of child processes created is

A
255-255

B
2554-2551

C
2557-2556

D
2559-2552

Solution

Value of x first decremented(--x)
while condition
fork calling
9
True
called
8
True
called
7
True
called
6
True
called
5
True
called
4
True
called
3
True
called
2
True
called (but break the while loop)

From the table it is clear that fork is called 8 times
Formula:
If n times fork is called the number of child processes created = 2n – 1
Calculation:
Since n = 8
number of child processes created = 28 - 1 = 255
Question 9
2 / -0.33

A system has 12 identical resources and Y processes competing for them. Each process can request at most 3 resources. Which one of the following values of Y could lead to a deadlock?

A
5

B
6

C
7

D
Both 6 and 7

Solution

Data:
Available identical Resources = R = 12
Max needs per process = 3
Concepts:
Deadlock can occur If any process gets available resource < needed (requested) resource
Max resource per process to be in deadlock = needed – 1 = 3 – 1 = 2
For Y process, max resource to be in deadlock = Y × 2 = 2Y
Condition for deadlock
2Y ≥ R
2Y ≥ 12
Y ≥ 6
values of Y could lead to a deadlock is 6 and 7
Question 10
2 / -0.33

At time 0, an OS with round-robin scheduling (time quantum = 5 units) has 5 processes P_1,P₂, P₃, P₄, P₅ with burst times 15, 10, 20, 5, 15 respectively. What is the average turn-around time?

A
40 units

B
46 units

C
56 units

D
65 units

Solution

Concept:
The round-robin (RR) scheduling algorithm is designed especially for timesharing systems.
It is similar to FCFS scheduling, but preemption is added to enable the system to switch between processes.
A small unit of time, called a time quantum or time slice, is defined.
The ready queue is treated as a circular queue. The CPU scheduler goes around the ready queue, allocating the CPU to each process for a time interval of up to 1-time quantum.
Time quantum = 5 units
Gantt Chart:
P₁
P₂
P₃
P₄
P₅
P₁
P₂
P₃
P₅
P₁
P₃
P₅
P₃
0 5 10 15 20 25 30 35 40 45 50 55 60 65
Calculation
Process
Arrival Time (AT)
Burst Time (BT)
Completion Time (CT)
Turnaround time =
CT – AT
P₁
0
15
50
50
P₂
0
10
35
35
P₃
0
20
65
65
P₄
0
5
20
20
P₅
0
15
45
45

Average turn around time = \(\frac{20 + 35 + 50 + 60 + 65}{5} = 46\)
Question 11
2 / -0.33

Which of the following is/are not true about the deadlock scheme?
I. In deadlock avoidance, number the resource uniquely and never requests a lower-numbered resource
II. In deadlock avoidance, the request for the resources is always granted if the resulting state is safe.
III. In deadlock prevention, never request a resource after releasing any resource.

A
Only II

B
Only II and III

C
Only I and III

D
Only I, II and III

Solution

Statement I: not true (False)
In deadlock prevention, number the resource uniquely and never requests a lower-numbered resource
Statement II: True
In deadlock avoidance, the request for the resources is always granted if the resulting state is safe.
Statement IIi: not true (False)
In deadlock prevention, the request can be made after releasing the resource
Statement I and III are not true
Question 12
2 / -0.33

Consider a counting semaphore S. The operation P(S) performs the operation S = S – 1 and operation V(S) performs the operation V = V + 1. During program execution, 13P and 5V operation is performed in the some order. Find the number of processes in a blocked state with initial counting semaphore value 1.

A
7-7

B
74-71

C
77-76

D
79-72

Solution

Concepts:
V(S): Signal will increment the semaphore variable, that is, S++.
P(S): Signal will decrement the semaphore variable., that is, S--.
Data:
Initial counting semaphore = x = 1
Signal operation = 11 V
Wait operation = 17 P
n process in blocked state
Final counting semaphore (F) = -n
Formula:
n = x + 13P + 5V
Calculation:
n = 1 + 13(-1) + 5(+1)
∴ n = -7
∴ number of process in blocked state is 7
Question 13
2 / -0.33

A computer uses 48 bits Virtual Address Space, 256 GB of physical memory with page size of 4 KB. The page table size is 4 bytes and if each entry fits in a single page the number of levels of paging required is _____.

A
4-4

B
44-41

C
47-46

D
49-42

Solution

Data:
Virtual address space (VAS) = 48 bits
Virtual memory (VA) = 2⁴⁸ Byte
Physical Memory (PA) = 256 GB
Page size (PS) = 4 KB = 2¹² B
Page Table size (PTS) = 4 bytes
Formula:
\({\rm{page\;table\;size\;}} = \frac{{{\rm{VA}}}}{{{\rm{PS}}}} \times {\rm{PTS\;}}\)
Calculation:
\({\rm{Level\;}}1:{\rm{\;page\;table\;size\;}} = \frac{{{2^{48}}}}{{{2^{12}}}} \times 4\;{\rm{byte}} = {2^{38}}\;{\rm{byte}} > {\rm{PTS}}\)
\({\rm{Level\;}}2:{\rm{page\;table\;size\;}} = \frac{{{2^{38}}}}{{{2^{12}}}} \times 4\;{\rm{byte}} = {2^{28}}\;{\rm{byte}} > {\rm{PTS}}\)
\({\rm{Level\;}}3:{\rm{page\;table\;size\;}} = \frac{{{2^{28}}}}{{{2^{12}}}} \times 4\;{\rm{byte}} = {2^{18}}\;{\rm{byte}} > {\rm{PTS}}\)
\({\rm{Level\;}}4:{\rm{page\;table\;size\;}} = \frac{{{2^{18}}}}{{{2^{12}}}} \times 4\;{\rm{byte}} = {2^8}\;{\rm{byte}} \le {\rm{PTS}}\)
The number of levels of paging required is 4.
Important Points:
If page table size (PTS) is less than or equal to page size, then page table can be stored in that level.

Question 14

2 / -0.33

Consider two processes P1 and P2 executing their respective codes initial values of global variables P1_inside and P2_inside are “false”.

while(true)

{

while(P1_inside);

P1 inside=true;

C.S. // critical section

P1 inside =false;

}

while(true)

{

while(P2_inside);

P2 inside=true;

C.S. // critical section

P2 inside=false;

}

Consider the below statements:

Mutual exclusion is satisfied
Mutual Exclusion is not satisfied
It is possible for deadlock

Which of the above are true?

Only 1

Only 2

Only 1 and 3

Only 2 and 3

Solution

P1 executes while and P2 executes while. Then by executing assignment statement, both P1 and P2 can enter the critical section. So mutual exclusion is not satisfied.

Question 15

2 / -0.33

	Allocation	Max	Available
	A B C D	A B C D	A B C D
P0	0 0 1 2	0 0 1 2	1 5 2 0
P1	1 0 0 0	1 7 5 0
P2	1 3 5 4	2 3 5 6
P3	0 6 3 2	0 6 5 2
P4	0 0 1 4	0 6 5 6

Which of the following is true for the above-given resource allocation table?

The system is in safe state

The system is in unsafe state

The system is in deadlock

Insufficient Information

Solution

Concepts:

If all processes able to terminate from a state than the state is safe.

In other words, if we can find a safe sequence (order of termination of the process) then the process is in a safe state.

Resources needed:

	Need
	A B C D
P0	0 0 0 0
P1	0 7 5 0
P2	1 0 0 2
P3	0 0 2 0
P4	0 6 4 2

Available <1, 5, 2, 0>

Sequence of safe state:

Since P0 doesn't needed any instance of resource, it will release it allocated resources

∴ Available = <1, 5, 2, 0> + <0, 0, 1, 2> = <1, 5, 3, 2>

P3 needs <0, 0, 2, 0>, is less than avaible, after P3 execution

∴ Available = <1, 5, 3, 2> + <0, 6, 3, 2> = <1, 11, 6, 4>

P2 needs <1, 0, 0, 2>, is less than avaible, after P2 execution

∴ Available = <1, 11, 6, 4> + <1, 3, 5, 4> = <2, 14, 11, 8>

P4 needs <0, 6, 4, 2>, is less than avaible, after P4 execution

∴ Available = <2, 14, 11, 8> + <0, 0, 1, 4> = <2, 14, 12, 12>

P1 needs <0, 7, 5, 0>, is less than avaible, after P1 execution

∴ Available = <2, 14, 12, 12> + <1, 0, 0, 0> = <3, 14, 12, 12>

Process are able to finish in this situation so system is in safe state, and one of the safe sequence is <P0, P3, P2, P4, P1>

Question 16
2 / -0.33

Consider a file system that uses inodes to represent files. Disk blocks are 12 KB in size, and a pointer to a disk block requires 8 bytes. This file system has 10 direct disk blocks, as well as single and double indirect disk blocks. What is the maximum size in GB (up to two decimals) of a file that can be stored in this file system?

A
27.01-27.02

B
27.014-27.021

C
27.017-27.026

D
27.019-27.022

Solution

Data:
In index node of a Unix- style file system,
Direct Block pointer = 10
Single Indirect Block pointer = 1
Double Indirect Block pointer = 1
Disk block address = disk block entries size = 8 B
Disk Block size = 12 KB
Calculation:
Number of entries in a block = \(\frac{disk\; block\; size}{disk\; block\; address\; size} = \frac{12\;K B}{8\;B} \) = 3 × 2⁹
File size = (10 direct + 1 singel indirect + 1 double indirect) × 12 KB
File size = (10 × 1 + 1 × 3 × 29 + 1 × 3 × 29 × 3 × 29 )× 12 KB
File size = (15 × 213 + 9 × 221 B + 27 × 2³⁰)
File size = 0.00011 GB + 0.017578 GB + 27 GB
∴ file size ≈ 27.02
Question 17
2 / -0.33

Consider a n-way set associative cache memory with 8 sets and 16 cache blocks (0-15) and a main memory with 512 blocks (0 - 511). What memory blocks will be present in the cache after the sequence 17, 4, 13, 105, 28, 29, 200, 408, 127, 205, 411, 251 if LRU is used as a cache replacement block?

A
17, 4, 13, 105, 28, 29, 200, 408, 127, 205, 411

B
17, 13, 105, 28, 29, 200, 408, 127, 205, 411, 251

C
17, 4, 105, 28, 29, 200, 408, 127, 205, 411, 251

D
4, 13, 105, 28, 29, 200, 408, 127, 205, 411, 251

Solution

Data:
number of sets = 8
cache blocks = 16
Formula:
memory block placed in cache = memory block%8
Calculation:
set associative = \(\frac{{16}}{8} = 2\)
∴ in one set two memory block can be placed
0
200
408
1
17
105
2

3
411
251
4
4
28
5
13 (Replace 13) with 205
29
6

7
127

4, 13, 105, 28, 29, 200, 408, 127, 205, 411, 251 is the correct answer:
Important Points:
LRU is used and hence 13 is replaced with 205
Question 18
2 / -0.33

Three CPU-bound tasks, with execution times of 15, 12 and 5 time units respectively arrive at times 0, t and 8, respectively. If the operating system implements a shortest remaining time first scheduling algorithm, what should be the value of t to have 4 context switches?
Ignore the context switches at time 0 and at the end.

A
0 < t < 3

B
t = 0

C
t < = 3

D
3 < t < 8

Solution

Process scheduling Algorithm: shortest remaining time first scheduling
Process Table:
Process
Arrival Time
Burst Time
P₀
0
15
P₁
t
12
P₂
8
5

Gantt Chart:
At t = 0:
P₁
P₁
P₂
P₀
0           8            12          17          32
1^st2^nd
2 context switches occur. Hence Option 2 and 3 is eliminated.
At t = 1:
P₀
P₁
P₂
P₁
P₀
0           1            8           13           18         32
              1^st 2^nd 3^rd          4^th
4 context switches occur.
Since P₂ and P₁ has the same burst time we can choose any process at t = 8
At t = 2:
P₀
P₁
P₂
P₁
P₀
0           2            8           13           19          32
1^st         2^nd         3^rd          4^th
4 context switches occur.
At = 5:
P₀
P₀
P₂
P₀
P₁
0           5            8 13           20          32
1^st         2^nd          3^rd
Only 3 context switches occur. Hence Option 4 is also eliminated.
Therefore, option 1 is the correct answer

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Process Id	Arrival Time (AT)	Burst Time (BT)	Completion Time (CT)	Turn Around Time (TAT)	Waiting Time (WT)
A	4	3	7	3	0
B	9	1	14	5	4
C	0	4	4	4	0
D	8	5	13	5	0

Value of x first decremented(--x)	while condition	fork calling
9	True	called
8	True	called
7	True	called
6	True	called
5	True	called
4	True	called
3	True	called
2	True	called (but break the while loop)

Process	Arrival Time (AT)	Burst Time (BT)	Completion Time (CT)	Turnaround time = CT – AT
P₁	0	15	50	50
P₂	0	10	35	35
P₃	0	20	65	65
P₄	0	5	20	20
P₅	0	15	45	45

0	200	408
1	17	105
2
3	411	251
4	4	28
5	13 (Replace 13) with 205	29
6
7	127

Operating System Test 1

Result

Operating System Test 1

Score

-

Rank

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SHARING IS CARING

Question 1

swapping a process to the disk

assigning ready process to the CPU

suspending some of the processes when the CPU load is high

bring processes from the disk to the main memory

Solution

Question 2

0.75

1

1.25

1.5

Solution

Question 3

89-89

894-891

897-896

899-892

Solution

Question 4

26

27

28

29

Solution

Question 5

S1 and S2 only

S2 and S3 only

S1 and S3 only

All are true

Solution

Question 6

32-32

324-321

327-326

329-322

Solution

Question 7

A,B,C,B,E,D,B,A,F

A,B,C,B,E,B,D,B,A,F

A,B,C,B,E,B,D,A,F

A,B,C,B,B,B,E,D,A,F

Solution

Question 8

255-255

2554-2551

2557-2556

2559-2552

Solution

Question 9

5

6

7

Both 6 and 7

Solution

Question 10

40 units

46 units

56 units

65 units

Solution

Question 11

Only II

Only II and III

Only I and III

Only I, II and III

Solution

Question 12

7-7

74-71

77-76

79-72

Solution

S₁ and S₂ only

S₂ and S₃ only

S₁ and S₃ only