Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
CUET PG MCA Previous Year Questions (PYQs)
CUET PG MCA Operating System PYQ
Consider a system with 1K pages and 512 frames and each page is of size 2 KB. How many bits are required to represent the virtual address space memory:
Go to Discussion
Solution
To determine the number of bits required to represent the virtual address space:- Number of pages = 1K = 210.
- Page size = 2 KB = 211 bytes.
- Virtual address space = Number of pages × Page size = 210×211=221 bytes.
- Bits required =21 bits.
How does the number of page frames affect and number of page faults for a given memory access pattern in FIFO page replacement algorithm?
Go to Discussion
Solution
On a system using simple segmentation, following is the segment table:
What is the physical address for the logical address 2, 212?
| Segment | Limit | Base |
| 0 | 500 | 1000 |
| 1 | 200 | 2000 |
| 2 | 300 | 2500 |
| 3 | 100 | 1700 |
Go to Discussion
Solution
Decreasing the RAM of a computer typically leads to which of the following outcomes
?
Go to Discussion
Solution
We can say that a schedule is conflict serializable?
(1) If a schedule T can be transformed into a serial schedule U by a series of swaps of conflicting operations.
(2) If a schedule T can be transformed into a serial schedule U by a series of swaps of nonconflicting operations.
(3) If a schedule T can be transformed into a non serial schedule U by a series of swaps of conflicting operations.
(4) If a schedule T can be transformed into a non serial schedule U by a series of swaps of non conflicting operations.
Go to Discussion
Solution
An operating system cotains 4 user processor each requiring 5 units of resource R. The minimum number ofrequired units of R such that no deadlock will every occur is
Go to Discussion
Solution
Minimum Number of Required Units of Resource R to Prevent Deadlock
Number of processes (P) = 4 Maximum demand per process = 5 units
Formula: R≥(P−1)×max demand+1
Substituting values:
R≥(4−1)×5+1=3×5+1=15+1=16
Minimum required units of resource R = 16
Number of processes (P) = 4 Maximum demand per process = 5 units
Formula: R≥(P−1)×max demand+1
Substituting values:
R≥(4−1)×5+1=3×5+1=15+1=16
Minimum required units of resource R = 16
The current allocation and Maximum requirement of different types of resources for four processes are given below:
Consider the following four statements.
(A) P2 → P4 → P1 → P3 is a safe sequence
(B) P4 → P2 → P1 → P3 is a safe sequence
(C) P4 → P2 → P3 → P1 is a safe sequence
(D) P1 → P4 → P2 → P3 is a safe sequence
Identify correct statements from the given options.
(A) P2 → P4 → P1 → P3 is a safe sequence
(B) P4 → P2 → P1 → P3 is a safe sequence
(C) P4 → P2 → P3 → P1 is a safe sequence
(D) P1 → P4 → P2 → P3 is a safe sequence
Identify correct statements from the given options.
Go to Discussion
Solution
Banker's Algorithm Solution
Step 1: Need Matrix
The Need Matrix is calculated as:
Need = Max - Allocation
| Process | R1 | R2 | R3 |
|---|---|---|---|
| P1 | 4 | 5 | 2 |
| P2 | 1 | 0 | 1 |
| P3 | 6 | 0 | 0 |
| P4 | 0 | 1 | 0 |
Step 2: Initial Available Resources
The Available resources at the start are:
- R1 = 4
- R2 = 4
- R3 = 5
Step 3: Evaluate Each Sequence
Sequence (A): P2 → P4 → P1 → P3
- P2: Need = [1, 0, 1], Available = [4, 4, 5]. Allocate resources. New Available = [7, 5, 7].
- P4: Need = [0, 1, 0], Available = [7, 5, 7]. Allocate resources. New Available = [10, 7, 9].
- P1: Need = [4, 5, 2], Available = [10, 7, 9]. Allocate resources. New Available = [14, 8, 11].
- P3: Need = [6, 0, 0], Available = [14, 8, 11]. Allocate resources. New Available = [17, 12, 14].
Result: Sequence (A) is valid.
Sequence (B): P2 → P1 → P3 → P4
- P2: Need = [1, 0, 1], Available = [4, 4, 5]. Allocate resources. New Available = [7, 5, 7].
- P1: Need = [4, 5, 2], Available = [7, 5, 7]. Allocate resources. New Available = [11, 6, 9].
- P3: Need = [6, 0, 0], Available = [11, 6, 9]. Allocate resources. New Available = [14, 10, 12].
- P4: Need = [0, 1, 0], Available = [14, 10, 12]. Allocate resources. New Available = [17, 12, 14].
Result: Sequence (B) is valid.
Sequence (C): P4 → P2 → P3 → P1
- P4: Need = [0, 1, 0], Available = [4, 4, 5]. Allocate resources. New Available = [7, 6, 7].
- P2: Need = [1, 0, 1], Available = [7, 6, 7]. Allocate resources. New Available = [10, 7, 9].
- P3: Need = [6, 0, 0], Available = [10, 7, 9]. Allocate resources. New Available = [14, 8, 11].
- P1: Need = [4, 5, 2], Available = [14, 8, 11]. Allocate resources. New Available = [17, 12, 14].
Result: Sequence (C) is valid.
Sequence (D): P1 → P4 → P2 → P3
- P1: Need = [4, 5, 2], Available = [4, 4, 5]. Cannot proceed as Need > Available.
Result: Sequence (D) is invalid.
Final Answer:
Safe Sequences: (A), (B), (C)
Invalid Sequence: (D)
How many child processes will be created by following fork() system call?
fork();
fork();
fork();
fork();
Go to Discussion
Solution
Given: 4 calls to fork().
Formula: Total processes created = 2^n
Calculation:
- Total processes =
2^4 = 16 - Child processes =
16 - 1 = 15
Answer: 15 child processes
Match List – I with List – II
Choose the correct answer from the options given below:
| List - I | List - II |
| (A) Critical Region | (I) Circular Wait |
| (B) Working Set | (II) Condition variable |
| (C) Deadlock | (III) Principle of locality |
| (D) Wait/Signal | (IV) Mutual Exclusion |
Go to Discussion
Solution
Matching List-I with List-II:
| List - I | List - II | Explanation |
|---|---|---|
| Critical Region | Mutual Exclusion (IV) | The critical region ensures mutual exclusion, allowing only one process to access a shared resource at a time. |
| Working Set | Principle of Locality (III) | The working set is a concept used in memory management, based on the principle of locality to predict the set of pages a process is likely to use. |
| Deadlock | Circular Wait (I) | One of the necessary conditions for a deadlock is circular wait, where processes are waiting for resources held by one another in a circular chain. |
| Wait/Signal | Condition Variable (II) | The wait and signal operations are synchronization mechanisms associated with condition variables in multi-threaded environments. |
Final Answer:
(A) → (IV), (B) → (III), (C) → (I), (D) → (II)
In a computer if the page fault service time is 10 ms and average memory access time is 30 ns. If one page faultis generated for every 106 memory accesses. What is the effective access time for the memory?
Go to Discussion
Solution
CUET PG MCA
Online Test Series,
Information About Examination,
Syllabus, Notification
and More.
View More
CUET PG MCA
Online Test Series,
Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
