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Department: Computer Science and Engineering B.Tech. CSE – INTERNAL ASSESSMENT – II – MAR 2023 2022-23 EVEN SEMESTER Date: (^) Session: Semester: IV Course Code: 20CS2036 Maximum Marks: 40 Course Name: OPERATING SYSTEMS Time: 2 Hours Course Outcomes for Assessment in this IA: COs COURSE OUTCOME CO3 analyze the thread context management, synchronization methods and various scheduling algorithms.. CO4 apply various memory management schemes especially paging and segmentation in real time applications. PART – A (20 Marks) Bloom’s Level Mapping COs Marks 1.a) Differentiate binary and counting semaphores. A Binary Semaphore is a semaphore whose integer value range over 0 and 1. A counting semaphore is a semaphore that has multiple values of the counter.
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b) List the necessary conditions for the deadlock prevention. mutual exclusion, hold and wait, no preemption, and circular wait.
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c) Sketch the schematic view of a monitor. A CO3 2 d) There are 3 processes P1, P2 and P3, and two types of resources R and R2. The resources are having 1 instance each. R1 is being used by P1, P2 is holding R2 and waiting for R1, P3 is waiting for R1 as well as R2. Based on the above scenario, create a resource allocation graph. C CO3 2 Reg. No.
e) i) Consider five philosophers who spend their lives thinking and eating. The philosophers share a circular table surrounded by five chairs, each belonging to one philosopher. In the center of the table is a bowl of rice, and the table is laid with five single chopsticks. When a philosopher thinks, she does not interact with her colleagues. From time to time, a philosopher gets hungry and tries to pick up the two chopsticks that are closest to her. A philosopher may pick up only one chopstick at a time. Obviously, she cannot pick up a chopstick that is already in the hand of a neighbor. When a hungry philosopher has both her chopsticks at the same time, she eats without releasing the chopsticks. When she is finished eating, she puts down both chopsticks and starts thinking again. Explain this using semaphore. Answer: do { wait (chopstick[i] ); wait (chopStick[ (i + 1) % 5] ); // eat signal (chopstick[i] ); signal (chopstick[ (i + 1) % 5] ); // think } while (TRUE);
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ii) Consider the following snap shot of a system: Allocation Max Available A B C D A B C D A B C D P0 2 0 0 1 4 2 1 2 3 3 2 1 P1 3 1 2 1 5 2 5 2 P2 2 1 0 3 2 3 1 6 P3 1 3 1 2 1 4 2 4 P4 1 4 3 2 3 6 6 5 Determine the following questions using the banker’s algorithm: a. Identify the content of the matrix need. 2 2 1 1 2 1 3 1 0 2 1 3 0 1 1 2 2 2 3 3 b. Illustrate the system is in a safe state or not. It is in safe State P0 P3 P4 P1 P
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PART- B (20 Marks) Bloom’s Level Mapping COs Marks 2.a) Name the two different types of fragmentations.
- Internal
- External
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b) Define critical section. Each process must ask permission to enter critical section in entry section, may follow critical section with exit section, then remainder section
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c) Compare and contrast logical and physical address. The logical address is rendered by CPU while a program is executed, whereas the physical address directs to a location in the memory unit. An CO4 2
Bloom’s Level-wise Mark Distribution: Remember Understand Apply Analyze Evaluate Create Total 5 15 16 2 - 2 40
Course Outcome wise Mark Distribution COs Remember Understand Apply Analyze Evaluate Create Total CO3 2 8 9 - - 2 21 CO4 3 7 7 2 - - 19 Total 5 15 16 2 - 2 40
