In this article, you will get answer for Operating System Fundamentals NPTEL Assignment Answers of Week 9 (2023)
Last Date for Week – 9 Due date: 2023-09-27, 23:59 IST
1 ] Given memory partitions of 100K, 200K, 300K, and 600K (in order), the First-fit memory management algorithm (search always starts from the at the beginning) places the following processes of size 212K, 417K, 112K, and 426K (in order). What is the amount of external fragmentation?
(A) 288K
(B) 200K
(C) 600K
(D) 359K
(E) 500K
2] Consider a logical address space of sixteen pages of 1024 words each. mapped onto a physical memory of sixty-four frames. How many bits are there in the logical address and physical address?
(A) 13 15
(B) 14 15
(C) 14 16
(D) 16 14
(E) 15 15
3 ] Assuming the page size is 1 KB. what are the page numbers and offsets for the logical address 4050 (provided as decimal numbers)?
(A) 3 978
(B) 2 2002
(C) 4 979
(D) 3 799
(E) 3 980
4 ] The objective of using the page table in a paging scheme is
(A) To store the actual data of the process
(B) To translate logical addresses to physical addresses
(C) To manage the CPU registers
(D) To perform dynamic memory allocation
(E) TO perform fixed memory allocation
5 ] Which of the following is responsible for converting logical addresses to physical addresses?
(A) Central Processing Unit
(B) Arithmetic Logic Unit
(C) Cache memory
(D) Memory Management Unit
(E) Input/Output (I/O) Controller
6 ] What is the purpose of the Segment-table base register (STBR)?
(A) To store the actual data of the process
(B) To hold the base addresses of different segments
(C) To manage the CPU registers
(D) To handle memory allocation for the operating system
(E) To reduce the memory requirements of large programs
7 ] What is the purpose of the Segment-table length register (STLR)?
(A) To store the actual data of the process
(B) To reduce the memory requirements of large programs
(C) To hold the length (size) of the Segment Table
(D) To manage the CPU registers
(E) To handle memory allocation for the operating system
8 ] What is the main object of using overlay techniques in main memory management?
(A) To handle memory allocation for the operating system
(B) To manage the CPU registers
(C) To reduce the memory requirements of large programs
(D) To hold the base addresses of different segments
(E) To speed up the execution of the process
9 ] The main advantage of dynamic address binding is
(A) Faster program loading
(B) Reduced memory fragmentation
(C) Early error detection
(D) Increased security
(E) More efficient memory utilization
10 ] Which of the statement is true?
(A) Logical address is generated by the CPU
(B) Physical address is decided after binding the logical address and the offset
(C) Physical address is obtained by adding the base address of the process to the logical address
(D) Logical and physical addresses are identical for compile-time and load-time address- binding schemes
(E) All of the above
11 ] Assume that a computer system uses a 32-bit virtual address.32-bit physical address. and has a page size of 4KB. The system uses a two-level paging scheme. The first-level page table has 1024 entries. and each second-level page table also has 1024 entries. The system uses a single MMU register to hold the first-level page table entry. What is the size of the total memory consumed by the first-level and second-level page tables?
(A)4 GB. and 4 MB
(B)4 MB. and 4 GB
(C)4 MB and 4 MB
(D)4 GB. and 4 GB
(E) 2 GB, and 2 GB
12 ] Assume that in a computer system. a program uses a library with various utility functions. The program uses five functions from the library. and the size of each function is 5MB. The size of the library is 100MB. The library is dynamically linked to the program. and dynamic loading is employed for loading the library components at runtime. Assume that the size of program data and code is 50MB. What is the amount of memory required by the program and the library if dynamic loading and dynamic linking are used?
(A) 5OMB
(B) 75MB
(C) 90MB
(D) 95MB
(E) 125MB
13 ] Assume that in a computer system uses a paging scheme with page size 2KB. The size of the
user process is 72500 bytes. What is the internal fragmentation?
(A) 1228 bytes
(B) 1024 bytes
(C) 2048 bytes
(D) 512 bytes
(E) 3072 bytes
14 ] Assume that a computer system uses m bits for logical address and n bits for physical address. If the size of the logical address space is 2m. and a page size is 2n bytes, how many bits are required to represent the page number and offset.
(A) (n – m) bits. and n bits
(B) n bits. and m bits
(C) (m – n) bits. and n bits
(D) (m – n) bits. and m bits
(E) m bits, and n bits
15 ] In the context of memory management process, compaction is a technique
(A) to divide physical memory into fixed-size blocks
(B) to handle page faults in virtual memory systems
(C) of converting logical addresses to physical addresses
(D) of removing fragmentation by rearranging memory contents
(E) of swapping out pages from the disk to free up memory
