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Test: Fragmentation - Computer Science Engineering (CSE) MCQ


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10 Questions MCQ Test Operating System - Test: Fragmentation

Test: Fragmentation for Computer Science Engineering (CSE) 2024 is part of Operating System preparation. The Test: Fragmentation questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Fragmentation MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Fragmentation below.
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Test: Fragmentation - Question 1

Internal fragmentation occurs when a

Detailed Solution for Test: Fragmentation - Question 1

Some of the memory allocated to process remain unused,which happens if a process is allocated more memory than it needed

Test: Fragmentation - Question 2

In External fragmentation,merging of free memory areas using boundary tags.which of the following statement stands TRUE.

  1. Boundary tags ,is a status descriptor for a memory area.
  2. It consists of an ordered pair giving allocation status of the area;whether it is free or allocated.
  3. Boundary tags are identical tags stored at the start and end of memory area.
  4. when an area of memory becomes free ,the kernel checks the boundary tags of its neighboring areas.

Detailed Solution for Test: Fragmentation - Question 2

Boundary tags ,is a status descriptor for a memory area. It consists of an ordered pair giving allocation status of the area;whether it is free or allocated. Boundary tags are identical tags stored at the start and end of memory area. when an area of memory becomes free ,the kernel checks the boundary tags of its neighboring areas.

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Test: Fragmentation - Question 3

Which of following statements stands true of a relation called 50-percent rule.

  1. the number of free area is half the number of allocated area ie. m=n/2
  2. it helps in estimating the size of the free list
  3. it also gives us method of estimating the free area in the memory at any time
  4. if sf is the average size of free area of memory,the total free memory is sf×n/2

Detailed Solution for Test: Fragmentation - Question 3

The number of free area is half the number of allocated area ie. m=n/2 it helps in estimating the size of the free list it also gives us method of estimating the free area in the memory at any time if sf is the average size of free area of memory,the total free memory is sf×n/2

Test: Fragmentation - Question 4

Which of the following statements stands true for memory compaction method.

  1. it involves movement of code and data in the memory.
  2. it is feasible only if computer system provides relocation register;the relocation can be achieved by simply changing the address in the relocation register
  3. it does not involves movement of code and data in the memory
  4. it does not involves use of relocation register
Detailed Solution for Test: Fragmentation - Question 4

Memory compaction is not simple as suggested.it involves movement of code and data in memoryin free list if process memory has free memory on either side of it.it needs to be relocated to execute correctly from the new memory area allocated to it.Relocation involves modification of all addresses used by a process,including address of heap-allocated data and address contained in the general purppose registers.it is feasible only if the computer system provides a relocation register;relocation can be achieved by simply changing the address in the relocation register.

Test: Fragmentation - Question 5

In ________ memory binding are changed in such manner that all free memory area can be merged to form a single free memory area.

Detailed Solution for Test: Fragmentation - Question 5

It is achieved by packing all allocated areas towards once end of the memory.

Test: Fragmentation - Question 6

Buddy system and power of 2 allocators leads to ___________

Detailed Solution for Test: Fragmentation - Question 6

It performs allocation of memory in blocks of a few standard sizes.this features leads to internal fragmentation because some memory in each allocated memory block may be wasted.

Test: Fragmentation - Question 7

Which of the following statement is true for buddy system allocators?

  1. Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.
  2. No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.
  3. When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2i for the smallest value of i such that 2i ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.
  4. When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2i.if the list is empty, it checks the lists for block of size 2i+1.it takes one block off this list and splits it into two halves of size 2i.it put one of these blocks into the free list of size 2i,and uses the other block to satisfy the request.
Detailed Solution for Test: Fragmentation - Question 7

Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation. When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2i.if the list is empty, it checks the lists for block of size 2i+1.it takes one block off this list and splits it into two halves of size 2i.it put one of these blocks into the free list of size 2i,and uses the other block to satisfy the request.

Test: Fragmentation - Question 8

Which of the following statement is true for Power-of-2 allocators?

  1. Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.
  2. No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.
  3. When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2i for the smallest value of i such that 2i ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.
  4. When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2i.if the list is empty, it checks the lists for block of size 2i+1.it takes one block off this list and splits it into two halves of size 2i.it put one of these blocks into the free list of size 2i,and uses the other block to satisfy the request.
Detailed Solution for Test: Fragmentation - Question 8

No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list. When a request is made for m bytes ,the allocator first check the free list containing blocks whose size is 2i for the smallest value of i such that 2i ≥ m. If the free list is empty, it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.

Test: Fragmentation - Question 9

The power of 2 allocators is faster than the Buddy System allocators.

Detailed Solution for Test: Fragmentation - Question 9

because it does not need to performs splitting and merging.

Test: Fragmentation - Question 10

The buddy and power of 2 allocators are faster than the first-fit ,best-fit, next-fit allocators

Detailed Solution for Test: Fragmentation - Question 10

because they avoid searches in free lists

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